Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 9 - Rotational Dynamics - Focus On Concepts - Page 242: 3


b) is correct.

Work Step by Step

Forces $F_1$ and $F_4$ pass through the point where the axis of rotation also passes, so they produce no torques. Take $l$ to be the distance from the axis of rotation to the point where forces $F_2$ and $F_3$ are applied and $\theta$ to be the angle $F_2$ makes with the door. Then we have $$\tau_2=Fl\sin\theta$$ $$\tau_3=Fl$$ Because $l\gt l\sin\theta$, $\tau_3\gt\tau_2$ So to rank the torques from largest to smallest, $\tau_3\gt\tau_2\gt\tau_1=\tau_4$
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