In physics and mechanics, torque is the rotational analogue of linear force.Serway, R. A. and Jewett, J. W. Jr. (2003). Physics for Scientists and Engineers. 6th ed. Brooks Cole. . It is also referred to as the moment of force (also abbreviated to moment). It describes the rate of change of angular momentum that would be imparted to an isolated body.
The concept originated with the studies by Archimedes of the usage of , which is reflected in his famous quote: "Give me a lever and a place to stand and I will move the Earth". 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. Torque is defined as the product of the magnitude of the perpendicular component of the force and the distance of the line of action of a force from the point around which it is being determined. The law of conservation of energy can also be used to understand torque. The symbol for torque is typically $\backslash boldsymbol\backslash tau$, the lowercase Greek alphabet tau. When being referred to as moment of force, it is commonly denoted by .
In three dimensions, the torque is a pseudovector; for point particles, it is given by the cross product of the Euclidean vector and the force vector. The magnitude of torque applied to 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:$$\backslash boldsymbol\; \backslash tau\; =\; \backslash mathbf\{r\}\backslash times\; \backslash mathbf\{F\}$$ $$\backslash tau\; =\; rF\backslash sin\; \backslash theta,$$
where
The SI units for torque is the newtonmetre (N⋅m). For more on the units of torque, see .
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. Physics for Engineering by Hendricks, Subramony, and Van Blerk, Chinappi page 148, Web link
In the UK and in US mechanical engineering, torque is referred to as moment of force, usually shortened to moment.Kane, T.R. Kane and D.A. Levinson (1985). Dynamics, Theory and Applications pp. 90–99: Free download . This terminology can be traced back to at least 1811 in Siméon Denis Poisson's Traité de mécanique. An English translation of Poisson's work appears in 1842.
More generally, the torque on a point particle (which has the position r in some reference frame) can be defined as the cross product:
$$\backslash boldsymbol\{\backslash tau\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{F\},$$
where F is the force acting on the particle. The magnitude τ of the torque is given by
$$\backslash tau\; =\; rF\backslash sin\backslash theta,$$
where F is the magnitude of the force applied, and θ is the angle between the position and force vectors. Alternatively,
$$\backslash tau\; =\; rF\_\{\backslash perp\},$$
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 righthand rule.
The net torque on a body determines the rate of change of the body's angular momentum,
$$\backslash boldsymbol\{\backslash tau\}\; =\; \backslash frac\{\backslash mathrm\{d\}\backslash mathbf\{L\}\}\{\backslash mathrm\{d\}t\}$$
where L is the angular momentum vector and t is time.
For the motion of a point particle,
$$\backslash mathbf\{L\}\; =\; I\backslash boldsymbol\{\backslash omega\},$$
where is the moment of inertia and ω is the orbital angular velocity pseudovector. It follows that
$$\backslash boldsymbol\{\backslash tau\}\_\{\backslash mathrm\{net\}\}\; =\; I\_1\backslash dot\{\backslash omega\_1\}\backslash hat\{\backslash boldsymbol\{e\_1\}\}\; +\; I\_2\backslash dot\{\backslash omega\_2\}\backslash hat\{\backslash boldsymbol\{e\_2\}\}\; +\; I\_3\backslash dot\{\backslash omega\_3\}\backslash hat\{\backslash boldsymbol\{e\_3\}\}\; +\; I\_1\backslash omega\_1\backslash frac\{d\backslash hat\{\backslash boldsymbol\{e\_1\}\}\}\{dt\}\; +\; I\_2\backslash omega\_2\backslash frac\{d\backslash hat\{\backslash boldsymbol\{e\_2\}\}\}\{dt\}\; +\; I\_3\backslash omega\_3\backslash frac\{d\backslash hat\{\backslash boldsymbol\{e\_3\}\}\}\{dt\}\; =\; I\backslash boldsymbol\backslash vec\{\backslash dot\{\backslash omega\}\}\; +\; \backslash boldsymbol\backslash vec\{\backslash omega\}\; \backslash times\; (I\backslash boldsymbol\backslash vec\{\backslash omega\})$$
using the derivative of a Unit vector is$$\{d\backslash boldsymbol\{\backslash hat\{e\_i\}\}\; \backslash over\; dt\}\; =\; \backslash vec\{\backslash boldsymbol\{\backslash omega\}\}\; \backslash times\; \backslash boldsymbol\{\backslash hat\{e\_i\}\}$$This equation is the rotational analogue of Newton'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 be$$\backslash boldsymbol\{\backslash tau\}\; =\; I\backslash boldsymbol\{\backslash alpha\}$$where $I\; =\; mr^2$ and $\backslash boldsymbol\backslash alpha\; =\; \backslash boldsymbol\backslash vec\backslash dot\backslash omega$.
