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 217: 47

Answer

$F=Mgtan(\frac{\theta}{2})$

Work Step by Step

We call the center of the wheel the axis of rotation. We know that the wheel will not rotate when the net torques are zero. We know that the torque due to friction is given by: $\tau = r Mg sin(.5\theta) cos(.5\theta)$ Where r is the radius of the wheel. We know that the force on the wheel is given by: $ Fcos^2(.5\theta)$ Setting these two expressions equal and solving for F gives: $F=Mgtan(\frac{\theta}{2})$
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