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Torque

From Wikipedia, the free encyclopedia
(Redirected fromRotatum)
Turning force around an axis
For other uses, seeTorque (disambiguation).
It has been suggested thatLine of action bemerged into this article. (Discuss) Proposed since February 2025.
Torque
Relationship betweenforceF, torqueτ,linear momentump, andangular momentumL in a system which has rotation constrained to only one plane (forces and moments due togravity andfriction not considered).
Common symbols
τ{\displaystyle \tau },M
SI unitN⋅m
Other units
pound-force-feet,lbf⋅inch, ozf⋅in
InSI base unitskg⋅m2⋅s−2
DimensionML2T−2{\displaystyle {\mathsf {M}}{\mathsf {L}}^{2}{\mathsf {T}}^{-2}}
Part of a series on
Classical mechanics
F=dpdt{\displaystyle {\textbf {F}}={\frac {d\mathbf {p} }{dt}}}

Inphysics andmechanics,torque is therotational analogue of linearforce.[1] It is also referred to as themoment of force (also abbreviated tomoment). The symbol for torque is typicallyτ{\displaystyle {\boldsymbol {\tau }}}, the lowercaseGreek lettertau. When being referred to asmoment of force, it is commonly denoted byM. Just as a linear force is a push or a pull applied to a body, a torque can be thought of as a twist applied to an object with respect to a chosen point; for example, driving ascrew uses torque to force it into an object, which is applied by thescrewdriver rotating around itsaxis to thedrives on the head.

History

[edit]
See also:Couple (mechanics)

The termtorque (fromLatintorquēre, 'to twist') is said to have been suggested byJames Thomson and appeared in print in April, 1884.[2][3][4] Usage is attested the same year bySilvanus P. Thompson in the first edition ofDynamo-Electric Machinery.[4] Thompson motivates the term as follows:[3]

Just as the Newtonian definition offorce is that which produces or tends to producemotion (along a line), sotorque may be defined as that which produces or tends to producetorsion (around an axis). It is better to use a term which treats this action as a single definite entity than to use terms like "couple" and "moment", which suggest more complex ideas. The single notion of a twist applied to turn a shaft is better than the more complex notion of applying a linear force (or a pair of forces) with a certain leverage.

Today, torque is referred to using different vocabulary depending on geographical location and field of study. This article follows the definition used in US physics in its usage of the wordtorque.[5]

In the UK and in USmechanical engineering, torque is referred to asmoment of force, usually shortened tomoment.[6] This terminology can be traced back to at least 1811 inSiméon Denis Poisson'sTraité de mécanique.[7] An English translation of Poisson's work appears in 1842.

Definition and relation to other physical quantities

[edit]
A particle is located at positionr relative to its axis of rotation. When a forceF is applied to the particle, only the perpendicular componentF⊥ produces a torque. This torqueτ =r ×F has magnitudeτ = |r| |F⊥| = |r| |F| sinθ and is directed outward from the page.

A force applied perpendicularly to a lever multiplied by its distance from thelever's fulcrum (the length of thelever arm) is its torque. Therefore, torque is defined as the product of the magnitude of the perpendicular component of the force and the distance of theline of action of a force from the point around which it is being determined. In three dimensions, the torque is apseudovector; forpoint particles, it is given by thecross product of thedisplacement vector and the force vector. The direction of the torque can be determined by using theright hand grip rule: if the fingers of the right hand are curled from the direction of the lever arm to the direction of the force, then the thumb points in the direction of the torque.[8] It follows that thetorque vector is perpendicular to both theposition andforce vectors and defines the plane in which the two vectors lie. The resultingtorque vector direction is determined by the right-hand rule. Therefore any force directed parallel to the particle's position vector does not produce a torque.[9][10] The magnitude of torque applied to arigid body depends on three quantities: the force applied, thelever arm vector[11] connecting the point about which the torque is being measured to the point of force application, and the angle between the force and lever arm vectors. In symbols:

τ=r×F⟹τ=rF⊥=rFsin⁡θ{\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} \implies \tau =rF_{\perp }=rF\sin \theta }

where

TheSI unit for torque is thenewton-metre (N⋅m). For more on the units of torque, see§ Units.

