College Physics (4th Edition)

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

Chapter 3 - Problems - Page 118: 112


(a) $\theta = 21.8^{\circ}$ (b) $a = 0.91~m/s^2$

Work Step by Step

(a) When the brick starts to slide, we can assume that the magnitude of the brick's weight directed down the slope is equal in magnitude to the maximum possible force of static friction at the angle $\theta$: $mg~sin~\theta = mg~cos~\theta~\mu_s$ $tan~\theta = \mu_s$ $\theta = tan^{-1}(\mu_s)$ $\theta = tan^{-1}(0.40)$ $\theta = 21.8^{\circ}$ (b) We can find the acceleration as the brick slides down the board: $\sum F = ma$ $mg~sin~\theta-mg~cos~\theta~\mu_k = ma$ $a = g~(sin~\theta-\mu_k~cos~\theta)$ $a = (9.80~m/s^2)~(sin~21.8^{\circ}-0.30~cos~21.8^{\circ})$ $a = 0.91~m/s^2$
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