## College Physics (4th Edition)

The spring is stretched a distance of $0.792~m$
We can consider the torques about the rotation axis at the corner. The object's weight exerts a clockwise torque, and the spring exerts a counterclockwise torque. Since the system is in equilibrium, the magnitude of these torques must be equal. Let $L$ be the length of the metal object. We can find the force $F$ exerted by the spring: $\tau_{ccw} = \tau_{cw}$ $L~F~sin~\theta = mg~(\frac{L}{2})~cos~\theta$ $F = (\frac{mg}{2})~cot~\theta$ $F = \frac{(53.0~kg)(9.80~m/s^2)}{2}~cot~50.0^{\circ}$ $F = 217.9~N$ We can find the distance the spring is stretched: $kx = F$ $x = \frac{F}{k}$ $x = \frac{217.9~N}{275~N/m}$ $x = 0.792~m$ The spring is stretched a distance of $0.792~m$.