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, $$\tau_{max}=(15N)\times(0.283m)=4.25N.m$$
Small 1582256016
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.