**Torque**

Torque is the result of applying a force to rotate an object around an axis. It is a measure of the tendency of a force to change the rotational motion of an object. In a previous section, angular acceleration was introduced, but the possible sources of this were not discussed. Applying a force to produce a torque is one way that angular acceleration can be applied. The SI unit of torque is the Newton-meter,

Torque depends on an applied force, but it also depends on the distance between the rotation axis and the location that the force is applied. This distance is called either the lever arm or moment arm. Torque is a vector quantity, but it is important to note that it does not point in the direction of the applied force, or along the moment arm. The torque vector points along the axis of rotation, in one of two directions. This means that the torque vector is always perpendicular to the applied force and the moment arm.

The formula for torque depends on the cross product of two vectors, which was presented in a previous section. The “cross” or “vector” product of two vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is written\(\overrightarrow{a}\times \overrightarrow{b}\). The result of the cross product is a vector value, equal to the magnitudes of the vectors multiplied by the sine of the angle between them θ, and multiplied by a unit vector that is perpendicular to both \(\overrightarrow{a}\) and\(\overrightarrow{b}\), which is given the label \(\widehat{n}\),

\(\overrightarrow{a}\times \overrightarrow{b}\,\,=\,\,|\overrightarrow{a}|\,|\overrightarrow{b}|\,\sin \,\theta \,\widehat{n}\)

= \(ab\,\sin \,\theta \,\widehat{n}\)

The vector can point in the positive or negative direction, and this direction can be found using the right-hand rule (the direction of your right thumb when curling your fingers from the first vector to the second vector).

The formula for the torque vector is \(\vec{\tau }\,=\,\vec{r}\,\times \,\vec{F}\).

The label used for torque is the Greek letter τ (“tau”). The vector \(\overrightarrow{F}\) is the applied force, and the vector \(\overrightarrow{r}\) is the vector drawn from the axis of rotation to the location where the force is applied. The magnitude of the vector \(\overrightarrow{r}\) is therefore the moment arm length. The torque vector is equal to \(\vec{\tau }\,=\,rF\,\sin \theta \,\hat{n}\).

The angle θ is the angle between the moment arm and the force. If this value is 90° (which is equal to π/2 radians), then the force is perpendicular to the moment arm, and the torque is, \(\overrightarrow{\tau }\,\,=\,\,r\,F\,\sin \,(\pi /2\,\,radians)\,\widehat{n}\) (Perpendicular force and Moment arm)

∴ \(\overrightarrow{\tau }\,\,\,=\,\,r\,F(1)\,\widehat{n}\)

∴ \(\overrightarrow{\tau }\,\,=\,\,r\,F\,\widehat{n}\)

So, when the moment arm and applied force are perpendicular, the sine term is equal to 1, and the magnitude of the torque is, \(\tau \,\,=\,\,|\overrightarrow{\tau }|\,\,=\,\,r\,F\) (Perpendicular force and moment arm).

The direction of the torque can be found using the right-hand rule, by curling the right hand from the moment arm vector to the force vector, and following the direction of the thumb. For an object that appears to be rotating clockwise, this means that the direction of the torque vector is down. For an object that appears to be rotating counter-clockwise, the torque vector points up.

In general, the tangential component of the applied force (F_{tan}) has a magnitude of F_{tan} = F sinθ.

The magnitude of the torque is therefore, τ = r F_{tan} (Force with any component perpendicular to the moment arm).

**How to find the Torque?**

**Problem: **In a hurry to catch a cab, you rush through a frictionless swinging door and onto the sidewalk. The force you exerted on the door was 50N, applied perpendicular to the plane of the door. The door is 1.0m wide. Assuming that you pushed the door at its edge, what was the torque on the swinging door (taking the hinge as the pivot point)?

**Solution: **The pivot point is at the hinges of the door, opposite to where you were pushing the door. The force you used was 50N, at a distance 1.0m from the pivot point. You hit the door perpendicular to its plane, so the angle between the door and the direction of force was 90 degrees. Since, τ = r x F = r F sinθ.

Then the torque on the door was:

τ = (1m) (50 N) sin 90°

τ = 50 Nm.