Nenek Ani
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Nenek Ani
Nenek Ani dirawat di rumah sakit. Menurut doktornya, asmanya sudah semakin teruk hingga perlu dipasangkan salur oksigen. Sudah beberapa hari dia tidak bercakap dan seperti orang koma. Dikira sudah menjelang ajal, anaknya memangilkan seorang Mudhin (tukang do'a) agar di doakan. Sedang asyik Pak Mudhin berdoa, tibatiba muka nenek Ani bertukar warna biru seolaholah tidak boleh bernafas. Tangannya menggigil. Dengan menggunakan bahasa isyarat nenek Ani minta diambilkan kertas dan alat tulis. Sisasisa tenaga yang ada digunakan oleh nenek Ani untuk menulis sesuatu dan memberi kertas tersebut kepada Pak Mudhin.
Sambil terus berdoa Pak Mudhin langsung menyimpan kertas tersebut tanpa membacanya kerana fikirannya dia tidak sanggup membaca surat wasiat tersebut didepan Ani. Tak lama kemudian nenek Ani meninggal dunia. Pada hari ketujuh meninggalnya nenek Ani, Pak Mudhin diundang untuk datang kerumah Ani.
Selesai memimpin do'a, Pak Mudhin berbicara, "Saudarasaudara sekalian, ini ada surat wasiat dari almarhum nenek Ani yang belum sempat saya sampaikan, yang saya pasti nasehat untuk anak cucunya semua. Mari kita samasama membaca suratnya".
Pak Mudhin membaca surat tersebut, yang ternyata berbunyi :
"Mudhin jangan berdiri di situ...! Jangan pijak saluran oksigenku"
Sambil terus berdoa Pak Mudhin langsung menyimpan kertas tersebut tanpa membacanya kerana fikirannya dia tidak sanggup membaca surat wasiat tersebut didepan Ani. Tak lama kemudian nenek Ani meninggal dunia. Pada hari ketujuh meninggalnya nenek Ani, Pak Mudhin diundang untuk datang kerumah Ani.
Selesai memimpin do'a, Pak Mudhin berbicara, "Saudarasaudara sekalian, ini ada surat wasiat dari almarhum nenek Ani yang belum sempat saya sampaikan, yang saya pasti nasehat untuk anak cucunya semua. Mari kita samasama membaca suratnya".
Pak Mudhin membaca surat tersebut, yang ternyata berbunyi :
"Mudhin jangan berdiri di situ...! Jangan pijak saluran oksigenku"
 bikinanu
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Re: Nenek Ani
kesian nenek Ani..
[spoiler]The Company Constitution.
Introduction
The company’s constitution was previously divided between the Memorandum of Association (external aspects) and the Articles of Association (internal aspects, including the balance of power between the board of directors and the general meeting) where this pattern has continued under Companies Acts. This two constitution is for defining what kind of company is and how its business and affairs are to be conducted. Without this constitution, a new company cannot be registered and established to do a legal business in Malaysia under the Companies Acts 1965. The important and relevant of this constitution is to carrying out issues related to the company and all the people who work directly or indirectly in the top management.
1.0 The Memorandum of Association
Memorandum of Association is a document that lists written constitution. Section 18 of Companies Acts 1965 must fulfill with this requirement for registering the company:
1.1 The name of company
1.2 Registered office
1.3 Objectives of the company
1.4 The limited liability of members
1.5 The authorized share capital
1.6 Name, address, work and signature for memorandum.
2.0 The Articles of Association
A document which content information related with administration affairs and management of the company. In addition, the articles of association should have the issue
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Companies Acts 1965
Section 18 of Companies Acts 1965
and transfer of shares, alteration of capital structure, calling general meetings and voting right, appointment, powers and proceedings of directors, dividend, accounts and the issue of notices. Limited company with allowed share by Company Act 1965 is not necessary to register their Article; this matter is provided in section 29(1). However, section 29(1) must read together with section 30(2) for those company that chose to not register the Articles, and Table A (a set of completed rules) will be automatically used. The other features in Articles of Association are:
2.1 The Article should be printed and divided with numbered spacing and signed by each signer in front of at least two witnesses. (Section 29(2))
2.2 Company have right to chose a matter in Article Association.
2.3 Non limited companies which have share capital declare the share amount and divided it to share unit.
2.4 Non limited company and limited company in guarantee to share and have to declare members that suggested.
The Memorandum ends with a declaration that the subscribers wish to form the company pursuant to the Memorandum and agree to take the number of shares opposite their names. For all companies, public as well as private, it is now only necessary to have two subscribers each of who signs the Memorandum. The subscriber’s signatures are witnessed, usually by a single witness who signs as such. The Memorandum is dated and is treated as deed executed under seal (J.M. Gullick, Company Law 13rd edition, 1987). The Articles may as a general rule is altered simply by passing a special resolution. Clauses which could be included in the articles may be placed in the memorandum in order to make it more difficult to alter them. The memorandum and the articles of association will be described in the next page with description and a relevant case in order to understand of application for incorporated company.
______________________________________________________________________________
Section 29(1), Company Act 1965
Section 29(2), Company Act 1965
Section 30(2), Company Act 1965
1.0 The Memorandum of Association.
1.1 The name of company.
In Malaysia, Suruhanjaya Syarikat Malaysia is the registry for a new company that doing business inside Malaysia.
The name of the company serves to identify it and to distinguish it from any other company. For this reason, company which might mislead the public will be change automatically by the registrar that has statutory powers of control over the choice of names. The choice of name must end with the word ‘Sendirian Berhad’ if it is a public company or ‘Berhad’ if it is private limited company, unless permitted to omit ‘Berhad’ from its name an unlimited company does not have to end with ‘Berhad’.
Other rules that will count by the registry for name registration are:
a) No company may have same name which is already exist.
b) No criminal offence or considered offensive.
c) In registrar opinion, the name is like have connection with the government or local authority.
1.1.1 Changing name – In section 23 of Companies Act 1965, a company may change its name by passing a special resolution and obtaining the registrar’s certificate of incorporation that has registered the company under new name. But before the name has been change it should be discuss and approve in the general meeting of the company. In case Oshkosh B’Gosh, the same limitations as above apply to adoption of a name by change of name as by incorporation of a new company. The registrar also can compel to change the name of company if the name same as or in the registrar opinion too like the other company name or the name give misleading an indication of its activities as to be likely to cause harm to the public.
1.1.2 Business name – Most company carry on a commercial activity. Hence the law on company names may give rise to business identity problems of two kinds.
First, a business or professional activity if carried on successfully soon acquires ‘goodwill’.
___________________________________________________________________________ Section 23, Company Act 1965
Customer or client that works with the company's name will bring to the attention of other people, anyone who hears the name of the company with good reputation will put confidence in the company while providing a high confidence to do business. However, if there is a competition that leads to the suppression of business caused by a competitor who has a similar company name, then legal action can be taken. For example in the case Ewing Vs Buttercup Margarine Co Ltd (1917), The plaintiff, a sole trader, had network of retail shops in Scotland and north of England, through which is sold margarine and tea. He traded ‘the Buttercup Dairy Co.’. The company was form to carry on a business of wholesale margarine distributors in England, it’s included a power to trade as retailers. The plaintiff obtained an injunction to prevent the defendant from trading under its own name on the grounds that there might be confusion especially as the plaintiff planned to extend his business to the south of England.
Second, the sole trader or a partnership can use a name that represents the company by mutual consent, however, the sole trader or a partnership cannot use the information held by the companies represented. sole trader or a partnership must use the original information is held as the company address, registration number or a telephone line so that the parent company did not receive overlapping claims and the errors made by a sole trader or a partnership.
1.1.3 Public company name  a public company must have a special name that ends with 'public limited company', where in Malaysia it is called a 'Sendirian Berhad'. This principle is buttressed by rules which:
a) Prohibit any person who is not a public company from carrying on trade, profession or business under a name which includes the public company designation.
b) A public company is guilty of an offence if, in circumstance which the fact that it is a public company is likely to be material to any person, it uses a name which may reasonable be expected to give the impression that it is a private company. (Company Law, 13rd Edition, by J.M Gullick 1987)
1.2 Registered office.
Section 119, Company Act 1965, (1) A company shall as from the day on which it begin to carry on business or as from the fourteenth day after the date of its incorporation, whichever is the earlier, have a registered office within Malaysia to which all communications and notices may be addressed and which shall be open and accessible to the public for not less than three hours during ordinary business hours on each day, Saturdays, weekly and public holidays excepted. (2) If default is made in complying with subsection (1) the company and every officer of the company who is in default shall be guilty of an offence against this Act.
1.3 Objectives of the company.
Objectives in the memorandum of association are concerned with the type of business. It will become the company's guidelines on the proper direction in matters involving. Ordinarily, a company will have many objects clause and the area to enable the company to operate in various business activities. However, if the company conducts a business activity that is unrelated to the purpose of its establishment as contained in the objects clause, then, the company has acted contrary to the objects clause and the association has committed an ultra vires.
1.3.1 Development of Object Clause
In determining the objects clause, it should be towards a power for companies to control the flow of capital is used more effectively. The shareholders will see the use of capital has been invested with the interest of the employer or employee. However, there are some cases that show the establishment clause of the object beyond the object itself, for example in the case of Bell House Ltd. Vs City Wall Properties Ltd (1966), the object clause was to carry on the business of builders and developers. The third paragraph was ‘to carry on any other trade or business whatsoever’ which can in the opinion of the board of directors be advantageously carried on by the company in connection of the above businesses or the general business of
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Section 119(1) of the Company Act 1965
Section 119(2) of the Company Act 1965
the company. The companies also have power to turn to account any of its property and acquired to contact with Swiss bankers who had money to lend on mortgage for property development. The defendant sought the help from plaintiff to introduce defendant to the Swiss bankers and in return for an introduction, fee £20,000 will given by defendant. They later refused to pay because the contract was ultra vires the company and void since its objects did not include mortgage broking. Judge has decided that the fee was binding. The directors did genuinely believe that the effecting of an introduction for fee was a business which could be advantageously carried on as ancillary to the general development of the company. This is what the object clause required and permitted. It was also a case of turning to account an asset of the company.
1.3.2 Alteration of objects
Changing the object clause is under Section 21 of Companies Act 1965, the company can alter as long as Companies Act 1965 allows to do so. The object clause can be altered under Section 28 of Companies Act 1965. According to Section 28(1) of Companies Act 1965, the alteration of object clause can be done by special resolution. Section 28(2) of Companies Act 1965 requires the company to give 21 day notice of meeting to all the members. Section 28(3) of Companies Act 1965 states that notice must be given to all members of company about the alteration of object clause. According to Section 152 of Companies Act 1965, there must be at least 75% majority of the members of company to agree to change the object clause. Section 28(8) of Companies Act 1965 states those 21 days after passing the special resolution, any members of company can apply to object. Section 28(5) of Companies Act 1965 states that there must be at least 10% of issued share capital if they want to object. Section 181 of Companies Act 1965 states any minority, which is less than 10%, can object under this section. Under Section 28(9) of Companies Act 1965, must lodge with SSM within 14 days. Section 28(10) of Companies Act 1965 states the alteration of object clause takes effect when the resolution has been lodged.
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Section 21 of Companies Act 1965
Section 28(1) (2) (3) (5) (8) (9) (10) of Companies Act 1965
Section 152 of Companies Act 1965
Section 181 of Companies Act 1965
1.3.3 Ultra Vires
Ultra Vires may be defined as a situation where the company has conducted business activities beyond the limit and the terms in the memorandum of association. However, a company that has made an ultra vires transaction or arrangement that is not contained in the memorandum of association shall remain valid. These are the effect of the amendment laws under section 20 (1) of the Companies Act 1965 which states: “no such act or acts of the company will be revoked only by reason of the fact that the company has no ability or authority to act or perform or take such conveyance or transfer of even more than what is stipulated in the memorandum of association”.
Case Ashbury Railway Carriage & Iron Co. Ltd. Vs Riche (1875), the company had an object clause which stated that its objects were to make and sell, or lend on hire, railway carriages and wagons and all kinds of railway plant, fittings, machinery and rolling stock and to carry on business as mechanical engineers. The company bought a concession to build a railway in Belgium, subcontracting the work to defendant. Later the company repudiated the contract. Judge was deciding that, constructing a railway was not within the company’s object so the company did not have capacity to enter into either the concession contract or the subcontract. The contract was void for ultra vires and so the defendant had no right to damages for breach. The members could not ratify it and the company could neither enforce the contract nor be forced into performing its obligations.
Ultra vires issue can be raised by the company or other shareholders through the provision of sections that have been made in the Companies Act 1965 section 20 (2) (a) (b) and (c). Explanation of the issue of ultra virus provided in section 20 (2) (a) which provides “the proceedings of any member, or if the company has issued debentures secured by floating charge over all or any of its assets by holders of any debentures or trustee for debenture holders to prevent the commission of any act or acts or the conveyance or transfer any
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Section 20 (1) of Companies Act 1965
Section 20 (2) (a) of Company Act 1965
property to or by the company”. This means that a trustee or member may take action to restrict or prevent the company from taking any action ultra vires.
Case Hawkesbury Development Co. Ltd Vs Landmark Finance Pty. Ltd. (1969), plaintiff is the holder of all shares in Landmark Finance. It has demanded that the court declare the Landmark production of some debentures to finance United Dominion Corporation (UDC) is invalid because it was ultra vires. Plaintiff also asked the court to prohibit enforcement of UDC of the debentures. However, plaintiff's application is not successful, and failed to obtain court approval to void the declaration of the UDC. This is because, the plaintiffs are shareholders of the Landmark Finance. The application should be made to Landmark instead of UDC Finance is a third party.
1.4 The limited liability of members.
This clause states that the liability of the members of the company is limited. In the case of a company limited by shares, the member is liable only to the amount unpaid on the shares taken by members. In the case of a company limited by guarantee the members are liable to the amount undertaken to be contributed by them to the assets of the company in the event of its being wound up. However, this clause is omitted from the memorandum of association of unlimited companies. Any alteration in the memorandum compelling a member to take up more shares, or which increases his liability, would be null and void. If a company carries on business for more than six months, less than 2 members in case of a private company each member aware of this fact, is liable for all the debts contracted by the company after the period of six months has elapsed.
1.5 The authorized share capital.
Authorized share capital is the ceiling of capital available for issue. For example, if the company’s authorized capital is RM100, 000.00 divided into 100,000 ordinary share of RM1.00, that means the company can only increase its paid up capital to a maximum of RM100, 000.00. The company may increase its authorized capital anytime by holding an EGM (Extraordinary General Meeting). A fee is payable to the Companies Commission of Malaysia for the increase of authorized share capital.
1.6 Name, address, position and signature for memorandum.
Normal procedures to complete a memorandum of association as have to fill up for more information in company structure. Example:
Picture 1
2.0 The Articles of Association.
Articles of association are a document that was touched on in the internal affairs of a company that it provides information on how companies should conduct business. For example, it touches on the relationship between the shareholders of each other shareholder, the withdrawal, confiscation and transfer of shares, transfer of the share capital, the notice of the proceedings, the particulars of the appointment, powers, action directors, accounts, audit, dividend, loans and winding. Every company, except a company limited by shares must register the articles of association to the Registrar of Companies. This is consistent with the requirements under Section 29 of the Companies Act 1965. This Act also provides a set of examples of articles of association in Table A of the Schedule 4. Any company is free to imitate some/all of them. If a company limited by shares were not registered with the Registrar of the article itself, then Table A will apply automatically. (Refer to appendix, Sample of Article of Association)
2.1 Effect of Memorandum and Articles of Association
Section 33 (1) of the Companies Act 1965 provides:
"subject to this Act, the memorandum and articles shall when registered, according to the company and its members as if the memorandum and articles, each signed and sealed by each member and contained covenants on the part of each member to comply with all provisions of the memorandum and articles ".
From the provisions of this section, the memorandum and articles on the registration would create the effect of the contract. The effects are:
2.1.1 Contractual effect between the companies and member.
2.1.2 Contractual effect between members (members inter se).
2.1.3 Relationships between companies and outsiders.
______________________________________________________________________________ Section 29 of the Companies Act 1965
Section 33 (1) of the Companies Act 1965
2.1.1 Contractual effect between the companies and member.
Effects arising from the contract provisions of section 33 (1) of the Companies Act 1965 allows companies to enforce the provisions of the articles and memorandum of association of members. In other words, companies can take action against the member to ensure that they comply with the provisions in the articles and memorandum if the member refuses to follow the provisions voluntarily.
In case Hickman (H) Vs Kent (1915), the facts are as follows: a provision in the articles of a company provide that, "if there are any differences that exist between the association and any of its members, then such dispute shall be referred to arbitration." plaintiffs (H), is a members in these companies have taken legal action against the company relating to a dispute concerning his dismissal from the company. This is a dispute between the H at the position as a member of the company. The company has relied on the relevant articles and requested that the trial is terminated. The court decided the company was entitled to terminate the trial because the article was a contract between his company and H, and such difference shall be referred to arbitration as provided by the articles of association. The proceeding would be stayed since the dispute (which related to matters affecting H as a member) must, in conformity with the articles, be submitted to arbitration. In reviewing other cases the court held that if the matter does not affect members generally, if it is something confined to one member in his personal capacity, it is not question of members right and obligations and is not subject to the rule.
However, there are a few cases the effect of a contract excludes the application of section 33 (1) of the Companies Act 1965 to be valid by the position of other members. For example, the position as director, promoter or lawyer. This is because there is among individuals or parties who registered as a member of the company and at the same time they hold certain positions within the company. If members want to enforce the provisions of the articles, giving the position of other members of the claim will not succeed.
In case Beattie Vs E & F Beattie Ltd (1938), the article also provided for arbitration on disputes. The company sued its managing directors to recover money improperly paid to him.
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Section 33 (1) of the Companies Act 1965
The managing director applied to have the action stayed as he contended that the dispute should be submitted to arbitration. The court judge that the claim was against the defendant as a director and the arbitration clause was limited to disputes with members as such. He could not rely on the arbitration clause. Provisions of section 33 (1), is enshrined in the company and its members only. It will not be valid to a third party even if the third party has held a position in the company and manage the affairs of the company. This matter can be looked in Eley Vs Positive Government Security Life Assurance Co. (1876), In this case, Eley is a drafter of articles and the articles of association have provided:
"that Eley is a lawyer and the company only can terminated him for misconduct"
Eley acting as company lawyer for some time though there is no separate employment contract signing. Eley also receive a distribution of shares in the company in return for work done during the establishment of the company. Eley is then fired by company as a company lawyer. Eley claimed the company had breached the contract. The Court held, the article does not give rights to the member if the member seeking to enforce their rights in a position other than a member. In this case, Eley try to enforce the provisions of articles that give it rights in the company's position as a lawyer. If Eley wish to do so, he should sign a contract other than as provided in the articles of association.
In case Re New British Iron Co, ex parte Beckwith (1898) showed about separated contract with the company. In this case, the articles provided that the remuneration of the directors should be £1,000 per annum to be divided between them as they saw fit. By accepting office the directors entered into separate contracts between themselves and the company but nothing more was said about their remuneration. When the company went into liquidation the directors claimed arrears of fees. The liquidator denied that there was any contract to pay the fees. The court judge, although the directors could not rely on the articles as a contract for the payment of their fees, they could refer to the articles to establish the amount payable under the separate contracts with the company made when they accepted office as directors.
2.1.2 Contractual effect between members (members inter se).
Every member is bound to the other members through an article. There's disagreement on how the law is. What seemed clear was that if the article provided to a member of a right alone so he could enforce those rights without the help of members of the company. It can also impose a contract on the members in their dealings with each other as illustrated by the case below. To avoid doubt and difficulty it is usual however to draft articles, especially on members right of first refusal of other members shares, so that each stage is dealing between the company and the members, to which Section 33(1) clearly applies. So that, a member who intends to transfer his shares must, if the articles so require, give notice of his intention to the company and the company must then give notice to other members that they have an option to take up his shares.
In the case of Rayfield v Hands (1960), articles of association of a company alone has allocated:
"Every member of the type that wishes to transfer shares shall notify the director and they will take the same share many of them with a reasonable price.”
Pursuant to the provisions of this article, the plaintiff was trying to force three directors of the company to buy its shares. But all three refused to buy its shares. The Court held that the relationship created by the articles that are binding and director of the company as between members and nonmembers and as between a member and director. With that the three directors are bound to purchase shares of the plaintiff.
The effect also taken to be more complex when a share holder is still want to hold it share but have to follow the rules or article. Take a look in Malaysian case, In the case of Malaysia, Wong Kim Fatt lwn Leong & Co. Sdn Bhd (1976), provisions of the articles of association of limited companies have been allocated as follows: holders of the 7 / 10 of the issued capital of the company may at any time make a request to the company to enforce the transfer of any shares certain that they are not held by the requester. In this case the company has 2 shareholders, A and B where A is a minority shareholder who holds 50,000 shares. B also holds 250,000 shares and the more he is the holder from 7 / 10 of its issued capital. B had made an application for get a share based on the provisions of the article, but a refuses to sell and claimed in court that he was entitled to continue holding its shares. Court decided that it is purely contractual obligations and the claim must be tied to contractual obligations he has received. The question of whether fair or not a provision of the article, is not relevant in determine whether the article could be enforced or not. Rights and dependents under article is a contractual obligation.
2.1.3 Relationships between companies and outsiders.
Articles do not involve contracts with outsiders. This is because outsiders are not members and cannot enforce any alleged rights derived from the memorandum or articles of the company. Look at the example of the case opposite the Raffles Hotel Ltd Malayan Banking Bhd (1996). Malayan Banking is the site of the lessor Raffles Hotel. Things in the article had stated that the lessor has power to appoint a director. Malayan Banking has appointed himself as the director led the Raffles Hotel has applied for a declaration that the appointment was illegal. The court ruled invalid because the appointment of Malayan Banking is not a member of the company and the power held within the article mentions only Malayan Banking authorities to appoint a director.
However, relations with people outside the company should be viewed in a broader context. There are also cases which do not deny the power of the article to be applied into the relationship. As an example the case of Southern foundries (1926) Ltd. Vs Shirlaw (1940) state that the article was to authorize the trustees of the late director to appoint a director of the company. The trustee had made an appointment but was challenged by several members. The Court held that the appointment of a trustee is legal. This action confirms the appointment may be more affected by the responsibility and it should be seen as justice for the rights of the more significant.
2.2 Alteration of the Articles.
Section 31 (1) provides:
'Subject to this act and to such terms in the memorandum, a company may be through a special resolution to amend or add to the articles'
Clear from the provisions of this section that company has the right to amend or add to the articles of association. However, an amendment to articles is only possible through a special resolution received 75% majority vote and the amendment is binding even though there are members who do not give consent. While section 31 (2) also provides for any modifications or additions to the articles of association shall be deemed valid as if it were originally contained in the articles of association. If all members of the company agree to an alteration without passing the general meeting or special resolution, this is still an effective under ‘assent principle’. It can be looked over Cane Vs Jones (1980) where the shareholders signed a written agreement by which the chairman was deprived of the casting vote at general meeting given to him by the articles. No special resolution was passed to make this alteration. Later one group of shareholders contended that the articles had not been altered and so the chairman still had casting vote since special resolution was the only permissible method of altering the articles. The court judge that a special resolution was the means by which majority of members could alter the articles so that all members become bound by the alteration. But it is also a principle of company law that all the members acting together maybe their unanimous agreement take a binding decision on any matter for which a majority of votes at a general meeting is required to bind a minority. The alteration was affective and the chairman had lost his casting vote.
This is about the winning of a majority in their voting, but if there have some unfair decision that make the minority felt injustice, section 181 (1) can be use to protect the rights.
