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# TORQUE

 Torque, also called moment or moment of force (see "Terminology" below), is the tendency of a force to rotate an object about an axis, [1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.In more basic terms, torque measures how hard something is rotated. For example, imagine a wrench or spanner trying to twist a nut or bolt. The amount of "twist" (torque) depends on how long the wrench is, how hard you push down on it, and how well you are pushing it in the correct direction. Relationship between force F, torque τ, and momentum vectors p and L in a system which has rotation constrained in one plane only. (Forces and moments due to gravity and friction not considered.)The terminology for this concept is not straightforward: In physics, it is usually called "torque", and in mechanical engineering, it is called "moment". [2] However, in mechanical engineering, the term "torque" means something different, [3] described below. In this article, the word "torque" is always used in the physics sense, synonymous with "moment" in engineering.The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.The magnitude of torque depends on three quantities: First, the force applied; second, the length of the lever arm [4] connecting the axis to the point of force application; and third, the angle between the two. In symbols: whereτ is the torque vector and τ is the magnitude of the torque, r is the lever arm vector (vector from the axis to the point of force application), and r is the length (or magnitude) of the lever arm vector, F is the force vector, and F is the magnitude of the force, × denotes the cross product, θ is the angle between the force vector and the lever arm vector. The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.The SI unit for torque is the newton meter (N·m). In Imperial and U.S. customary units, it is measured in foot pounds (ft·lbf) (also known as 'pound feet') and for smaller measurement of torque: inch pounds (in·lbf) or even inch ounces (in·ozf)

# TORQUE

Relationship between force (F), torque (τ), and momentum vectors (p and L) in a system which has rotation constrained in one plane only. (Forces and moments due to gravity and friction not shown.)

Torque, also called moment or moment of force (see "Terminology" below), is the tendency of a force to rotate an object about an axis,[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.

In more basic terms, torque measures how hard something is twisted. For example, imagine a wrench or spanner trying to twist a nut or bolt. The amount of "twist" (torque) depends on how long the wrench is, how hard you push down on it, and how well you are pushing it in the correct direction.

The terminology for this concept is not straightforward: In physics, it is usually called "torque", and in mechanical engineering, it is called "moment".[2] However, in mechanical engineering, the term "torque" means something different,[3] described below. In this article, the word "torque" is always used in the physics sense, synonymous with "moment" in engineering.

The symbol for torque is typically $\tau\,\!$, the Greek letter tau. When it is called moment, it is commonly denoted $M\,\!$

The magnitude of torque depends on three quantities: First, the force applied; second, the length of the lever arm[4] connecting the axis to the point of force application; and third, the angle between the two. In symbols:

$\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!$
$\tau = rF\sin \theta\,\!$

where

$\boldsymbol{\tau}\,\!$ is the torque vector and $\mathbf{\tau}\,\!$ is the magnitude of the torque,
$\mathbf{r}\,\!$ is the lever arm vector (vector from the axis to the point of force application), and $r\,\!$ is the length (or magnitude) of the lever arm vector,
$\mathbf{F}\!$ is the force vector, and $F\,\!$ is the magnitude of the force,
$\times\,\!$ denotes the cross product,
$\theta\,\!$ is the angle between the force vector and the lever arm vector.

The length of the lever arm is particularly important; choosing this length appropriately lies behind the operation of levers, pulleys, gears, and most other simple machines involving a mechanical advantage.

The SI unit for torque is the newton meter (N·m). In Imperial and U.S. customary units, it is measured in foot pounds (ft·lbf) (also known as 'pound feet') and for smaller measurement of torque: inch pounds (in·lbf) or even inch ounces (in·ozf).

Terminology

In mechanical engineering (unlike physics), the terms "torque" and "moment" are not interchangeable. "Moment" is the general term for the tendency of one or more applied forces to rotate an object about an axis (the concept which in physics is called torque).[3] "Torque" is a special case of this: If the applied force vectors add to zero (i.e., their "resultant" is zero), then the forces are called a "couple" and their moment is called a "torque".[3]

For example, a rotational force down a shaft, such as a turning screw-driver, forms a couple, so the resulting moment is called a "torque". By contrast, a lateral force on a beam produces a moment (called a bending moment), but since the net force is nonzero, this bending moment is not called a "torque".

This article follows physics terminology by calling all moments by the term "torque", whether or not they are associated with a couple.

## History

The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.

## Explanation

The force applied 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 meters from the fulcrum, for example, exerts the same torque as one newton applied six meters from the fulcrum. This assumes the force is in a direction at right angles to the straight lever. The direction of the torque can be determined by using the right hand grip rule: curl the fingers of your right hand to indicate the direction of rotation, and stick your thumb out so it is aligned with the axis of rotation. Your thumb points in the direction of the torque vector.[5]

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

$\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$

where

r is the particle's position vector relative to the fulcrum
F is the force acting on the particle.

