Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 116: 66

Answer

$\theta_{max}=42^o$

Work Step by Step

According to Newton's 3rd Law of Motion, if $F$ is the force exerted on each crutch by the ground, there has to be a $P$ force in equal magnitude but in opposite direction exerted on the ground by the crutch. We will be concerned with force $P$ here, as portrayed in the free-body diagram below. $\vec{P}$, in the diagram, is analyzed $\vec{P}\cos\theta$ and $\vec{P}\sin\theta$. - $\vec{P}\sin\theta$ is the force that causes the crutch to slip, if it overcomes frictional force $f_s^{max}$. Since we are finding $\theta_{max}$ so that the crutch does not slip, we take that $$P\sin\theta_{max}=f_s^{max}=\mu_sF_N$$ - On the vertical sides, we have $\vec{P}\cos\theta$ and $\vec{F}_N$ in opposite directions. Since there is no vertical movement, we have $$F_N=P\cos\theta$$ Therefore, $$P\sin\theta_{max}=\mu_sP\cos\theta_{max}$$ $$\frac{\sin\theta_{max}}{\cos\theta_{max}}=\tan\theta_{max}=\mu_s=0.9$$ $$\theta_{max}=42^o$$
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