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Section 31 (1) of the Companies Act 1965
Section 31 (2) of the Companies Act 1965
Section 181 (1) of the Companies Act 1965
This section states in paragraphs 1 and 2 are split into several branches of a particular section. They are:
Section 181 (1): Any member or holder of debentures of the company, the minister may apply to the court for an order under this section on the ground:
(a): that the company's affairs are conducted or the powers of directors are carried out in a manner oppressive to one or more members or holders of debentures including himself or herself or does not take into account their interests as members, shareholders or holders of debentures of the company, or
Section 181 (2): If the court considers the application of one of the reasons could be proved:
(a): directing or preventing any act or cancel or vary any transaction or resolution;
(b): manage the affairs of the company's actions in the future;
(c): provides for the purchase of shares or debentures of the company by other members or debenture holders of the company or the company itself;
(d): in case of purchase of shares by the company, it provides for deduction of the share capital of the company, or
(e): provides that the company can be deposed.
2.2.1 Limits for the amendment article
There are some acts that limit the amendments in the article. The Acts are:
Section 33 (3): Any amendment which requires stockholders to take the shares or increase their liability to contribute to the capital of the company is not allowed. Unless the shareholder got written consent from members before and after the amendment of the articles.
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Section 33 (3) of the Companies Act 1965
Section 181 (1) (a) of the Companies Act 1965
Section 181 (2) (a) (b) (c) (d) (e) of the Companies Act 1965
Section 181 (4): Shareholders or debenture holders are allowed to apply for a remedy on the grounds of oppression prevailing in the company. Upon application, the court has the power to issue such an order to amend the memorandum or articles of association. If the court had issued a revised order of the articles, the company cannot make changes other than those permitted by the court.
Section 65 (1): For a company having a share capital divided into several classes of shares and have different rights between the classes of shares, the class of rights amendment is not allowed except in the manner provided for in clause variation, the clause that provides a way to amend the right according to the classes of shares. Freedom to amend the articles is limited when several classes of shares issued and any amendments to the rights granted to the class of shares must follow the procedures set out in clause variations.
The principle of ‘Majority Rule’: When a company wants to amend the articles of association, the principle of 'majority rule' will apply. Through this principle, amendments to the articles could be enforced upon approval of 3 / 4 or 75% majority of the company and the decision is binding on the minority. In this case, the majority of members can add to or amend the provisions of the article. However, to prevent abuse of this principle, amendments to articles by the majority has a limit of the amendment must be made by Bona Fide, which is in good faith for the benefit of the company as a whole. Therefore, the amendment of articles by the majority cannot be enforced if it is not made in good faith for the good of the company as a whole.
Bona Fide can be looked over case of Allen Vs Gold Reefs of West Africa Ltd (1900), under the articles in their original form the company had a lien on ‘all shares (not being fully paid shares) of members’. All the members held partly paid shares, to which the lien attached, but only Z also held fully paid shares to which the lien in that form did
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Section 65 (1) of the Companies Act 1965
Section 181 (4) of the Companies Act 1965
not attach. A special resolution was passed to remove from the articles the words ‘not being paid shares’. As a result the company would then have a lien on the fully paid shares of Z to secure the amount owing on his partly paid shares. The position of other members remained the same since the lien in its original form already attached to their partly paid shares. The executors of Z (who had died) objected that the alteration was discriminatory since it only altered the position of one member (Z) and that it was retrospective in its effect since Z had acquired his fully paid shares (free of the lien) before he was allotted partly paid shares and so incurred the debt for which the lien, extended to his fully paid shares, was security. The judge held that, the alteration was for the benefit of the company as a whole and could apply to any member who might hold fully paid shares. As regards the retrospective effect a member held his shares subject to the memorandum and articles of association and his shareholder’s right were therefore subject to any alteration which might be made affecting them. The alteration was therefore valid.
2.3 Methods of internal affairs (the principle of agency in the company)
This principle highlights the rights agents appointed to represent something or someone in the business of the company. Powers of agents is divided into three, namely:
2.3.1 Express Authority
Real power is something that is expressly given to these agents. For example, someone who had been appointed by the company and given a power to issuing and receiving payments on behalf of the company's debt or borrow money and give security against the debt. The responsibility is clearly within the jurisdiction of a director of the company.
2.3.2 Implied Authority
Implied powers not expressly mentioned, and it is subject to state authority. It exists in two states, namely:
i) Power to make matters incidental to the matters expressly authorized.
ii) Power to do things normally done by an agent in such a position.
For example, an agent has been appointed to a personal manager to a company, it is given real power. However, when the agent advertising to call the candidates to attend an interview and took a successful candidate to work in a company then it is binds to the company, this task is a task usually done by a manager.
2.3.3 Ostensible Authority
One situation where the law will infer the creation of an agency by implied appointment is when the principal, by his words or conduct, creates an inference among third parties that the agent has the actual authority to contract even though no actual authority was actually given to him. This situation may arise when the principal allows the agent to order goods and services on his behalf and he proceeds to pay for them. In such a scenario, if the agent contracts within the limits of the ostensible (apparent) authority, although without any actual authority, the principal will be bound to third parties by his agent’s acts.
In case Chew Hock San Vs Connaught Housing Development SDN. BHD (1985), defendant is the developer. The plaintiff entered into an agreement to buy several houses a particular store from the defendant. The plaintiff has to pay the deposit to the clerk of the defendant to the book shop though at the time of booking that has not been opened. Defendant refused to comply with the contract and the plaintiff's claim for specific performance. The Court ruled that the clerk has no authority to take actions such deposits, and therefore there is no representation from the company to be bound by the acts the defendant. Thus, plaintiff has failed to claim its actions.
However, there is an exception in this ostensible authorities to block the company were causing the lack of power and the doctrine of estopel. Circumstances in which the doctrine of estopel / restrictions apply are:
1) There is a representation by the company to a state that the agent has the authority to take action in question.
 Representation of this can arise in several ways: first, an agent with apparent authority to do all such acts as agent in the usual position. Second, a principal by his actions in a certain time period would represent an implicit: his agent is authorized to sign contracts on behalf of himself.
 In case of Freeman & Lockyer Vs Buckhurst Park Properties (Mangal) Ltd. (1964), A property development company which was established by two men and one named Kapoor. Kapoor acted as managing director although never elected to that position, but the other directors are satisfied with the way management Kapoor. Then, Kapoor had hired the plaintiffs to post an architect and land surveyor. Company refuses to pay the plaintiff as alleged Kapoor has no such power. The Court ruled, Kapoor has ostensible authority, the board of directors know that Kapoor has acted as managing director and appoint an agent for the company. With the permission and that they were, he has the power to enter into ordinary contracts signed by a manager. The plaintiff was motivated by a sense of trust that Kapoor has been authorized by the representation to perform services for the company. So, the company has estopel from denying the power of Kapoor and companies have to pay the plaintiff.
2) Representations must be made by a person having authority to make representations on behalf of the company.
 To make these companies are bound, these representations must be made by a person who has actual authority to make such representations.
 In case Armagas Ltd. Vs Mundogas (1986), Mundogas is the owner of "Ocean Frost 'is a shipping company. A consultation has been done to sell the ships to the Armada Group and it was carried out by representatives of the parties, Magelssen. Representatives of the Armada Group only wants to hire the vessel for a period of 3 years and Magelssen not have the power to hire only sales. Magelssen agree to the lease agreement with a false representation that he has the power to enter into the lease contract. Armagas Ltd was incorporated to a ship and Magelssen had completed the charter agreement for the Mundogas for 3 years. To cheat Mundogas, Magelssen has provided the rental agreement for 12 months for the ship of Armagas Ltd. Thereafter, Mundogas decided not to proceed with charter breach caused Armagas claimed promise. Armagas Magelssen believes is authorized to manage the rental of the mundogas. The Court ruled that the application is rejected because Magelssen not have real power to complete the rental for 3 years. Magelssen has no authority to make representations that he was authorized to do so. Misrepresentation made to the powers is not binding Mundogas.
3) People who want to enforce the contract against the company must have relied on these representations.
 Estopel would only arise if the contractor is dependent on the representation that the agent has the authority to complete the transaction for the parties. If the contractor does not know the agent has the power, then he would not rely on the
representation and estopel will not arise. Thus, knowledge of the power of agents is necessary to avoid problems arising if there fraud or offense committed without the knowledge of real company.
2.4 Duties as responsible from company to agent
Duties of an Agent to his Principal:
1. To obey the principal’s instructions
2. In the absence of instructions from the principal. To act according to the prevailing customs of that business
3. To exercise care and diligence in carrying out his work and to use such skill as he possesses
4. To render proper accounts when required
5. To pay to his principal all sums received on his behalf
6. To communicate with the principal
7. Not to let his own interest conflict with his duty
8. Not to make secret profits out of the performance of his duty.
9. Not to disclose confidential information or documents entrusted to him by the principal
10. Not to delegate his authority
Duties of Principal to his Agent
1. To pay the agent the commission or other agreed remuneration unless the agency relationship is gratuitous
2. Not to willfully prevent or hinder the agent from earning his commission
3. To indemnify and reimburse the agent for acts done in the exercise of his duties
2.5 Methods of internal affairs (Indoor Management Rule)
If an agent had apparent authority to do something, then someone who is dealing with a company is entitled to presume that all matters of internal management and procedures set by the company have been complied with. Indoor management rule sometimes can be known as the rule of Turquand case. It is necessary to begin with the case itself, Royal British Bank Vs Turquand (1856), under the company’s deed of settlement (corresponding with the modern articles) the directors had power to borrow only such sums as had been authorized by general resolution of the company. A debenture for £2,000 had been issued under the seal, signed by two directors and by the secretary. The resolution passed in general meeting to sanction this borrowing did not specify the amount which might be borrowed. It was therefore no authority to the directors to borrow anything. The bank had not seen the resolution and had no knowledge of its inadequacy. But it was held to be entitled to assume that the authority prescribed by the articles, which requirement it had constructive notice, had been effectually given by ordinary resolution, in example by internal procedure of which the bank had neither actual nor constructive notice.
As an 'outsider', the bank is not able to find out whether ordinary resolution was passed because of this information does not typically file in the registry. In this case, the bank has the right to assume that a resolution has been approved legally. The principle of 'internal affairs rule' that sets a third party dealing with the right to make assumptions (if they have no knowledge that the company's internal regulations required by the articles of association have been complied with).
2.5.1 Exclusion
There are limitations on the ability of a party to rely on the internal management rules. First, if the contracting party knew or should have known that the agent has no power, then the contracting parties cannot rely on this method. Typically, these cases occur when the contract is "insider” of his position, should have known about the disability. For example in case Howard Vs Patent Ivory Manufacturing Co. (1888), Director of the company has lent money to companies based on the guarantee of the debentures. Article states that the company can only borrow up to the limit and has exceeded a certain limit. The director wants to enforce the debentures. The Court held that it refuses to allow them to do so. All this is because, as the directors they know or should know about borrowing limits. Thus, they cannot rely on the internal management rules. Second, if there are situations where a reasonable person would make the suspect feeling the power of the agent. So the contracting parties must make reasonable inquiries. If these questions indicate the agent has no authority, contracting parties cannot rely on internal management tool to help him. In case Morris Vs Kanssen (1946) lord of house has saying that:
“Person contracting with a company and dealing in good faith may assume that acts within its constitution and powers have been properly and duly performed and are not bound to inquire whether acts of internal management have been regular”
Later, the judge has pointed out the rationale of the rule ‘The wheels of business will not go smoothly round unless it may be assumed that is in order which appears to be in order’. Third, if the review of the memorandum or articles of association of the company clearly shows that the agent's authority is limited, so the contract cannot rely on the internal management rules.
Lawton and BrownieWilkinson LJJ delivered concurring opinions, “the presumption of regularity cannot be relied on by ‘inside’. In example, persons who by virtue of their position in the company are in a position to know whether or not the internal regulation have been observed”. (page 211, Cases and Material in Company Law, L.S. Seally [1985])
For the conclusion, these two documents are important for determining the direction of the company. This document has clearly touched a lot about management and regulatory compliance by directors, shareholders and even employees who worked in the company. The importance of the two documents is to bring the birth of new companies registered to continue to drive the mission and vision for the establishment of better economic importance.
The end
SOALAN ASSIGNMENT:
a) Nyatakan sebabsebab yang diberikan oleh mahkamah untuk justifikasi
mengenai penerimaan dengan surat dihantar melalui pos?
b) Pada tahun lepas, Linda telah meninggalkan sekolah semasa beliau berumur 16 tahun. Dia telah berkerja dalam bidang hotel sebagai pelatih(trainee) pembantu dapur. Pendapatan Linda adalah 50 pounds seminggu, dan seperti yang diperuntukkan oleh hotel dia harus memberi notis selama 3 bulan jika ingin menamatkan pekerjaannya.
Barubaru ini Linda telah bersetuju untuk membeli motosikal “ Osaka” sebab beliau ingin meluangkan masa bersama kawan baiknya Malcolm, yang turut berminat terhadap motosikal. Linda juga telah menandatangani perjanjian bertulis untuk membeli syer ( satu perempat) yang dikenali sebagai syer “Dingo”.
Linda kini telah ditawarkan pekerjaan sebagai tukang masak di restauran dengan gaji
sebanyak 100 pounds seminggu, dengan syarat beliau boleh memulakan
tugasnya dengan sertamerta. Dia telah gagal untuk membayar untuk harga motosikal
dan bahagian dalam syer Dingo.
Nasihatkan Linda berdasarkan Akta Kontrak yang relevan dan keskes yang
berkaitan/diputuskan oleh mahkamah
Jawapan soalan a).
Undangundang kontrak dalam seksyen 4(2) Akta Kontrak 1950 ada menyatakan bahawa komunikasi penerimaan adalah lengkap terhadap pencadang apabila komunikasi penerimaan itu telah dimasukkan kedalam peti surat untuk perjalanan surat pengirim kepadanya, di mana ia adalah di luar kuasa penerima. Kes Entores Ltd. lawan Miles Far East corporation (1955) menyatakan bahawa, “apabila suatu kontrak dibuat melalui pos, undangundang dalam semua negeri Common Law jelas bahawa penerimaan adalah lengkap selepas surat itu dimasukkan ke dalam peti surat dan itulah tempatnya kontrak telah dibuat”.
Undangundang ini juga adalah untuk mewujudkan kesahihan terhadap kontrak sama ada boleh dikuatkuasakan oleh undangundang atau pun tidak. Kesahihan undangundang itu perlu bagi mengesahkan sesuatu kontrak sedang berjalan atau pun tidak, kerana pencadang boleh membuat pembatalan kontrak sebelum tarikh luput kontrak tersebut tanpa mengetahui jika terdapat penerimaan yang sedang dalam perjalanan. Sebagai contoh Kes Byrne lawan Van Tienhoven, defendan telah menawarkan akan menjual 1000 kotak tin bersadur timah kepada plaintif melalui surat pada 1 Oktober, pada 8 Oktober defendan telah membuat pembatalan cadangan yang dibuat pada 1 Oktober. Surat pembatalan tiba kepada plaintif pada 20 Oktober. Namun plaintif telah menghantar utusan penerimaannya pada 11 Oktober dan mengesahkannya pula pada 15 Oktober. Mahkamah memutuskan bahawa kontrak telah berjalan kerana pembatalan tidak berkuatkuasa sehingga 20 Oktober, iaitu apabila sampai ke pengetahuan plaintif sedangkan plaintif telah pun menerima cadangan tersebut pada 15 Oktober.
Bagi penerima pula, sesuatu kontrak melalui komunikasi penerimaan pos akan berlaku apabila pencadang mendapat maklum balas positif untuk melanjutkan sesuatu kontrak yang dicadangkan. Mengambil contoh dalam kes Ignatius lawan Bell (1913), plaintif ingin menuntut pelaksanaan spesifik mengenai perjanjian opsyen yang mengikut plaintif memberinya pilihan membeli hakhak defendan keatas sebidang tanah. Plaintif menghantar notis penerimaannya melalui surat berdaftar pada 16 Ogos, tetapi surat tidak sampai kepada pengetahuan defendan sehingga petang 25 Ogos kerana defendan tiada di rumah. Mahkamah telah memutuskan bahawa opsyen tersebut adalah sah dipergunakan oleh plaintif apabila surat tersebut sudah diposkan pada 16 Ogos. Pengesahan kontrak tersebut telah berjalan dikuatkan lagi dengan niat penerima iaitu plaintif untuk berkontrak dengan mendaftar surat tersebut.
Walaubagaimana pun, sesuatu kontrak yang melalui komunikasi penerimaan pos mempunyai akta tersendiri dan cara tertentu untuk dibatalkan. Seksyen 5(2) Akta Kontrak 1950 menyatakan sesuatu penerimaan boleh dibatalkan pada bilabila masa sebelum komunikasi penerimaan lengkap terhadap penerima. Antara cara pembatalan kontrak melalui pos adalah:
a) Mengkomunikasikan notis pembatalan,
b) Luputnya waktu yang munasabah,
c) Kemungkiran penerima berkenaan syarat kontrak
d) Meninggal dunia ataupun sakit otak.
Kita mengambil contoh pembatalan luputnya waktu yang munasabah. Kes yang berkaitan menjelaskan keadaan ini adalah Ramsgate Victoria Hotel Co. Ltd lawan Montefiore (1866), Defendan telah memohon suatu saham dalam syarikat hotel tersebut pada bulan Jun. Tetapi hanya bulan November, syarikat tersebut memberitahu sahamsaham itu telah diagihkan dan diberi. Maka, pemegang saham perlu menjelaskan baki harga belian dimana defendan enggan membayarnya. Mahkamah telah memutuskan bahawa keengganan defendan menjelaskan baki harga belian tersebut sah kerana maklumat saham tersebut sepatutnya diketahui oleh defendan dalam masa yang munasabah bukan mengambil masa antara Jun dan November.
Mahkamah menetapkan justifikasi undangundang seperti ini adalah untuk menjaga hak penerima kontrak selain daripada hak pencadang. Kontrak melalui pos adalah suatu kontrak yang memerlukan persefahaman antara keduadua pihak, kerana ia melibatkan masa dan tindak balas daripada pencadang mahu pun penerima. Namun, hak antara penerima dan pencadang cadangan ini mempunyai kriteria tertentu untuk dipatuhi supaya keduadua belah pihak mendapat hak keadilan yang seimbang.
\
Jawapan soalan b).
Akta Dewasa 1971 menyatakan bahawa, seseorang dikatakan dewasa apabila mencapai umur 18 tahun dan layak untuk membuat kontrak. Namun, umur dewasa boleh berbeza untuk tujuan lain sebagaimana ditetapkan oleh undangundang lain. Contohnya, untuk tujuan mengundi dalam pilihan raya negeri, umur dewasa adalah 21 tahun sementara untuk kontrak perantisan dan ahli keanggotaan kesatuan sekerja adalah 16 tahun. Ini bermakna Linda yang berumur 16 tahun boleh membuat kontrak untuk tujuan pekerjaan.
Isu.
1. Adakah Linda perlu memberi notis jika ingin menamatkan pekerjaannya?
2. Adakah sah untuk Linda memasuki kontrak pembelian motosikal dan syer “Dingo”?
3. Bolehkah Linda menerima tawaran pekerjaan sebagai tukang masak?
4. Kegagalan Linda membayar harga motosikal dan syer “Dingo”.
Penyelesaian.
Isu 1.
Ya. Linda perlu memberi notis jika ingin berhenti daripada kerjanya sebagai pelatih pembantu dapur. Ini kerana pihak hotel telah memperuntukkan notis tersebut sebagai peraturan pekerja yang ingin berhenti.
Isu 2.
Kanakkanak tidak boleh memasuki kontrak pembelian motosikal mahupun pembelian untuk syer kerana kontrak tersebut tidak termasuk didalam senarai kekecualian kepada kanakkanak. Tetapi dalam kes Linda, dia telah berniat untuk memasuki kontrak kerana mempunyai kerja sebagai pelatih pembantu dapur dengan pendapatan £50 seminggu. Seksyen 69 (d) menyatakan, kanakkanak akan bertanggungan hanya sekiranya mempunyai harta. Sebagai contoh dalam kes Doyle lawan White City Stadium Ltd (1935), seorang kanakkanak yang merupakan peninju professional telah bersetuju untuk mematuhi peraturan Lembaga Kawalan Tinju British(LKTB) dalam urusan profesionnya ketika menerima lesen dari LKTB. Sejumlah wang telah dipegang oleh LKTB seperti yang diperuntukkan dalam peraturan LKTB kerana kanakkanak tersebut telah melanggar peraturan yang ditetapkan. Mahkamah memutuskan kontrak diantara kanakkanak tersebut dengan LKTB adalah sah dan mengikat. Kes ini serupa dengan kesahihan Linda dengan niat untuk berkontrak dan mengikat janji dalam pembelian motosikal serta pembelian syer yang dikenali sebagai syer “Dingo” dan Linda adalah terikat dengan perjanjian komersial tersebut.
Isu 3.
Linda boleh menerima tawaran tersebut yang menawarkan gaji £100 seminggu. Ini kerana pekerjaan tersebut boleh membantu Linda untuk memenuhi dan menanggung keperluan hidupnya. Walaubagaimana pun, Linda perlu mengemukakan notis selama 3 bulan kepada majikan terdahulu yang merupakan syarat yang telah diperuntukkan sebelum dia menerima pekerjaan sebagai tukang masak di restoran.
Isu 4.
Kegagalan Linda untuk membayar harga motosikal dan bahagian dalam syer Dingo merupakan suatu kesalahan kerana perjanjian yang telah dibuat mesti dipatuhi manakala perjanjian tersebut adalah sah dan mengikat dengan niat serta boleh dikuatkuasakan selari dengan peruntukan seksyen 2(h) Akta Kontrak 1950 iaitu sesuatu perjanjian yang boleh dikuatkuasakan oleh undangundang dan seksyen 10(1) Akta Kontrak 1950 yang menyatakan, sesuatu perjanjian adalah kontrak jika dibuat atas kerelaan bebas pihakpihak yang layak berkontrak, untuk sesuatu balasan yang sah, dan dengan suatu tujuan yang sah, dan tidak ditetapkan dengan nyata dibawah peruntukan akta ini bahawa ianya batal. Dalam kes Koh Kia Hong lawan Guo Enterprise Pte. Ltd (1990), pihak plaintif membuat tawaran untuk membeli hartanah yang diiklankan oleh defendan. Pihak plaintif telah menyerahkan deposit kepada pihak defendan. Satu nota juga telah ditandatangani oleh keduadua pihak dan resit rasmi telah dikeluarkan oleh defendan. Mahkamah memutuskan bahawa keduadua pihak mempunyai niat untuk mengikat diri dalam satu kontrak. Kes tersebut sama dengan tindakan Linda yang telah bersetuju untuk membeli sebuah motosikal dan turut menandatangani perjanjian bertulis untuk membeli syer Dingo. Kegagalan Linda untuk membayar seperti yang telah ditetapkan dalam
perjanjian yang dibuatnya membolehkan Linda dikenakan tindakan yang sah dan didakwa di mahkamah.
Rumusan.
1. Linda perlu menilai kemampuannya terlebih dahulu sebelum ingin memasuki pelbagai kontrak serta melihat had keperluannya sesuai dengan umur.
2. Linda perlu bertanggungjawab dan menanggung semua tindakan mahkamah atas perjanjian yang telah dibuat.
3. Linda tidak sepatutnya meninggalkan persekolahan terlalu awal kerana dia perlu menghabiskan pelajaran. Ini kerana pelajaran yang tinggi boleh membantu Linda mendapatkan pekerjaan yang lebih baik.[/spoiler]
[spoiler]The Company Constitution.
Introduction
The company’s constitution was previously divided between the Memorandum of Association (external aspects) and the Articles of Association (internal aspects, including the balance of power between the board of directors and the general meeting) where this pattern has continued under Companies Acts. This two constitution is for defining what kind of company is and how its business and affairs are to be conducted. Without this constitution, a new company cannot be registered and established to do a legal business in Malaysia under the Companies Acts 1965. The important and relevant of this constitution is to carrying out issues related to the company and all the people who work directly or indirectly in the top management.
