Question

1. lever arm you open a door by pushing at a right angle to the door. your friend pushes at the same place, but at an angle of 55° from the perpendicular. if both you and your friend exert the same torque on the door, how do the forces you and your friend applied compare?

Explanation:

Step1: Recall Torque Formula

Torque \(\tau\) is given by \(\tau = rF\sin\theta\), where \(r\) is the distance from the pivot (constant here), \(F\) is the force, and \(\theta\) is the angle between the position vector and the force vector.
For you: \(\theta_1 = 90^\circ\), so \(\tau_1 = rF_1\sin90^\circ = rF_1\) (since \(\sin90^\circ = 1\)).
For your friend: \(\theta_2 = 55^\circ\), so \(\tau_2 = rF_2\sin55^\circ\).

Step2: Set Torques Equal

Given \(\tau_1 = \tau_2\), so \(rF_1 = rF_2\sin55^\circ\).
Cancel \(r\) from both sides: \(F_1 = F_2\sin55^\circ\).
Solve for \(F_2\): \(F_2 = \frac{F_1}{\sin55^\circ}\).
Calculate \(\sin55^\circ \approx 0.8192\), so \(F_2 \approx \frac{F_1}{0.8192} \approx 1.22F_1\).

Answer:

Your friend's force (\(F_2\)) is approximately \(\frac{1}{\sin55^\circ}\) (or about 1.22 times) your force (\(F_1\)), i.e., \(F_{\text{friend}}=\frac{F_{\text{you}}}{\sin55^\circ}\) (or \(F_{\text{friend}}\approx1.22F_{\text{you}}\)).