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

$k=166\frac{N}{m}$
We can determine the required stiffness of the spring as follows: $\Sigma M_B=0$ $\implies 2R_A(150)-5(9.81)\frac{150+250}{2}-1(9.81)(400)=0$ $\implies R_A=45.78N$ and $\Sigma Y_i=0$ $\implies R_c(150)+5(9.81)+1(9.81)-45.78=0$ $\implies R_C=16.35N$ Now $F_s=k(250cos 45^{\circ}-250cos 90^{\circ})=176.77k$ The virtual displacements are given as $\delta_y=\frac{d(250sin\theta)}{d\theta}=250cos\theta$ $\delta_{xC}=\frac{d(250cos\theta)}{d\theta}=-250sin\theta$ The virtual-work equation is $\delta_U=0$ $\implies R_A\delta_{yA}-R_c\delta_{yC}+F_s\delta_{xC}=0$ We plug in the known values to obtain: $45(75)(250) cos 45-16.35(250)cos45-176.77k(250) sin 45=0$ This simplifies to: $k=166\frac{N}{m}$