## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 12 - Exercises and Problems - Page 216: 33

#### Answer

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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