Relationship with the angular momentum

[edit]

The net torque on a body determines the rate of change of the body'sangular momentum,

τ=dLdt{\displaystyle {\boldsymbol {\tau }}={\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}}

whereL is the angular momentum vector andt is time. For the motion of a point particle,

L=Iω,{\displaystyle \mathbf {L} =I{\boldsymbol {\omega }},}

whereI=mr2{\textstyle I=mr^{2}} is themoment of inertia andω is the orbitalangular velocity pseudovector. It follows that

τnet=I1ω1˙e1^+I2ω2˙e2^+I3ω3˙e3^+I1ω1de1^dt+I2ω2de2^dt+I3ω3de3^dt=Iω˙+ω×(Iω){\displaystyle {\boldsymbol {\tau }}_{\mathrm {net} }=I_{1}{\dot {\omega _{1}}}{\hat {\boldsymbol {e_{1}}}}+I_{2}{\dot {\omega _{2}}}{\hat {\boldsymbol {e_{2}}}}+I_{3}{\dot {\omega _{3}}}{\hat {\boldsymbol {e_{3}}}}+I_{1}\omega _{1}{\frac {d{\hat {\boldsymbol {e_{1}}}}}{dt}}+I_{2}\omega _{2}{\frac {d{\hat {\boldsymbol {e_{2}}}}}{dt}}+I_{3}\omega _{3}{\frac {d{\hat {\boldsymbol {e_{3}}}}}{dt}}=I{\boldsymbol {\dot {\omega }}}+{\boldsymbol {\omega }}\times (I{\boldsymbol {\omega }})}

using the derivative of avector isdei^dt=ω×ei^{\displaystyle {d{\boldsymbol {\hat {e_{i}}}} \over dt}={\boldsymbol {\omega }}\times {\boldsymbol {\hat {e_{i}}}}}This equation is the rotational analogue ofNewton's second law for point particles, and is valid for any type of trajectory. In some simple cases like a rotating disc, where only the moment of inertia on rotating axis is, the rotational Newton's second law can beτ=Iα{\displaystyle {\boldsymbol {\tau }}=I{\boldsymbol {\alpha }}}whereα=ω˙{\displaystyle {\boldsymbol {\alpha }}={\dot {\boldsymbol {\omega }}}}.

Proof of the equivalence of definitions

[edit]

The definition of angular momentum for a single point particle is:L=r×p{\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }wherep is the particle'slinear momentum andr is the position vector from the origin. The time-derivative of this is:

dLdt=r×dpdt+drdt×p.{\displaystyle {\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}=\mathbf {r} \times {\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}+{\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}\times \mathbf {p} .}

This result can easily be proven by splitting the vectors into components and applying theproduct rule. But because the rate of change of linear momentum is forceF{\textstyle \mathbf {F} } and the rate of change of position is velocityv{\textstyle \mathbf {v} },

dLdt=r×F+v×p{\displaystyle {\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}=\mathbf {r} \times \mathbf {F} +\mathbf {v} \times \mathbf {p} }

The cross product of momentump{\displaystyle \mathbf {p} } with its associated velocityv{\displaystyle \mathbf {v} } is zero because velocity and momentum are parallel, so the second term vanishes. Therefore, torque on a particle isequal to thefirst derivative of its angular momentum with respect to time. If multiple forces are applied, accordingNewton's second law it follows thatdLdt=r×Fnet=τnet.{\displaystyle {\frac {\mathrm {d} \mathbf {L} }{\mathrm {d} t}}=\mathbf {r} \times \mathbf {F} _{\mathrm {net} }={\boldsymbol {\tau }}_{\mathrm {net} }.}

This is a general proof for point particles, but it can be generalized to a system of point particles by applying the above proof to each of the point particles and then summing over all the point particles. Similarly, the proof can be generalized to a continuous mass by applying the above proof to each point within the mass, and thenintegrating over the entire mass.

