College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 8 - Multiple-Choice Questions - Page 310: 9


The correct answer is: (c) $(\frac{mg}{2})~sin~\theta$

Work Step by Step

In general, the torque is $\tau = r\times F$, where $r$ is the smallest distance between the force vector and the axis, and $F$ is the perpendicular component of the force vector. To hold the bar in equilibrium, the force F must produce a torque that is equal in magnitude to the torque produced by gravity. Let $L$ be the length of the bar. Note that we can consider that the force of gravity is applied at the midpoint of the bar. We can find the magnitude of the torque produced by gravity: $\tau_g = mg~(\frac{L}{2})~sin~\theta$ We can equate the magnitudes of the torques to find $F$: $L\times F = mg~(\frac{L}{2})~sin~\theta$ $F = (\frac{mg}{2})~sin~\theta$ The correct answer is: (c) $(\frac{mg}{2})~sin~\theta$
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