The torque on a body determines the rate of change of its angular momentum,

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

where

L is the angular momentum vector
t is time.

As can be seen from either of these relationships, torque is a vector, which points along the axis of the rotation it would tend to cause.

### Proof of the equivalence of definitions

The definition of angular momentum for a single particle is:

$\mathbf{L} = \mathbf{r} \times \mathbf{p}$

where "×" indicates the vector cross product and p is the particle's linear momentum. The time-derivative of this is:

$\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}$

This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv, we can see that:

$\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} + \mathbf{v} \times m\mathbf{v}$

But the cross product of any vector with itself is zero, so the second term vanishes. Hence with the definition of force F = ma, (Newton's 2nd law) we obtain:

$\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}$

And by definition, torque τ = r×F.

Note that there is a hidden assumption that mass is constant — this is quite valid in non-relativistic mechanics. Also, total (summed) forces and torques have been used — it perhaps would have been more rigorous to write:

 $\frac{d\mathbf{L}}{dt}$ $= \boldsymbol{\tau}_{tot}$ $= \sum_{i} \mathbf{\tau}_i$

## Units

Torque has dimensions of force times distance. Official SI literature suggests using the unit "Newton meter" (N m) or "Joule per radian".[6] The unit Newton meter is properly denoted "N m" or "N·m", but not other combinations[7] (this avoids ambiguity—for example, mN is the symbol for millinewtons, nm is the symbol for nanometers, etc.)

The joule, which is the SI unit for energy or work, is dimensionally equivalent to a N m, but this unit is not used for torque. Energy and torque are entirely different concepts, so the practice of using different unit names for them helps avoid mistakes and misunderstandings.[6] The dimensional equivalence of these units, of course, is not simply a coincidence: a torque of 1 N m applied through a full revolution will require an energy of exactly 2π joules. Mathematically,

$E= \tau \theta\$

where

E is the energy
τ is torque
θ is the angle moved, in radians.

(This equation motivates the alternate unit name of "Joules per radian".)[6]

Other non-SI units of torque include "pound-force-feet" or "foot-pounds-force" or "inch-pounds-force" or "ounce-force-inches" or "meter-kilograms-force" or "kilogrammeter" (kgm). For all these units, the word "force" is often left out,[8] for example abbreviating "pound-force-foot" to simply "pound-foot". (In this case, it would be implicit that the "pound" is pound-force and not pound-mass.)

## Special cases and other facts

### Moment arm formula

Moment arm diagram

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

$|\tau| = (\textrm{moment\ arm}) (\textrm{force}).$

The construction of the "moment arm" is shown in the figure below, 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:

$|\tau| = (\textrm{distance\ to\ center}) (\textrm{force}).$

For example, if a person places a force of 10 N on a spanner (wrench) which is 0.5 m long, the torque will be 5 N m, assuming that the person pulls the spanner by applying force perpendicular to the spanner.

### Force at an angle

If a force of magnitude F is at an angle θ from the displacement arm of length r (and within the plane perpendicular to the rotation axis), then from the definition of cross product, the magnitude of the torque arising is:

τ = rFsinθ

### 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, we use three equations.

### Torque as a function of time

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.

Torque is the time-derivative of angular momentum, just as force is the time derivative of linear momentum:

$\boldsymbol{\tau} ={\mathrm{d}\mathbf{L} \over \mathrm{d}t} \,\!$

where

L is angular momentum.

Angular momentum on a rigid body can be written in terms of its moment of inertia $\boldsymbol I \,\!$ and its angular velocity $\boldsymbol{\omega}$:

$\mathbf{L}=I\,\boldsymbol{\omega} \,\!$

so if $\boldsymbol I \,\!$ is constant,

$\boldsymbol{\tau}=I{\mathrm{d}\boldsymbol{\omega} \over \mathrm{d}t}=I\boldsymbol{\alpha} \,\!$

where α is angular acceleration, a quantity usually measured in radians per second squared.

## Machine torque

Torque is part of the basic specification of an engine: the power output of an engine is expressed as its torque multiplied by its rotational speed of the axis. 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). The varying torque output over that range can be measured with a dynamometer, and shown as a torque curve. The peak of that torque curve occurs somewhat below the overall power peak. The torque peak cannot, by definition, appear at higher rpm than the power peak.

Understanding the relationship between torque, power and engine speed is vital in automotive engineering, concerned as it is with transmitting power from the engine through the drive train to the wheels. Power is typically a function of torque and engine speed. The gearing of the drive train must be chosen appropriately to make the most of the motor's torque characteristics.

