University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 5 - Applying Newton's Laws - Problems - Exercises - Page 162: 5.25

Answer

$\theta = 50.2^{\circ}$

Work Step by Step

We can let the friction force directed up the incline be equal in magnitude to the component of the person's weight directed down the incline. $mg~sin(\theta) = mg~cos(\theta)~\mu_s$ $tan(\theta) = \mu_s$ $\theta = arctan(\mu_s) = arctan(1.20)$ $\theta = 50.2^{\circ}$
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