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For other uses, see Torque (disambiguation).

In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. It represents the capability of a force to produce change in the rotational motion of the body. The concept originated with the studies by Archimedes of the usage of levers. Just as a linear force is a push or a pull, a torque can be thought of as a twist to an object around a specific axis. Torque is defined as the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation. The symbol for torque is typically τ {\displaystyle {\boldsymbol {\tau }}} , the lowercase Greek letter tau. When being referred to as moment of force, it is commonly denoted byM.

Torque
Relationship between force F, torque τ, linear momentum p, and angular momentum L in a system which has rotation constrained to only one plane (forces and moments due to gravity and friction not considered).
Common symbols
τ {\displaystyle \tau } , M
SI unitN⋅m
Other units
pound-force-feet, lbf⋅inch, ozf⋅in
In SI base unitskg⋅m2⋅s−2
DimensionM L2T−2

In three dimensions, the torque is a pseudovector; for point particles, it is given by the cross product of the position vector (distance vector) and the force vector. The magnitude of torque of a rigid body depends on three quantities: the force applied, the lever arm vector 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 {\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} \,\!}
τ = r F sin θ {\displaystyle \tau =\|\mathbf {r} \|\,\|\mathbf {F} \|\sin \theta \,\!}

where

  • τ {\displaystyle {\boldsymbol {\tau }}} is the torque vector and τ {\displaystyle \tau } is the magnitude of the torque,
  • r {\displaystyle \mathbf {r} } is the position vector (a vector from the point about which the torque is being measured to the point where the force is applied),
  • F {\displaystyle \mathbf {F} } is the force vector,
  • × {\displaystyle \times } denotes the cross product, which produces a vector that is perpendicular to bothr andF following the right-hand rule,
  • θ {\displaystyle \theta } is the angle between the force vector and the lever arm vector.

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

Contents

The term torque (from Latin torquēre "to twist") is said to have been suggested by James Thomson and appeared in print in April, 1884. Usage is attested the same year by Silvanus P. Thompson in the first edition of Dynamo-Electric Machinery. Thompson motivates the term as follows:

"Just as the Newtonian definition of force is that which produces or tends to produce motion (along a line), so torque may be defined as that which produces or tends to produce torsion (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 word torque. In the UK and in US mechanical engineering, torque is referred to as moment of force, usually shortened to moment. These terms are interchangeable in US physics and UK physics terminology, unlike in US mechanical engineering, where the term torque is used for the closely related "resultant moment of a couple".[contradictory]

Torque and moment in the US mechanical engineering terminology

In US mechanical engineering, torque is defined mathematically as the rate of change of angular momentum of an object (in physics it is called "net torque"). The definition of torque states that one or both of the angular velocity or the moment of inertia of an object are changing. Moment is the general term used for the tendency of one or more applied forces to rotate an object about an axis, but not necessarily to change the angular momentum of the object (the concept which is called torque in physics). For example, a rotational force applied to a shaft causing acceleration, such as a drill bit accelerating from rest, results in a moment called a torque. By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the angular momentum of the beam is not changing, this bending moment is not called a torque. Similarly with any force couple on an object that has no change to its angular momentum, such moment is also not called a torque.

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 torqueτ = r × F has magnitudeτ = |r| |F| = |r| |F| sin θ and is directed outward from the page.

A force applied perpendicularly 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 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.

More generally, the torque on a point particle (which has the position r in some reference frame) can be defined as the cross product:

τ = r × F , {\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} ,}

where F is the force acting on the particle. The magnitude τ of the torque is given by

τ = r F sin θ , {\displaystyle \tau =rF\sin \theta ,}

where F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,

τ = r F , {\displaystyle \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.

It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors. Conversely, the torque vector defines the plane in which the position and force vectors lie. The resulting torque vector direction is determined by the right-hand rule.

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

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

where L is the angular momentum vector and t is time.

