Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 12 - Exercises and Problems - Page 216: 33


The proof is below.

Work Step by Step

The ladder is at static equilibrium, so the net torques and the net forces are 0. Thus, we use the fact that the x and y forces are 0 to find: $F_{n1}\mu_1=F_{n2}$ $mg = F_{n1}+F_{n2}\mu_2$ We know that the torques sum to zero, so using the bottom of the latter as the axis of rotation, we find: $\frac{mgsin(90+\phi)}{2}=-\mu_2F_{n2}sin(90-\phi)-F{n2}sin(-\phi+180)$ Combining these equations, we obtain: $\phi = tan^{-1}(\frac{1-\mu_1\mu_2}{2\mu_1})$
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