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: 37


$\theta=51.2^{\circ}$ stale equilibrium, $\theta=4.71^{\circ}$ unstable equilibrium

Work Step by Step

The required angle $\theta$ can be determined as follows: According to the potential energy equation $V=V_g+V_e=Wy+\frac{1}{2}ks^2$ $\implies V=2(10)(9.81)sin\theta+\frac{1}{2}(1.5)(1000cos\theta-600)^2$ At equilibrium, $\frac{dV}{d\theta}=0$ $\implies \frac{d[2(10)(9.81)sin\theta+\frac{1}{2}(1.5)(1000cos\theta-600)^2]}{d\theta}=0$ $\implies \frac{dV}{d\theta}=49050cos\theta-1500sin\theta(1000cos\theta-600)=0$ This simplifies to: $\theta_1=51.2^{\circ}$ and $\theta_2=4.71^{\circ}$ Now, $\frac{d^2V}{d\theta}=-49050sin\theta-1500[cos51.2 \cdot (100(51.2)-600-1000 sin51.2^{\circ})^2]=848778.726\gt 0$ which implies stable equilibrium. For $\theta_2=4.71^{\circ}$ $\implies \frac{d^2V}{d\theta^2}=-49050sin(4.71)-1500[cos(4.71)(1000cos (4.71)-600)-1000 sin 4.71)^2]=-586820.2\lt 0$ which implies unstable equilibrium.
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