College Physics (4th Edition)

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

Chapter 2 - Problems - Page 68: 70

Answer

$\theta = 16.7^{\circ}$

Work Step by Step

Let $\theta$ be the angle between the ramp and the ground when the box starts to slide. At that angle, the force of friction exerted on the box is equal in magnitude to the component of the box's weight directed down the ramp. $mg~sin~\theta = F_N~\mu_s$ $mg~sin~\theta = mg~cos~\theta~\mu_s$ $tan~\theta = \mu_s$ $\theta = tan^{-1}~\mu_s$ $\theta = tan^{-1}~0.30$ $\theta = 16.7^{\circ}$
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