Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 9 - Static Equilibrium; Elasticity and Fracture - General Problems - Page 258: 71

Answer

$\theta = 45^{\circ}$.

Work Step by Step

Consider the beam on the left. There is a normal force from the floor pushing up. There is gravity pulling down, at the center of mass. There is a frictional force from the floor, acting toward the left, of maximum magnitude $\mu_sF_N=\mu_s mg$. For the beam to be in equilibrium, the normal force balances the weight of the beam. The net torque must be zero. By taking torques about the top of the beam, we manage to ignore the force from the other beam, since r = 0. Let clockwise torque be in the positive direction. $$\Sigma \tau=0=F_NL cos \theta-mg(\frac{L}{2})cos \theta- F_{friction}L sin \theta $$ $$0=mgL cos \theta-mg(\frac{L}{2})cos \theta- \mu_s mg L sin \theta $$ $$0=cos \theta- (\frac{1}{2})cos \theta- \mu_s sin \theta $$ $$0=cos \theta- (\frac{1}{2})cos \theta-\frac{1}{2} sin \theta $$ $$sin \theta=cos \theta$$ $$tan \theta=1$$ $$\theta = 45^{\circ}$$
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