$$\backslash frac\{\backslash mathrm\{d\}\backslash mathbf\{L\}\}\{\backslash mathrm\{d\}t\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash frac\{\backslash mathrm\{d\}\backslash mathbf\{p\}\}\{\backslash mathrm\{d\}t\}\; +\; \backslash frac\{\backslash mathrm\{d\}\backslash mathbf\{r\}\}\{\backslash mathrm\{d\}t\}\; \backslash times\; \backslash mathbf\{p\}.$$
This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definition of force $\backslash mathbf\{F\}\; =\; \backslash frac\{\backslash mathrm\{d\}\backslash mathbf\{p\}\}\{\backslash mathrm\{d\}t\}$ (whether or not mass is constant) and the definition of velocity $\backslash frac\{\backslash mathrm\{d\}\backslash mathbf\{r\}\}\{\backslash mathrm\{d\}t\}\; =\; \backslash mathbf\{v\}$
$$\backslash frac\{\backslash mathrm\{d\}\backslash mathbf\{L\}\}\{\backslash mathrm\{d\}t\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{F\}\; +\; \backslash mathbf\{v\}\; \backslash times\; \backslash mathbf\{p\}.$$
The cross product of momentum $\backslash mathbf\{p\}$ with its associated velocity $\backslash mathbf\{v\}$ 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 reads , and it follows that $$\backslash frac\{\backslash mathrm\{d\}\backslash mathbf\{L\}\}\{\backslash mathrm\{d\}t\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{F\}\_\{\backslash mathrm\{net\}\}\; =\; \backslash boldsymbol\{\backslash tau\}\_\{\backslash mathrm\{net\}\}.$$
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.
The traditional imperial and U.S. customary units for torque are the pound foot (lbfft), or for small values the pound inch (lbfin). In the US, torque is most commonly referred to as the footpound (denoted as either lbft or ftlb) and the inchpound (denoted as inlb). Demonstration that, as in most US industrial settings, the torque ranges are given in ftlb rather than lbfft.
$$\backslash tau\; =\; (\backslash text\{moment\; arm\})\; (\backslash text\{force\}).$$
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 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:
$$\backslash tau\; =\; (\backslash text\{distance\; to\; centre\})\; (\backslash text\{force\}).$$
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.
and 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 steamengines and electric motors can start heavy loads from zero rpm without a clutch.
$$W\; =\; \backslash int\_\{\backslash theta\_1\}^\{\backslash theta\_2\}\; \backslash tau\backslash \; \backslash mathrm\{d\}\backslash theta,$$
where τ is torque, and θ_{1} and θ_{2} represent (respectively) the initial and final of the body.
