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: 27


Unstable at $\theta=34.6^{\circ}$; stable at $\theta=145^{\circ}$

Work Step by Step

We can determine the required positions for equilibrium and the stability at these positions as follows: As given that, $V=12sin2\theta+15cos\theta$ But for equilibrium, $\frac{dV}{d\theta}=0$ $\implies \frac{d(12sin2\theta+15cos\theta)}{d\theta}= 24cos2\theta-15sin\theta=0$eq(1) $\implies cos2\theta=\frac{5}{8}sin\theta$ This simplifies to: $\theta_1=34.6^{\circ}$ and $\theta_2=145^{\circ}$ To check the stability at the above two positions, we take the second derivative of eq(1) $\frac{d^2(12sin2\theta+15cos\theta)}{d\theta^2}=-48sin2\theta-15cos\theta$ Now, we plug in the values of $\theta_1$ and $\theta_2$ into the above equation For $\theta_1=34.6^{\circ}$ $\implies \frac{d^2V}{d\theta^2}=-48sin(2(34.6^{\circ}))-15cos 34.6^{\circ}=-57.22\lt 0$ which means that it is in unstable equilibrium at this position. For $\theta_2=145^{\circ}$ $\frac{d^2V}{d\theta^2}=-48sin(2(145^{\circ}))-15cos145^{\circ}=56.21\gt 0$ which means, there is a stable equilibrium at this position.
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