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 609: 38

Answer

$k=157N/m$ stable at $\theta=60^{\circ}$

Work Step by Step

We can determine the required stiffness of the spring as follows: According to the potential energy equation $V=V_g+V_e=W_1y_1+W_2y_2+\frac{1}{2}ks^2$ $\implies V=8(9.81)(0.75)cos\theta+8(9.81)(2.25)cos\theta+\frac{1}{2}k(3cos\theta-1)^2$ $\implies V=235.44cos\theta+\frac{1}{2}k(3cos\theta-1)^2$ At equilibrium $\frac{dV}{d\theta}=0$ $\implies \frac{d(235.44cos\theta+\frac{1}{2}k(3cos\theta-1)^2)}{d\theta}=-235.44sin\theta-3ksin\theta(3cos\theta-1)=0$ It is given that $\theta=60^{\circ}$ $\implies -235.44sin60-3ksin60(3cos60-1)=0$ This simplifies to: $k=157N/m$
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