The length of a bicycle pedal arm is 0.152m, and a downward force of 111N is applied to the pedal by the rider. What is the magnitude of the torque about the pedal arm’s pivot when the arm is at angle (a) 30^o , (b) 90^o , and (c) 180^o with the vertical?

1. The length of pedal arm is, $r=0.152m$.
2. The downward force is, $F=111N$
3. ${\theta }_1={30}^o$
4. ${\theta }_2={90}^o$
5. ${\theta }_3={180}^o$
   

Torqueis a turning action on a body about a rotation axis due to a force. If force is applied at a point, then total torque is the cross product of radial vector and force exerted on the body. The magnitude of torque is$\mathit{\tau }\mathbf{=}\mathit{r}\mathit{F}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

$\begin{array}{rcl}\mathit{\tau }& \mathbf{=}& \mathit{r}\mathbf{\times }\mathit{F}\\ & \mathbf{=}& \mathit{r}\mathit{F}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\end{array}$

According to the formula, we can calculate the magnitude of torque for${\theta }_1={30}^o$

$$\begin{gathered} \tau =r\times F \\ \tau =rF\sin \theta . \end{gathered}$$

Substitute all the value in the above equation.

$\begin{array}{rcl}\tau & =& 0.152m\times 111N\times \sin 30\\ & =& 8.44N.m\\ \tau & \approx & 8.4N.m\end{array}$

Hence the torque is, 8.4N.m .

According to formula we can calculate magnitude of torque for ${\theta }_2={90}^o$

$$\begin{gathered} \tau =r\times F \\ \tau =rF\sin \theta \end{gathered}$$

Substitute all the value in the above equation.

$\begin{array}{rcl}\tau & =& 0.152m\times 111N\times \sin 90\\ & =& 16.90N.m\\ \tau & \approx & 17N.m\end{array}$

Hence the torque is, 17N.m .

According to formula we can calculate magnitude of torque for${\theta }_3={180}^o$

$$\begin{gathered} \tau =r\times F \\ \tau =rF\sin \theta \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} \tau =0.152m\times 111N\times \sin 180 \\ \tau =0N.m \end{gathered}$$

Hence the torque is, 0N.m.