$$W\; =\; \backslash int\_\{s\_1\}^\{s\_2\}\; \backslash mathbf\{F\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash mathbf\{s\}$$
However, the infinitesimal linear displacement $\backslash mathrm\{d\}\backslash mathbf\{s\}$ is related to a corresponding angular displacement $\backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}$ and the radius vector $\backslash mathbf\{r\}$ as
$$\backslash mathrm\{d\}\backslash mathbf\{s\}\; =\; \backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}\backslash times\backslash mathbf\{r\}$$
Substitution in the above expression for work gives $$W\; =\; \backslash int\_\{s\_1\}^\{s\_2\}\; \backslash mathbf\{F\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}\; \backslash times\; \backslash mathbf\{r\}$$
The expression $\backslash mathbf\{F\}\backslash cdot\backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}\backslash times\backslash mathbf\{r\}$ is a scalar triple product given by $\backslash left\backslash mathbf\{F\}\backslash ,\backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}\backslash ,\backslash mathbf\{r\}\backslash right$. An alternate expression for the same scalar triple product is
$$\backslash left\backslash mathbf\{F\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{F\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}$$
But as per the definition of torque,
$$\backslash boldsymbol\{\backslash tau\}\; =\; \backslash mathbf\{r\}\; \backslash times\; \backslash mathbf\{F\}$$
Corresponding substitution in the expression of work gives
$$W\; =\; \backslash int\_\{s\_1\}^\{s\_2\}\; \backslash boldsymbol\{\backslash tau\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}$$
Since the parameter of integration has been changed from linear displacement to angular displacement, the limits of the integration also change correspondingly, giving
$$W\; =\; \backslash int\_\{\backslash theta\; \_1\}^\{\backslash theta\; \_2\}\; \backslash boldsymbol\{\backslash tau\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}$$
If the torque and the angular displacement are in the same direction, then the scalar product reduces to a product of magnitudes; i.e., $\backslash boldsymbol\{\backslash tau\}\backslash cdot\; \backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}\; =\; \backslash left\backslash boldsymbol\{\backslash tau\}\backslash right\; \backslash left\; \backslash mathrm\{d\}\backslash boldsymbol\{\backslash theta\}\backslash right\backslash cos\; 0\; =\; \backslash tau\; \backslash ,\; \backslash mathrm\{d\}\backslash theta$ giving
$$W\; =\; \backslash int\_\{\backslash theta\; \_1\}^\{\backslash theta\; \_2\}\; \backslash tau\; \backslash ,\; \backslash mathrm\{d\}\backslash theta$$
It follows from the work–energy principle that W also represents the change in the rotational kinetic energy E_{r} of the body, given by
$$E\_\{\backslash mathrm\{r\}\}\; =\; \backslash tfrac\{1\}\{2\}I\backslash omega^2,$$
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
$$P\; =\; \backslash boldsymbol\{\backslash tau\}\; \backslash cdot\; \backslash boldsymbol\{\backslash omega\},$$
where P is power, τ is torque, ω is the angular velocity, and $\backslash cdot$ represents the scalar 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).
In practice, this relationship can be observed in : Bicycles are typically composed of two road wheels, front and rear gears (referred to as sprockets) meshing with a bicycle chain, and a derailleur gears if the bicycle's transmission system allows multiple gear ratios to be used (i.e. multispeed bicycle), all of which attached to the bicycle 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 angular speed (i.e. the number of pedal revolutions per minute times 2 π) and the torque at the Axle 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, 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 multispeed bicycles, angular speed of the road wheels is decreased while the torque is increased, product of which (i.e. power) does not change.
For SI units, the unit of power is the watt, the unit of torque is the newtonmetre and the unit of angular speed is the radian per second (not rpm and not revolutions per second).
The unit newtonmetre is dimensionally equivalent to the joule, which is the unit of energy. 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 newtonmetre and the joule may be applied in the former, but not in the latter case. This problem is addressed in , which treats the radian as a base unit rather than as a dimensionless unit.
$$P\; =\; \backslash tau\; \backslash cdot\; 2\; \backslash pi\; \backslash cdot\; \backslash nu$$
Showing units:
$$P\; \_\{\backslash rm\; W\}\; =\; \backslash tau\; \_\{\backslash rm\; N\; \{\backslash cdot\}\; m\}\; \backslash cdot\; 2\; \backslash pi\; \_\{\backslash rm\; rad/rev\}\; \backslash cdot\; \backslash nu\; \_\{\backslash rm\; rev/s\}$$
Dividing by 60 seconds per minute gives us the following.
$$P\; \_\{\backslash rm\; W\}\; =\; \backslash frac\{\; \backslash tau\; \_\{\backslash rm\; N\; \{\backslash cdot\}\; m\}\; \backslash cdot\; 2\; \backslash pi\; \_\{\backslash rm\; rad/rev\}\; \backslash cdot\; \backslash nu\; \_\{\backslash rm\; rev/min\}\; \}\; \{\backslash rm\; 60\; ~s/min\}$$
where rotational speed is in revolutions per minute (rpm, rev/min).
