Physics (10th Edition)

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

Chapter 9 - Rotational Dynamics - Problems - Page 243: 6



Work Step by Step

We have $$\tau=Fl$$ Because $F$ is constant here, the change in torque depends on the change in the lever arm. In a square, the diagonal is the longest line. So the lever arm $l$ is maximum when the lever arm is half this diagonal. In other words, $$l_{max}=\frac{1}{2}\sqrt{0.4^2+0.4^2}=0.283m$$ Therefore, $$\tau_{max}=(15N)\times(0.283m)=4.25N.m$$
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