1.0 The Memorandum of Association
Memorandum of Association is a document that lists written constitution. Section 18 of Companies Acts 1965 must fulfill with this requirement for registering the company:
1.1 The name of company
1.2 Registered office
1.3 Objectives of the company
1.4 The limited liability of members
1.5 The authorized share capital
1.6 Name, address, work and signature for memorandum.
2.0 The Articles of Association
A document which content information related with administration affairs and management of the company. In addition, the articles of association should have the issue
______________________________________________________________________________
Companies Acts 1965
Section 18 of Companies Acts 1965
and transfer of shares, alteration of capital structure, calling general meetings and voting right, appointment, powers and proceedings of directors, dividend, accounts and the issue of notices. Limited company with allowed share by Company Act 1965 is not necessary to register their Article; this matter is provided in section 29(1). However, section 29(1) must read together with section 30(2) for those company that chose to not register the Articles, and Table A (a set of completed rules) will be automatically used. The other features in Articles of Association are:
2.1 The Article should be printed and divided with numbered spacing and signed by each signer in front of at least two witnesses. (Section 29(2))
2.2 Company have right to chose a matter in Article Association.
2.3 Non limited companies which have share capital declare the share amount and divided it to share unit.
2.4 Non limited company and limited company in guarantee to share and have to declare members that suggested.
The Memorandum ends with a declaration that the subscribers wish to form the company pursuant to the Memorandum and agree to take the number of shares opposite their names. For all companies, public as well as private, it is now only necessary to have two subscribers each of who signs the Memorandum. The subscriber’s signatures are witnessed, usually by a single witness who signs as such. The Memorandum is dated and is treated as deed executed under seal (J.M. Gullick, Company Law 13rd edition, 1987). The Articles may as a general rule is altered simply by passing a special resolution. Clauses which could be included in the articles may be placed in the memorandum in order to make it more difficult to alter them. The memorandum and the articles of association will be described in the next page with description and a relevant case in order to understand of application for incorporated company.
______________________________________________________________________________
Section 29(1), Company Act 1965
Section 29(2), Company Act 1965
Section 30(2), Company Act 1965
1.0 The Memorandum of Association.
1.1 The name of company.
In Malaysia, Suruhanjaya Syarikat Malaysia is the registry for a new company that doing business inside Malaysia.
The name of the company serves to identify it and to distinguish it from any other company. For this reason, company which might mislead the public will be change automatically by the registrar that has statutory powers of control over the choice of names. The choice of name must end with the word ‘Sendirian Berhad’ if it is a public company or ‘Berhad’ if it is private limited company, unless permitted to omit ‘Berhad’ from its name an unlimited company does not have to end with ‘Berhad’.
Other rules that will count by the registry for name registration are:
a) No company may have same name which is already exist.
b) No criminal offence or considered offensive.
c) In registrar opinion, the name is like have connection with the government or local authority.
1.1.1 Changing name – In section 23 of Companies Act 1965, a company may change its name by passing a special resolution and obtaining the registrar’s certificate of incorporation that has registered the company under new name. But before the name has been change it should be discuss and approve in the general meeting of the company. In case Oshkosh B’Gosh, the same limitations as above apply to adoption of a name by change of name as by incorporation of a new company. The registrar also can compel to change the name of company if the name same as or in the registrar opinion too like the other company name or the name give misleading an indication of its activities as to be likely to cause harm to the public.
1.1.2 Business name – Most company carry on a commercial activity. Hence the law on company names may give rise to business identity problems of two kinds.
First, a business or professional activity if carried on successfully soon acquires ‘goodwill’.
___________________________________________________________________________ Section 23, Company Act 1965
Customer or client that works with the company's name will bring to the attention of other people, anyone who hears the name of the company with good reputation will put confidence in the company while providing a high confidence to do business. However, if there is a competition that leads to the suppression of business caused by a competitor who has a similar company name, then legal action can be taken. For example in the case Ewing Vs Buttercup Margarine Co Ltd (1917), The plaintiff, a sole trader, had network of retail shops in Scotland and north of England, through which is sold margarine and tea. He traded ‘the Buttercup Dairy Co.’. The company was form to carry on a business of wholesale margarine distributors in England, it’s included a power to trade as retailers. The plaintiff obtained an injunction to prevent the defendant from trading under its own name on the grounds that there might be confusion especially as the plaintiff planned to extend his business to the south of England.
Second, the sole trader or a partnership can use a name that represents the company by mutual consent, however, the sole trader or a partnership cannot use the information held by the companies represented. sole trader or a partnership must use the original information is held as the company address, registration number or a telephone line so that the parent company did not receive overlapping claims and the errors made by a sole trader or a partnership.
1.1.3 Public company name  a public company must have a special name that ends with 'public limited company', where in Malaysia it is called a 'Sendirian Berhad'. This principle is buttressed by rules which:
a) Prohibit any person who is not a public company from carrying on trade, profession or business under a name which includes the public company designation.
b) A public company is guilty of an offence if, in circumstance which the fact that it is a public company is likely to be material to any person, it uses a name which may reasonable be expected to give the impression that it is a private company. (Company Law, 13rd Edition, by J.M Gullick 1987)
1.2 Registered office.
Section 119, Company Act 1965, (1) A company shall as from the day on which it begin to carry on business or as from the fourteenth day after the date of its incorporation, whichever is the earlier, have a registered office within Malaysia to which all communications and notices may be addressed and which shall be open and accessible to the public for not less than three hours during ordinary business hours on each day, Saturdays, weekly and public holidays excepted. (2) If default is made in complying with subsection (1) the company and every officer of the company who is in default shall be guilty of an offence against this Act.
1.3 Objectives of the company.
Objectives in the memorandum of association are concerned with the type of business. It will become the company's guidelines on the proper direction in matters involving. Ordinarily, a company will have many objects clause and the area to enable the company to operate in various business activities. However, if the company conducts a business activity that is unrelated to the purpose of its establishment as contained in the objects clause, then, the company has acted contrary to the objects clause and the association has committed an ultra vires.
1.3.1 Development of Object Clause
In determining the objects clause, it should be towards a power for companies to control the flow of capital is used more effectively. The shareholders will see the use of capital has been invested with the interest of the employer or employee. However, there are some cases that show the establishment clause of the object beyond the object itself, for example in the case of Bell House Ltd. Vs City Wall Properties Ltd (1966), the object clause was to carry on the business of builders and developers. The third paragraph was ‘to carry on any other trade or business whatsoever’ which can in the opinion of the board of directors be advantageously carried on by the company in connection of the above businesses or the general business of
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Section 119(1) of the Company Act 1965
Section 119(2) of the Company Act 1965
the company. The companies also have power to turn to account any of its property and acquired to contact with Swiss bankers who had money to lend on mortgage for property development. The defendant sought the help from plaintiff to introduce defendant to the Swiss bankers and in return for an introduction, fee £20,000 will given by defendant. They later refused to pay because the contract was ultra vires the company and void since its objects did not include mortgage broking. Judge has decided that the fee was binding. The directors did genuinely believe that the effecting of an introduction for fee was a business which could be advantageously carried on as ancillary to the general development of the company. This is what the object clause required and permitted. It was also a case of turning to account an asset of the company.
1.3.2 Alteration of objects
Changing the object clause is under Section 21 of Companies Act 1965, the company can alter as long as Companies Act 1965 allows to do so. The object clause can be altered under Section 28 of Companies Act 1965. According to Section 28(1) of Companies Act 1965, the alteration of object clause can be done by special resolution. Section 28(2) of Companies Act 1965 requires the company to give 21 day notice of meeting to all the members. Section 28(3) of Companies Act 1965 states that notice must be given to all members of company about the alteration of object clause. According to Section 152 of Companies Act 1965, there must be at least 75% majority of the members of company to agree to change the object clause. Section 28(8) of Companies Act 1965 states those 21 days after passing the special resolution, any members of company can apply to object. Section 28(5) of Companies Act 1965 states that there must be at least 10% of issued share capital if they want to object. Section 181 of Companies Act 1965 states any minority, which is less than 10%, can object under this section. Under Section 28(9) of Companies Act 1965, must lodge with SSM within 14 days. Section 28(10) of Companies Act 1965 states the alteration of object clause takes effect when the resolution has been lodged.
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Section 21 of Companies Act 1965
Section 28(1) (2) (3) (5) (8) (9) (10) of Companies Act 1965
Section 152 of Companies Act 1965
Section 181 of Companies Act 1965
1.3.3 Ultra Vires
Ultra Vires may be defined as a situation where the company has conducted business activities beyond the limit and the terms in the memorandum of association. However, a company that has made an ultra vires transaction or arrangement that is not contained in the memorandum of association shall remain valid. These are the effect of the amendment laws under section 20 (1) of the Companies Act 1965 which states: “no such act or acts of the company will be revoked only by reason of the fact that the company has no ability or authority to act or perform or take such conveyance or transfer of even more than what is stipulated in the memorandum of association”.
Case Ashbury Railway Carriage & Iron Co. Ltd. Vs Riche (1875), the company had an object clause which stated that its objects were to make and sell, or lend on hire, railway carriages and wagons and all kinds of railway plant, fittings, machinery and rolling stock and to carry on business as mechanical engineers. The company bought a concession to build a railway in Belgium, subcontracting the work to defendant. Later the company repudiated the contract. Judge was deciding that, constructing a railway was not within the company’s object so the company did not have capacity to enter into either the concession contract or the subcontract. The contract was void for ultra vires and so the defendant had no right to damages for breach. The members could not ratify it and the company could neither enforce the contract nor be forced into performing its obligations.
Ultra vires issue can be raised by the company or other shareholders through the provision of sections that have been made in the Companies Act 1965 section 20 (2) (a) (b) and (c). Explanation of the issue of ultra virus provided in section 20 (2) (a) which provides “the proceedings of any member, or if the company has issued debentures secured by floating charge over all or any of its assets by holders of any debentures or trustee for debenture holders to prevent the commission of any act or acts or the conveyance or transfer any
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Section 20 (1) of Companies Act 1965
Section 20 (2) (a) of Company Act 1965
property to or by the company”. This means that a trustee or member may take action to restrict or prevent the company from taking any action ultra vires.
Case Hawkesbury Development Co. Ltd Vs Landmark Finance Pty. Ltd. (1969), plaintiff is the holder of all shares in Landmark Finance. It has demanded that the court declare the Landmark production of some debentures to finance United Dominion Corporation (UDC) is invalid because it was ultra vires. Plaintiff also asked the court to prohibit enforcement of UDC of the debentures. However, plaintiff's application is not successful, and failed to obtain court approval to void the declaration of the UDC. This is because, the plaintiffs are shareholders of the Landmark Finance. The application should be made to Landmark instead of UDC Finance is a third party.
1.4 The limited liability of members.
This clause states that the liability of the members of the company is limited. In the case of a company limited by shares, the member is liable only to the amount unpaid on the shares taken by members. In the case of a company limited by guarantee the members are liable to the amount undertaken to be contributed by them to the assets of the company in the event of its being wound up. However, this clause is omitted from the memorandum of association of unlimited companies. Any alteration in the memorandum compelling a member to take up more shares, or which increases his liability, would be null and void. If a company carries on business for more than six months, less than 2 members in case of a private company each member aware of this fact, is liable for all the debts contracted by the company after the period of six months has elapsed.
1.5 The authorized share capital.
Authorized share capital is the ceiling of capital available for issue. For example, if the company’s authorized capital is RM100, 000.00 divided into 100,000 ordinary share of RM1.00, that means the company can only increase its paid up capital to a maximum of RM100, 000.00. The company may increase its authorized capital anytime by holding an EGM (Extraordinary General Meeting). A fee is payable to the Companies Commission of Malaysia for the increase of authorized share capital.
1.6 Name, address, position and signature for memorandum.
Normal procedures to complete a memorandum of association as have to fill up for more information in company structure. Example:
Picture 1
2.0 The Articles of Association.
Articles of association are a document that was touched on in the internal affairs of a company that it provides information on how companies should conduct business. For example, it touches on the relationship between the shareholders of each other shareholder, the withdrawal, confiscation and transfer of shares, transfer of the share capital, the notice of the proceedings, the particulars of the appointment, powers, action directors, accounts, audit, dividend, loans and winding. Every company, except a company limited by shares must register the articles of association to the Registrar of Companies. This is consistent with the requirements under Section 29 of the Companies Act 1965. This Act also provides a set of examples of articles of association in Table A of the Schedule 4. Any company is free to imitate some/all of them. If a company limited by shares were not registered with the Registrar of the article itself, then Table A will apply automatically. (Refer to appendix, Sample of Article of Association)
2.1 Effect of Memorandum and Articles of Association
Section 33 (1) of the Companies Act 1965 provides:
"subject to this Act, the memorandum and articles shall when registered, according to the company and its members as if the memorandum and articles, each signed and sealed by each member and contained covenants on the part of each member to comply with all provisions of the memorandum and articles ".
From the provisions of this section, the memorandum and articles on the registration would create the effect of the contract. The effects are:
2.1.1 Contractual effect between the companies and member.
2.1.2 Contractual effect between members (members inter se).
2.1.3 Relationships between companies and outsiders.
______________________________________________________________________________ Section 29 of the Companies Act 1965
Section 33 (1) of the Companies Act 1965
2.1.1 Contractual effect between the companies and member.
Effects arising from the contract provisions of section 33 (1) of the Companies Act 1965 allows companies to enforce the provisions of the articles and memorandum of association of members. In other words, companies can take action against the member to ensure that they comply with the provisions in the articles and memorandum if the member refuses to follow the provisions voluntarily.
In case Hickman (H) Vs Kent (1915), the facts are as follows: a provision in the articles of a company provide that, "if there are any differences that exist between the association and any of its members, then such dispute shall be referred to arbitration." plaintiffs (H), is a members in these companies have taken legal action against the company relating to a dispute concerning his dismissal from the company. This is a dispute between the H at the position as a member of the company. The company has relied on the relevant articles and requested that the trial is terminated. The court decided the company was entitled to terminate the trial because the article was a contract between his company and H, and such difference shall be referred to arbitration as provided by the articles of association. The proceeding would be stayed since the dispute (which related to matters affecting H as a member) must, in conformity with the articles, be submitted to arbitration. In reviewing other cases the court held that if the matter does not affect members generally, if it is something confined to one member in his personal capacity, it is not question of members right and obligations and is not subject to the rule.
However, there are a few cases the effect of a contract excludes the application of section 33 (1) of the Companies Act 1965 to be valid by the position of other members. For example, the position as director, promoter or lawyer. This is because there is among individuals or parties who registered as a member of the company and at the same time they hold certain positions within the company. If members want to enforce the provisions of the articles, giving the position of other members of the claim will not succeed.
In case Beattie Vs E & F Beattie Ltd (1938), the article also provided for arbitration on disputes. The company sued its managing directors to recover money improperly paid to him.
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Section 33 (1) of the Companies Act 1965
The managing director applied to have the action stayed as he contended that the dispute should be submitted to arbitration. The court judge that the claim was against the defendant as a director and the arbitration clause was limited to disputes with members as such. He could not rely on the arbitration clause. Provisions of section 33 (1), is enshrined in the company and its members only. It will not be valid to a third party even if the third party has held a position in the company and manage the affairs of the company. This matter can be looked in Eley Vs Positive Government Security Life Assurance Co. (1876), In this case, Eley is a drafter of articles and the articles of association have provided:
"that Eley is a lawyer and the company only can terminated him for misconduct"
Eley acting as company lawyer for some time though there is no separate employment contract signing. Eley also receive a distribution of shares in the company in return for work done during the establishment of the company. Eley is then fired by company as a company lawyer. Eley claimed the company had breached the contract. The Court held, the article does not give rights to the member if the member seeking to enforce their rights in a position other than a member. In this case, Eley try to enforce the provisions of articles that give it rights in the company's position as a lawyer. If Eley wish to do so, he should sign a contract other than as provided in the articles of association.
In case Re New British Iron Co, ex parte Beckwith (1898) showed about separated contract with the company. In this case, the articles provided that the remuneration of the directors should be £1,000 per annum to be divided between them as they saw fit. By accepting office the directors entered into separate contracts between themselves and the company but nothing more was said about their remuneration. When the company went into liquidation the directors claimed arrears of fees. The liquidator denied that there was any contract to pay the fees. The court judge, although the directors could not rely on the articles as a contract for the payment of their fees, they could refer to the articles to establish the amount payable under the separate contracts with the company made when they accepted office as directors.
2.1.2 Contractual effect between members (members inter se).
Every member is bound to the other members through an article. There's disagreement on how the law is. What seemed clear was that if the article provided to a member of a right alone so he could enforce those rights without the help of members of the company. It can also impose a contract on the members in their dealings with each other as illustrated by the case below. To avoid doubt and difficulty it is usual however to draft articles, especially on members right of first refusal of other members shares, so that each stage is dealing between the company and the members, to which Section 33(1) clearly applies. So that, a member who intends to transfer his shares must, if the articles so require, give notice of his intention to the company and the company must then give notice to other members that they have an option to take up his shares.
In the case of Rayfield v Hands (1960), articles of association of a company alone has allocated:
"Every member of the type that wishes to transfer shares shall notify the director and they will take the same share many of them with a reasonable price.”
Pursuant to the provisions of this article, the plaintiff was trying to force three directors of the company to buy its shares. But all three refused to buy its shares. The Court held that the relationship created by the articles that are binding and director of the company as between members and nonmembers and as between a member and director. With that the three directors are bound to purchase shares of the plaintiff.
The effect also taken to be more complex when a share holder is still want to hold it share but have to follow the rules or article. Take a look in Malaysian case, In the case of Malaysia, Wong Kim Fatt lwn Leong & Co. Sdn Bhd (1976), provisions of the articles of association of limited companies have been allocated as follows: holders of the 7 / 10 of the issued capital of the company may at any time make a request to the company to enforce the transfer of any shares certain that they are not held by the requester. In this case the company has 2 shareholders, A and B where A is a minority shareholder who holds 50,000 shares. B also holds 250,000 shares and the more he is the holder from 7 / 10 of its issued capital. B had made an application for get a share based on the provisions of the article, but a refuses to sell and claimed in court that he was entitled to continue holding its shares. Court decided that it is purely contractual obligations and the claim must be tied to contractual obligations he has received. The question of whether fair or not a provision of the article, is not relevant in determine whether the article could be enforced or not. Rights and dependents under article is a contractual obligation.
2.1.3 Relationships between companies and outsiders.
Articles do not involve contracts with outsiders. This is because outsiders are not members and cannot enforce any alleged rights derived from the memorandum or articles of the company. Look at the example of the case opposite the Raffles Hotel Ltd Malayan Banking Bhd (1996). Malayan Banking is the site of the lessor Raffles Hotel. Things in the article had stated that the lessor has power to appoint a director. Malayan Banking has appointed himself as the director led the Raffles Hotel has applied for a declaration that the appointment was illegal. The court ruled invalid because the appointment of Malayan Banking is not a member of the company and the power held within the article mentions only Malayan Banking authorities to appoint a director.
However, relations with people outside the company should be viewed in a broader context. There are also cases which do not deny the power of the article to be applied into the relationship. As an example the case of Southern foundries (1926) Ltd. Vs Shirlaw (1940) state that the article was to authorize the trustees of the late director to appoint a director of the company. The trustee had made an appointment but was challenged by several members. The Court held that the appointment of a trustee is legal. This action confirms the appointment may be more affected by the responsibility and it should be seen as justice for the rights of the more significant.
2.2 Alteration of the Articles.
Section 31 (1) provides:
'Subject to this act and to such terms in the memorandum, a company may be through a special resolution to amend or add to the articles'
Clear from the provisions of this section that company has the right to amend or add to the articles of association. However, an amendment to articles is only possible through a special resolution received 75% majority vote and the amendment is binding even though there are members who do not give consent. While section 31 (2) also provides for any modifications or additions to the articles of association shall be deemed valid as if it were originally contained in the articles of association. If all members of the company agree to an alteration without passing the general meeting or special resolution, this is still an effective under ‘assent principle’. It can be looked over Cane Vs Jones (1980) where the shareholders signed a written agreement by which the chairman was deprived of the casting vote at general meeting given to him by the articles. No special resolution was passed to make this alteration. Later one group of shareholders contended that the articles had not been altered and so the chairman still had casting vote since special resolution was the only permissible method of altering the articles. The court judge that a special resolution was the means by which majority of members could alter the articles so that all members become bound by the alteration. But it is also a principle of company law that all the members acting together maybe their unanimous agreement take a binding decision on any matter for which a majority of votes at a general meeting is required to bind a minority. The alteration was affective and the chairman had lost his casting vote.
This is about the winning of a majority in their voting, but if there have some unfair decision that make the minority felt injustice, section 181 (1) can be use to protect the rights.
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Section 31 (1) of the Companies Act 1965
Section 31 (2) of the Companies Act 1965
Section 181 (1) of the Companies Act 1965
This section states in paragraphs 1 and 2 are split into several branches of a particular section. They are:
Section 181 (1): Any member or holder of debentures of the company, the minister may apply to the court for an order under this section on the ground:
(a): that the company's affairs are conducted or the powers of directors are carried out in a manner oppressive to one or more members or holders of debentures including himself or herself or does not take into account their interests as members, shareholders or holders of debentures of the company, or
Section 181 (2): If the court considers the application of one of the reasons could be proved:
(a): directing or preventing any act or cancel or vary any transaction or resolution;
(b): manage the affairs of the company's actions in the future;
(c): provides for the purchase of shares or debentures of the company by other members or debenture holders of the company or the company itself;
(d): in case of purchase of shares by the company, it provides for deduction of the share capital of the company, or
(e): provides that the company can be deposed.
2.2.1 Limits for the amendment article
There are some acts that limit the amendments in the article. The Acts are:
Section 33 (3): Any amendment which requires stockholders to take the shares or increase their liability to contribute to the capital of the company is not allowed. Unless the shareholder got written consent from members before and after the amendment of the articles.
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Section 33 (3) of the Companies Act 1965
Section 181 (1) (a) of the Companies Act 1965
Section 181 (2) (a) (b) (c) (d) (e) of the Companies Act 1965
Section 181 (4): Shareholders or debenture holders are allowed to apply for a remedy on the grounds of oppression prevailing in the company. Upon application, the court has the power to issue such an order to amend the memorandum or articles of association. If the court had issued a revised order of the articles, the company cannot make changes other than those permitted by the court.
Section 65 (1): For a company having a share capital divided into several classes of shares and have different rights between the classes of shares, the class of rights amendment is not allowed except in the manner provided for in clause variation, the clause that provides a way to amend the right according to the classes of shares. Freedom to amend the articles is limited when several classes of shares issued and any amendments to the rights granted to the class of shares must follow the procedures set out in clause variations.
The principle of ‘Majority Rule’: When a company wants to amend the articles of association, the principle of 'majority rule' will apply. Through this principle, amendments to the articles could be enforced upon approval of 3 / 4 or 75% majority of the company and the decision is binding on the minority. In this case, the majority of members can add to or amend the provisions of the article. However, to prevent abuse of this principle, amendments to articles by the majority has a limit of the amendment must be made by Bona Fide, which is in good faith for the benefit of the company as a whole. Therefore, the amendment of articles by the majority cannot be enforced if it is not made in good faith for the good of the company as a whole.
