Answer
$176N/m$
Work Step by Step
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:
$K=176N/m$
