Torque EX 4

The pedal length is equal to the distance of the point where the force is applied from the axis of rotation (arm of force). We can use three different methods to determine the torque:

1. Shift the vector of force ${\vec {F}}$ and the vector of the arm of force ${\vec {r}}$ so that they are "tail-to-tail" and then use the rules of geometry to find the smaller angle $\theta$ between them. With this angle we can use the formula $\tau =Fr\sin {\theta }$ and the right-hand rule to determine the direction of the torque.
2. We can determine the component of the force perpendicular to ${\vec {r}}$ and then use the formula $\tau =rF\perp$ and the right-hand rule to determine the direction of the torque.
3. Lastly, we can also determine the perpendicular distance of the line of force from the axis and then use the formula $\tau =Fr\perp$ and the right-hand rule to determine the direction of the torque.

The diagram below shows all three methods so that you can choose which one of them suits you.

Note that the direction of the torque is clockwise (the sign of all torques should be negative) since the force makes the pedal turn clockwise.

Angle	Method 1	Method 2	Method 3
0°	$\tau =(0.20)(120)\sin(90^{\circ }+0^{\circ })=24Nm$	$\tau =(0.20)(120\times \cos {0^{\circ }})=24Nm$	$\tau =(0.20\times \cos {0^{\circ }})(120)=24Nm$
30°	$\tau =(0.20)(120)\sin(90^{\circ }+30^{\circ })=21Nm$	$\tau =(0.20)(120\times \cos {30^{\circ }})=21Nm$	$\tau =(0.20\times \cos {30^{\circ }})(120)=21Nm$
45°	$\tau =(0.20)(120)\sin(90^{\circ }+45^{\circ })=17Nm$	$\tau =(0.20)(120\times \cos {45^{\circ }})=17Nm$	$\tau =(0.20\times \cos {45^{\circ }})(120)=17Nm$
60°	$\tau =(0.20)(120)\sin(90^{\circ }+60^{\circ })=12Nm$	$\tau =(0.20)(120\times \cos {60^{\circ }})=12Nm$	$\tau =(0.20\times \cos {60^{\circ }})(120)=12Nm$