Bona Fide can be looked over case of Allen Vs Gold Reefs of West Africa Ltd (1900), under the articles in their original form the company had a lien on ‘all shares (not being fully paid shares) of members’. All the members held partly paid shares, to which the lien attached, but only Z also held fully paid shares to which the lien in that form did
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Section 65 (1) of the Companies Act 1965
Section 181 (4) of the Companies Act 1965
not attach. A special resolution was passed to remove from the articles the words ‘not being paid shares’. As a result the company would then have a lien on the fully paid shares of Z to secure the amount owing on his partly paid shares. The position of other members remained the same since the lien in its original form already attached to their partly paid shares. The executors of Z (who had died) objected that the alteration was discriminatory since it only altered the position of one member (Z) and that it was retrospective in its effect since Z had acquired his fully paid shares (free of the lien) before he was allotted partly paid shares and so incurred the debt for which the lien, extended to his fully paid shares, was security. The judge held that, the alteration was for the benefit of the company as a whole and could apply to any member who might hold fully paid shares. As regards the retrospective effect a member held his shares subject to the memorandum and articles of association and his shareholder’s right were therefore subject to any alteration which might be made affecting them. The alteration was therefore valid.
2.3 Methods of internal affairs (the principle of agency in the company)
This principle highlights the rights agents appointed to represent something or someone in the business of the company. Powers of agents is divided into three, namely:
2.3.1 Express Authority
Real power is something that is expressly given to these agents. For example, someone who had been appointed by the company and given a power to issuing and receiving payments on behalf of the company's debt or borrow money and give security against the debt. The responsibility is clearly within the jurisdiction of a director of the company.
2.3.2 Implied Authority
Implied powers not expressly mentioned, and it is subject to state authority. It exists in two states, namely:
i) Power to make matters incidental to the matters expressly authorized.
ii) Power to do things normally done by an agent in such a position.
For example, an agent has been appointed to a personal manager to a company, it is given real power. However, when the agent advertising to call the candidates to attend an interview and took a successful candidate to work in a company then it is binds to the company, this task is a task usually done by a manager.
2.3.3 Ostensible Authority
One situation where the law will infer the creation of an agency by implied appointment is when the principal, by his words or conduct, creates an inference among third parties that the agent has the actual authority to contract even though no actual authority was actually given to him. This situation may arise when the principal allows the agent to order goods and services on his behalf and he proceeds to pay for them. In such a scenario, if the agent contracts within the limits of the ostensible (apparent) authority, although without any actual authority, the principal will be bound to third parties by his agent’s acts.
In case Chew Hock San Vs Connaught Housing Development SDN. BHD (1985), defendant is the developer. The plaintiff entered into an agreement to buy several houses a particular store from the defendant. The plaintiff has to pay the deposit to the clerk of the defendant to the book shop though at the time of booking that has not been opened. Defendant refused to comply with the contract and the plaintiff's claim for specific performance. The Court ruled that the clerk has no authority to take actions such deposits, and therefore there is no representation from the company to be bound by the acts the defendant. Thus, plaintiff has failed to claim its actions.
However, there is an exception in this ostensible authorities to block the company were causing the lack of power and the doctrine of estopel. Circumstances in which the doctrine of estopel / restrictions apply are:
1) There is a representation by the company to a state that the agent has the authority to take action in question.
 Representation of this can arise in several ways: first, an agent with apparent authority to do all such acts as agent in the usual position. Second, a principal by his actions in a certain time period would represent an implicit: his agent is authorized to sign contracts on behalf of himself.
 In case of Freeman & Lockyer Vs Buckhurst Park Properties (Mangal) Ltd. (1964), A property development company which was established by two men and one named Kapoor. Kapoor acted as managing director although never elected to that position, but the other directors are satisfied with the way management Kapoor. Then, Kapoor had hired the plaintiffs to post an architect and land surveyor. Company refuses to pay the plaintiff as alleged Kapoor has no such power. The Court ruled, Kapoor has ostensible authority, the board of directors know that Kapoor has acted as managing director and appoint an agent for the company. With the permission and that they were, he has the power to enter into ordinary contracts signed by a manager. The plaintiff was motivated by a sense of trust that Kapoor has been authorized by the representation to perform services for the company. So, the company has estopel from denying the power of Kapoor and companies have to pay the plaintiff.
2) Representations must be made by a person having authority to make representations on behalf of the company.
 To make these companies are bound, these representations must be made by a person who has actual authority to make such representations.
 In case Armagas Ltd. Vs Mundogas (1986), Mundogas is the owner of "Ocean Frost 'is a shipping company. A consultation has been done to sell the ships to the Armada Group and it was carried out by representatives of the parties, Magelssen. Representatives of the Armada Group only wants to hire the vessel for a period of 3 years and Magelssen not have the power to hire only sales. Magelssen agree to the lease agreement with a false representation that he has the power to enter into the lease contract. Armagas Ltd was incorporated to a ship and Magelssen had completed the charter agreement for the Mundogas for 3 years. To cheat Mundogas, Magelssen has provided the rental agreement for 12 months for the ship of Armagas Ltd. Thereafter, Mundogas decided not to proceed with charter breach caused Armagas claimed promise. Armagas Magelssen believes is authorized to manage the rental of the mundogas. The Court ruled that the application is rejected because Magelssen not have real power to complete the rental for 3 years. Magelssen has no authority to make representations that he was authorized to do so. Misrepresentation made to the powers is not binding Mundogas.
3) People who want to enforce the contract against the company must have relied on these representations.
 Estopel would only arise if the contractor is dependent on the representation that the agent has the authority to complete the transaction for the parties. If the contractor does not know the agent has the power, then he would not rely on the
representation and estopel will not arise. Thus, knowledge of the power of agents is necessary to avoid problems arising if there fraud or offense committed without the knowledge of real company.
2.4 Duties as responsible from company to agent
Duties of an Agent to his Principal:
1. To obey the principal’s instructions
2. In the absence of instructions from the principal. To act according to the prevailing customs of that business
3. To exercise care and diligence in carrying out his work and to use such skill as he possesses
4. To render proper accounts when required
5. To pay to his principal all sums received on his behalf
6. To communicate with the principal
7. Not to let his own interest conflict with his duty
8. Not to make secret profits out of the performance of his duty.
9. Not to disclose confidential information or documents entrusted to him by the principal
10. Not to delegate his authority
Duties of Principal to his Agent
1. To pay the agent the commission or other agreed remuneration unless the agency relationship is gratuitous
2. Not to willfully prevent or hinder the agent from earning his commission
3. To indemnify and reimburse the agent for acts done in the exercise of his duties
2.5 Methods of internal affairs (Indoor Management Rule)
If an agent had apparent authority to do something, then someone who is dealing with a company is entitled to presume that all matters of internal management and procedures set by the company have been complied with. Indoor management rule sometimes can be known as the rule of Turquand case. It is necessary to begin with the case itself, Royal British Bank Vs Turquand (1856), under the company’s deed of settlement (corresponding with the modern articles) the directors had power to borrow only such sums as had been authorized by general resolution of the company. A debenture for £2,000 had been issued under the seal, signed by two directors and by the secretary. The resolution passed in general meeting to sanction this borrowing did not specify the amount which might be borrowed. It was therefore no authority to the directors to borrow anything. The bank had not seen the resolution and had no knowledge of its inadequacy. But it was held to be entitled to assume that the authority prescribed by the articles, which requirement it had constructive notice, had been effectually given by ordinary resolution, in example by internal procedure of which the bank had neither actual nor constructive notice.
As an 'outsider', the bank is not able to find out whether ordinary resolution was passed because of this information does not typically file in the registry. In this case, the bank has the right to assume that a resolution has been approved legally. The principle of 'internal affairs rule' that sets a third party dealing with the right to make assumptions (if they have no knowledge that the company's internal regulations required by the articles of association have been complied with).
2.5.1 Exclusion
There are limitations on the ability of a party to rely on the internal management rules. First, if the contracting party knew or should have known that the agent has no power, then the contracting parties cannot rely on this method. Typically, these cases occur when the contract is "insider” of his position, should have known about the disability. For example in case Howard Vs Patent Ivory Manufacturing Co. (1888), Director of the company has lent money to companies based on the guarantee of the debentures. Article states that the company can only borrow up to the limit and has exceeded a certain limit. The director wants to enforce the debentures. The Court held that it refuses to allow them to do so. All this is because, as the directors they know or should know about borrowing limits. Thus, they cannot rely on the internal management rules. Second, if there are situations where a reasonable person would make the suspect feeling the power of the agent. So the contracting parties must make reasonable inquiries. If these questions indicate the agent has no authority, contracting parties cannot rely on internal management tool to help him. In case Morris Vs Kanssen (1946) lord of house has saying that:
“Person contracting with a company and dealing in good faith may assume that acts within its constitution and powers have been properly and duly performed and are not bound to inquire whether acts of internal management have been regular”
Later, the judge has pointed out the rationale of the rule ‘The wheels of business will not go smoothly round unless it may be assumed that is in order which appears to be in order’. Third, if the review of the memorandum or articles of association of the company clearly shows that the agent's authority is limited, so the contract cannot rely on the internal management rules.
Lawton and BrownieWilkinson LJJ delivered concurring opinions, “the presumption of regularity cannot be relied on by ‘inside’. In example, persons who by virtue of their position in the company are in a position to know whether or not the internal regulation have been observed”. (page 211, Cases and Material in Company Law, L.S. Seally [1985])
For the conclusion, these two documents are important for determining the direction of the company. This document has clearly touched a lot about management and regulatory compliance by directors, shareholders and even employees who worked in the company. The importance of the two documents is to bring the birth of new companies registered to continue to drive the mission and vision for the establishment of better economic importance.
The end
SOALAN ASSIGNMENT:
a) Nyatakan sebabsebab yang diberikan oleh mahkamah untuk justifikasi
mengenai penerimaan dengan surat dihantar melalui pos?
b) Pada tahun lepas, Linda telah meninggalkan sekolah semasa beliau berumur 16 tahun. Dia telah berkerja dalam bidang hotel sebagai pelatih(trainee) pembantu dapur. Pendapatan Linda adalah 50 pounds seminggu, dan seperti yang diperuntukkan oleh hotel dia harus memberi notis selama 3 bulan jika ingin menamatkan pekerjaannya.
Barubaru ini Linda telah bersetuju untuk membeli motosikal “ Osaka” sebab beliau ingin meluangkan masa bersama kawan baiknya Malcolm, yang turut berminat terhadap motosikal. Linda juga telah menandatangani perjanjian bertulis untuk membeli syer ( satu perempat) yang dikenali sebagai syer “Dingo”.
Linda kini telah ditawarkan pekerjaan sebagai tukang masak di restauran dengan gaji
sebanyak 100 pounds seminggu, dengan syarat beliau boleh memulakan
tugasnya dengan sertamerta. Dia telah gagal untuk membayar untuk harga motosikal
dan bahagian dalam syer Dingo.
Nasihatkan Linda berdasarkan Akta Kontrak yang relevan dan keskes yang
berkaitan/diputuskan oleh mahkamah
Jawapan soalan a).
Undangundang kontrak dalam seksyen 4(2) Akta Kontrak 1950 ada menyatakan bahawa komunikasi penerimaan adalah lengkap terhadap pencadang apabila komunikasi penerimaan itu telah dimasukkan kedalam peti surat untuk perjalanan surat pengirim kepadanya, di mana ia adalah di luar kuasa penerima. Kes Entores Ltd. lawan Miles Far East corporation (1955) menyatakan bahawa, “apabila suatu kontrak dibuat melalui pos, undangundang dalam semua negeri Common Law jelas bahawa penerimaan adalah lengkap selepas surat itu dimasukkan ke dalam peti surat dan itulah tempatnya kontrak telah dibuat”.
Undangundang ini juga adalah untuk mewujudkan kesahihan terhadap kontrak sama ada boleh dikuatkuasakan oleh undangundang atau pun tidak. Kesahihan undangundang itu perlu bagi mengesahkan sesuatu kontrak sedang berjalan atau pun tidak, kerana pencadang boleh membuat pembatalan kontrak sebelum tarikh luput kontrak tersebut tanpa mengetahui jika terdapat penerimaan yang sedang dalam perjalanan. Sebagai contoh Kes Byrne lawan Van Tienhoven, defendan telah menawarkan akan menjual 1000 kotak tin bersadur timah kepada plaintif melalui surat pada 1 Oktober, pada 8 Oktober defendan telah membuat pembatalan cadangan yang dibuat pada 1 Oktober. Surat pembatalan tiba kepada plaintif pada 20 Oktober. Namun plaintif telah menghantar utusan penerimaannya pada 11 Oktober dan mengesahkannya pula pada 15 Oktober. Mahkamah memutuskan bahawa kontrak telah berjalan kerana pembatalan tidak berkuatkuasa sehingga 20 Oktober, iaitu apabila sampai ke pengetahuan plaintif sedangkan plaintif telah pun menerima cadangan tersebut pada 15 Oktober.
Bagi penerima pula, sesuatu kontrak melalui komunikasi penerimaan pos akan berlaku apabila pencadang mendapat maklum balas positif untuk melanjutkan sesuatu kontrak yang dicadangkan. Mengambil contoh dalam kes Ignatius lawan Bell (1913), plaintif ingin menuntut pelaksanaan spesifik mengenai perjanjian opsyen yang mengikut plaintif memberinya pilihan membeli hakhak defendan keatas sebidang tanah. Plaintif menghantar notis penerimaannya melalui surat berdaftar pada 16 Ogos, tetapi surat tidak sampai kepada pengetahuan defendan sehingga petang 25 Ogos kerana defendan tiada di rumah. Mahkamah telah memutuskan bahawa opsyen tersebut adalah sah dipergunakan oleh plaintif apabila surat tersebut sudah diposkan pada 16 Ogos. Pengesahan kontrak tersebut telah berjalan dikuatkan lagi dengan niat penerima iaitu plaintif untuk berkontrak dengan mendaftar surat tersebut.
Walaubagaimana pun, sesuatu kontrak yang melalui komunikasi penerimaan pos mempunyai akta tersendiri dan cara tertentu untuk dibatalkan. Seksyen 5(2) Akta Kontrak 1950 menyatakan sesuatu penerimaan boleh dibatalkan pada bilabila masa sebelum komunikasi penerimaan lengkap terhadap penerima. Antara cara pembatalan kontrak melalui pos adalah:
a) Mengkomunikasikan notis pembatalan,
b) Luputnya waktu yang munasabah,
c) Kemungkiran penerima berkenaan syarat kontrak
d) Meninggal dunia ataupun sakit otak.
Kita mengambil contoh pembatalan luputnya waktu yang munasabah. Kes yang berkaitan menjelaskan keadaan ini adalah Ramsgate Victoria Hotel Co. Ltd lawan Montefiore (1866), Defendan telah memohon suatu saham dalam syarikat hotel tersebut pada bulan Jun. Tetapi hanya bulan November, syarikat tersebut memberitahu sahamsaham itu telah diagihkan dan diberi. Maka, pemegang saham perlu menjelaskan baki harga belian dimana defendan enggan membayarnya. Mahkamah telah memutuskan bahawa keengganan defendan menjelaskan baki harga belian tersebut sah kerana maklumat saham tersebut sepatutnya diketahui oleh defendan dalam masa yang munasabah bukan mengambil masa antara Jun dan November.
Mahkamah menetapkan justifikasi undangundang seperti ini adalah untuk menjaga hak penerima kontrak selain daripada hak pencadang. Kontrak melalui pos adalah suatu kontrak yang memerlukan persefahaman antara keduadua pihak, kerana ia melibatkan masa dan tindak balas daripada pencadang mahu pun penerima. Namun, hak antara penerima dan pencadang cadangan ini mempunyai kriteria tertentu untuk dipatuhi supaya keduadua belah pihak mendapat hak keadilan yang seimbang.
\
Jawapan soalan b).
Akta Dewasa 1971 menyatakan bahawa, seseorang dikatakan dewasa apabila mencapai umur 18 tahun dan layak untuk membuat kontrak. Namun, umur dewasa boleh berbeza untuk tujuan lain sebagaimana ditetapkan oleh undangundang lain. Contohnya, untuk tujuan mengundi dalam pilihan raya negeri, umur dewasa adalah 21 tahun sementara untuk kontrak perantisan dan ahli keanggotaan kesatuan sekerja adalah 16 tahun. Ini bermakna Linda yang berumur 16 tahun boleh membuat kontrak untuk tujuan pekerjaan.
Isu.
1. Adakah Linda perlu memberi notis jika ingin menamatkan pekerjaannya?
2. Adakah sah untuk Linda memasuki kontrak pembelian motosikal dan syer “Dingo”?
3. Bolehkah Linda menerima tawaran pekerjaan sebagai tukang masak?
4. Kegagalan Linda membayar harga motosikal dan syer “Dingo”.
Penyelesaian.
Isu 1.
Ya. Linda perlu memberi notis jika ingin berhenti daripada kerjanya sebagai pelatih pembantu dapur. Ini kerana pihak hotel telah memperuntukkan notis tersebut sebagai peraturan pekerja yang ingin berhenti.
Isu 2.
Kanakkanak tidak boleh memasuki kontrak pembelian motosikal mahupun pembelian untuk syer kerana kontrak tersebut tidak termasuk didalam senarai kekecualian kepada kanakkanak. Tetapi dalam kes Linda, dia telah berniat untuk memasuki kontrak kerana mempunyai kerja sebagai pelatih pembantu dapur dengan pendapatan £50 seminggu. Seksyen 69 (d) menyatakan, kanakkanak akan bertanggungan hanya sekiranya mempunyai harta. Sebagai contoh dalam kes Doyle lawan White City Stadium Ltd (1935), seorang kanakkanak yang merupakan peninju professional telah bersetuju untuk mematuhi peraturan Lembaga Kawalan Tinju British(LKTB) dalam urusan profesionnya ketika menerima lesen dari LKTB. Sejumlah wang telah dipegang oleh LKTB seperti yang diperuntukkan dalam peraturan LKTB kerana kanakkanak tersebut telah melanggar peraturan yang ditetapkan. Mahkamah memutuskan kontrak diantara kanakkanak tersebut dengan LKTB adalah sah dan mengikat. Kes ini serupa dengan kesahihan Linda dengan niat untuk berkontrak dan mengikat janji dalam pembelian motosikal serta pembelian syer yang dikenali sebagai syer “Dingo” dan Linda adalah terikat dengan perjanjian komersial tersebut.
Isu 3.
Linda boleh menerima tawaran tersebut yang menawarkan gaji £100 seminggu. Ini kerana pekerjaan tersebut boleh membantu Linda untuk memenuhi dan menanggung keperluan hidupnya. Walaubagaimana pun, Linda perlu mengemukakan notis selama 3 bulan kepada majikan terdahulu yang merupakan syarat yang telah diperuntukkan sebelum dia menerima pekerjaan sebagai tukang masak di restoran.
Isu 4.
Kegagalan Linda untuk membayar harga motosikal dan bahagian dalam syer Dingo merupakan suatu kesalahan kerana perjanjian yang telah dibuat mesti dipatuhi manakala perjanjian tersebut adalah sah dan mengikat dengan niat serta boleh dikuatkuasakan selari dengan peruntukan seksyen 2(h) Akta Kontrak 1950 iaitu sesuatu perjanjian yang boleh dikuatkuasakan oleh undangundang dan seksyen 10(1) Akta Kontrak 1950 yang menyatakan, sesuatu perjanjian adalah kontrak jika dibuat atas kerelaan bebas pihakpihak yang layak berkontrak, untuk sesuatu balasan yang sah, dan dengan suatu tujuan yang sah, dan tidak ditetapkan dengan nyata dibawah peruntukan akta ini bahawa ianya batal. Dalam kes Koh Kia Hong lawan Guo Enterprise Pte. Ltd (1990), pihak plaintif membuat tawaran untuk membeli hartanah yang diiklankan oleh defendan. Pihak plaintif telah menyerahkan deposit kepada pihak defendan. Satu nota juga telah ditandatangani oleh keduadua pihak dan resit rasmi telah dikeluarkan oleh defendan. Mahkamah memutuskan bahawa keduadua pihak mempunyai niat untuk mengikat diri dalam satu kontrak. Kes tersebut sama dengan tindakan Linda yang telah bersetuju untuk membeli sebuah motosikal dan turut menandatangani perjanjian bertulis untuk membeli syer Dingo. Kegagalan Linda untuk membayar seperti yang telah ditetapkan dalam
perjanjian yang dibuatnya membolehkan Linda dikenakan tindakan yang sah dan didakwa di mahkamah.
Rumusan.
1. Linda perlu menilai kemampuannya terlebih dahulu sebelum ingin memasuki pelbagai kontrak serta melihat had keperluannya sesuai dengan umur.
2. Linda perlu bertanggungjawab dan menanggung semua tindakan mahkamah atas perjanjian yang telah dibuat.
3. Linda tidak sepatutnya meninggalkan persekolahan terlalu awal kerana dia perlu menghabiskan pelajaran. Ini kerana pelajaran yang tinggi boleh membantu Linda mendapatkan pekerjaan yang lebih baik.[/spoiler]

rockoz  Teman Setia SF
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Re: Nenek Ani
budu baini si Muhidin... [spoiler]kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
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kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:[/spoiler]
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:[/spoiler]
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[spoiler]Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:[/spoiler]
[spoiler]Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:[/spoiler]

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Re: Nenek Ani
[spoiler]kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:[/spoiler]
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:[/spoiler]
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[spoiler]The Company Constitution.
Introduction
The company’s constitution was previously divided between the Memorandum of Association (external aspects) and the Articles of Association (internal aspects, including the balance of power between the board of directors and the general meeting) where this pattern has continued under Companies Acts. This two constitution is for defining what kind of company is and how its business and affairs are to be conducted. Without this constitution, a new company cannot be registered and established to do a legal business in Malaysia under the Companies Acts 1965. The important and relevant of this constitution is to carrying out issues related to the company and all the people who work directly or indirectly in the top management.
1.0 The Memorandum of Association
Memorandum of Association is a document that lists written constitution. Section 18 of Companies Acts 1965 must fulfill with this requirement for registering the company:
1.1 The name of company
1.2 Registered office
1.3 Objectives of the company
1.4 The limited liability of members
1.5 The authorized share capital
1.6 Name, address, work and signature for memorandum.
2.0 The Articles of Association
A document which content information related with administration affairs and management of the company. In addition, the articles of association should have the issue
______________________________________________________________________________
Companies Acts 1965
Section 18 of Companies Acts 1965
and transfer of shares, alteration of capital structure, calling general meetings and voting right, appointment, powers and proceedings of directors, dividend, accounts and the issue of notices. Limited company with allowed share by Company Act 1965 is not necessary to register their Article; this matter is provided in section 29(1). However, section 29(1) must read together with section 30(2) for those company that chose to not register the Articles, and Table A (a set of completed rules) will be automatically used. The other features in Articles of Association are:
2.1 The Article should be printed and divided with numbered spacing and signed by each signer in front of at least two witnesses. (Section 29(2))
2.2 Company have right to chose a matter in Article Association.
2.3 Non limited companies which have share capital declare the share amount and divided it to share unit.
2.4 Non limited company and limited company in guarantee to share and have to declare members that suggested.
The Memorandum ends with a declaration that the subscribers wish to form the company pursuant to the Memorandum and agree to take the number of shares opposite their names. For all companies, public as well as private, it is now only necessary to have two subscribers each of who signs the Memorandum. The subscriber’s signatures are witnessed, usually by a single witness who signs as such. The Memorandum is dated and is treated as deed executed under seal (J.M. Gullick, Company Law 13rd edition, 1987). The Articles may as a general rule is altered simply by passing a special resolution. Clauses which could be included in the articles may be placed in the memorandum in order to make it more difficult to alter them. The memorandum and the articles of association will be described in the next page with description and a relevant case in order to understand of application for incorporated company.
______________________________________________________________________________
Section 29(1), Company Act 1965
Section 29(2), Company Act 1965
Section 30(2), Company Act 1965
1.0 The Memorandum of Association.
1.1 The name of company.
In Malaysia, Suruhanjaya Syarikat Malaysia is the registry for a new company that doing business inside Malaysia.
The name of the company serves to identify it and to distinguish it from any other company. For this reason, company which might mislead the public will be change automatically by the registrar that has statutory powers of control over the choice of names. The choice of name must end with the word ‘Sendirian Berhad’ if it is a public company or ‘Berhad’ if it is private limited company, unless permitted to omit ‘Berhad’ from its name an unlimited company does not have to end with ‘Berhad’.
