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 607: 29


$\theta=38.7^{\circ}$ unstable, $\theta=90^{\circ}$ stable, $\theta=141^{\circ}$ unstable

Work Step by Step

We can determine the required equilibrium positions and stability at these positions as follows: As given that, $V=10cos2\theta+25sin\theta$ At equilibrium, $\frac{dV}{d\theta}=0$ $\implies \frac{d(10cos2\theta+25sin\theta)}{d\theta}=-20sin2\theta+25cos\theta=0$eq(1) This simplifies to: $\theta_1=38.7^{\circ}$ $\theta_2=90^{\circ}$ and $\theta_3=141^{\circ}$ Now, we take the second derivative of eq(1) and plug in these values, one-by-one to obtain: For $\theta_1=38.7^{\circ}$ $\frac{d^2V}{d\theta^2}=-40cos(2(38.7^{\circ}))-25sin(38.7^{\circ})=-24.375\lt 0$ which means unstable equilibrium. For $\theta_2=90^{\circ}$ $\frac{d^2V}{d\theta^2}=-40cos(2(90))-25sin90=15\gt 0$ which means stable equilibrium. For $\theta_3=141^{\circ}$ $\frac{d^2V}{d^2\theta}=-40cos(2(141))-25sin141=-24.375\lt 0$ which means unstable equilibrium.
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