Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 12 - Exercises and Problems - Page 225: 59

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: $\mu \gt \frac{tan\theta}{2+tan^2\theta}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.