For the motion of a point particle,

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

whereI is the moment of inertia and ω is the orbital angular velocity pseudovector. It follows that

τ n e t = d L d t = d ( I ω ) d t = I d ω d t + d I d t ω = I α + d ( m r 2 ) d t ω = I α + 2 r p | | ω , {\displaystyle {\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}}+{\frac {\mathrm {d} I}{\mathrm {d} t}}{\boldsymbol {\omega }}=I{\boldsymbol {\alpha }}+{\frac {\mathrm {d} (mr^{2})}{\mathrm {d} t}}{\boldsymbol {\omega }}=I{\boldsymbol {\alpha }}+2rp_{||}{\boldsymbol {\omega }},}

where α is the angular acceleration of the particle, and p|| is the radial component of its linear momentum. This equation is the rotational analogue of Newton's Second Law for point particles, and is valid for any type of trajectory. Note that although force and acceleration are always parallel and directly proportional, the torque τ need not be parallel or directly proportional to the angular acceleration α. This arises from the fact that although mass is always conserved, the moment of inertia in general is not.

Proof of the equivalence of definitions

The definition of angular momentum for a single point particle is:

L = r × p {\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }
where p is the particle's linear momentum and r is the position vector from the origin. The time-derivative of this is:
d L d t = r × d p d t + d r d t × 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 the product rule. Now using the definition of force F = d p d t {\textstyle \mathbf {F} ={\frac {\mathrm {d} \mathbf {p} }{\mathrm {d} t}}} (whether or not mass is constant) and the definition of velocity d r d t = v {\textstyle {\frac {\mathrm {d} \mathbf {r} }{\mathrm {d} t}}=\mathbf {v} }

d L d t = 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 momentum p {\displaystyle \mathbf {p} } with its associated velocity v {\displaystyle \mathbf {v} } is zero because velocity and momentum are parallel, so the second term vanishes.

By definition, torque τ = r × F. Therefore, torque on a particle is equal to the first derivative of its angular momentum with respect to time.

If multiple forces are applied, Newton's second law instead readsFnet = ma, and it follows that

d L d t = r × F n e t = τ n e t . {\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.

The proof 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 then integrating over the entire mass.

Torque has the dimension of force times distance, symbolicallyT−2L2M. Although those fundamental dimensions are the same as that for energy or work, official SI literature suggests using the unit newton metre (N⋅m) and never the joule. The unit newton metre is properly denoted N⋅m.

The traditional Imperial and U.S. customary units for torque are the pound foot (lbf-ft), or for small values the pound inch (lbf-in). In the US, torque is most commonly referred to as the foot-pound (denoted as either lb-ft or ft-lb) and the inch-pound (denoted as in-lb). 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).

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:

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 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 three-dimensional 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:

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 exactly 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 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 two-dimensional 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 two-dimensions, three equations are used.

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 current-carrying loop in a uniform magnetic field is the same regardless of the point of reference. If the net force F {\displaystyle \mathbf {F} } is not zero, and τ 1 {\displaystyle {\boldsymbol {\tau }}_{1}} is the torque measured from r 1 {\displaystyle \mathbf {r} _{1}} , then the torque measured from r 2 {\displaystyle \mathbf {r} _{2}} is

τ 2 = τ 1 + ( r 1 r 2 ) × F {\displaystyle {\boldsymbol {\tau }}_{2}={\boldsymbol {\tau }}_{1}+(\mathbf {r} _{1}-\mathbf {r} _{2})\times \mathbf {F} }
Torque curve of a motorcycle ("BMW K 1200 R 2005"). The horizontal axis shows 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 forms 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. Internal-combustion engines produce useful torque only over a limited range of rotational speeds (typically from around 1,000–6,000 rpm for a small car). One can measure the varying torque output over that range with a dynamometer, and show it as a torque curve.

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 and electric motors can start heavy loads from zero rpm without a clutch.

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, the work W can be expressed as

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

where τ is torque, and θ1 and θ2 represent (respectively) the initial and final angular positions of the body.

