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

$176N/m$
We can find the required stiffness of the spring as follows: $F_{spring}=K\Delta x$ $\implies F_{spring}=K(3.61-2)=1.61KN$ Now $\Sigma F_x=0$ $\implies -F_{CD}cos(45)+F_{spring} cos(33.7)=0$ $\implies F_{CD}=1.89KN$ and $\Sigma F_y=0$ $\implies F_{CD}sin (45)+F_{spring} sin (33.7)-W=0$ We plug in the known values in the above equation to obtain: $1.89K.sin(45)+1.61K.sin(33.7)-392.4=0$ This simplifies to: $K=176N/m$