Other rules that will count by the registry for name registration are:
a) No company may have same name which is already exist.
b) No criminal offence or considered offensive.
c) In registrar opinion, the name is like have connection with the government or local authority.
1.1.1 Changing name – In section 23 of Companies Act 1965, a company may change its name by passing a special resolution and obtaining the registrar’s certificate of incorporation that has registered the company under new name. But before the name has been change it should be discuss and approve in the general meeting of the company. In case Oshkosh B’Gosh, the same limitations as above apply to adoption of a name by change of name as by incorporation of a new company. The registrar also can compel to change the name of company if the name same as or in the registrar opinion too like the other company name or the name give misleading an indication of its activities as to be likely to cause harm to the public.
1.1.2 Business name – Most company carry on a commercial activity. Hence the law on company names may give rise to business identity problems of two kinds.
First, a business or professional activity if carried on successfully soon acquires ‘goodwill’.
___________________________________________________________________________ Section 23, Company Act 1965
Customer or client that works with the company's name will bring to the attention of other people, anyone who hears the name of the company with good reputation will put confidence in the company while providing a high confidence to do business. However, if there is a competition that leads to the suppression of business caused by a competitor who has a similar company name, then legal action can be taken. For example in the case Ewing Vs Buttercup Margarine Co Ltd (1917), The plaintiff, a sole trader, had network of retail shops in Scotland and north of England, through which is sold margarine and tea. He traded ‘the Buttercup Dairy Co.’. The company was form to carry on a business of wholesale margarine distributors in England, it’s included a power to trade as retailers. The plaintiff obtained an injunction to prevent the defendant from trading under its own name on the grounds that there might be confusion especially as the plaintiff planned to extend his business to the south of England.
Second, the sole trader or a partnership can use a name that represents the company by mutual consent, however, the sole trader or a partnership cannot use the information held by the companies represented. sole trader or a partnership must use the original information is held as the company address, registration number or a telephone line so that the parent company did not receive overlapping claims and the errors made by a sole trader or a partnership.
1.1.3 Public company name  a public company must have a special name that ends with 'public limited company', where in Malaysia it is called a 'Sendirian Berhad'. This principle is buttressed by rules which:
a) Prohibit any person who is not a public company from carrying on trade, profession or business under a name which includes the public company designation.
b) A public company is guilty of an offence if, in circumstance which the fact that it is a public company is likely to be material to any person, it uses a name which may reasonable be expected to give the impression that it is a private company. (Company Law, 13rd Edition, by J.M Gullick 1987)
1.2 Registered office.
Section 119, Company Act 1965, (1) A company shall as from the day on which it begin to carry on business or as from the fourteenth day after the date of its incorporation, whichever is the earlier, have a registered office within Malaysia to which all communications and notices may be addressed and which shall be open and accessible to the public for not less than three hours during ordinary business hours on each day, Saturdays, weekly and public holidays excepted. (2) If default is made in complying with subsection (1) the company and every officer of the company who is in default shall be guilty of an offence against this Act.
1.3 Objectives of the company.
Objectives in the memorandum of association are concerned with the type of business. It will become the company's guidelines on the proper direction in matters involving. Ordinarily, a company will have many objects clause and the area to enable the company to operate in various business activities. However, if the company conducts a business activity that is unrelated to the purpose of its establishment as contained in the objects clause, then, the company has acted contrary to the objects clause and the association has committed an ultra vires.
1.3.1 Development of Object Clause
In determining the objects clause, it should be towards a power for companies to control the flow of capital is used more effectively. The shareholders will see the use of capital has been invested with the interest of the employer or employee. However, there are some cases that show the establishment clause of the object beyond the object itself, for example in the case of Bell House Ltd. Vs City Wall Properties Ltd (1966), the object clause was to carry on the business of builders and developers. The third paragraph was ‘to carry on any other trade or business whatsoever’ which can in the opinion of the board of directors be advantageously carried on by the company in connection of the above businesses or the general business of
___________________________________________________________________________
Section 119(1) of the Company Act 1965
Section 119(2) of the Company Act 1965
the company. The companies also have power to turn to account any of its property and acquired to contact with Swiss bankers who had money to lend on mortgage for property development. The defendant sought the help from plaintiff to introduce defendant to the Swiss bankers and in return for an introduction, fee £20,000 will given by defendant. They later refused to pay because the contract was ultra vires the company and void since its objects did not include mortgage broking. Judge has decided that the fee was binding. The directors did genuinely believe that the effecting of an introduction for fee was a business which could be advantageously carried on as ancillary to the general development of the company. This is what the object clause required and permitted. It was also a case of turning to account an asset of the company.
1.3.2 Alteration of objects
Changing the object clause is under Section 21 of Companies Act 1965, the company can alter as long as Companies Act 1965 allows to do so. The object clause can be altered under Section 28 of Companies Act 1965. According to Section 28(1) of Companies Act 1965, the alteration of object clause can be done by special resolution. Section 28(2) of Companies Act 1965 requires the company to give 21 day notice of meeting to all the members. Section 28(3) of Companies Act 1965 states that notice must be given to all members of company about the alteration of object clause. According to Section 152 of Companies Act 1965, there must be at least 75% majority of the members of company to agree to change the object clause. Section 28(8) of Companies Act 1965 states those 21 days after passing the special resolution, any members of company can apply to object. Section 28(5) of Companies Act 1965 states that there must be at least 10% of issued share capital if they want to object. Section 181 of Companies Act 1965 states any minority, which is less than 10%, can object under this section. Under Section 28(9) of Companies Act 1965, must lodge with SSM within 14 days. Section 28(10) of Companies Act 1965 states the alteration of object clause takes effect when the resolution has been lodged.
___________________________________________________________________________
Section 21 of Companies Act 1965
Section 28(1) (2) (3) (5) (8) (9) (10) of Companies Act 1965
Section 152 of Companies Act 1965
Section 181 of Companies Act 1965
1.3.3 Ultra Vires
Ultra Vires may be defined as a situation where the company has conducted business activities beyond the limit and the terms in the memorandum of association. However, a company that has made an ultra vires transaction or arrangement that is not contained in the memorandum of association shall remain valid. These are the effect of the amendment laws under section 20 (1) of the Companies Act 1965 which states: “no such act or acts of the company will be revoked only by reason of the fact that the company has no ability or authority to act or perform or take such conveyance or transfer of even more than what is stipulated in the memorandum of association”.
Case Ashbury Railway Carriage & Iron Co. Ltd. Vs Riche (1875), the company had an object clause which stated that its objects were to make and sell, or lend on hire, railway carriages and wagons and all kinds of railway plant, fittings, machinery and rolling stock and to carry on business as mechanical engineers. The company bought a concession to build a railway in Belgium, subcontracting the work to defendant. Later the company repudiated the contract. Judge was deciding that, constructing a railway was not within the company’s object so the company did not have capacity to enter into either the concession contract or the subcontract. The contract was void for ultra vires and so the defendant had no right to damages for breach. The members could not ratify it and the company could neither enforce the contract nor be forced into performing its obligations.
Ultra vires issue can be raised by the company or other shareholders through the provision of sections that have been made in the Companies Act 1965 section 20 (2) (a) (b) and (c). Explanation of the issue of ultra virus provided in section 20 (2) (a) which provides “the proceedings of any member, or if the company has issued debentures secured by floating charge over all or any of its assets by holders of any debentures or trustee for debenture holders to prevent the commission of any act or acts or the conveyance or transfer any
___________________________________________________________________________
Section 20 (1) of Companies Act 1965
Section 20 (2) (a) of Company Act 1965
property to or by the company”. This means that a trustee or member may take action to restrict or prevent the company from taking any action ultra vires.
Case Hawkesbury Development Co. Ltd Vs Landmark Finance Pty. Ltd. (1969), plaintiff is the holder of all shares in Landmark Finance. It has demanded that the court declare the Landmark production of some debentures to finance United Dominion Corporation (UDC) is invalid because it was ultra vires. Plaintiff also asked the court to prohibit enforcement of UDC of the debentures. However, plaintiff's application is not successful, and failed to obtain court approval to void the declaration of the UDC. This is because, the plaintiffs are shareholders of the Landmark Finance. The application should be made to Landmark instead of UDC Finance is a third party.
1.4 The limited liability of members.
This clause states that the liability of the members of the company is limited. In the case of a company limited by shares, the member is liable only to the amount unpaid on the shares taken by members. In the case of a company limited by guarantee the members are liable to the amount undertaken to be contributed by them to the assets of the company in the event of its being wound up. However, this clause is omitted from the memorandum of association of unlimited companies. Any alteration in the memorandum compelling a member to take up more shares, or which increases his liability, would be null and void. If a company carries on business for more than six months, less than 2 members in case of a private company each member aware of this fact, is liable for all the debts contracted by the company after the period of six months has elapsed.
1.5 The authorized share capital.
Authorized share capital is the ceiling of capital available for issue. For example, if the company’s authorized capital is RM100, 000.00 divided into 100,000 ordinary share of RM1.00, that means the company can only increase its paid up capital to a maximum of RM100, 000.00. The company may increase its authorized capital anytime by holding an EGM (Extraordinary General Meeting). A fee is payable to the Companies Commission of Malaysia for the increase of authorized share capital.
1.6 Name, address, position and signature for memorandum.
Normal procedures to complete a memorandum of association as have to fill up for more information in company structure. Example:
Picture 1
2.0 The Articles of Association.
Articles of association are a document that was touched on in the internal affairs of a company that it provides information on how companies should conduct business. For example, it touches on the relationship between the shareholders of each other shareholder, the withdrawal, confiscation and transfer of shares, transfer of the share capital, the notice of the proceedings, the particulars of the appointment, powers, action directors, accounts, audit, dividend, loans and winding. Every company, except a company limited by shares must register the articles of association to the Registrar of Companies. This is consistent with the requirements under Section 29 of the Companies Act 1965. This Act also provides a set of examples of articles of association in Table A of the Schedule 4. Any company is free to imitate some/all of them. If a company limited by shares were not registered with the Registrar of the article itself, then Table A will apply automatically. (Refer to appendix, Sample of Article of Association)
2.1 Effect of Memorandum and Articles of Association
Section 33 (1) of the Companies Act 1965 provides:
"subject to this Act, the memorandum and articles shall when registered, according to the company and its members as if the memorandum and articles, each signed and sealed by each member and contained covenants on the part of each member to comply with all provisions of the memorandum and articles ".
From the provisions of this section, the memorandum and articles on the registration would create the effect of the contract. The effects are:
2.1.1 Contractual effect between the companies and member.
2.1.2 Contractual effect between members (members inter se).
2.1.3 Relationships between companies and outsiders.
______________________________________________________________________________ Section 29 of the Companies Act 1965
Section 33 (1) of the Companies Act 1965
2.1.1 Contractual effect between the companies and member.
Effects arising from the contract provisions of section 33 (1) of the Companies Act 1965 allows companies to enforce the provisions of the articles and memorandum of association of members. In other words, companies can take action against the member to ensure that they comply with the provisions in the articles and memorandum if the member refuses to follow the provisions voluntarily.
In case Hickman (H) Vs Kent (1915), the facts are as follows: a provision in the articles of a company provide that, "if there are any differences that exist between the association and any of its members, then such dispute shall be referred to arbitration." plaintiffs (H), is a members in these companies have taken legal action against the company relating to a dispute concerning his dismissal from the company. This is a dispute between the H at the position as a member of the company. The company has relied on the relevant articles and requested that the trial is terminated. The court decided the company was entitled to terminate the trial because the article was a contract between his company and H, and such difference shall be referred to arbitration as provided by the articles of association. The proceeding would be stayed since the dispute (which related to matters affecting H as a member) must, in conformity with the articles, be submitted to arbitration. In reviewing other cases the court held that if the matter does not affect members generally, if it is something confined to one member in his personal capacity, it is not question of members right and obligations and is not subject to the rule.
However, there are a few cases the effect of a contract excludes the application of section 33 (1) of the Companies Act 1965 to be valid by the position of other members. For example, the position as director, promoter or lawyer. This is because there is among individuals or parties who registered as a member of the company and at the same time they hold certain positions within the company. If members want to enforce the provisions of the articles, giving the position of other members of the claim will not succeed.
In case Beattie Vs E & F Beattie Ltd (1938), the article also provided for arbitration on disputes. The company sued its managing directors to recover money improperly paid to him.
___________________________________________________________________________
Section 33 (1) of the Companies Act 1965
The managing director applied to have the action stayed as he contended that the dispute should be submitted to arbitration. The court judge that the claim was against the defendant as a director and the arbitration clause was limited to disputes with members as such. He could not rely on the arbitration clause. Provisions of section 33 (1), is enshrined in the company and its members only. It will not be valid to a third party even if the third party has held a position in the company and manage the affairs of the company. This matter can be looked in Eley Vs Positive Government Security Life Assurance Co. (1876), In this case, Eley is a drafter of articles and the articles of association have provided:
"that Eley is a lawyer and the company only can terminated him for misconduct"
Eley acting as company lawyer for some time though there is no separate employment contract signing. Eley also receive a distribution of shares in the company in return for work done during the establishment of the company. Eley is then fired by company as a company lawyer. Eley claimed the company had breached the contract. The Court held, the article does not give rights to the member if the member seeking to enforce their rights in a position other than a member. In this case, Eley try to enforce the provisions of articles that give it rights in the company's position as a lawyer. If Eley wish to do so, he should sign a contract other than as provided in the articles of association.
In case Re New British Iron Co, ex parte Beckwith (1898) showed about separated contract with the company. In this case, the articles provided that the remuneration of the directors should be £1,000 per annum to be divided between them as they saw fit. By accepting office the directors entered into separate contracts between themselves and the company but nothing more was said about their remuneration. When the company went into liquidation the directors claimed arrears of fees. The liquidator denied that there was any contract to pay the fees. The court judge, although the directors could not rely on the articles as a contract for the payment of their fees, they could refer to the articles to establish the amount payable under the separate contracts with the company made when they accepted office as directors.
2.1.2 Contractual effect between members (members inter se).
Every member is bound to the other members through an article. There's disagreement on how the law is. What seemed clear was that if the article provided to a member of a right alone so he could enforce those rights without the help of members of the company. It can also impose a contract on the members in their dealings with each other as illustrated by the case below. To avoid doubt and difficulty it is usual however to draft articles, especially on members right of first refusal of other members shares, so that each stage is dealing between the company and the members, to which Section 33(1) clearly applies. So that, a member who intends to transfer his shares must, if the articles so require, give notice of his intention to the company and the company must then give notice to other members that they have an option to take up his shares.
In the case of Rayfield v Hands (1960), articles of association of a company alone has allocated:
"Every member of the type that wishes to transfer shares shall notify the director and they will take the same share many of them with a reasonable price.”
Pursuant to the provisions of this article, the plaintiff was trying to force three directors of the company to buy its shares. But all three refused to buy its shares. The Court held that the relationship created by the articles that are binding and director of the company as between members and nonmembers and as between a member and director. With that the three directors are bound to purchase shares of the plaintiff.
The effect also taken to be more complex when a share holder is still want to hold it share but have to follow the rules or article. Take a look in Malaysian case, In the case of Malaysia, Wong Kim Fatt lwn Leong & Co. Sdn Bhd (1976), provisions of the articles of association of limited companies have been allocated as follows: holders of the 7 / 10 of the issued capital of the company may at any time make a request to the company to enforce the transfer of any shares certain that they are not held by the requester. In this case the company has 2 shareholders, A and B where A is a minority shareholder who holds 50,000 shares. B also holds 250,000 shares and the more he is the holder from 7 / 10 of its issued capital. B had made an application for get a share based on the provisions of the article, but a refuses to sell and claimed in court that he was entitled to continue holding its shares. Court decided that it is purely contractual obligations and the claim must be tied to contractual obligations he has received. The question of whether fair or not a provision of the article, is not relevant in determine whether the article could be enforced or not. Rights and dependents under article is a contractual obligation.
2.1.3 Relationships between companies and outsiders.
Articles do not involve contracts with outsiders. This is because outsiders are not members and cannot enforce any alleged rights derived from the memorandum or articles of the company. Look at the example of the case opposite the Raffles Hotel Ltd Malayan Banking Bhd (1996). Malayan Banking is the site of the lessor Raffles Hotel. Things in the article had stated that the lessor has power to appoint a director. Malayan Banking has appointed himself as the director led the Raffles Hotel has applied for a declaration that the appointment was illegal. The court ruled invalid because the appointment of Malayan Banking is not a member of the company and the power held within the article mentions only Malayan Banking authorities to appoint a director.
However, relations with people outside the company should be viewed in a broader context. There are also cases which do not deny the power of the article to be applied into the relationship. As an example the case of Southern foundries (1926) Ltd. Vs Shirlaw (1940) state that the article was to authorize the trustees of the late director to appoint a director of the company. The trustee had made an appointment but was challenged by several members. The Court held that the appointment of a trustee is legal. This action confirms the appointment may be more affected by the responsibility and it should be seen as justice for the rights of the more significant.
2.2 Alteration of the Articles.
Section 31 (1) provides:
'Subject to this act and to such terms in the memorandum, a company may be through a special resolution to amend or add to the articles'
Clear from the provisions of this section that company has the right to amend or add to the articles of association. However, an amendment to articles is only possible through a special resolution received 75% majority vote and the amendment is binding even though there are members who do not give consent. While section 31 (2) also provides for any modifications or additions to the articles of association shall be deemed valid as if it were originally contained in the articles of association. If all members of the company agree to an alteration without passing the general meeting or special resolution, this is still an effective under ‘assent principle’. It can be looked over Cane Vs Jones (1980) where the shareholders signed a written agreement by which the chairman was deprived of the casting vote at general meeting given to him by the articles. No special resolution was passed to make this alteration. Later one group of shareholders contended that the articles had not been altered and so the chairman still had casting vote since special resolution was the only permissible method of altering the articles. The court judge that a special resolution was the means by which majority of members could alter the articles so that all members become bound by the alteration. But it is also a principle of company law that all the members acting together maybe their unanimous agreement take a binding decision on any matter for which a majority of votes at a general meeting is required to bind a minority. The alteration was affective and the chairman had lost his casting vote.
This is about the winning of a majority in their voting, but if there have some unfair decision that make the minority felt injustice, section 181 (1) can be use to protect the rights.
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Section 31 (1) of the Companies Act 1965
Section 31 (2) of the Companies Act 1965
Section 181 (1) of the Companies Act 1965
This section states in paragraphs 1 and 2 are split into several branches of a particular section. They are:
Section 181 (1): Any member or holder of debentures of the company, the minister may apply to the court for an order under this section on the ground:
(a): that the company's affairs are conducted or the powers of directors are carried out in a manner oppressive to one or more members or holders of debentures including himself or herself or does not take into account their interests as members, shareholders or holders of debentures of the company, or
Section 181 (2): If the court considers the application of one of the reasons could be proved:
(a): directing or preventing any act or cancel or vary any transaction or resolution;
(b): manage the affairs of the company's actions in the future;
(c): provides for the purchase of shares or debentures of the company by other members or debenture holders of the company or the company itself;
(d): in case of purchase of shares by the company, it provides for deduction of the share capital of the company, or
(e): provides that the company can be deposed.
2.2.1 Limits for the amendment article
There are some acts that limit the amendments in the article. The Acts are:
Section 33 (3): Any amendment which requires stockholders to take the shares or increase their liability to contribute to the capital of the company is not allowed. Unless the shareholder got written consent from members before and after the amendment of the articles.
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Section 33 (3) of the Companies Act 1965
Section 181 (1) (a) of the Companies Act 1965
Section 181 (2) (a) (b) (c) (d) (e) of the Companies Act 1965
Section 181 (4): Shareholders or debenture holders are allowed to apply for a remedy on the grounds of oppression prevailing in the company. Upon application, the court has the power to issue such an order to amend the memorandum or articles of association. If the court had issued a revised order of the articles, the company cannot make changes other than those permitted by the court.
Section 65 (1): For a company having a share capital divided into several classes of shares and have different rights between the classes of shares, the class of rights amendment is not allowed except in the manner provided for in clause variation, the clause that provides a way to amend the right according to the classes of shares. Freedom to amend the articles is limited when several classes of shares issued and any amendments to the rights granted to the class of shares must follow the procedures set out in clause variations.
The principle of ‘Majority Rule’: When a company wants to amend the articles of association, the principle of 'majority rule' will apply. Through this principle, amendments to the articles could be enforced upon approval of 3 / 4 or 75% majority of the company and the decision is binding on the minority. In this case, the majority of members can add to or amend the provisions of the article. However, to prevent abuse of this principle, amendments to articles by the majority has a limit of the amendment must be made by Bona Fide, which is in good faith for the benefit of the company as a whole. Therefore, the amendment of articles by the majority cannot be enforced if it is not made in good faith for the good of the company as a whole.
Bona Fide can be looked over case of Allen Vs Gold Reefs of West Africa Ltd (1900), under the articles in their original form the company had a lien on ‘all shares (not being fully paid shares) of members’. All the members held partly paid shares, to which the lien attached, but only Z also held fully paid shares to which the lien in that form did
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Section 65 (1) of the Companies Act 1965
Section 181 (4) of the Companies Act 1965
not attach. A special resolution was passed to remove from the articles the words ‘not being paid shares’. As a result the company would then have a lien on the fully paid shares of Z to secure the amount owing on his partly paid shares. The position of other members remained the same since the lien in its original form already attached to their partly paid shares. The executors of Z (who had died) objected that the alteration was discriminatory since it only altered the position of one member (Z) and that it was retrospective in its effect since Z had acquired his fully paid shares (free of the lien) before he was allotted partly paid shares and so incurred the debt for which the lien, extended to his fully paid shares, was security. The judge held that, the alteration was for the benefit of the company as a whole and could apply to any member who might hold fully paid shares. As regards the retrospective effect a member held his shares subject to the memorandum and articles of association and his shareholder’s right were therefore subject to any alteration which might be made affecting them. The alteration was therefore valid.
2.3 Methods of internal affairs (the principle of agency in the company)
This principle highlights the rights agents appointed to represent something or someone in the business of the company. Powers of agents is divided into three, namely:
2.3.1 Express Authority
Real power is something that is expressly given to these agents. For example, someone who had been appointed by the company and given a power to issuing and receiving payments on behalf of the company's debt or borrow money and give security against the debt. The responsibility is clearly within the jurisdiction of a director of the company.
2.3.2 Implied Authority
Implied powers not expressly mentioned, and it is subject to state authority. It exists in two states, namely:
i) Power to make matters incidental to the matters expressly authorized.
ii) Power to do things normally done by an agent in such a position.
For example, an agent has been appointed to a personal manager to a company, it is given real power. However, when the agent advertising to call the candidates to attend an interview and took a successful candidate to work in a company then it is binds to the company, this task is a task usually done by a manager.
2.3.3 Ostensible Authority
One situation where the law will infer the creation of an agency by implied appointment is when the principal, by his words or conduct, creates an inference among third parties that the agent has the actual authority to contract even though no actual authority was actually given to him. This situation may arise when the principal allows the agent to order goods and services on his behalf and he proceeds to pay for them. In such a scenario, if the agent contracts within the limits of the ostensible (apparent) authority, although without any actual authority, the principal will be bound to third parties by his agent’s acts.
In case Chew Hock San Vs Connaught Housing Development SDN. BHD (1985), defendant is the developer. The plaintiff entered into an agreement to buy several houses a particular store from the defendant. The plaintiff has to pay the deposit to the clerk of the defendant to the book shop though at the time of booking that has not been opened. Defendant refused to comply with the contract and the plaintiff's claim for specific performance. The Court ruled that the clerk has no authority to take actions such deposits, and therefore there is no representation from the company to be bound by the acts the defendant. Thus, plaintiff has failed to claim its actions.