Derivatives of torque

[edit]

Inphysics,rotatum is the derivative oftorque with respect totime[12]

P=dτdt,{\displaystyle \mathbf {P} ={\frac {\mathrm {d} {\boldsymbol {\tau }}}{\mathrm {d} t}},}

whereτ is torque.

This word is derived from theLatin wordrotātus meaning 'to rotate', but the termrotatum is not universally recognized but is commonly used. There is not a universally accepted lexicon to indicate the successive derivatives of rotatum, even if sometimes various proposals have been made.

Using the cross product definition of torque, an alternative expression for rotatum is:

P=r×dFdt+drdt×F.{\displaystyle \mathbf {P} =\mathbf {r} \times {\frac {\mathrm {d} \mathbf {F} }{\mathrm {d} t}}+{\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}\times \mathbf {F} .}

Because the rate of change of force is yankY{\textstyle \mathbf {Y} } and the rate of change of position is velocityv{\textstyle \mathbf {v} }, the expression can be further simplified to:

P=r×Y+v×F.{\displaystyle \mathbf {P} =\mathbf {r} \times \mathbf {Y} +\mathbf {v} \times \mathbf {F} .}

Relationship with power and energy

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The law ofconservation of energy can also be used to understand torque. If aforce is allowed to act through a distance, it is doingmechanical work. Similarly, if torque is allowed to act through an angular displacement, it is doing work. Mathematically, for rotation about a fixed axis through thecenter of mass, the workW can be expressed as

W=∫θ1θ2τ dθ,{\displaystyle W=\int _{\theta _{1}}^{\theta _{2}}\tau \ \mathrm {d} \theta ,}

whereτ is torque, andθ1 andθ2 represent (respectively) the initial and finalangular positions of the body.[13]

It follows from thework–energy principle thatW also represents the change in therotational kinetic energyEr of the body, given by

Er=12Iω2,{\displaystyle E_{\mathrm {r} }={\tfrac {1}{2}}I\omega ^{2},}

whereI is themoment of inertia of the body andω is itsangular speed.[13]

Power is the work per unittime, given by

P=τ⋅ω,{\displaystyle P={\boldsymbol {\tau }}\cdot {\boldsymbol {\omega }},}

whereP is power,τ is torque,ω is theangular velocity, and⋅{\displaystyle \cdot } represents thescalar product.

Algebraically, the equation may be rearranged to compute torque for a given angular speed and power output. 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).

Proof

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The work done by a variable force acting over a finite linear displacements{\displaystyle s} is given by integrating the force with respect to an elemental linear displacementds{\displaystyle \mathrm {d} \mathbf {s} }

W=∫s1s2F⋅ds{\displaystyle W=\int _{s_{1}}^{s_{2}}\mathbf {F} \cdot \mathrm {d} \mathbf {s} }

However, the infinitesimal linear displacementds{\displaystyle \mathrm {d} \mathbf {s} } is related to a corresponding angular displacementdθ{\displaystyle \mathrm {d} {\boldsymbol {\theta }}} and the radius vectorr{\displaystyle \mathbf {r} } as

ds=dθ×r{\displaystyle \mathrm {d} \mathbf {s} =\mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} }

Substitution in the above expression for work, , givesW=∫s1s2F⋅dθ×r{\displaystyle W=\int _{s_{1}}^{s_{2}}\mathbf {F} \cdot \mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} }

The expression inside the integral is ascalar triple productF⋅dθ×r=r×F⋅dθ{\displaystyle \mathbf {F} \cdot \mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} =\mathbf {r} \times \mathbf {F} \cdot \mathrm {d} {\boldsymbol {\theta }}}, but as per the definition of torque, and since the parameter of integration has been changed from linear displacement to angular displacement, the equation becomes

W=∫θ1θ2τ⋅dθ{\displaystyle W=\int _{\theta _{1}}^{\theta _{2}}{\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}}