Proof

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

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

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

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

Substitution in the above expression for work gives

W = s 1 s 2 F d θ × r {\displaystyle W=\int _{s_{1}}^{s_{2}}\mathbf {F} \cdot \mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} }

The expression F d θ × r {\displaystyle \mathbf {F} \cdot \mathrm {d} {\boldsymbol {\theta }}\times \mathbf {r} } is a scalar triple product given by [ F d θ r ] {\displaystyle \left[\mathbf {F} \,\mathrm {d} {\boldsymbol {\theta }}\,\mathbf {r} \right]} . An alternate expression for the same scalar triple product is

[ F d θ r ] = r × F d θ {\displaystyle \left[\mathbf {F} \,\mathrm {d} {\boldsymbol {\theta }}\,\mathbf {r} \right]=\mathbf {r} \times \mathbf {F} \cdot \mathrm {d} {\boldsymbol {\theta }}}

But as per the definition of torque,

τ = r × F {\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} }

Corresponding substitution in the expression of work gives,

W = s 1 s 2 τ d θ {\displaystyle W=\int _{s_{1}}^{s_{2}}{\boldsymbol {\tau }}\cdot \mathrm {d} {\boldsymbol {\theta }}}

Since the parameter of integration has been changed from linear displacement to angular displacement, the limits of the integration also change correspondingly, giving

W = θ 1 θ 2 τ d θ {\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 d θ = | τ | | d θ | cos 0 = τ d θ {\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 = θ 1 θ 2 τ d θ {\displaystyle W=\int _{\theta _{1}}^{\theta _{2}}\tau \,\mathrm {d} \theta }

It follows from the work–energy principle that W also represents the change in the rotational kinetic energy Er of the body, given by

E r = 1 2 I ω 2 , {\displaystyle E_{\mathrm {r} }={\tfrac {1}{2}}I\omega ^{2},}

where I is the moment of inertia of the body and ω is its angular speed.

Power is the work per unit time, given by

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

where P is power, τ is torque, ω is the angular velocity, and {\displaystyle \cdot } represents the scalar product.

Algebraically, the equation may be rearranged to compute torque for a given angular speed and 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 bicycles: Bicycles are typically composed of two road wheels, front and rear gears (referred to as sprockets) meshing with a circular chain, and a derailleur mechanism if the bicycle's transmission system allows multiple gear ratios to be used (i.e. multi-speed bicycle), all of which attached to the frame. A cyclist, the person who rides the bicycle, provides the input power by turning pedals, thereby cranking the front sprocket (commonly referred to as chainring). The input power provided by the cyclist is equal to the product of cadence (i.e. the number of pedal revolutions per minute) and the torque on spindle of the bicycle's crankset. The bicycle's drivetrain transmits the input power to the road wheel, which in turn conveys the received power to the road as the output power of the bicycle. Depending on the gear ratio of the bicycle, a (torque, rpm)input pair is converted to a (torque, rpm)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.

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. 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 in orientational analysis which treats radians as a base unit rather than a dimensionless unit.

Conversion to other units

A conversion factor may be necessary when using different units of power or torque. 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. In the following formulas, P is power, τ is torque, and ν (Greek letter nu) is rotational speed.

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

Showing units:

P ( W ) = τ ( N m ) 2 π ( r a d / r e v ) ν ( r e v / s e c ) {\displaystyle P({\rm {W}})=\tau {\rm {(N\cdot m)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu {\rm {(rev/sec)}}}

Dividing by 60 seconds per minute gives us the following.

P ( W ) = τ ( N m ) 2 π ( r a d / r e v ) ν ( r p m ) 60 {\displaystyle P({\rm {W}})={\frac {\tau {\rm {(N\cdot m)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu {\rm {(rpm)}}}{60}}}

where rotational speed is in revolutions per minute (rpm).

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

P ( h p ) = τ ( l b f f t ) 2 π ( r a d / r e v ) ν ( r p m ) 33 , 000 . {\displaystyle P({\rm {hp}})={\frac {\tau {\rm {(lbf\cdot ft)}}\cdot 2\pi {\rm {(rad/rev)}}\cdot \nu ({\rm {rpm}})}{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

For a rotating object, the linear distance covered at the circumference 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 of power:

power = force linear distance time = ( torque r ) ( 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 radius r and time t 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 2 π 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 2 π rotational speed ft lbf min horsepower 33 , 000 ft lbf min torque RPM 5 , 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}}}

because 5252.113122 33 , 000 2 π . {\displaystyle 5252.113122\approx {\frac {33,000}{2\pi }}.\,}

