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

$2.03m$
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: $\implies s=2.03m$