If the torque and the angular displacement are in the same direction, then the scalar product reduces to a product of magnitudes; i.e.,τ⋅dθ=|τ||dθ|cos⁡0=τdθ{\displaystyle {\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}=\left|{\boldsymbol {\tau }}\right|\left|\mathrm {d} {\boldsymbol {\theta }}\right|\cos 0=\tau \,\mathrm {d} \theta } giving

W=∫θ1θ2τdθ{\displaystyle W=\int _{\theta _{1}}^{\theta _{2}}\tau \,\mathrm {d} \theta }

Principle of moments

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The principle of moments, also known asVarignon's theorem (not to be confused with thegeometrical theorem of the same name) states that the resultant torques due to several forces applied to about a point is equal to the sum of the contributing torques:

τ=r1×F1+r2×F2+…+rN×FN.{\displaystyle \tau =\mathbf {r} _{1}\times \mathbf {F} _{1}+\mathbf {r} _{2}\times \mathbf {F} _{2}+\ldots +\mathbf {r} _{N}\times \mathbf {F} _{N}.}

From this it follows that the torques resulting from N number of forces acting around a pivot on an object are balanced when

r1×F1+r2×F2+…+rN×FN=0.{\displaystyle \mathbf {r} _{1}\times \mathbf {F} _{1}+\mathbf {r} _{2}\times \mathbf {F} _{2}+\ldots +\mathbf {r} _{N}\times \mathbf {F} _{N}=\mathbf {0} .}

Units

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Torque has thedimension of force timesdistance, symbolicallyT−2L2M and those fundamental dimensions are the same as that forenergy orwork. OfficialSI literature indicatesnewton-metre, is properly denoted N⋅m, as the unit for torque; although this isdimensionally equivalent to thejoule, which is not used for torque.[14][15] In the case of torque, the unit is assigned to avector, whereas forenergy, it is assigned to ascalar. This means that the dimensional equivalence of the newton-metre and the joule may be applied in the former but not in the latter case. This problem is addressed inorientational analysis, which treats the radian as a base unit rather than as a dimensionless unit.[16]

The traditional imperial units for torque are thepound foot (lbf-ft), or, for small values, the pound inch (lbf-in). In the US, torque is most commonly referred to as thefoot-pound (denoted as either lb-ft or ft-lb) and theinch-pound (denoted as in-lb).[17][18] Practitioners depend on context and the hyphen in the abbreviation to know that these refer to torque and not to energy or moment of mass (as the symbolism ft-lb would properly imply).

Conversion to other units

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A conversion factor may be necessary when using different units of power or torque. For example, ifrotational speed (unit: revolution per minute or second) is used in place of angular speed (unit: radian per second), we must multiply by 2π radians per revolution. In the following formulas,P is power,τ is torque, andν (Greek letter nu) is rotational speed.

P=τ⋅2π⋅ν{\displaystyle P=\tau \cdot 2\pi \cdot \nu }

Showing units:

PW=τN⋅m⋅2πrad/rev⋅νrev/s{\displaystyle P_{\rm {W}}=\tau _{\rm {N{\cdot }m}}\cdot 2\pi _{\rm {rad/rev}}\cdot \nu _{\rm {rev/s}}}

Dividing by 60 seconds per minute gives us the following.

PW=τN⋅m⋅2πrad/rev⋅νrev/min60 s/min{\displaystyle P_{\rm {W}}={\frac {\tau _{\rm {N{\cdot }m}}\cdot 2\pi _{\rm {rad/rev}}\cdot \nu _{\rm {rev/min}}}{\rm {60~s/min}}}}

where rotational speed is in revolutions per minute (rpm, rev/min).

