Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 11 - Virtual Work - Section 11.7 - Stability of Equilibrium Configuration - Problems - Page 608: 36


$\theta=12.1^{\circ}$, unstable equilibrium

Work Step by Step

We can determine the required angle $\theta$ as follows: According to the potential energy equation $V=V_{g,bar}-V_{g,block}=2W_{bar}y_1-W_{block}y_2$ $\implies V=2(3)(250)sin\theta-7(1500-1000cos\theta)$ At equilibrium $\frac{dV}{d\theta}=0$ $\implies \frac{d[2(3)(250)sin\theta-7(1500-1000cos\theta)]}{d\theta}=1500cos\theta-700sin\theta=0$ This simplifies to: $\theta=12.095^{\circ}\approx12.1^{\circ}$ Now, $\frac{d^2V}{d\theta^2}=-1500sin\theta-700cos\theta$ For $\theta=12.095^{\circ}$ $\implies \frac{d^2V}{d\theta^2}=-1500sin(12.095^{\circ})-700cos(12.095^{\circ})=-7158.9\lt 0$ which implies unstable equilibrium.
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