College Physics (4th Edition)

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

Chapter 8 - Problems - Page 318: 94

Answer

$\tau = 98~N \cdot m$

Work Step by Step

We can find the torque due to the weight about the horizontal axis: $\tau = r~F$ $\tau = r~mg$ $\tau = (1.0~m)(10.0~kg)(9.80~m/s^2)$ $\tau = 98~N \cdot m$
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