However, there is an exception in this ostensible authorities to block the company were causing the lack of power and the doctrine of estopel. Circumstances in which the doctrine of estopel / restrictions apply are:
1) There is a representation by the company to a state that the agent has the authority to take action in question.
 Representation of this can arise in several ways: first, an agent with apparent authority to do all such acts as agent in the usual position. Second, a principal by his actions in a certain time period would represent an implicit: his agent is authorized to sign contracts on behalf of himself.
 In case of Freeman & Lockyer Vs Buckhurst Park Properties (Mangal) Ltd. (1964), A property development company which was established by two men and one named Kapoor. Kapoor acted as managing director although never elected to that position, but the other directors are satisfied with the way management Kapoor. Then, Kapoor had hired the plaintiffs to post an architect and land surveyor. Company refuses to pay the plaintiff as alleged Kapoor has no such power. The Court ruled, Kapoor has ostensible authority, the board of directors know that Kapoor has acted as managing director and appoint an agent for the company. With the permission and that they were, he has the power to enter into ordinary contracts signed by a manager. The plaintiff was motivated by a sense of trust that Kapoor has been authorized by the representation to perform services for the company. So, the company has estopel from denying the power of Kapoor and companies have to pay the plaintiff.
2) Representations must be made by a person having authority to make representations on behalf of the company.
 To make these companies are bound, these representations must be made by a person who has actual authority to make such representations.
 In case Armagas Ltd. Vs Mundogas (1986), Mundogas is the owner of "Ocean Frost 'is a shipping company. A consultation has been done to sell the ships to the Armada Group and it was carried out by representatives of the parties, Magelssen. Representatives of the Armada Group only wants to hire the vessel for a period of 3 years and Magelssen not have the power to hire only sales. Magelssen agree to the lease agreement with a false representation that he has the power to enter into the lease contract. Armagas Ltd was incorporated to a ship and Magelssen had completed the charter agreement for the Mundogas for 3 years. To cheat Mundogas, Magelssen has provided the rental agreement for 12 months for the ship of Armagas Ltd. Thereafter, Mundogas decided not to proceed with charter breach caused Armagas claimed promise. Armagas Magelssen believes is authorized to manage the rental of the mundogas. The Court ruled that the application is rejected because Magelssen not have real power to complete the rental for 3 years. Magelssen has no authority to make representations that he was authorized to do so. Misrepresentation made to the powers is not binding Mundogas.
3) People who want to enforce the contract against the company must have relied on these representations.
 Estopel would only arise if the contractor is dependent on the representation that the agent has the authority to complete the transaction for the parties. If the contractor does not know the agent has the power, then he would not rely on the
representation and estopel will not arise. Thus, knowledge of the power of agents is necessary to avoid problems arising if there fraud or offense committed without the knowledge of real company.
2.4 Duties as responsible from company to agent
Duties of an Agent to his Principal:
1. To obey the principal’s instructions
2. In the absence of instructions from the principal. To act according to the prevailing customs of that business
3. To exercise care and diligence in carrying out his work and to use such skill as he possesses
4. To render proper accounts when required
5. To pay to his principal all sums received on his behalf
6. To communicate with the principal
7. Not to let his own interest conflict with his duty
8. Not to make secret profits out of the performance of his duty.
9. Not to disclose confidential information or documents entrusted to him by the principal
10. Not to delegate his authority
Duties of Principal to his Agent
1. To pay the agent the commission or other agreed remuneration unless the agency relationship is gratuitous
2. Not to willfully prevent or hinder the agent from earning his commission
3. To indemnify and reimburse the agent for acts done in the exercise of his duties
2.5 Methods of internal affairs (Indoor Management Rule)
If an agent had apparent authority to do something, then someone who is dealing with a company is entitled to presume that all matters of internal management and procedures set by the company have been complied with. Indoor management rule sometimes can be known as the rule of Turquand case. It is necessary to begin with the case itself, Royal British Bank Vs Turquand (1856), under the company’s deed of settlement (corresponding with the modern articles) the directors had power to borrow only such sums as had been authorized by general resolution of the company. A debenture for £2,000 had been issued under the seal, signed by two directors and by the secretary. The resolution passed in general meeting to sanction this borrowing did not specify the amount which might be borrowed. It was therefore no authority to the directors to borrow anything. The bank had not seen the resolution and had no knowledge of its inadequacy. But it was held to be entitled to assume that the authority prescribed by the articles, which requirement it had constructive notice, had been effectually given by ordinary resolution, in example by internal procedure of which the bank had neither actual nor constructive notice.
As an 'outsider', the bank is not able to find out whether ordinary resolution was passed because of this information does not typically file in the registry. In this case, the bank has the right to assume that a resolution has been approved legally. The principle of 'internal affairs rule' that sets a third party dealing with the right to make assumptions (if they have no knowledge that the company's internal regulations required by the articles of association have been complied with).
2.5.1 Exclusion
There are limitations on the ability of a party to rely on the internal management rules. First, if the contracting party knew or should have known that the agent has no power, then the contracting parties cannot rely on this method. Typically, these cases occur when the contract is "insider” of his position, should have known about the disability. For example in case Howard Vs Patent Ivory Manufacturing Co. (1888), Director of the company has lent money to companies based on the guarantee of the debentures. Article states that the company can only borrow up to the limit and has exceeded a certain limit. The director wants to enforce the debentures. The Court held that it refuses to allow them to do so. All this is because, as the directors they know or should know about borrowing limits. Thus, they cannot rely on the internal management rules. Second, if there are situations where a reasonable person would make the suspect feeling the power of the agent. So the contracting parties must make reasonable inquiries. If these questions indicate the agent has no authority, contracting parties cannot rely on internal management tool to help him. In case Morris Vs Kanssen (1946) lord of house has saying that:
“Person contracting with a company and dealing in good faith may assume that acts within its constitution and powers have been properly and duly performed and are not bound to inquire whether acts of internal management have been regular”
Later, the judge has pointed out the rationale of the rule ‘The wheels of business will not go smoothly round unless it may be assumed that is in order which appears to be in order’. Third, if the review of the memorandum or articles of association of the company clearly shows that the agent's authority is limited, so the contract cannot rely on the internal management rules.
Lawton and BrownieWilkinson LJJ delivered concurring opinions, “the presumption of regularity cannot be relied on by ‘inside’. In example, persons who by virtue of their position in the company are in a position to know whether or not the internal regulation have been observed”. (page 211, Cases and Material in Company Law, L.S. Seally [1985])
For the conclusion, these two documents are important for determining the direction of the company. This document has clearly touched a lot about management and regulatory compliance by directors, shareholders and even employees who worked in the company. The importance of the two documents is to bring the birth of new companies registered to continue to drive the mission and vision for the establishment of better economic importance.
The end
SOALAN ASSIGNMENT:
a) Nyatakan sebabsebab yang diberikan oleh mahkamah untuk justifikasi
mengenai penerimaan dengan surat dihantar melalui pos?
b) Pada tahun lepas, Linda telah meninggalkan sekolah semasa beliau berumur 16 tahun. Dia telah berkerja dalam bidang hotel sebagai pelatih(trainee) pembantu dapur. Pendapatan Linda adalah 50 pounds seminggu, dan seperti yang diperuntukkan oleh hotel dia harus memberi notis selama 3 bulan jika ingin menamatkan pekerjaannya.
Barubaru ini Linda telah bersetuju untuk membeli motosikal “ Osaka” sebab beliau ingin meluangkan masa bersama kawan baiknya Malcolm, yang turut berminat terhadap motosikal. Linda juga telah menandatangani perjanjian bertulis untuk membeli syer ( satu perempat) yang dikenali sebagai syer “Dingo”.
Linda kini telah ditawarkan pekerjaan sebagai tukang masak di restauran dengan gaji
sebanyak 100 pounds seminggu, dengan syarat beliau boleh memulakan
tugasnya dengan sertamerta. Dia telah gagal untuk membayar untuk harga motosikal
dan bahagian dalam syer Dingo.
Nasihatkan Linda berdasarkan Akta Kontrak yang relevan dan keskes yang
berkaitan/diputuskan oleh mahkamah
Jawapan soalan a).
Undangundang kontrak dalam seksyen 4(2) Akta Kontrak 1950 ada menyatakan bahawa komunikasi penerimaan adalah lengkap terhadap pencadang apabila komunikasi penerimaan itu telah dimasukkan kedalam peti surat untuk perjalanan surat pengirim kepadanya, di mana ia adalah di luar kuasa penerima. Kes Entores Ltd. lawan Miles Far East corporation (1955) menyatakan bahawa, “apabila suatu kontrak dibuat melalui pos, undangundang dalam semua negeri Common Law jelas bahawa penerimaan adalah lengkap selepas surat itu dimasukkan ke dalam peti surat dan itulah tempatnya kontrak telah dibuat”.
Undangundang ini juga adalah untuk mewujudkan kesahihan terhadap kontrak sama ada boleh dikuatkuasakan oleh undangundang atau pun tidak. Kesahihan undangundang itu perlu bagi mengesahkan sesuatu kontrak sedang berjalan atau pun tidak, kerana pencadang boleh membuat pembatalan kontrak sebelum tarikh luput kontrak tersebut tanpa mengetahui jika terdapat penerimaan yang sedang dalam perjalanan. Sebagai contoh Kes Byrne lawan Van Tienhoven, defendan telah menawarkan akan menjual 1000 kotak tin bersadur timah kepada plaintif melalui surat pada 1 Oktober, pada 8 Oktober defendan telah membuat pembatalan cadangan yang dibuat pada 1 Oktober. Surat pembatalan tiba kepada plaintif pada 20 Oktober. Namun plaintif telah menghantar utusan penerimaannya pada 11 Oktober dan mengesahkannya pula pada 15 Oktober. Mahkamah memutuskan bahawa kontrak telah berjalan kerana pembatalan tidak berkuatkuasa sehingga 20 Oktober, iaitu apabila sampai ke pengetahuan plaintif sedangkan plaintif telah pun menerima cadangan tersebut pada 15 Oktober.
Bagi penerima pula, sesuatu kontrak melalui komunikasi penerimaan pos akan berlaku apabila pencadang mendapat maklum balas positif untuk melanjutkan sesuatu kontrak yang dicadangkan. Mengambil contoh dalam kes Ignatius lawan Bell (1913), plaintif ingin menuntut pelaksanaan spesifik mengenai perjanjian opsyen yang mengikut plaintif memberinya pilihan membeli hakhak defendan keatas sebidang tanah. Plaintif menghantar notis penerimaannya melalui surat berdaftar pada 16 Ogos, tetapi surat tidak sampai kepada pengetahuan defendan sehingga petang 25 Ogos kerana defendan tiada di rumah. Mahkamah telah memutuskan bahawa opsyen tersebut adalah sah dipergunakan oleh plaintif apabila surat tersebut sudah diposkan pada 16 Ogos. Pengesahan kontrak tersebut telah berjalan dikuatkan lagi dengan niat penerima iaitu plaintif untuk berkontrak dengan mendaftar surat tersebut.
Walaubagaimana pun, sesuatu kontrak yang melalui komunikasi penerimaan pos mempunyai akta tersendiri dan cara tertentu untuk dibatalkan. Seksyen 5(2) Akta Kontrak 1950 menyatakan sesuatu penerimaan boleh dibatalkan pada bilabila masa sebelum komunikasi penerimaan lengkap terhadap penerima. Antara cara pembatalan kontrak melalui pos adalah:
a) Mengkomunikasikan notis pembatalan,
b) Luputnya waktu yang munasabah,
c) Kemungkiran penerima berkenaan syarat kontrak
d) Meninggal dunia ataupun sakit otak.
Kita mengambil contoh pembatalan luputnya waktu yang munasabah. Kes yang berkaitan menjelaskan keadaan ini adalah Ramsgate Victoria Hotel Co. Ltd lawan Montefiore (1866), Defendan telah memohon suatu saham dalam syarikat hotel tersebut pada bulan Jun. Tetapi hanya bulan November, syarikat tersebut memberitahu sahamsaham itu telah diagihkan dan diberi. Maka, pemegang saham perlu menjelaskan baki harga belian dimana defendan enggan membayarnya. Mahkamah telah memutuskan bahawa keengganan defendan menjelaskan baki harga belian tersebut sah kerana maklumat saham tersebut sepatutnya diketahui oleh defendan dalam masa yang munasabah bukan mengambil masa antara Jun dan November.
Mahkamah menetapkan justifikasi undangundang seperti ini adalah untuk menjaga hak penerima kontrak selain daripada hak pencadang. Kontrak melalui pos adalah suatu kontrak yang memerlukan persefahaman antara keduadua pihak, kerana ia melibatkan masa dan tindak balas daripada pencadang mahu pun penerima. Namun, hak antara penerima dan pencadang cadangan ini mempunyai kriteria tertentu untuk dipatuhi supaya keduadua belah pihak mendapat hak keadilan yang seimbang.
\
Jawapan soalan b).
Akta Dewasa 1971 menyatakan bahawa, seseorang dikatakan dewasa apabila mencapai umur 18 tahun dan layak untuk membuat kontrak. Namun, umur dewasa boleh berbeza untuk tujuan lain sebagaimana ditetapkan oleh undangundang lain. Contohnya, untuk tujuan mengundi dalam pilihan raya negeri, umur dewasa adalah 21 tahun sementara untuk kontrak perantisan dan ahli keanggotaan kesatuan sekerja adalah 16 tahun. Ini bermakna Linda yang berumur 16 tahun boleh membuat kontrak untuk tujuan pekerjaan.
Isu.
1. Adakah Linda perlu memberi notis jika ingin menamatkan pekerjaannya?
2. Adakah sah untuk Linda memasuki kontrak pembelian motosikal dan syer “Dingo”?
3. Bolehkah Linda menerima tawaran pekerjaan sebagai tukang masak?
4. Kegagalan Linda membayar harga motosikal dan syer “Dingo”.
Penyelesaian.
Isu 1.
Ya. Linda perlu memberi notis jika ingin berhenti daripada kerjanya sebagai pelatih pembantu dapur. Ini kerana pihak hotel telah memperuntukkan notis tersebut sebagai peraturan pekerja yang ingin berhenti.
Isu 2.
Kanakkanak tidak boleh memasuki kontrak pembelian motosikal mahupun pembelian untuk syer kerana kontrak tersebut tidak termasuk didalam senarai kekecualian kepada kanakkanak. Tetapi dalam kes Linda, dia telah berniat untuk memasuki kontrak kerana mempunyai kerja sebagai pelatih pembantu dapur dengan pendapatan £50 seminggu. Seksyen 69 (d) menyatakan, kanakkanak akan bertanggungan hanya sekiranya mempunyai harta. Sebagai contoh dalam kes Doyle lawan White City Stadium Ltd (1935), seorang kanakkanak yang merupakan peninju professional telah bersetuju untuk mematuhi peraturan Lembaga Kawalan Tinju British(LKTB) dalam urusan profesionnya ketika menerima lesen dari LKTB. Sejumlah wang telah dipegang oleh LKTB seperti yang diperuntukkan dalam peraturan LKTB kerana kanakkanak tersebut telah melanggar peraturan yang ditetapkan. Mahkamah memutuskan kontrak diantara kanakkanak tersebut dengan LKTB adalah sah dan mengikat. Kes ini serupa dengan kesahihan Linda dengan niat untuk berkontrak dan mengikat janji dalam pembelian motosikal serta pembelian syer yang dikenali sebagai syer “Dingo” dan Linda adalah terikat dengan perjanjian komersial tersebut.
Isu 3.
Linda boleh menerima tawaran tersebut yang menawarkan gaji £100 seminggu. Ini kerana pekerjaan tersebut boleh membantu Linda untuk memenuhi dan menanggung keperluan hidupnya. Walaubagaimana pun, Linda perlu mengemukakan notis selama 3 bulan kepada majikan terdahulu yang merupakan syarat yang telah diperuntukkan sebelum dia menerima pekerjaan sebagai tukang masak di restoran.
Isu 4.
Kegagalan Linda untuk membayar harga motosikal dan bahagian dalam syer Dingo merupakan suatu kesalahan kerana perjanjian yang telah dibuat mesti dipatuhi manakala perjanjian tersebut adalah sah dan mengikat dengan niat serta boleh dikuatkuasakan selari dengan peruntukan seksyen 2(h) Akta Kontrak 1950 iaitu sesuatu perjanjian yang boleh dikuatkuasakan oleh undangundang dan seksyen 10(1) Akta Kontrak 1950 yang menyatakan, sesuatu perjanjian adalah kontrak jika dibuat atas kerelaan bebas pihakpihak yang layak berkontrak, untuk sesuatu balasan yang sah, dan dengan suatu tujuan yang sah, dan tidak ditetapkan dengan nyata dibawah peruntukan akta ini bahawa ianya batal. Dalam kes Koh Kia Hong lawan Guo Enterprise Pte. Ltd (1990), pihak plaintif membuat tawaran untuk membeli hartanah yang diiklankan oleh defendan. Pihak plaintif telah menyerahkan deposit kepada pihak defendan. Satu nota juga telah ditandatangani oleh keduadua pihak dan resit rasmi telah dikeluarkan oleh defendan. Mahkamah memutuskan bahawa keduadua pihak mempunyai niat untuk mengikat diri dalam satu kontrak. Kes tersebut sama dengan tindakan Linda yang telah bersetuju untuk membeli sebuah motosikal dan turut menandatangani perjanjian bertulis untuk membeli syer Dingo. Kegagalan Linda untuk membayar seperti yang telah ditetapkan dalam
perjanjian yang dibuatnya membolehkan Linda dikenakan tindakan yang sah dan didakwa di mahkamah.
Rumusan.
1. Linda perlu menilai kemampuannya terlebih dahulu sebelum ingin memasuki pelbagai kontrak serta melihat had keperluannya sesuai dengan umur.
2. Linda perlu bertanggungjawab dan menanggung semua tindakan mahkamah atas perjanjian yang telah dibuat.
3. Linda tidak sepatutnya meninggalkan persekolahan terlalu awal kerana dia perlu menghabiskan pelajaran. Ini kerana pelajaran yang tinggi boleh membantu Linda mendapatkan pekerjaan yang lebih baik.[/spoiler]
[spoiler]The Company Constitution.
Introduction
The company’s constitution was previously divided between the Memorandum of Association (external aspects) and the Articles of Association (internal aspects, including the balance of power between the board of directors and the general meeting) where this pattern has continued under Companies Acts. This two constitution is for defining what kind of company is and how its business and affairs are to be conducted. Without this constitution, a new company cannot be registered and established to do a legal business in Malaysia under the Companies Acts 1965. The important and relevant of this constitution is to carrying out issues related to the company and all the people who work directly or indirectly in the top management.
1.0 The Memorandum of Association
Memorandum of Association is a document that lists written constitution. Section 18 of Companies Acts 1965 must fulfill with this requirement for registering the company:
1.1 The name of company
1.2 Registered office
1.3 Objectives of the company
1.4 The limited liability of members
1.5 The authorized share capital
1.6 Name, address, work and signature for memorandum.
2.0 The Articles of Association
A document which content information related with administration affairs and management of the company. In addition, the articles of association should have the issue
______________________________________________________________________________
Companies Acts 1965
Section 18 of Companies Acts 1965
and transfer of shares, alteration of capital structure, calling general meetings and voting right, appointment, powers and proceedings of directors, dividend, accounts and the issue of notices. Limited company with allowed share by Company Act 1965 is not necessary to register their Article; this matter is provided in section 29(1). However, section 29(1) must read together with section 30(2) for those company that chose to not register the Articles, and Table A (a set of completed rules) will be automatically used. The other features in Articles of Association are:
2.1 The Article should be printed and divided with numbered spacing and signed by each signer in front of at least two witnesses. (Section 29(2))
2.2 Company have right to chose a matter in Article Association.
2.3 Non limited companies which have share capital declare the share amount and divided it to share unit.
2.4 Non limited company and limited company in guarantee to share and have to declare members that suggested.
The Memorandum ends with a declaration that the subscribers wish to form the company pursuant to the Memorandum and agree to take the number of shares opposite their names. For all companies, public as well as private, it is now only necessary to have two subscribers each of who signs the Memorandum. The subscriber’s signatures are witnessed, usually by a single witness who signs as such. The Memorandum is dated and is treated as deed executed under seal (J.M. Gullick, Company Law 13rd edition, 1987). The Articles may as a general rule is altered simply by passing a special resolution. Clauses which could be included in the articles may be placed in the memorandum in order to make it more difficult to alter them. The memorandum and the articles of association will be described in the next page with description and a relevant case in order to understand of application for incorporated company.
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Section 29(1), Company Act 1965
Section 29(2), Company Act 1965
Section 30(2), Company Act 1965
1.0 The Memorandum of Association.
1.1 The name of company.
In Malaysia, Suruhanjaya Syarikat Malaysia is the registry for a new company that doing business inside Malaysia.
The name of the company serves to identify it and to distinguish it from any other company. For this reason, company which might mislead the public will be change automatically by the registrar that has statutory powers of control over the choice of names. The choice of name must end with the word ‘Sendirian Berhad’ if it is a public company or ‘Berhad’ if it is private limited company, unless permitted to omit ‘Berhad’ from its name an unlimited company does not have to end with ‘Berhad’.
Other rules that will count by the registry for name registration are:
a) No company may have same name which is already exist.
b) No criminal offence or considered offensive.
c) In registrar opinion, the name is like have connection with the government or local authority.
1.1.1 Changing name – In section 23 of Companies Act 1965, a company may change its name by passing a special resolution and obtaining the registrar’s certificate of incorporation that has registered the company under new name. But before the name has been change it should be discuss and approve in the general meeting of the company. In case Oshkosh B’Gosh, the same limitations as above apply to adoption of a name by change of name as by incorporation of a new company. The registrar also can compel to change the name of company if the name same as or in the registrar opinion too like the other company name or the name give misleading an indication of its activities as to be likely to cause harm to the public.
1.1.2 Business name – Most company carry on a commercial activity. Hence the law on company names may give rise to business identity problems of two kinds.
First, a business or professional activity if carried on successfully soon acquires ‘goodwill’.
___________________________________________________________________________ Section 23, Company Act 1965
Customer or client that works with the company's name will bring to the attention of other people, anyone who hears the name of the company with good reputation will put confidence in the company while providing a high confidence to do business. However, if there is a competition that leads to the suppression of business caused by a competitor who has a similar company name, then legal action can be taken. For example in the case Ewing Vs Buttercup Margarine Co Ltd (1917), The plaintiff, a sole trader, had network of retail shops in Scotland and north of England, through which is sold margarine and tea. He traded ‘the Buttercup Dairy Co.’. The company was form to carry on a business of wholesale margarine distributors in England, it’s included a power to trade as retailers. The plaintiff obtained an injunction to prevent the defendant from trading under its own name on the grounds that there might be confusion especially as the plaintiff planned to extend his business to the south of England.
Second, the sole trader or a partnership can use a name that represents the company by mutual consent, however, the sole trader or a partnership cannot use the information held by the companies represented. sole trader or a partnership must use the original information is held as the company address, registration number or a telephone line so that the parent company did not receive overlapping claims and the errors made by a sole trader or a partnership.
1.1.3 Public company name  a public company must have a special name that ends with 'public limited company', where in Malaysia it is called a 'Sendirian Berhad'. This principle is buttressed by rules which:
a) Prohibit any person who is not a public company from carrying on trade, profession or business under a name which includes the public company designation.
b) A public company is guilty of an offence if, in circumstance which the fact that it is a public company is likely to be material to any person, it uses a name which may reasonable be expected to give the impression that it is a private company. (Company Law, 13rd Edition, by J.M Gullick 1987)
1.2 Registered office.
Section 119, Company Act 1965, (1) A company shall as from the day on which it begin to carry on business or as from the fourteenth day after the date of its incorporation, whichever is the earlier, have a registered office within Malaysia to which all communications and notices may be addressed and which shall be open and accessible to the public for not less than three hours during ordinary business hours on each day, Saturdays, weekly and public holidays excepted. (2) If default is made in complying with subsection (1) the company and every officer of the company who is in default shall be guilty of an offence against this Act.