Some people (e.g., American automotive engineers) usehorsepower (mechanical) for power, foot-pounds (lbf⋅ft) for torque and rpm for rotational speed. This results in the formula changing to:

Php=τlbf⋅ft⋅2πrad/rev⋅νrev/min33,000.{\displaystyle P_{\rm {hp}}={\frac {\tau _{\rm {lbf{\cdot }ft}}\cdot 2\pi _{\rm {rad/rev}}\cdot \nu _{\rm {rev/min}}}{33,000}}.}

The constant below (in foot-pounds per minute) changes with the definition of the horsepower; for example, using metric horsepower, it becomes approximately 32,550.

The use of other units (e.g.,BTU per hour for power) would require a different custom conversion factor.

Derivation

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For a rotating object, thelinear distance covered at thecircumference of rotation is the product of the radius with the angle covered. That is: linear distance = radius × angular distance. And by definition, linear distance = linear speed × time = radius × angular speed × time.

By the definition of torque: torque = radius × force. We can rearrange this to determine force = torque ÷ radius. These two values can be substituted into the definition ofpower:

power=force⋅linear distancetime=(torquer)⋅(r⋅angular speed⋅t)t=torque⋅angular speed.{\displaystyle {\begin{aligned}{\text{power}}&={\frac {{\text{force}}\cdot {\text{linear distance}}}{\text{time}}}\\[6pt]&={\frac {\left({\dfrac {\text{torque}}{r}}\right)\cdot (r\cdot {\text{angular speed}}\cdot t)}{t}}\\[6pt]&={\text{torque}}\cdot {\text{angular speed}}.\end{aligned}}}

The radiusr and timet have dropped out of the equation. However, angular speed must be in radians per unit of time, 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:

power=torque⋅2π⋅rotational speed.{\displaystyle {\text{power}}={\text{torque}}\cdot 2\pi \cdot {\text{rotational speed}}.\,}

If torque is in newton-metres and rotational speed in revolutions per second, the above equation gives power in newton-metres per second or watts. If Imperial units are used, and if torque is in pounds-force feet and rotational speed in revolutions per minute, the above equation gives power in foot pounds-force per minute. The horsepower form of the equation is then derived by applying the conversion factor 33,000 ft⋅lbf/min per horsepower:

power=torque⋅2π⋅rotational speed⋅ft⋅lbfmin⋅horsepower33,000⋅ft⋅lbfmin≈torque⋅RPM5,252{\displaystyle {\begin{aligned}{\text{power}}&={\text{torque}}\cdot 2\pi \cdot {\text{rotational speed}}\cdot {\frac {{\text{ft}}{\cdot }{\text{lbf}}}{\text{min}}}\cdot {\frac {\text{horsepower}}{33,000\cdot {\frac {{\text{ft}}\cdot {\text{lbf}}}{\text{min}}}}}\\[6pt]&\approx {\frac {{\text{torque}}\cdot {\text{RPM}}}{5,252}}\end{aligned}}}

because5252.113122≈33,0002π.{\displaystyle 5252.113122\approx {\frac {33,000}{2\pi }}.\,}

Special cases and other facts

[edit]

Moment arm formula

[edit]
Moment arm diagram

A very useful special case, often given as the definition of torque in fields other than physics, is as follows:

τ=(moment arm)(force).{\displaystyle \tau =({\text{moment arm}})({\text{force}}).}

The construction of the "moment arm" is shown in the figure to the right, along with the vectorsr andF 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 three-dimensional cases. If the force is perpendicular to the displacement vectorr, 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:

τ=(distance to centre)(force).{\displaystyle \tau =({\text{distance to centre}})({\text{force}}).}

For example, if a person places a force of 10 N at the terminal end of a wrench that is 0.5 m long (or a force of 10 N acting 0.5 m from the twist point of a wrench of any length), the torque will be 5 N⋅m – assuming that the person moves the wrench by applying force in the plane of movement and perpendicular to the wrench.

The torque caused by the two opposing forcesFg and −Fg causes a change in the angular momentumL in the direction of that torque. This causes the top toprecess.

