Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems - Page 223: 27

Answer

$\Sigma \tau = mg \mathcal{l}_2-mg \mathcal{l}_1$ $clockwise$.

Work Step by Step

There is a torque due to the weight on the left that tends to accelerate the rod counterclockwise and a clockwise torque due to the weight on the right. Let’s call clockwise the positive direction. $$\Sigma \tau = mg \mathcal{l}_2-mg \mathcal{l}_1$$ The net torque is clockwise.
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