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 resultant torques due to several forces applied to about a point is equal to the sum of the contributing torques:

τ = r 1 × F 1 + r 2 × F 2 + + r N × F N . {\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 two forces acting around a pivot on an object are balanced when

r 1 × F 1 + r 2 × F 2 = 0 . {\displaystyle \mathbf {r} _{1}\times \mathbf {F} _{1}+\mathbf {r} _{2}\times \mathbf {F} _{2}=\mathbf {0} .}
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 or gear box. Such a mechanism multiplies torque, as rotation rate is reduced.

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Torque Article Talk Language Watch Edit For other uses see Torque disambiguation In physics and mechanics torque is the rotational equivalent of linear force 1 It is also referred to as the moment moment of force rotational force or turning effect depending on the field of study It represents the capability of a force to produce change in the rotational motion of the body The concept originated with the studies by Archimedes of the usage of levers Just as a linear force is a push or a pull a torque can be thought of as a twist to an object around a specific axis Torque is defined as the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation The symbol for torque is typically t displaystyle boldsymbol tau the lowercase Greek letter tau When being referred to as moment of force it is commonly denoted by M TorqueRelationship between force F torque t linear momentum p and angular momentum L in a system which has rotation constrained to only one plane forces and moments due to gravity and friction not considered Common symbolst displaystyle tau MSI unitN mOther unitspound force feet lbf inch ozf inIn SI base unitskg m2 s 2DimensionM L2T 2 In three dimensions the torque is a pseudovector for point particles it is given by the cross product of the position vector distance vector and the force vector The magnitude of torque of a rigid body depends on three quantities the force applied the lever arm vector 2 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 t r F displaystyle boldsymbol tau mathbf r times mathbf F t r F sin 8 displaystyle tau mathbf r mathbf F sin theta where t displaystyle boldsymbol tau is the torque vector and t displaystyle tau is the magnitude of the torque r displaystyle mathbf r is the position vector a vector from the point about which the torque is being measured to the point where the force is applied F displaystyle mathbf F is the force vector displaystyle times denotes the cross product which produces a vector that is perpendicular to both r and F following the right hand rule 8 displaystyle theta is the angle between the force vector and the lever arm vector The SI unit for torque is the newton metre N m For more on the units of torque see Units Contents 1 Defining terminology 1 1 Torque and moment in the US mechanical engineering terminology 2 Definition and relation to angular momentum 2 1 Proof of the equivalence of definitions 3 Units 4 Special cases and other facts 4 1 Moment arm formula 4 2 Static equilibrium 4 3 Net force versus torque 5 Machine torque 6 Relationship between torque power and energy 6 1 Proof 6 2 Conversion to other units 6 3 Derivation 7 Principle of moments 8 Torque multiplier 9 See also 10 References 11 External linksDefining terminology EditSee also Couple mechanics The term torque from Latin torquere to twist is said to have been suggested by James Thomson and appeared in print in April 1884 3 4 5 Usage is attested the same year by Silvanus P Thompson in the first edition of Dynamo Electric Machinery 5 Thompson motivates the term as follows 4 Just as the Newtonian definition of force is that which produces or tends to produce motion along a line so torque may be defined as that which produces or tends to produce torsion 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 word torque 6 In the UK and in US mechanical engineering torque is referred to as moment of force usually shortened to moment 7 These terms are interchangeable in US physics 6 and UK physics terminology unlike in US mechanical engineering where the term torque is used for the closely related resultant moment of a couple 7 contradictory Torque and moment in the US mechanical engineering terminology Edit In US mechanical engineering torque is defined mathematically as the rate of change of angular momentum of an object in physics it is called net torque The definition of torque states that one or both of the angular velocity or the moment of inertia of an object are changing Moment is the general term used for the tendency of one or more applied forces to rotate an object about an axis but not necessarily to change the angular momentum of the object the concept which is called torque in physics 7 For example a rotational force applied to a shaft causing acceleration such as a drill bit accelerating from rest results in a moment called a torque By contrast a lateral force