College Physics (4th Edition)

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

Chapter 8 - Problems - Page 312: 21


(a) $\tau = 58.5~N\cdot m$ (b) $\tau = 39.9~N\cdot m$ (c) $\tau = 0$

Work Step by Step

The magnitude of the torque can be expressed as $\tau = r~F~sin~\theta$, where $r$ is the displacement between the rotation axis and the point where the force is applied, $F$ is the force, and $\theta$ is the angle between the force vector and the displacement vector $r$. We can find the magnitude of the torque for each situation: (a) $\tau = r~ F~sin~\theta$ $\tau = (1.26~m)(46.4~N)~sin~90^{\circ}$ $\tau = 58.5~N\cdot m$ (b) $\tau = r~ F~sin~\theta$ $\tau = (1.26~m)(46.4~N)~sin~43.0^{\circ}$ $\tau = 39.9~N\cdot m$ (c) $\tau = r~ F~sin~\theta$ $\tau = (1.26~m)(46.4~N)~sin~0^{\circ}$ $\tau = 0$
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