Answer
The spring is stretched a distance of $0.792~m$
Work Step by Step
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 $0.792~m$.