Static equilibrium

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For an object to be instatic equilibrium, not only must the sum of the forces be zero, but also the sum of the torques (moments) about any point. For a two-dimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations:ΣH = 0 andΣV = 0, and the torque a third equation:Στ = 0. That is, to solvestatically determinate equilibrium problems in two-dimensions, three equations are used.

Net force versus torque

[edit]

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 current-carrying loop in a uniform magnetic field is the same regardless of the point of reference. If the net forceF{\displaystyle \mathbf {F} } is not zero, andτ1{\displaystyle {\boldsymbol {\tau }}_{1}} is the torque measured fromr1{\displaystyle \mathbf {r} _{1}}, then the torque measured fromr2{\displaystyle \mathbf {r} _{2}} isτ2=τ1+(r2−r1)×F{\displaystyle {\boldsymbol {\tau }}_{2}={\boldsymbol {\tau }}_{1}+(\mathbf {r} _{2}-\mathbf {r} _{1})\times \mathbf {F} }

Machine torque

[edit]
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis shows therotational speed (inrpm) that thecrankshaft is turning, and the vertical axis is the torque (innewton-metres) that the engine is capable of providing at that speed.

Torque forms part of the basic specification of anengine: thepower output of an engine is expressed as its torque multiplied by the angular speed of the drive shaft.Internal-combustion engines produce useful torque only over a limited range ofrotational speeds (typically from around 1,000–6,000 rpm for a small car). One can measure the varying torque output over that range with adynamometer, and show it as a torque curve.Steam engines andelectric 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 and electric motors can start heavy loads from zero rpm without aclutch.

In practice, the relationship between power and torque can be observed inbicycles: Bicycles are typically composed of two road wheels, front and rear gears (referred to assprockets) meshing with achain, and aderailleur mechanism if the bicycle's transmission system allows multiple gear ratios to be used (i.e.multi-speed bicycle), all of which attached to theframe. Acyclist, the person who rides the bicycle, provides the input power by turning pedals, therebycranking the front sprocket (commonly referred to aschainring). The input power provided by the cyclist is equal to the product of angular speed (i.e. the number of pedal revolutions per minute times 2π) and the torque at thespindle of the bicycle'scrankset. The bicycle'sdrivetrain transmits the input power to the roadwheel, which in turn conveys the received power to the road as the output power of the bicycle. Depending on thegear ratio of the bicycle, a (torque, angular speed)input pair is converted to a (torque, angular speed)output pair. By using a larger rear gear, or by switching to a lower gear in multi-speed bicycles,angular speed of the road wheels is decreased while the torque is increased, product of which (i.e. power) does not change.

Torque multiplier

[edit]
Main article:Torque multiplier

Torque can be multiplied via three methods: by locating the fulcrum such that the length of a lever is increased; by using a longer lever; or by the use of a speed-reducing gearset orgear box. Such a mechanism multiplies torque, as rotation rate is reduced.

See also

[edit]

