Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 351: 65

Answer

$30.6 \;\rm N$

Work Step by Step

The pole is in equilibrium which means that the net force exerted on it and the net torque exerted on it are zeros. We can see that the pole is uniform and so do the sign. From the geometry of the given figure, we can see that the angle $\theta$ is given by $$\theta=\tan^{-1}\left[ \dfrac{250}{200}\right]=\color{blue}{51.3^\circ}$$ Now we need to find the torque around the left end of the pole the one that touches the wall. We choose counterclockwise to be the positive torque direction. $$\sum \tau=F_{nV}(0)+F_{nH}(0)-\dfrac{m_{\rm sign}g}{2}(0.8)-m_{\rm pole}g(1.0)-\dfrac{m_{\rm sign}g}{2}(2.0)+T\sin\theta (2)+T\cos\theta(2)(0)=0$$ We know that the horizontal component of the tension force is doing zero torque on our pole since it is parallel to it. $$ - 0.4m_{\rm sign}g -m_{\rm pole}g - m_{\rm sign}g +2T\sin\theta =0$$ Hence, $$ -m_{\rm pole}g +2T\sin\theta =1.4 m_{\rm sign}g$$ $$ m_{\rm sign}= \dfrac{ -m_{\rm pole}g +2T\sin\theta }{1.4 g}$$ The maximum mass makes the tension maximized. Thus, plugging the known; $$ m_{\rm sign}= \dfrac{ (-5)(9.8) +(2)(300)\sin51.3^\circ}{1.4 (9.8)}$$ $$ m_{\rm sign}=\color{red}{\bf 30.6}\;\rm N$$
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