on a beam produces a moment called a bending moment but since the angular momentum of the beam is not changing this bending moment is not called a torque Similarly with any force couple on an object that has no change to its angular momentum such moment is also not called a torque Definition and relation to angular momentum Edit 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 t r F has magnitude t r F r F sin 8 and is directed outward from the page A force applied perpendicularly 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 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 More generally the torque on a point particle which has the position r in some reference frame can be defined as the cross product t r F displaystyle boldsymbol tau mathbf r times mathbf F where F is the force acting on the particle The magnitude t of the torque is given by t r F sin 8 displaystyle tau rF sin theta where F is the magnitude of the force applied and 8 is the angle between the position and force vectors Alternatively t r F displaystyle 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 9 10 It follows from the properties of the cross product that the torque vector is perpendicular to both the position and force vectors Conversely the torque vector defines the plane in which the position and force vectors lie The resulting torque vector direction is determined by the right hand rule 9 The net torque on a body determines the rate of change of the body s angular momentum t d L d t displaystyle boldsymbol tau frac mathrm d mathbf L mathrm d t where L is the angular momentum vector and t is time For the motion of a point particle L I w displaystyle mathbf L I boldsymbol omega where I is the moment of inertia and w is the orbital angular velocity pseudovector It follows that t n e t d L d t d I w d t I d w d t d I d t w I a d m r 2 d t w I a 2 r p w displaystyle 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 frac mathrm d I mathrm d t boldsymbol omega I boldsymbol alpha frac mathrm d mr 2 mathrm d t boldsymbol omega I boldsymbol alpha 2rp boldsymbol omega where a is the angular acceleration of the particle and p is the radial component of its linear momentum This equation is the rotational analogue of Newton s Second Law for point particles and is valid for any type of trajectory Note that although force and acceleration are always parallel and directly proportional the torque t need not be parallel or directly proportional to the angular acceleration a This arises from the fact that although mass is always conserved the moment of inertia in general is not 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 where p is the particle s linear momentum and r is the position vector from the origin The time derivative of this is d L d t r d p d t d r d t 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 the product rule Now using the definition of force F d p d t textstyle mathbf F frac mathrm d mathbf p mathrm d t whether or not mass is constant and the definition of velocity d r d t v textstyle frac mathrm d mathbf r mathrm d t mathbf v d L d t 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 momentum p displaystyle mathbf p with its associated velocity v displaystyle mathbf v is zero because velocity and momentum are parallel so the second term vanishes By definition torque t r F Therefore torque on a particle is equal to the first derivative of its angular momentum with respect to time If multiple forces are applied Newton s second law instead reads Fnet ma and it follows thatd L d t r F n e t t n e t 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 The proof 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 then integrating over the entire mass Units EditTorque has the dimension of force times distance symbolically T 2L 2M Although those fundamental dimensions are the same as that for energy or work official SI literature suggests using the unit newton metre N m and never the joule 11 12 The unit newton metre is properly denoted N m 12 The traditional Imperial and U S customary units for torque are the pound foot lbf ft or for small values the pound inch lbf in In the US torque is most commonly referred to as the foot pound denoted as either lb ft or ft lb and the inch pound denoted as in lb 13 14 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 Special cases and other facts EditMoment 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 t 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 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 three dimensional 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 t 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 exactly 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 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 Edit 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 two dimensional situation with horizontal and vertical forces the sum of the forces requirement is two equations SH 0 and SV 0 and the torque a third equation St 