References

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  1. ^Serway, R. A. and Jewett, J. W. Jr. (2003).Physics for Scientists and Engineers. 6th ed. Brooks Cole.ISBN 0-534-40842-7.
  2. ^Thomson, James; Larmor, Joseph (1912).Collected Papers in Physics and Engineering. University Press. p. civ.
  3. ^abThompson, Silvanus Phillips (1893).Dynamo-electric machinery: A Manual For Students Of Electrotechnics (4th ed.). New York, Harvard publishing co. p. 108.
  4. ^ab"torque".Oxford English Dictionary. 1933.
  5. ^Physics for Engineering by Hendricks, Subramony, and Van Blerk, Chinappi page 148,Web linkArchived 2017-07-11 at theWayback Machine
  6. ^Kane, T.R. Kane and D.A. Levinson (1985).Dynamics, Theory and Applications pp. 90–99:Free downloadArchived 2015-06-19 at theWayback Machine.
  7. ^Poisson, Siméon-Denis (1811).Traité de mécanique, tome premier. p. 67.
  8. ^"Right Hand Rule for Torque".Archived from the original on 2007-08-19. Retrieved2007-09-08.
  9. ^Halliday, David; Resnick, Robert (1970).Fundamentals of Physics. John Wiley & Sons. pp. 184–85.
  10. ^Knight, Randall; Jones, Brian; Field, Stuart (2016).College Physics: A Strategic Approach (3rd technology update ed.). Boston: Pearson. p. 199.ISBN 9780134143323.OCLC 922464227.
  11. ^Tipler, Paul (2004).Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics (5th ed.). W. H. Freeman.ISBN 0-7167-0809-4.
  12. ^Kumar, Shitij; Savur, Celal; Sahin, Ferat (2021)."Survey of Human–Robot Collaboration in Industrial Settings: Awareness, Intelligence, and Compliance".IEEE Transactions on Systems, Man, and Cybernetics: Systems.51:280–297.doi:10.1109/TSMC.2020.3041231.
  13. ^abKleppner, Daniel; Kolenkow, Robert (1973).An Introduction to Mechanics. McGraw-Hill. pp. 267–268.ISBN 9780070350489.
  14. ^From theofficial SI websiteArchived 2021-04-19 at theWayback Machine, The International System of Units – 9th edition – Text in English Section 2.3.4: "For example, the quantity torque is the cross product of a position vector and a force vector. The SI unit is newton-metre. Even though torque has the same dimension as energy (SI unit joule), the joule is never used for expressing torque."
  15. ^"SI brochure Ed. 9, Section 2.3.4"(PDF). Bureau International des Poids et Mesures. 2019.Archived(PDF) from the original on 2020-07-26. Retrieved2020-05-29.
  16. ^Page, Chester H. (1979). "Rebuttal to de Boer's 'Group properties of quantities and units'".American Journal of Physics.47 (9): 820.Bibcode:1979AmJPh..47..820P.doi:10.1119/1.11704.
  17. ^"Dial Torque Wrenches from Grainger". Grainger. 2020. Demonstration that, as in most US industrial settings, the torque ranges are given in ft-lb rather than lbf-ft.
  18. ^Erjavec, Jack (22 January 2010).Manual Transmissions & Transaxles: Classroom manual. Cengage Learning. p. 38.ISBN 978-1-4354-3933-7.

External links

[edit]
Look uptorque in Wiktionary, the free dictionary.
Wikimedia Commons has media related toTorque.
Authority control databases: NationalEdit this at Wikidata
Linear/translational quantitiesAngular/rotational quantities
Dimensions1LL2Dimensions1θθ2
Ttime:t
s		absement:A
m s		Ttime:t
s
1distance:d,position:r,s,x,displacement
m		area:A
m2		1angle:θ,angular displacement:θ
rad		solid angle:Ω
rad2, sr
T−1frequency:f
s−1,Hz		speed:v,velocity:v
m s−1		kinematic viscosity:ν,
specific angular momentum: h
m2 s−1		T−1frequency:f,rotational speed:n,rotational velocity:n
s−1,Hz		angular speed:ω,angular velocity:ω
rad s−1
T−2acceleration:a
m s−2		T−2rotational acceleration
s−2		angular acceleration:α
rad s−2
T−3jerk:j
m s−3		T−3angular jerk:ζ
rad s−3
Mmass:m
kg		weighted position:M ⟨x⟩ = ∑mxmoment of inertia: I
kg m2		ML
MT−1Mass flow rate:m˙{\displaystyle {\dot {m}}}
kg s−1		momentum:p,impulse:J
kg m s−1,N s		action:𝒮,actergy:ℵ
kg m2 s−1,J s		MLT−1angular momentum:L,angular impulse:ΔL
kg m rad s−1
MT−2force:F,weight:Fg
kg m s−2,N		energy:E,work:W,Lagrangian:L
kg m2 s−2,J		MLT−2torque:τ,moment:M
kg m rad s−2,N m
MT−3yank:Y
kg m s−3, N s−1		power:P
kg m2 s−3, W		MLT−3rotatum:P
kg m rad s−3, N m s−1
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