1.3 Objectives of the company.
Objectives in the memorandum of association are concerned with the type of business. It will become the company's guidelines on the proper direction in matters involving. Ordinarily, a company will have many objects clause and the area to enable the company to operate in various business activities. However, if the company conducts a business activity that is unrelated to the purpose of its establishment as contained in the objects clause, then, the company has acted contrary to the objects clause and the association has committed an ultra vires.
1.3.1 Development of Object Clause
In determining the objects clause, it should be towards a power for companies to control the flow of capital is used more effectively. The shareholders will see the use of capital has been invested with the interest of the employer or employee. However, there are some cases that show the establishment clause of the object beyond the object itself, for example in the case of Bell House Ltd. Vs City Wall Properties Ltd (1966), the object clause was to carry on the business of builders and developers. The third paragraph was ‘to carry on any other trade or business whatsoever’ which can in the opinion of the board of directors be advantageously carried on by the company in connection of the above businesses or the general business of
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Section 119(1) of the Company Act 1965
Section 119(2) of the Company Act 1965
the company. The companies also have power to turn to account any of its property and acquired to contact with Swiss bankers who had money to lend on mortgage for property development. The defendant sought the help from plaintiff to introduce defendant to the Swiss bankers and in return for an introduction, fee £20,000 will given by defendant. They later refused to pay because the contract was ultra vires the company and void since its objects did not include mortgage broking. Judge has decided that the fee was binding. The directors did genuinely believe that the effecting of an introduction for fee was a business which could be advantageously carried on as ancillary to the general development of the company. This is what the object clause required and permitted. It was also a case of turning to account an asset of the company.
1.3.2 Alteration of objects
Changing the object clause is under Section 21 of Companies Act 1965, the company can alter as long as Companies Act 1965 allows to do so. The object clause can be altered under Section 28 of Companies Act 1965. According to Section 28(1) of Companies Act 1965, the alteration of object clause can be done by special resolution. Section 28(2) of Companies Act 1965 requires the company to give 21 day notice of meeting to all the members. Section 28(3) of Companies Act 1965 states that notice must be given to all members of company about the alteration of object clause. According to Section 152 of Companies Act 1965, there must be at least 75% majority of the members of company to agree to change the object clause. Section 28(8) of Companies Act 1965 states those 21 days after passing the special resolution, any members of company can apply to object. Section 28(5) of Companies Act 1965 states that there must be at least 10% of issued share capital if they want to object. Section 181 of Companies Act 1965 states any minority, which is less than 10%, can object under this section. Under Section 28(9) of Companies Act 1965, must lodge with SSM within 14 days. Section 28(10) of Companies Act 1965 states the alteration of object clause takes effect when the resolution has been lodged.
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Section 21 of Companies Act 1965
Section 28(1) (2) (3) (5) (8) (9) (10) of Companies Act 1965
Section 152 of Companies Act 1965
Section 181 of Companies Act 1965
1.3.3 Ultra Vires
Ultra Vires may be defined as a situation where the company has conducted business activities beyond the limit and the terms in the memorandum of association. However, a company that has made an ultra vires transaction or arrangement that is not contained in the memorandum of association shall remain valid. These are the effect of the amendment laws under section 20 (1) of the Companies Act 1965 which states: “no such act or acts of the company will be revoked only by reason of the fact that the company has no ability or authority to act or perform or take such conveyance or transfer of even more than what is stipulated in the memorandum of association”.
Case Ashbury Railway Carriage & Iron Co. Ltd. Vs Riche (1875), the company had an object clause which stated that its objects were to make and sell, or lend on hire, railway carriages and wagons and all kinds of railway plant, fittings, machinery and rolling stock and to carry on business as mechanical engineers. The company bought a concession to build a railway in Belgium, subcontracting the work to defendant. Later the company repudiated the contract. Judge was deciding that, constructing a railway was not within the company’s object so the company did not have capacity to enter into either the concession contract or the subcontract. The contract was void for ultra vires and so the defendant had no right to damages for breach. The members could not ratify it and the company could neither enforce the contract nor be forced into performing its obligations.
Ultra vires issue can be raised by the company or other shareholders through the provision of sections that have been made in the Companies Act 1965 section 20 (2) (a) (b) and (c). Explanation of the issue of ultra virus provided in section 20 (2) (a) which provides “the proceedings of any member, or if the company has issued debentures secured by floating charge over all or any of its assets by holders of any debentures or trustee for debenture holders to prevent the commission of any act or acts or the conveyance or transfer any
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Section 20 (1) of Companies Act 1965
Section 20 (2) (a) of Company Act 1965
property to or by the company”. This means that a trustee or member may take action to restrict or prevent the company from taking any action ultra vires.
Case Hawkesbury Development Co. Ltd Vs Landmark Finance Pty. Ltd. (1969), plaintiff is the holder of all shares in Landmark Finance. It has demanded that the court declare the Landmark production of some debentures to finance United Dominion Corporation (UDC) is invalid because it was ultra vires. Plaintiff also asked the court to prohibit enforcement of UDC of the debentures. However, plaintiff's application is not successful, and failed to obtain court approval to void the declaration of the UDC. This is because, the plaintiffs are shareholders of the Landmark Finance. The application should be made to Landmark instead of UDC Finance is a third party.
1.4 The limited liability of members.
This clause states that the liability of the members of the company is limited. In the case of a company limited by shares, the member is liable only to the amount unpaid on the shares taken by members. In the case of a company limited by guarantee the members are liable to the amount undertaken to be contributed by them to the assets of the company in the event of its being wound up. However, this clause is omitted from the memorandum of association of unlimited companies. Any alteration in the memorandum compelling a member to take up more shares, or which increases his liability, would be null and void. If a company carries on business for more than six months, less than 2 members in case of a private company each member aware of this fact, is liable for all the debts contracted by the company after the period of six months has elapsed.
1.5 The authorized share capital.
Authorized share capital is the ceiling of capital available for issue. For example, if the company’s authorized capital is RM100, 000.00 divided into 100,000 ordinary share of RM1.00, that means the company can only increase its paid up capital to a maximum of RM100, 000.00. The company may increase its authorized capital anytime by holding an EGM (Extraordinary General Meeting). A fee is payable to the Companies Commission of Malaysia for the increase of authorized share capital.
1.6 Name, address, position and signature for memorandum.
Normal procedures to complete a memorandum of association as have to fill up for more information in company structure. Example:
Picture 1
2.0 The Articles of Association.
Articles of association are a document that was touched on in the internal affairs of a company that it provides information on how companies should conduct business. For example, it touches on the relationship between the shareholders of each other shareholder, the withdrawal, confiscation and transfer of shares, transfer of the share capital, the notice of the proceedings, the particulars of the appointment, powers, action directors, accounts, audit, dividend, loans and winding. Every company, except a company limited by shares must register the articles of association to the Registrar of Companies. This is consistent with the requirements under Section 29 of the Companies Act 1965. This Act also provides a set of examples of articles of association in Table A of the Schedule 4. Any company is free to imitate some/all of them. If a company limited by shares were not registered with the Registrar of the article itself, then Table A will apply automatically. (Refer to appendix, Sample of Article of Association)
2.1 Effect of Memorandum and Articles of Association
Section 33 (1) of the Companies Act 1965 provides:
"subject to this Act, the memorandum and articles shall when registered, according to the company and its members as if the memorandum and articles, each signed and sealed by each member and contained covenants on the part of each member to comply with all provisions of the memorandum and articles ".
From the provisions of this section, the memorandum and articles on the registration would create the effect of the contract. The effects are:
2.1.1 Contractual effect between the companies and member.
2.1.2 Contractual effect between members (members inter se).
2.1.3 Relationships between companies and outsiders.
______________________________________________________________________________ Section 29 of the Companies Act 1965
Section 33 (1) of the Companies Act 1965
2.1.1 Contractual effect between the companies and member.
Effects arising from the contract provisions of section 33 (1) of the Companies Act 1965 allows companies to enforce the provisions of the articles and memorandum of association of members. In other words, companies can take action against the member to ensure that they comply with the provisions in the articles and memorandum if the member refuses to follow the provisions voluntarily.
In case Hickman (H) Vs Kent (1915), the facts are as follows: a provision in the articles of a company provide that, "if there are any differences that exist between the association and any of its members, then such dispute shall be referred to arbitration." plaintiffs (H), is a members in these companies have taken legal action against the company relating to a dispute concerning his dismissal from the company. This is a dispute between the H at the position as a member of the company. The company has relied on the relevant articles and requested that the trial is terminated. The court decided the company was entitled to terminate the trial because the article was a contract between his company and H, and such difference shall be referred to arbitration as provided by the articles of association. The proceeding would be stayed since the dispute (which related to matters affecting H as a member) must, in conformity with the articles, be submitted to arbitration. In reviewing other cases the court held that if the matter does not affect members generally, if it is something confined to one member in his personal capacity, it is not question of members right and obligations and is not subject to the rule.
However, there are a few cases the effect of a contract excludes the application of section 33 (1) of the Companies Act 1965 to be valid by the position of other members. For example, the position as director, promoter or lawyer. This is because there is among individuals or parties who registered as a member of the company and at the same time they hold certain positions within the company. If members want to enforce the provisions of the articles, giving the position of other members of the claim will not succeed.
In case Beattie Vs E & F Beattie Ltd (1938), the article also provided for arbitration on disputes. The company sued its managing directors to recover money improperly paid to him.
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Section 33 (1) of the Companies Act 1965
The managing director applied to have the action stayed as he contended that the dispute should be submitted to arbitration. The court judge that the claim was against the defendant as a director and the arbitration clause was limited to disputes with members as such. He could not rely on the arbitration clause. Provisions of section 33 (1), is enshrined in the company and its members only. It will not be valid to a third party even if the third party has held a position in the company and manage the affairs of the company. This matter can be looked in Eley Vs Positive Government Security Life Assurance Co. (1876), In this case, Eley is a drafter of articles and the articles of association have provided:
"that Eley is a lawyer and the company only can terminated him for misconduct"
Eley acting as company lawyer for some time though there is no separate employment contract signing. Eley also receive a distribution of shares in the company in return for work done during the establishment of the company. Eley is then fired by company as a company lawyer. Eley claimed the company had breached the contract. The Court held, the article does not give rights to the member if the member seeking to enforce their rights in a position other than a member. In this case, Eley try to enforce the provisions of articles that give it rights in the company's position as a lawyer. If Eley wish to do so, he should sign a contract other than as provided in the articles of association.
In case Re New British Iron Co, ex parte Beckwith (1898) showed about separated contract with the company. In this case, the articles provided that the remuneration of the directors should be £1,000 per annum to be divided between them as they saw fit. By accepting office the directors entered into separate contracts between themselves and the company but nothing more was said about their remuneration. When the company went into liquidation the directors claimed arrears of fees. The liquidator denied that there was any contract to pay the fees. The court judge, although the directors could not rely on the articles as a contract for the payment of their fees, they could refer to the articles to establish the amount payable under the separate contracts with the company made when they accepted office as directors.
2.1.2 Contractual effect between members (members inter se).
Every member is bound to the other members through an article. There's disagreement on how the law is. What seemed clear was that if the article provided to a member of a right alone so he could enforce those rights without the help of members of the company. It can also impose a contract on the members in their dealings with each other as illustrated by the case below. To avoid doubt and difficulty it is usual however to draft articles, especially on members right of first refusal of other members shares, so that each stage is dealing between the company and the members, to which Section 33(1) clearly applies. So that, a member who intends to transfer his shares must, if the articles so require, give notice of his intention to the company and the company must then give notice to other members that they have an option to take up his shares.
In the case of Rayfield v Hands (1960), articles of association of a company alone has allocated:
"Every member of the type that wishes to transfer shares shall notify the director and they will take the same share many of them with a reasonable price.”
Pursuant to the provisions of this article, the plaintiff was trying to force three directors of the company to buy its shares. But all three refused to buy its shares. The Court held that the relationship created by the articles that are binding and director of the company as between members and nonmembers and as between a member and director. With that the three directors are bound to purchase shares of the plaintiff.
The effect also taken to be more complex when a share holder is still want to hold it share but have to follow the rules or article. Take a look in Malaysian case, In the case of Malaysia, Wong Kim Fatt lwn Leong & Co. Sdn Bhd (1976), provisions of the articles of association of limited companies have been allocated as follows: holders of the 7 / 10 of the issued capital of the company may at any time make a request to the company to enforce the transfer of any shares certain that they are not held by the requester. In this case the company has 2 shareholders, A and B where A is a minority shareholder who holds 50,000 shares. B also holds 250,000 shares and the more he is the holder from 7 / 10 of its issued capital. B had made an application for get a share based on the provisions of the article, but a refuses to sell and claimed in court that he was entitled to continue holding its shares. Court decided that it is purely contractual obligations and the claim must be tied to contractual obligations he has received. The question of whether fair or not a provision of the article, is not relevant in determine whether the article could be enforced or not. Rights and dependents under article is a contractual obligation.
2.1.3 Relationships between companies and outsiders.
Articles do not involve contracts with outsiders. This is because outsiders are not members and cannot enforce any alleged rights derived from the memorandum or articles of the company. Look at the example of the case opposite the Raffles Hotel Ltd Malayan Banking Bhd (1996). Malayan Banking is the site of the lessor Raffles Hotel. Things in the article had stated that the lessor has power to appoint a director. Malayan Banking has appointed himself as the director led the Raffles Hotel has applied for a declaration that the appointment was illegal. The court ruled invalid because the appointment of Malayan Banking is not a member of the company and the power held within the article mentions only Malayan Banking authorities to appoint a director.
However, relations with people outside the company should be viewed in a broader context. There are also cases which do not deny the power of the article to be applied into the relationship. As an example the case of Southern foundries (1926) Ltd. Vs Shirlaw (1940) state that the article was to authorize the trustees of the late director to appoint a director of the company. The trustee had made an appointment but was challenged by several members. The Court held that the appointment of a trustee is legal. This action confirms the appointment may be more affected by the responsibility and it should be seen as justice for the rights of the more significant.
2.2 Alteration of the Articles.
Section 31 (1) provides:
'Subject to this act and to such terms in the memorandum, a company may be through a special resolution to amend or add to the articles'
Clear from the provisions of this section that company has the right to amend or add to the articles of association. However, an amendment to articles is only possible through a special resolution received 75% majority vote and the amendment is binding even though there are members who do not give consent. While section 31 (2) also provides for any modifications or additions to the articles of association shall be deemed valid as if it were originally contained in the articles of association. If all members of the company agree to an alteration without passing the general meeting or special resolution, this is still an effective under ‘assent principle’. It can be looked over Cane Vs Jones (1980) where the shareholders signed a written agreement by which the chairman was deprived of the casting vote at general meeting given to him by the articles. No special resolution was passed to make this alteration. Later one group of shareholders contended that the articles had not been altered and so the chairman still had casting vote since special resolution was the only permissible method of altering the articles. The court judge that a special resolution was the means by which majority of members could alter the articles so that all members become bound by the alteration. But it is also a principle of company law that all the members acting together maybe their unanimous agreement take a binding decision on any matter for which a majority of votes at a general meeting is required to bind a minority. The alteration was affective and the chairman had lost his casting vote.
This is about the winning of a majority in their voting, but if there have some unfair decision that make the minority felt injustice, section 181 (1) can be use to protect the rights.
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Section 31 (1) of the Companies Act 1965
Section 31 (2) of the Companies Act 1965
Section 181 (1) of the Companies Act 1965
This section states in paragraphs 1 and 2 are split into several branches of a particular section. They are:
Section 181 (1): Any member or holder of debentures of the company, the minister may apply to the court for an order under this section on the ground:
(a): that the company's affairs are conducted or the powers of directors are carried out in a manner oppressive to one or more members or holders of debentures including himself or herself or does not take into account their interests as members, shareholders or holders of debentures of the company, or
Section 181 (2): If the court considers the application of one of the reasons could be proved:
(a): directing or preventing any act or cancel or vary any transaction or resolution;
(b): manage the affairs of the company's actions in the future;
(c): provides for the purchase of shares or debentures of the company by other members or debenture holders of the company or the company itself;
(d): in case of purchase of shares by the company, it provides for deduction of the share capital of the company, or
(e): provides that the company can be deposed.
2.2.1 Limits for the amendment article
There are some acts that limit the amendments in the article. The Acts are:
Section 33 (3): Any amendment which requires stockholders to take the shares or increase their liability to contribute to the capital of the company is not allowed. Unless the shareholder got written consent from members before and after the amendment of the articles.
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Section 33 (3) of the Companies Act 1965
Section 181 (1) (a) of the Companies Act 1965
Section 181 (2) (a) (b) (c) (d) (e) of the Companies Act 1965
Section 181 (4): Shareholders or debenture holders are allowed to apply for a remedy on the grounds of oppression prevailing in the company. Upon application, the court has the power to issue such an order to amend the memorandum or articles of association. If the court had issued a revised order of the articles, the company cannot make changes other than those permitted by the court.
Section 65 (1): For a company having a share capital divided into several classes of shares and have different rights between the classes of shares, the class of rights amendment is not allowed except in the manner provided for in clause variation, the clause that provides a way to amend the right according to the classes of shares. Freedom to amend the articles is limited when several classes of shares issued and any amendments to the rights granted to the class of shares must follow the procedures set out in clause variations.
The principle of ‘Majority Rule’: When a company wants to amend the articles of association, the principle of 'majority rule' will apply. Through this principle, amendments to the articles could be enforced upon approval of 3 / 4 or 75% majority of the company and the decision is binding on the minority. In this case, the majority of members can add to or amend the provisions of the article. However, to prevent abuse of this principle, amendments to articles by the majority has a limit of the amendment must be made by Bona Fide, which is in good faith for the benefit of the company as a whole. Therefore, the amendment of articles by the majority cannot be enforced if it is not made in good faith for the good of the company as a whole.
Bona Fide can be looked over case of Allen Vs Gold Reefs of West Africa Ltd (1900), under the articles in their original form the company had a lien on ‘all shares (not being fully paid shares) of members’. All the members held partly paid shares, to which the lien attached, but only Z also held fully paid shares to which the lien in that form did
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Section 65 (1) of the Companies Act 1965
Section 181 (4) of the Companies Act 1965
not attach. A special resolution was passed to remove from the articles the words ‘not being paid shares’. As a result the company would then have a lien on the fully paid shares of Z to secure the amount owing on his partly paid shares. The position of other members remained the same since the lien in its original form already attached to their partly paid shares. The executors of Z (who had died) objected that the alteration was discriminatory since it only altered the position of one member (Z) and that it was retrospective in its effect since Z had acquired his fully paid shares (free of the lien) before he was allotted partly paid shares and so incurred the debt for which the lien, extended to his fully paid shares, was security. The judge held that, the alteration was for the benefit of the company as a whole and could apply to any member who might hold fully paid shares. As regards the retrospective effect a member held his shares subject to the memorandum and articles of association and his shareholder’s right were therefore subject to any alteration which might be made affecting them. The alteration was therefore valid.
2.3 Methods of internal affairs (the principle of agency in the company)
This principle highlights the rights agents appointed to represent something or someone in the business of the company. Powers of agents is divided into three, namely:
2.3.1 Express Authority
Real power is something that is expressly given to these agents. For example, someone who had been appointed by the company and given a power to issuing and receiving payments on behalf of the company's debt or borrow money and give security against the debt. The responsibility is clearly within the jurisdiction of a director of the company.
2.3.2 Implied Authority
Implied powers not expressly mentioned, and it is subject to state authority. It exists in two states, namely:
i) Power to make matters incidental to the matters expressly authorized.
ii) Power to do things normally done by an agent in such a position.
For example, an agent has been appointed to a personal manager to a company, it is given real power. However, when the agent advertising to call the candidates to attend an interview and took a successful candidate to work in a company then it is binds to the company, this task is a task usually done by a manager.
2.3.3 Ostensible Authority
One situation where the law will infer the creation of an agency by implied appointment is when the principal, by his words or conduct, creates an inference among third parties that the agent has the actual authority to contract even though no actual authority was actually given to him. This situation may arise when the principal allows the agent to order goods and services on his behalf and he proceeds to pay for them. In such a scenario, if the agent contracts within the limits of the ostensible (apparent) authority, although without any actual authority, the principal will be bound to third parties by his agent’s acts.
In case Chew Hock San Vs Connaught Housing Development SDN. BHD (1985), defendant is the developer. The plaintiff entered into an agreement to buy several houses a particular store from the defendant. The plaintiff has to pay the deposit to the clerk of the defendant to the book shop though at the time of booking that has not been opened. Defendant refused to comply with the contract and the plaintiff's claim for specific performance. The Court ruled that the clerk has no authority to take actions such deposits, and therefore there is no representation from the company to be bound by the acts the defendant. Thus, plaintiff has failed to claim its actions.
However, there is an exception in this ostensible authorities to block the company were causing the lack of power and the doctrine of estopel. Circumstances in which the doctrine of estopel / restrictions apply are:
1) There is a representation by the company to a state that the agent has the authority to take action in question.
 Representation of this can arise in several ways: first, an agent with apparent authority to do all such acts as agent in the usual position. Second, a principal by his actions in a certain time period would represent an implicit: his agent is authorized to sign contracts on behalf of himself.
 In case of Freeman & Lockyer Vs Buckhurst Park Properties (Mangal) Ltd. (1964), A property development company which was established by two men and one named Kapoor. Kapoor acted as managing director although never elected to that position, but the other directors are satisfied with the way management Kapoor. Then, Kapoor had hired the plaintiffs to post an architect and land surveyor. Company refuses to pay the plaintiff as alleged Kapoor has no such power. The Court ruled, Kapoor has ostensible authority, the board of directors know that Kapoor has acted as managing director and appoint an agent for the company. With the permission and that they were, he has the power to enter into ordinary contracts signed by a manager. The plaintiff was motivated by a sense of trust that Kapoor has been authorized by the representation to perform services for the company. So, the company has estopel from denying the power of Kapoor and companies have to pay the plaintiff.
2) Representations must be made by a person having authority to make representations on behalf of the company.
 To make these companies are bound, these representations must be made by a person who has actual authority to make such representations.
 In case Armagas Ltd. Vs Mundogas (1986), Mundogas is the owner of "Ocean Frost 'is a shipping company. A consultation has been done to sell the ships to the Armada Group and it was carried out by representatives of the parties, Magelssen. Representatives of the Armada Group only wants to hire the vessel for a period of 3 years and Magelssen not have the power to hire only sales. Magelssen agree to the lease agreement with a false representation that he has the power to enter into the lease contract. Armagas Ltd was incorporated to a ship and Magelssen had completed the charter agreement for the Mundogas for 3 years. To cheat Mundogas, Magelssen has provided the rental agreement for 12 months for the ship of Armagas Ltd. Thereafter, Mundogas decided not to proceed with charter breach caused Armagas claimed promise. Armagas Magelssen believes is authorized to manage the rental of the mundogas. The Court ruled that the application is rejected because Magelssen not have real power to complete the rental for 3 years. Magelssen has no authority to make representations that he was authorized to do so. Misrepresentation made to the powers is not binding Mundogas.
3) People who want to enforce the contract against the company must have relied on these representations.
 Estopel would only arise if the contractor is dependent on the representation that the agent has the authority to complete the transaction for the parties. If the contractor does not know the agent has the power, then he would not rely on the
representation and estopel will not arise. Thus, knowledge of the power of agents is necessary to avoid problems arising if there fraud or offense committed without the knowledge of real company.
