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 215: 26



Work Step by Step

As we know that $N=Tsin(\theta)$ and the force of friction is given as $f=\mu N$ $\implies f=\mu Tsin(\theta)$ Now we can find the torque as $fR-Tcos(\theta)(\frac{1}{2}R)=0$ This simplifies to: $\mu sin(\theta)-cos(\theta)(\frac{1}{2})=0$ $\implies \mu=\frac{1}{2tan(\theta)}=\frac{1}{2tan(30)}=0.87$
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