Question

The length of a bicycle pedal arm is 0.152 m, and a downward force of 111 N 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°, (b) 90°, and (c) 180° with the vertical?

Answer 1

Answer:

8.436 Nm

16.872 Nm

0

Explanation:

When we multiply the force with the distance at which the force is being applied we get torque

$\tau=Frsin\theta$

where

F = Force = 111 N

r = Distance = 0.152 m

$\theta$ = Angle at which force is applied

When, $\theta=30^{\circ}$

$\tau=111\times 0.152sin30\\\Rightarrow \tau=8.436\ Nm$

The torque is 8.436 Nm

When, $\theta=90^{\circ}$

$\tau=111\times 0.152sin90\\\Rightarrow \tau=16.872\ Nm$

The torque is 16.872 Nm

When, $\theta=180^{\circ}$

$\tau=111\times 0.152sin180\\\Rightarrow \tau=0\ Nm$

The torque is 0