0 That is to solve statically 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 force F displaystyle mathbf F is not zero and t 1 displaystyle boldsymbol tau 1 is the torque measured from r 1 displaystyle mathbf r 1 then the torque measured from r 2 displaystyle mathbf r 2 ist 2 t 1 r 1 r 2 F displaystyle boldsymbol tau 2 boldsymbol tau 1 mathbf r 1 mathbf r 2 times mathbf F Machine torque Edit Torque curve of a motorcycle BMW K 1200 R 2005 The horizontal axis shows 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 forms 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 Internal combustion engines produce useful torque only over a limited range of rotational speeds typically from around 1 000 6 000 rpm for a small car One can measure the varying torque output over that range with a dynamometer and show it as a torque curve 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 and electric motors can start heavy loads from zero rpm without a clutch Relationship between torque power and energy EditIf 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 the work W can be expressed as W 8 1 8 2 t d 8 displaystyle W int theta 1 theta 2 tau mathrm d theta where t is torque and 81 and 82 represent respectively the initial and final angular positions of the body 15 Proof Edit The work done by a variable force acting over a finite linear displacement s displaystyle s is given by integrating the force with respect to an elemental linear displacement d s displaystyle mathrm d mathbf s W s 1 s 2 F d s displaystyle W int s 1 s 2 mathbf F cdot mathrm d mathbf s However the infinitesimal linear displacement d s displaystyle mathrm d mathbf s is related to a corresponding angular displacement d 8 displaystyle mathrm d boldsymbol theta and the radius vector r displaystyle mathbf r as d s d 8 r displaystyle mathrm d mathbf s mathrm d boldsymbol theta times mathbf r Substitution in the above expression for work gives W s 1 s 2 F d 8 r displaystyle W int s 1 s 2 mathbf F cdot mathrm d boldsymbol theta times mathbf r The expression F d 8 r displaystyle mathbf F cdot mathrm d boldsymbol theta times mathbf r is a scalar triple product given by F d 8 r displaystyle left mathbf F mathrm d boldsymbol theta mathbf r right An alternate expression for the same scalar triple product is F d 8 r r F d 8 displaystyle left mathbf F mathrm d boldsymbol theta mathbf r right mathbf r times mathbf F cdot mathrm d boldsymbol theta But as per the definition of torque t r F displaystyle boldsymbol tau mathbf r times mathbf F Corresponding substitution in the expression of work gives W s 1 s 2 t d 8 displaystyle W int s 1 s 2 boldsymbol tau cdot mathrm d boldsymbol theta Since the parameter of integration has been changed from linear displacement to angular displacement the limits of the integration also change correspondingly giving W 8 1 8 2 t d 8 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 t d 8 t d 8 cos 0 t d 8 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 8 1 8 2 t d 8 displaystyle W int theta 1 theta 2 tau mathrm d theta It follows from the work energy principle that W also represents the change in the rotational kinetic energy Er of the body given by E r 1 2 I w 2 displaystyle E mathrm r tfrac 1 2 I omega 2 where I is the moment of inertia of the body and w is its angular speed 15 Power is the work per unit time given by P t w displaystyle P boldsymbol tau cdot boldsymbol omega where P is power t is torque w is the angular velocity and displaystyle cdot represents the scalar product Algebraically the equation may be rearranged to compute torque for a given angular speed and 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 bicycles Bicycles are typically composed of two road wheels front and rear gears referred to as sprockets meshing with a circular chain and a derailleur mechanism if the bicycle s transmission system allows multiple gear ratios to be used i e multi speed bicycle all of which attached to the frame A cyclist the person who rides the bicycle provides the input power by turning pedals thereby cranking the front sprocket commonly referred to as chainring The input power provided by the cyclist is equal to the product of cadence i e the number of pedal revolutions per minute and the torque on spindle of the bicycle s crankset The bicycle s drivetrain transmits the input power to the road wheel which in turn conveys the received power to the road as the output power of the bicycle Depending on the gear ratio of the bicycle a torque rpm input pair is converted to a torque rpm 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 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 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 in orientational analysis which treats radians as a base unit rather than a dimensionless unit 16 Conversion to other units Edit A conversion factor may be necessary when using different units of power or torque For example if rotational speed revolutions per time is used in place of angular