Answer
$\tau_{net} = 5.0~N~m$
Work Step by Step
We can find the magnitude of the net torque about the axle as:
$\tau_{net} = \sum \tau$
$\tau_{net} = r_1~F_1 + r_2~F_2$
$\tau_{net} = (0)(50~N)+(0.10~m)(50~N)$
$\tau_{net} = 5.0~N~m$
Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)
by Knight, Randall D.
Published by
Pearson
ISBN 10:
0321740904
ISBN 13:
978-0-32174-090-8
Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 349: 19
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