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 - Problems - Page 255: 30

Answer

$\theta$ must be at least $\arctan\frac{1}{(2\mu_s)}$

Work Step by Step

We sum the forces vertically to relate the normal force to the mass of the ladder: $(\uparrow +) \sum \overrightarrow{F}_{y} =0 $ $-mg+F_N=0$ $F_N=mg$ Then, we sum the torques about the top of the ladder and solve for $\theta$: $(+ \circlearrowleft) \sum \tau=0$ $-\cos\theta*F_N*l-\sin\theta*\mu_s*F_N*l+mg\cos\theta*1/2*l=0$ $-\cos\theta*F_N-\sin\theta*\mu_s*F_N+mg\cos\theta*1/2=0$ $-\cos\theta*F_N-\sin\theta*\mu_s*F_N+mg\cos\theta*1/2=0$ $-\cos\theta*F_N-\sin\theta*\mu_s*F_N+mg\cos\theta*1/2=0$ $\cos\theta(-F_N+mg*1/2)-\sin\theta*\mu_s*F_N=0$ $\cos\theta(-F_N+mg*1/2)=\sin\theta*\mu_s*F_N$ $\frac{-F_N+mg*1/2)}{\mu_s*F_N}=\frac{\sin\theta}{\cos\theta}$ $\frac{-mg+mg*1/2)}{\mu_s*mg}=\tan\theta$ $\frac{-1+1/2}{\mu_s}=\tan\theta$ $\frac{1}{2\mu_s}=\tan\theta$ $\theta=\arctan1/(2\mu_s)$ Therefore, $\theta$ must be at least $\arctan\frac{1}{(2\mu_s)}$.
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