Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 9 - Rotational Dynamics - Problems - Page 245: 20



Work Step by Step

We have $$f_s=\mu_sF_N$$ The normal force $F_N$ here is the y-component of the force the ground exerts on the ladder $G$, so $$f_s=\mu_sG_y$$ Because the ladder does not slip horizontally, $f_s$ balances $G_x$, so $$f_s=G_x$$ $$\mu_sG_y=G_x$$ $$\mu_s=\frac{G_x}{G_y}=\frac{727N}{1230N}=0.591$$
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