Questions: A bicycle pedal is pushed straight downwards by a foot with a 12 Newton force. The shaft of the pedal is 20 cm long. If the shaft is π / 3 radians past horizontal, what is the magnitude of the torque about the point where the shaft is attached to the bicycle? Nm

Solution Steps

• Force applied, \( F = 12 \, \text{N} \)
• Length of the pedal shaft, \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)
• Angle past horizontal, \( \theta = \frac{\pi}{3} \, \text{radians} \)

Torque (\( \tau \)) is calculated using the formula: \[ \tau = r \cdot F \cdot \sin(\theta) \] where \( r \) is the distance from the pivot point to where the force is applied, \( F \) is the force, and \( \theta \) is the angle between the force vector and the lever arm.

Calculate \(\sin\left(\frac{\pi}{3}\right)\): \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

Substitute the known values into the torque formula: \[ \tau = 0.2 \, \text{m} \cdot 12 \, \text{N} \cdot \frac{\sqrt{3}}{2} \]

Calculate the torque: \[ \tau = 0.2 \cdot 12 \cdot \frac{\sqrt{3}}{2} = 1.2 \cdot \frac{\sqrt{3}}{2} = 1.2 \cdot 0.866 \approx 1.0392 \, \text{Nm} \]

Final Answer