2.4 Duties as responsible from company to agent
Duties of an Agent to his Principal:
1. To obey the principal’s instructions
2. In the absence of instructions from the principal. To act according to the prevailing customs of that business
3. To exercise care and diligence in carrying out his work and to use such skill as he possesses
4. To render proper accounts when required
5. To pay to his principal all sums received on his behalf
6. To communicate with the principal
7. Not to let his own interest conflict with his duty
8. Not to make secret profits out of the performance of his duty.
9. Not to disclose confidential information or documents entrusted to him by the principal
10. Not to delegate his authority
Duties of Principal to his Agent
1. To pay the agent the commission or other agreed remuneration unless the agency relationship is gratuitous
2. Not to willfully prevent or hinder the agent from earning his commission
3. To indemnify and reimburse the agent for acts done in the exercise of his duties
2.5 Methods of internal affairs (Indoor Management Rule)
If an agent had apparent authority to do something, then someone who is dealing with a company is entitled to presume that all matters of internal management and procedures set by the company have been complied with. Indoor management rule sometimes can be known as the rule of Turquand case. It is necessary to begin with the case itself, Royal British Bank Vs Turquand (1856), under the company’s deed of settlement (corresponding with the modern articles) the directors had power to borrow only such sums as had been authorized by general resolution of the company. A debenture for £2,000 had been issued under the seal, signed by two directors and by the secretary. The resolution passed in general meeting to sanction this borrowing did not specify the amount which might be borrowed. It was therefore no authority to the directors to borrow anything. The bank had not seen the resolution and had no knowledge of its inadequacy. But it was held to be entitled to assume that the authority prescribed by the articles, which requirement it had constructive notice, had been effectually given by ordinary resolution, in example by internal procedure of which the bank had neither actual nor constructive notice.
As an 'outsider', the bank is not able to find out whether ordinary resolution was passed because of this information does not typically file in the registry. In this case, the bank has the right to assume that a resolution has been approved legally. The principle of 'internal affairs rule' that sets a third party dealing with the right to make assumptions (if they have no knowledge that the company's internal regulations required by the articles of association have been complied with).
2.5.1 Exclusion
There are limitations on the ability of a party to rely on the internal management rules. First, if the contracting party knew or should have known that the agent has no power, then the contracting parties cannot rely on this method. Typically, these cases occur when the contract is "insider” of his position, should have known about the disability. For example in case Howard Vs Patent Ivory Manufacturing Co. (1888), Director of the company has lent money to companies based on the guarantee of the debentures. Article states that the company can only borrow up to the limit and has exceeded a certain limit. The director wants to enforce the debentures. The Court held that it refuses to allow them to do so. All this is because, as the directors they know or should know about borrowing limits. Thus, they cannot rely on the internal management rules. Second, if there are situations where a reasonable person would make the suspect feeling the power of the agent. So the contracting parties must make reasonable inquiries. If these questions indicate the agent has no authority, contracting parties cannot rely on internal management tool to help him. In case Morris Vs Kanssen (1946) lord of house has saying that:
“Person contracting with a company and dealing in good faith may assume that acts within its constitution and powers have been properly and duly performed and are not bound to inquire whether acts of internal management have been regular”
Later, the judge has pointed out the rationale of the rule ‘The wheels of business will not go smoothly round unless it may be assumed that is in order which appears to be in order’. Third, if the review of the memorandum or articles of association of the company clearly shows that the agent's authority is limited, so the contract cannot rely on the internal management rules.
Lawton and BrownieWilkinson LJJ delivered concurring opinions, “the presumption of regularity cannot be relied on by ‘inside’. In example, persons who by virtue of their position in the company are in a position to know whether or not the internal regulation have been observed”. (page 211, Cases and Material in Company Law, L.S. Seally [1985])
For the conclusion, these two documents are important for determining the direction of the company. This document has clearly touched a lot about management and regulatory compliance by directors, shareholders and even employees who worked in the company. The importance of the two documents is to bring the birth of new companies registered to continue to drive the mission and vision for the establishment of better economic importance.
The end
SOALAN ASSIGNMENT:
a) Nyatakan sebabsebab yang diberikan oleh mahkamah untuk justifikasi
mengenai penerimaan dengan surat dihantar melalui pos?
b) Pada tahun lepas, Linda telah meninggalkan sekolah semasa beliau berumur 16 tahun. Dia telah berkerja dalam bidang hotel sebagai pelatih(trainee) pembantu dapur. Pendapatan Linda adalah 50 pounds seminggu, dan seperti yang diperuntukkan oleh hotel dia harus memberi notis selama 3 bulan jika ingin menamatkan pekerjaannya.
Barubaru ini Linda telah bersetuju untuk membeli motosikal “ Osaka” sebab beliau ingin meluangkan masa bersama kawan baiknya Malcolm, yang turut berminat terhadap motosikal. Linda juga telah menandatangani perjanjian bertulis untuk membeli syer ( satu perempat) yang dikenali sebagai syer “Dingo”.
Linda kini telah ditawarkan pekerjaan sebagai tukang masak di restauran dengan gaji
sebanyak 100 pounds seminggu, dengan syarat beliau boleh memulakan
tugasnya dengan sertamerta. Dia telah gagal untuk membayar untuk harga motosikal
dan bahagian dalam syer Dingo.
Nasihatkan Linda berdasarkan Akta Kontrak yang relevan dan keskes yang
berkaitan/diputuskan oleh mahkamah
Jawapan soalan a).
Undangundang kontrak dalam seksyen 4(2) Akta Kontrak 1950 ada menyatakan bahawa komunikasi penerimaan adalah lengkap terhadap pencadang apabila komunikasi penerimaan itu telah dimasukkan kedalam peti surat untuk perjalanan surat pengirim kepadanya, di mana ia adalah di luar kuasa penerima. Kes Entores Ltd. lawan Miles Far East corporation (1955) menyatakan bahawa, “apabila suatu kontrak dibuat melalui pos, undangundang dalam semua negeri Common Law jelas bahawa penerimaan adalah lengkap selepas surat itu dimasukkan ke dalam peti surat dan itulah tempatnya kontrak telah dibuat”.
Undangundang ini juga adalah untuk mewujudkan kesahihan terhadap kontrak sama ada boleh dikuatkuasakan oleh undangundang atau pun tidak. Kesahihan undangundang itu perlu bagi mengesahkan sesuatu kontrak sedang berjalan atau pun tidak, kerana pencadang boleh membuat pembatalan kontrak sebelum tarikh luput kontrak tersebut tanpa mengetahui jika terdapat penerimaan yang sedang dalam perjalanan. Sebagai contoh Kes Byrne lawan Van Tienhoven, defendan telah menawarkan akan menjual 1000 kotak tin bersadur timah kepada plaintif melalui surat pada 1 Oktober, pada 8 Oktober defendan telah membuat pembatalan cadangan yang dibuat pada 1 Oktober. Surat pembatalan tiba kepada plaintif pada 20 Oktober. Namun plaintif telah menghantar utusan penerimaannya pada 11 Oktober dan mengesahkannya pula pada 15 Oktober. Mahkamah memutuskan bahawa kontrak telah berjalan kerana pembatalan tidak berkuatkuasa sehingga 20 Oktober, iaitu apabila sampai ke pengetahuan plaintif sedangkan plaintif telah pun menerima cadangan tersebut pada 15 Oktober.
Bagi penerima pula, sesuatu kontrak melalui komunikasi penerimaan pos akan berlaku apabila pencadang mendapat maklum balas positif untuk melanjutkan sesuatu kontrak yang dicadangkan. Mengambil contoh dalam kes Ignatius lawan Bell (1913), plaintif ingin menuntut pelaksanaan spesifik mengenai perjanjian opsyen yang mengikut plaintif memberinya pilihan membeli hakhak defendan keatas sebidang tanah. Plaintif menghantar notis penerimaannya melalui surat berdaftar pada 16 Ogos, tetapi surat tidak sampai kepada pengetahuan defendan sehingga petang 25 Ogos kerana defendan tiada di rumah. Mahkamah telah memutuskan bahawa opsyen tersebut adalah sah dipergunakan oleh plaintif apabila surat tersebut sudah diposkan pada 16 Ogos. Pengesahan kontrak tersebut telah berjalan dikuatkan lagi dengan niat penerima iaitu plaintif untuk berkontrak dengan mendaftar surat tersebut.
Walaubagaimana pun, sesuatu kontrak yang melalui komunikasi penerimaan pos mempunyai akta tersendiri dan cara tertentu untuk dibatalkan. Seksyen 5(2) Akta Kontrak 1950 menyatakan sesuatu penerimaan boleh dibatalkan pada bilabila masa sebelum komunikasi penerimaan lengkap terhadap penerima. Antara cara pembatalan kontrak melalui pos adalah:
a) Mengkomunikasikan notis pembatalan,
b) Luputnya waktu yang munasabah,
c) Kemungkiran penerima berkenaan syarat kontrak
d) Meninggal dunia ataupun sakit otak.
Kita mengambil contoh pembatalan luputnya waktu yang munasabah. Kes yang berkaitan menjelaskan keadaan ini adalah Ramsgate Victoria Hotel Co. Ltd lawan Montefiore (1866), Defendan telah memohon suatu saham dalam syarikat hotel tersebut pada bulan Jun. Tetapi hanya bulan November, syarikat tersebut memberitahu sahamsaham itu telah diagihkan dan diberi. Maka, pemegang saham perlu menjelaskan baki harga belian dimana defendan enggan membayarnya. Mahkamah telah memutuskan bahawa keengganan defendan menjelaskan baki harga belian tersebut sah kerana maklumat saham tersebut sepatutnya diketahui oleh defendan dalam masa yang munasabah bukan mengambil masa antara Jun dan November.
Mahkamah menetapkan justifikasi undangundang seperti ini adalah untuk menjaga hak penerima kontrak selain daripada hak pencadang. Kontrak melalui pos adalah suatu kontrak yang memerlukan persefahaman antara keduadua pihak, kerana ia melibatkan masa dan tindak balas daripada pencadang mahu pun penerima. Namun, hak antara penerima dan pencadang cadangan ini mempunyai kriteria tertentu untuk dipatuhi supaya keduadua belah pihak mendapat hak keadilan yang seimbang.
\
Jawapan soalan b).
Akta Dewasa 1971 menyatakan bahawa, seseorang dikatakan dewasa apabila mencapai umur 18 tahun dan layak untuk membuat kontrak. Namun, umur dewasa boleh berbeza untuk tujuan lain sebagaimana ditetapkan oleh undangundang lain. Contohnya, untuk tujuan mengundi dalam pilihan raya negeri, umur dewasa adalah 21 tahun sementara untuk kontrak perantisan dan ahli keanggotaan kesatuan sekerja adalah 16 tahun. Ini bermakna Linda yang berumur 16 tahun boleh membuat kontrak untuk tujuan pekerjaan.
Isu.
1. Adakah Linda perlu memberi notis jika ingin menamatkan pekerjaannya?
2. Adakah sah untuk Linda memasuki kontrak pembelian motosikal dan syer “Dingo”?
3. Bolehkah Linda menerima tawaran pekerjaan sebagai tukang masak?
4. Kegagalan Linda membayar harga motosikal dan syer “Dingo”.
Penyelesaian.
Isu 1.
Ya. Linda perlu memberi notis jika ingin berhenti daripada kerjanya sebagai pelatih pembantu dapur. Ini kerana pihak hotel telah memperuntukkan notis tersebut sebagai peraturan pekerja yang ingin berhenti.
Isu 2.
Kanakkanak tidak boleh memasuki kontrak pembelian motosikal mahupun pembelian untuk syer kerana kontrak tersebut tidak termasuk didalam senarai kekecualian kepada kanakkanak. Tetapi dalam kes Linda, dia telah berniat untuk memasuki kontrak kerana mempunyai kerja sebagai pelatih pembantu dapur dengan pendapatan £50 seminggu. Seksyen 69 (d) menyatakan, kanakkanak akan bertanggungan hanya sekiranya mempunyai harta. Sebagai contoh dalam kes Doyle lawan White City Stadium Ltd (1935), seorang kanakkanak yang merupakan peninju professional telah bersetuju untuk mematuhi peraturan Lembaga Kawalan Tinju British(LKTB) dalam urusan profesionnya ketika menerima lesen dari LKTB. Sejumlah wang telah dipegang oleh LKTB seperti yang diperuntukkan dalam peraturan LKTB kerana kanakkanak tersebut telah melanggar peraturan yang ditetapkan. Mahkamah memutuskan kontrak diantara kanakkanak tersebut dengan LKTB adalah sah dan mengikat. Kes ini serupa dengan kesahihan Linda dengan niat untuk berkontrak dan mengikat janji dalam pembelian motosikal serta pembelian syer yang dikenali sebagai syer “Dingo” dan Linda adalah terikat dengan perjanjian komersial tersebut.
Isu 3.
Linda boleh menerima tawaran tersebut yang menawarkan gaji £100 seminggu. Ini kerana pekerjaan tersebut boleh membantu Linda untuk memenuhi dan menanggung keperluan hidupnya. Walaubagaimana pun, Linda perlu mengemukakan notis selama 3 bulan kepada majikan terdahulu yang merupakan syarat yang telah diperuntukkan sebelum dia menerima pekerjaan sebagai tukang masak di restoran.
Isu 4.
Kegagalan Linda untuk membayar harga motosikal dan bahagian dalam syer Dingo merupakan suatu kesalahan kerana perjanjian yang telah dibuat mesti dipatuhi manakala perjanjian tersebut adalah sah dan mengikat dengan niat serta boleh dikuatkuasakan selari dengan peruntukan seksyen 2(h) Akta Kontrak 1950 iaitu sesuatu perjanjian yang boleh dikuatkuasakan oleh undangundang dan seksyen 10(1) Akta Kontrak 1950 yang menyatakan, sesuatu perjanjian adalah kontrak jika dibuat atas kerelaan bebas pihakpihak yang layak berkontrak, untuk sesuatu balasan yang sah, dan dengan suatu tujuan yang sah, dan tidak ditetapkan dengan nyata dibawah peruntukan akta ini bahawa ianya batal. Dalam kes Koh Kia Hong lawan Guo Enterprise Pte. Ltd (1990), pihak plaintif membuat tawaran untuk membeli hartanah yang diiklankan oleh defendan. Pihak plaintif telah menyerahkan deposit kepada pihak defendan. Satu nota juga telah ditandatangani oleh keduadua pihak dan resit rasmi telah dikeluarkan oleh defendan. Mahkamah memutuskan bahawa keduadua pihak mempunyai niat untuk mengikat diri dalam satu kontrak. Kes tersebut sama dengan tindakan Linda yang telah bersetuju untuk membeli sebuah motosikal dan turut menandatangani perjanjian bertulis untuk membeli syer Dingo. Kegagalan Linda untuk membayar seperti yang telah ditetapkan dalam
perjanjian yang dibuatnya membolehkan Linda dikenakan tindakan yang sah dan didakwa di mahkamah.
Rumusan.
1. Linda perlu menilai kemampuannya terlebih dahulu sebelum ingin memasuki pelbagai kontrak serta melihat had keperluannya sesuai dengan umur.
2. Linda perlu bertanggungjawab dan menanggung semua tindakan mahkamah atas perjanjian yang telah dibuat.
3. Linda tidak sepatutnya meninggalkan persekolahan terlalu awal kerana dia perlu menghabiskan pelajaran. Ini kerana pelajaran yang tinggi boleh membantu Linda mendapatkan pekerjaan yang lebih baik.[/spoiler]

rockoz  Teman Setia SF
 Posts: 2039
 Joined: Mon Apr 05, 2010 10:51 pm
 Credit in hand: 49.81
 Bank: 25,556.84
Re: Nenek Ani
Faham sudah sekarang [spoiler]kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:[/spoiler]
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:
\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,
If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}
because 5252.113122... = \frac {33000} {2 \pi}. \,
Principle of moments
The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:
(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).
Torque multiplier
A torque multiplier is a gear box, which works on the principle of epicyclic gearing. The given load at the input gets multiplied as per the multiplication factor and transmitted to the output, thereby achieving greater load with minimal effort.
sumber wikipediaGuntap...
kalau ikut ni penerangan pening jg tu kepala mo pikir...sinang cerita torque kekuatan tu injin tarik tu badan kereta
Torque, also called moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The terminology for this concept is not straightforward: In the US, in physics it is usually called "torque" and in mechanical engineering it is called "moment".[2] However outside the US this varies. In the UK for instance, most physicists will use the term "moment". In mechanical engineering, the term "torque" means something different,[3] described below. In this article the word "torque" is always used to mean the same as "moment".
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm[4] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:
\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!
\tau = rF\sin \theta\,\!
where
τ is the torque vector and τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied), and r is the length (or magnitude) of the lever arm vector,
F is the force vector, and F is the magnitude of the force,
× denotes the cross product,
θ is the angle between the force vector and the lever arm vector.
The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.
The SI unit for torque is the newton metre (N·m). For more on the units of torque, see below.
Contents
[hide]
1 Terminology
2 History
3 Definition and relation to angular momentum
3.1 Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
4 Units
5 Special cases and other facts
5.1 Moment arm formula
5.2 Static equilibrium
5.3 Net force versus torque
6 Machine torque
7 Relationship between torque, power and energy
7.1 Conversion to other units
7.2 Derivation
8 Principle of moments
9 Torque multiplier
10 See also
11 References
12 External links
Terminology
See also: Couple (mechanics)
In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]
For example, a rotational force down a shaft, such as a turning screwdriver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".
This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.
History
The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.
Definition and relation to angular momentum
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = r F⊥ = r F sinθ and is directed outward from the page.
Torque is defined about a point not specifically about axis as mentioned in several books.
A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque.[5]
More generally, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:
\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F},
where r is the particle's position vector relative to the fulcrum, and F is the force acting on the particle. The magnitude τ of the torque is given by
\tau = rF\sin\theta,\!
where r is the distance from the axis of rotation to the particle, F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
\tau = rF_{\perp},
where F⊥ is the amount of force directed perpendicularly to the position of the particle. Any force directed parallel to the particle's position vector does not produce a torque.[6]
It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. It points along the axis of rotation, and its direction is determined by the righthand rule.[6]
The unbalanced torque on a body along axis of rotation determines the rate of change of the body's angular momentum,
\boldsymbol{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}
where L is the angular momentum vector and t is time. If multiple torques are acting on the body, it is instead the net torque which determines the rate of change of the angular momentum:
\boldsymbol{\tau}_1 + \cdots + \boldsymbol{\tau}_n = \boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}.
For rotation about a fixed axis,
\mathbf{L} = I\boldsymbol{\omega},
where I is the moment of inertia and ω is the angular velocity. It follows that
\boldsymbol{\tau}_{\mathrm{net}} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}(I\boldsymbol{\omega})}{\mathrm{d}t} = I\frac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t} = I\boldsymbol{\alpha},
where α is the angular acceleration of the body, measured in rad·s−2.This equation has limitation that torque equation is to be only written about instantaneous axis of rotation or center of mass for any type of motioneither motion is pure translation,pure rotation or mixed motion.I=Moment of inertia about point about which torque is written(either about instantaneous axis of rotation or center of mass only). If body is in translatory equilibrium then torque equation is same about all points in plane of motion.
Proof of the equivalence of definitions for a fixed instantaneous centre of rotation
The definition of angular momentum for a single particle is:
\mathbf{L} = \mathbf{r} \times \mathbf{p}
where "×" indicates the vector cross product and p is the particle's linear momentum. The timederivative of this is:
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}.
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv (if mass is constant),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}.
The cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma (Newton's 2nd law),
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}.
Then by definition, torque τ = r × F.
If multiple forces are applied, Newton's second law instead reads Fnet = ma, and it follows that
\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}_{\mathrm{net}} = \boldsymbol{\tau}_{\mathrm{net}}.
The proof relies on the assumption that mass is constant; this is valid only in nonrelativistic systems in which no mass is being ejected.
Units
Torque has dimensions of force times distance. Official SI literature suggests using the unit newton metre (N·m) or the unit joule per radian.[7] The unit newton metre is properly denoted N·m or N m.[8] This avoids ambiguity with mN, millinewtons.
The joule, which is the SI unit for energy or work, is dimensionally equivalent to a newton metre, but it is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[7] The dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,
E= \tau \theta\
where E is the energy, τ is magnitude of the torque, and θ is the angle moved (in radians). This equation motivates the alternate unit name joules per radian.[7]
In British unit, "poundforcefeet" (lbf x ft), "footpoundsforce", "inchpoundsforce", "ounceforceinches" (oz x in) are used, and other nonSI units of torque includes "metrekilogramsforce". For all these units, the word "force" is often left out,[9] for example abbreviating "poundforcefoot" to simply "poundfoot" (in this case, it would be implicit that the "pound" is poundforce and not poundmass).
Sometimes one may see torque given units that don't dimensionally make sense. For example: g x cm . In these units, g should be understood as the force given by the weight of 1 gram at the surface of the earth. The surface of the earth is understood to have an average acceleration of gravity (approx. 9.80665 m/sec2).
Special cases and other facts
Moment arm formula
Moment arm diagram
A very useful special case, often given as the definition of torque in fields other than physics, is as follows:
\tau = (\textrm{moment\ arm}) (\textrm{force}).
The construction of the "moment arm" is shown in the figure to the right, along with the vectors r and F mentioned above. The problem with this definition is that it does not give the direction of the torque but only the magnitude, and hence it is difficult to use in threedimensional cases. If the force is perpendicular to the displacement vector r, the moment arm will be equal to the distance to the centre, and torque will be a maximum for the given force. The equation for the magnitude of a torque, arising from a perpendicular force:
\tau = (\textrm{distance\ to\ centre}) (\textrm{force}).
For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.
The torque caused by the two opposing forces Fg and −Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess.
Static equilibrium
For an object to be in static equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a twodimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: ΣH = 0 and ΣV = 0, and the torque a third equation: Στ = 0. That is, to solve statically determinate equilibrium problems in twodimensions, we use three equations.
Net force versus torque
When the net force on the system is zero, the torque measured from any point in space is the same. For example, the torque on a currentcarrying loop in a uniform magnetic field is the same regardless of your point of reference. If the net force \mathbf{F} is not zero, and \mathbf{\tau}_1 is the torque measured from \mathbf{r}_1, then the torque measured from \mathbf{r}_2 is \mathbf{\tau}_2 = \mathbf{\tau}_1 + (\mathbf{r}_1  \mathbf{r}_2) \times \mathbf{F}
Machine torque
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis is the speed (in rpm) that the crankshaft is turning, and the vertical axis is the torque (in Newton metres) that the engine is capable of providing at that speed.
Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. Internalcombustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.
Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics. Power at the drive wheels is equal to engine power less mechanical losses regardless of any gearing between the engine and drive wheels.
Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.
Relationship between torque, power and energy
If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Mathematically, for rotation about a fixed axis through the center of mass,
W = \int_{\theta_1}^{\theta_2} \tau\ \mathrm{d}\theta,
where W is work, τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.[10] It follows from the workenergy theorem that W also represents the change in the rotational kinetic energy Krot of the body, given by
K_{\mathrm{rot}} = \tfrac{1}{2}I\omega^2,
where I is the moment of inertia of the body and ω is its angular speed.[10]
Power is the work per unit time, given by
P = \boldsymbol{\tau} \cdot \boldsymbol{\omega},
where P is power, τ is torque, ω is the angular velocity, and · represents the scalar product.
Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed – not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed – not on the resulting acceleration, if any).
In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.
Consistent units must be used. For metric SI units power is watts, torque is newton metres and angular speed is radians per second (not rpm and not revolutions per second).
Also, the unit newton metre is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.
Conversion to other units
A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.
\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed}
Adding units:
\mbox{power (W)} = \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rps)}
Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.
\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}
where rotational speed is in revolutions per minute (rpm).
Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, footpounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:
\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.
The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.
Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.
Derivation
For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.
By the definition of torque: torque=force × radius. We can rearrange this to determine force=torque ÷ radius. These two values can be substituted into the definition of power:
\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{\displaystyle{r}}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.
The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by 2π in the above derivation to give:[/spoiler]
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