## Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson

# Chapter 11 - Virtual Work - Section 11.7 - Stability of Equilibrium Configuration - Problems - Page 608: 35

#### Answer

$k=2.81lb/ft$

#### Work Step by Step

We can determine the required stiffness of the springs as follows: According to the potential energy equation $V=V_g+V_e=Wy+\frac{1}{2}ks^2$ $\implies V=2(8)sin\theta +\frac{1}{2}k(4cos\theta-1)^2$ At equilibrium $\frac{dV}{d\theta}=0$ $\frac{d[2(8)sin\theta +\frac{1}{2}k(4cos\theta-1)^2)]}{d\theta}=16cos\theta-4ksin\theta(4cos\theta-2)=0$ We plug in $\theta=30^{\circ}$ $\implies 16cos30-4ksin30(4cos30-2)=0$ This simplifies to: $k=2.81lb/ft$

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