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

Published by Pearson

# Chapter 12 - Exercises and Problems - Page 218: 53

#### Answer

$\mu \gt \frac{tan\theta}{2+tan^2\theta}$

#### Work Step by Step

We call the point where the string attaches to the bar the axis of rotation. Thus, we find where the net torques will be zero to find when the bar will not slip. $\frac{L}{2}Mgsin\theta=\mu L F_n$ We now must find the value of the normal force, using the fact that the bar is not moving: $F_n = Tsin\theta+mgcos\theta$ We use the fact that the bar is not moving in the other direction to find tension: $Tcos\theta+\mu F_n = mgsin\theta$ $T =\frac{-\mu F_n + mgsin\theta}{cos\theta}$ We substitute this into the equation for the normal force to find: $F_n =(-\mu F_n + mgsin\theta) tan\theta+mgcos\theta$ From the first equation, we know that $Mgsin\theta = 2\mu F_n$, so it follows: $F_n =(-\mu F_n + 2\mu F_n) tan\theta+mgcos\theta$ $F_n =(\mu F_n ) tan\theta+mgcos\theta$ We also know that $mgcos\theta=\frac{2\mu F_ncos\theta}{sin\theta}$, so we find: $F_n =(\mu F_n ) tan\theta+\frac{2\mu F_ncos\theta}{sin\theta}$ $1=(\mu ) tan\theta+\frac{2\mu cos\theta}{sin\theta}$ Using simplification, we find: $\mu \gt \frac{tan\theta}{2+tan^2\theta}$

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