speed radians per time we multiply by a factor of 2p radians per revolution In the following formulas P is power t is torque and n Greek letter nu is rotational speed P t 2 p n displaystyle P tau cdot 2 pi cdot nu Showing units P W t N m 2 p r a d r e v n r e v s e c displaystyle P rm W tau rm N cdot m cdot 2 pi rm rad rev cdot nu rm rev sec Dividing by 60 seconds per minute gives us the following P W t N m 2 p r a d r e v n r p m 60 displaystyle P rm W frac tau rm N cdot m cdot 2 pi rm rad rev cdot nu rm rpm 60 where rotational speed is in revolutions per minute rpm Some people e g American automotive engineers use horsepower mechanical for power foot pounds lbf ft for torque and rpm for rotational speed This results in the formula changing to P h p t l b f f t 2 p r a d r e v n r p m 33 000 displaystyle P rm hp frac tau rm lbf cdot ft cdot 2 pi rm rad rev cdot nu rm rpm 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 Edit For a rotating object the linear distance covered at the circumference 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 of power power force linear distance time torque r r angular speed t t torque angular speed displaystyle begin aligned text power amp frac text force cdot text linear distance text time 6pt amp frac left dfrac text torque r right cdot r cdot text angular speed cdot t t 6pt amp text torque cdot text angular speed end aligned The radius r and time t 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 2p in the above derivation to give power torque 2 p 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 2 p rotational speed ft lbf min horsepower 33 000 ft lbf min torque RPM 5 252 displaystyle begin aligned text power amp 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 amp approx frac text torque cdot text RPM 5 252 end aligned because 5252 113122 33 000 2 p displaystyle 5252 113122 approx frac 33 000 2 pi Principle of moments EditThe principle of moments also known as Varignon s theorem not to be confused with the geometrical 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 t r 1 F 1 r 2 F 2 r N F N 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 two forces acting around a pivot on an object are balanced when r 1 F 1 r 2 F 2 0 displaystyle mathbf r 1 times mathbf F 1 mathbf r 2 times mathbf F 2 mathbf 0 Torque multiplier EditMain 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 or gear box Such a mechanism multiplies torque as rotation rate is reduced See also EditMoment Conversion of units Friction torque Mechanical equilibrium Rigid body dynamics Statics Torque converter Torque limiter Torque screwdriver Torque tester Torque wrench Torsion mechanics References Edit Serway R A and Jewett Jr J W 2003 Physics for Scientists and Engineers 6th Ed Brooks Cole ISBN 0 534 40842 7 Tipler Paul 2004 Physics for Scientists and Engineers Mechanics Oscillations and Waves Thermodynamics 5th ed W H Freeman ISBN 0 7167 0809 4 Thomson James Larmor Joseph 1912 Collected Papers in Physics and Engineering University Press p civ a b Thompson Silvanus Phillips 1893 Dynamo electric machinery A Manual For Students Of Electrotechnics 4th ed New York Harvard publishing co p 108 a b torque Oxford English Dictionary 1933 a b Physics for Engineering by Hendricks Subramony and Van Blerk Chinappi page 148 Web link Archived 2017 07 11 at the Wayback Machine a b c Kane T R Kane and D A Levinson 1985 Dynamics Theory and Applications pp 90 99 Free download Archived 2015 06 19 at the Wayback Machine Right Hand Rule for Torque Archived from the original on 2007 08 19 Retrieved 2007 09 08 a b Halliday David Resnick Robert 1970 Fundamentals of Physics John Wiley amp Sons Inc pp 184 85 Knight Randall Jones Brian Field Stuart 2016 College Physics A Strategic Approach Jones Brian 1960 Field Stuart 1958 Third edition technology update ed Boston Pearson p 199 ISBN 9780134143323 OCLC 922464227 From the official SI website Archived 2021 04 19 at the Wayback 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 a b 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 Retrieved 2020 05 29 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 Erjavec Jack 22 January 2010 Manual Transmissions amp Transaxles Classroom manual p 38 ISBN 978 1 4354 3933 7 a b Kleppner Daniel Kolenkow Robert 1973 An Introduction to Mechanics McGraw Hill pp 267 68 ISBN 9780070350489 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 External links EditLook up torque in Wiktionary the free dictionary Wikimedia Commons has media related to Torque Horsepower and Torque An article showing how power torque and gearing affect a vehicle s performance Torque vs Horsepower Yet Another Argument An automotive perspective Torque and Angular Momentum in Circular Motion on Project PHYSNET An interactive simulation of torque Torque Unit Converter A feel for torque An order of magnitude interactive Retrieved from https en wikipedia org w index php title Torque amp oldid 1091038191, wikipedia, 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