Question

The length of a bicycle pedal arm is 0.177 m, and a downward force of 145 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 an angle of (a)48.6°, (b) 90°, and (c) 180° with the vertical?

Answer 1

Answer:

(a) 19.25 N-m

(b) 25.67 N-m

(c) 0 N-m

Explanation:

Given:

Length of the pedal arm (L) = 0.177 m

Downward force $(|\vec{F}|)$ = 145 N

Magnitude of torque is given by the formula:

$T=FL\sin\theta$

Where, $\theta\to angle\ between\ F\ and\ L$

(a)

Given:

$\theta=48.6$°

Therefore, torque is given as:

$T=FL\sin\theta\\\\T=(145\ N)(0.177\ m)(\sin(48.6))\\\\T=19.25\ Nm$

Therefore, the torque is 19.25 N-m.

(b)

Given:

$\theta=90$°

Therefore, torque is given as:

$T=FL\sin\theta\\\\T=(145\ N)(0.177\ m)(\sin(90))\\\\T=25.67\ Nm$

Therefore, the torque is 25.67 N-m.

(c)

Given:

$\theta=180$°

Therefore, torque is given as:

$T=FL\sin\theta\\\\T=(145\ N)(0.177\ m)(\sin(180))\\\\T=0\ Nm$

Therefore, the torque is 0 N-m.