Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 11 - Virtual Work - Section 11.3 - Principle of Virtual Work for a System of Connected Rigid Bodies - Problems - Page 595: 18

Answer

$\theta=41.2^{\circ}$

Work Step by Step

We can determine the required angle $\theta$ as follows: $F_s=k(600sin\theta-600sin15)-(200sin\theta-200sin15)$ $\implies F_s=6000(sin\theta-sin15)$ The virtual displacements are given as $\delta_{yA}=\frac{d(200sin\theta)}{d\theta}=200cos\theta$ $\delta_{yB}=\frac{d(600sin\theta)}{d\theta}=200cos\theta$ and $\delta_{yC}=\frac{d(1600sin\theta)}{d\theta}=1600cos\theta$ Now, according to the virtual-work equation $\delta U=0$ $\implies P\delta_{yC}+F_s(\delta_{yB}-\delta_{yB})=0$ We plug in the known values to obtain: $600(1600cos\theta)-6000(sin\theta-sin15)(600cos\theta-200cos\theta)=0$ This simplifies to: $\theta=41.2^{\circ}$
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