Question

1. you stand on a bicycle pedal, as shown in figure 7. your mass is 65 kg. if the pedal makes an angle of 35° above the horizontal and the pedal is 18 cm from the center of the chain ring, how much torque would you exert?

Explanation:

Step1: Recall Torque Formula

Torque \(\tau\) is calculated by \(\tau = rF\sin\theta\), where \(r\) is the distance from the pivot, \(F\) is the force, and \(\theta\) is the angle between \(r\) and \(F\). Here, the force \(F\) is the weight, so \(F = mg\), with \(m = 65\space kg\), \(g = 9.8\space m/s^2\), \(r = 0.18\space m\) (converted from 18 cm), and \(\theta = 90^\circ - 35^\circ = 55^\circ\) (since the pedal is \(35^\circ\) above horizontal, the angle between \(r\) and the weight force is \(90^\circ - 35^\circ\)).

Step2: Calculate Force \(F\)

\(F = mg = 65 \times 9.8 = 637\space N\)

Step3: Calculate Torque \(\tau\)

Substitute \(r = 0.18\space m\), \(F = 637\space N\), and \(\theta = 55^\circ\) into the torque formula:
\(\tau = 0.18 \times 637 \times \sin(55^\circ)\)
First, \(\sin(55^\circ) \approx 0.8192\)
Then, \(\tau \approx 0.18 \times 637 \times 0.8192\)
\(0.18 \times 637 = 114.66\)
\(114.66 \times 0.8192 \approx 93.9\space N\cdot m\)

Answer:

Approximately \(93.9\space N\cdot m\) (or more precisely, around \(94\space N\cdot m\) depending on rounding during steps)