Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 18

Answer

The net torque about the axle is $-0.40~N~m$

Work Step by Step

We can find the net torque about the axle. Note that clockwise is the negative direction. Let $F_1 = 30~N$ and let $F_2 = 20~N$. $\tau_{net} = \sum \tau$ $\tau_{net} = r_1~F_1 + r_2~F_2$ $\tau_{net} = -(0.040~m)(30~N)+(0.040~m)(20~N)$ $\tau_{net} = -0.40~N~m$ The net torque about the axle is $-0.40~N~m$.
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