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Wize AP Physics C: Mechanics Textbook > Unit 5: Rotation (14-20%)

Torque

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Torque


Torque is simply tendency of rotation. It is a rotational equivalent of force and is defined as:


τ=∣r⃗ × F⃗∣=rFsin⁡θ\boxed{\tau=\left|\vec{r}\ \times\ \vec{F}\right|=r F\sin\theta}

  • where rris the magnitude of vector r⃗\vec{r}which is defined from center of rotation (also known as the reference point) to the point of exertion of force (also known as point of action).
  • θ\theta is the angle between r⃗\vec{r} and F⃗\vec{F}
  • unit for torque is N.mN.m







Watch Out!
Torque is a vector so it has magnitude and direction. In two dimensions, torque is positive if it generates a counter-clockwise rotation and it is negative if it generates a clockwise rotation.

Exam Tip
If there are more than one force acting on an object, net torque of the system would be the sum of all torque vector corresponding to different forces. Make sure to add them as Vectors!!

Example:





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Wize Tip
Forces cause NO torque when they are:
  1. Parallel to the axis of rotation,(sin⁡θ=0)(\sin\theta=0)
  2. Acting at the center of rotation (r=0)




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Example: Rotating a Door about its Hinge


You and your friend are trying to push open a gate that has a single hinge on one side. The gate starts opening if you apply a force of 60N perpendicular to the gate, at a point that is 60cm horizontally away from the hinge, parallel to the ground. If your friend is only able to push the gate with a force of 40N perpendicular to the door, where should they apply the force in order to open the gate?

Solution:

The same amount of torque is required to open the gate.

You

Since you apply a force of 60N perpendicular to the gate, the angle between r⃗\vec r and F⃗\vec F is 90°90\degree :
∣τ⃗∣=rFsin⁡(90°)|\vec \tau|=r F \sin(90\degree)
∣τ⃗∣=(0.60)(60)sin⁡(90°)|\vec \tau|=(0.60)(60)\sin(90\degree)
∣∣τ⃗∣∣=36 N.m||\vec \tau||=36 \ N.m

Your friend

∣τ⃗∣=rFsin⁡(90°)|\vec \tau|= r F\sin(90\degree)
36=r(40)sin⁡(90°)36= r(40)\sin(90\degree)
r=3640=0.9mr=\dfrac{36}{40}=0.9 m

Therefore, your friend should push the door at a distance of 90 cm away from the hinge, parallel to the ground.

Seven forces are applied to a beam supported at the center as shown. Rank the torques produced by each force about the hinge from greatest to least, labeling the forces from 1-7 starting left to right (2F forces make an angle of 45 °45\ \degreerespect to the beam)