Answer
$2.03m$
Work Step by Step
The required length can be determined as follows:
$W=mg$
$W=40(9.81)=392.4N$
and $F_{spring}=K\Delta x=180(3.61-s)$
The sum of forces in the $x$ direction is
$\Sigma F_x=0$
$\implies -F_{CD} cos(45)+F_{spring}cos(33.7)=0$...eq(1)
and the sum of forces in the $y$ direction is
$\Sigma F_y=0$
$\implies F_{CD} sin (45)+F_{spring} sin (33.7)-W=0$.....eq(2)
Adding eq(1) and eq(2), we obtain:
$F_{spring} sin (33.7)+F_{spring} cos (33.7)=392.4$
This simplifies to:
$F_{spring}=283N$
Now $F_{spring}=K\Delta x$
$\implies F_{spring}=180(3.61-s)$
This simplifies to:
$\implies s=2.03m$