Steam engines and electric motors tend to produce maximum torque close to zero rpm, with the torque diminishing as rotational speed rises (due to increasing friction and other constraints). Reciprocating steam engines can start heavy loads from zero RPM without a clutch.

Torque is also the easiest way to explain mechanical advantage in just about every simple machine.[citation needed]

## Relationship between torque, power and energy

If a force is allowed to act through a distance, it is doing mechanical work. Similarly, if torque is allowed to act through a rotational distance, it is doing work. Power is the work per unit time. However, time and rotational distance are related by the angular speed where each revolution results in the circumference of the circle being travelled by the force that is generating the torque. The power injected by the applied torque may be calculated as:

$\mbox{Power}=\mbox{torque} \cdot \mbox{angular speed} \,$

On the right hand side, this is a scalar product of two vectors, giving a scalar on the left hand side of the equation. Mathematically, the equation may be rearranged to compute torque for a given power output. Note that the power injected by the torque depends only on the instantaneous angular speed - not on whether the angular speed increases, decreases, or remains constant while the torque is being applied (this is equivalent to the linear case where the power injected by a force depends only on the instantaneous speed - not on the resulting acceleration, if any).

In practice, this relationship can be observed in power stations which are connected to a large electrical power grid. In such an arrangement, the generator's angular speed is fixed by the grid's frequency, and the power output of the plant is determined by the torque applied to the generator's axis of rotation.

Consistent units must be used. For metric SI units power is watts, torque is newton meters and angular speed is radians per second (not rpm and not revolutions per second).

Also, the unit newton meter is dimensionally equivalent to the joule, which is the unit of energy. However, in the case of torque, the unit is assigned to a vector, whereas for energy, it is assigned to a scalar.

### Conversion to other units

For different units of power, torque, or angular speed, a conversion factor must be inserted into the equation. Also, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), a conversion factor of must be added because there are radians in a revolution:

$\mbox{power} = \mbox{torque} \times 2 \pi \times \mbox{rotational speed} \,$,

where rotational speed is in revolutions per unit time.

Useful formula in SI units:

$\mbox{power (kW)} = \frac{ \mbox{torque (N}\cdot\mbox{m)} \times 2 \pi \times \mbox{rotational speed (rpm)}} {60000}$

where 60,000 comes from 60 seconds per minute times 1000 watts per kilowatt.

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

$\mbox{power (hp)} = \frac{ \mbox{torque(lbf}\cdot\mbox{ft)} \times 2 \pi \times \mbox{rotational speed (rpm)} }{33000}.$

The constant below in, ft·lbf./min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.

Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.

### Derivation

For a rotating object, the linear distance covered at the circumference in a radian of rotation is the product of the radius with the angular speed. That is: linear speed = radius × angular speed. By definition, linear distance=linear speed × time=radius × angular speed × time.

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

$\mbox{power} = \frac{\mbox{force} \times \mbox{linear distance}}{\mbox{time}}=\frac{\left(\frac{\mbox{torque}}{r}\right) \times (r \times \mbox{angular speed} \times t)} {t} = \mbox{torque} \times \mbox{angular speed}.$

The radius r and time t have dropped out of the equation. However angular speed must be in radians, by the assumed direct relationship between linear speed and angular speed at the beginning of the derivation. If the rotational speed is measured in revolutions per unit of time, the linear speed and distance are increased proportionately by in the above derivation to give:

$\mbox{power}=\mbox{torque} \times 2 \pi \times \mbox{rotational speed}. \,$

If torque is in lbf·ft and rotational speed in revolutions per minute, the above equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33,000 ft·lbf/min per horsepower:

$\mbox{power} = \mbox{torque } \times\ 2 \pi\ \times \mbox{ rotational speed} \cdot \frac{\mbox{ft}\cdot\mbox{lbf}}{\mbox{min}} \times \frac{\mbox{horsepower}}{33000 \cdot \frac{\mbox{ft }\cdot\mbox{ lbf}}{\mbox{min}} } \approx \frac {\mbox{torque} \times \mbox{RPM}}{5252}$

because $5252.113122... = \frac {33,000} {2 \pi}. \,$

## Principle of Moments

The Principle of Moments, also known as Varignon's theorem (not to be confused with the geometrical theorem of the same name) states that the sum of torques due to several forces applied to a single point is equal to the torque due to the sum (resultant) of the forces. Mathematically, this follows from:

$(\mathbf{r}\times\mathbf{F}_1) + (\mathbf{r}\times\mathbf{F}_2) + \cdots = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2 + \cdots).$

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