Some people (e.g., American automotive engineers) use horsepower (mechanical) for power, footpounds (lbf⋅ft) for torque and rpm for rotational speed. This results in the formula changing to:
$$P\; \_\{\backslash rm\; hp\}\; =\; \backslash frac\{\; \backslash tau\; \_\{\backslash rm\; lbf\; \{\backslash cdot\}\; ft\}\; \backslash cdot\; 2\; \backslash pi\; \_\{\backslash rm\; rad/rev\}\; \backslash cdot\; \backslash nu\; \_\{\backslash rm\; rev/min\}\}\; \{33,000\}.$$
The constant below (in footpounds 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.
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:
$$\backslash begin\{align\}\; \backslash text\{power\}\; \&\; =\; \backslash frac\{\backslash text\{force\}\; \backslash cdot\; \backslash text\{linear\; distance\}\}\{\backslash text\{time\}\}\; \backslash \backslash 6pt\; \&\; =\; \backslash frac\{\backslash left(\backslash dfrac\{\backslash text\{torque\}\}\; r\; \backslash right)\; \backslash cdot\; (r\; \backslash cdot\; \backslash text\{angular\; speed\}\; \backslash cdot\; t)\}\; t\; \backslash \backslash 6pt\; \&\; =\; \backslash text\{torque\}\; \backslash cdot\; \backslash text\{angular\; speed\}.\; \backslash end\{align\}$$
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 2 in the above derivation to give:
$$\backslash text\{power\}\; =\; \backslash text\{torque\}\; \backslash cdot\; 2\; \backslash pi\; \backslash cdot\; \backslash text\{rotational\; speed\}.\; \backslash ,$$
If torque is in newtonmetres and rotational speed in revolutions per second, the above equation gives power in newtonmetres per second or watts. If Imperial units are used, and if torque is in poundsforce feet and rotational speed in revolutions per minute, the above equation gives power in foot poundsforce per minute. The horsepower form of the equation is then derived by applying the conversion factor 33,000 ft⋅lbf/min per horsepower:
$$\backslash begin\{align\}\; \backslash text\{power\}\; \&\; =\; \backslash text\{torque\}\; \backslash cdot\; 2\; \backslash pi\; \backslash cdot\; \backslash text\{rotational\; speed\}\; \backslash cdot\; \backslash frac\{\backslash text\{ft\}\{\backslash cdot\}\backslash text\{lbf\}\}\{\backslash text\{min\}\}\; \backslash cdot\; \backslash frac\{\backslash text\{horsepower\}\}\{33,000\; \backslash cdot\; \backslash frac\{\backslash text\{ft\}\backslash cdot\backslash text\{lbf\}\}\{\backslash text\{min\}\}\}\; \backslash \backslash 6pt\; \&\; \backslash approx\; \backslash frac\; \{\backslash text\{torque\}\; \backslash cdot\; \backslash text\{RPM\}\}\{5,252\}\; \backslash end\{align\}$$
because $5252.113122\; \backslash approx\; \backslash frac\; \{33,000\}\; \{2\; \backslash pi\}.\; \backslash ,$
$$\backslash tau\; =\; \backslash mathbf\{r\}\_1\backslash times\backslash mathbf\{F\}\_1\; +\; \backslash mathbf\{r\}\_2\backslash times\backslash mathbf\{F\}\_2\; +\; \backslash ldots\; +\; \backslash mathbf\{r\}\_N\backslash times\backslash mathbf\{F\}\_N.$$
From this it follows that the torques resulting from two forces acting around a pivot on an object are balanced when
$$\backslash mathbf\{r\}\_1\backslash times\backslash mathbf\{F\}\_1\; +\; \backslash mathbf\{r\}\_2\backslash times\backslash mathbf\{F\}\_2\; =\; \backslash mathbf\{0\}.$$

