Question

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?

Short answer

1. The magnitude of the torque about the pedal arm’s pivot when the arm is at angle 30.0^o is 8.4N.m.
2. The magnitude of the torque about the pedal arm’s pivot when the arm is at angle 90.0^o is 17N.m.
3. The magnitude of the torque about the pedal arm’s pivot when the arm is at angle 180.0^o is 0N.m.

Step-by-step solution

Understanding the given information

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. ${\theta }_3={180}^o$
   

Concept and formula used in the given question

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$\mathit{\tau }\mathbf{=}\mathit{r}\mathit{F}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

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

(a) Calculation for the magnitude of the torque about the pedal arm's pivot when the arm is at angle 30o with the vertical

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 .

(b) Calculation for the magitude of the torque about the pedal arm's pivot when the arm is at angle 90o with the vertical. 

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 .

(c) Calculation for the magnitude of the torque about the pedal arm's pivot when the arm is at angle 180o with the vertical.

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.