Let the length of the wire be $l$ and the maximum angle of displacement be $\theta_0$. When the load is released from angle $\theta_0$, it falls through a vertical height $h = l - l\cos\theta_0 = l(1 - \cos\theta_0)$. By conservation of mechanical energy, the kinetic energy at the lowest (equilibrium) position is equal to the loss in potential energy: $\frac{1}{2}mv^2 = mgh = mgl(1 - \cos\theta_0)$ $v^2 = 2gl(1 - \cos\theta_0)$ At the equilibrium position, the tension $T$ is maximum and provides the required centripetal force: $T - mg = \frac{mv^2}{l}$ $T = mg + \frac{m}{l}[2gl(1 - \cos\theta_0)]$ $T = mg + 2mg(1 - \cos\theta_0) = mg(3 - 2\cos\theta_0)$ Given maximum tension $T_{\text{max}} = 2940 \text{ N}$, mass $m = 150 \text{ kg}$, and taking acceleration due to gravity $g = 9.8 \text{ m/s}^2$: $2940 = 150 \times 9.8 \times (3 - 2\cos\theta_0)$ $2940 = 1470(3 - 2\cos\theta_0)$ $2 = 3 - 2\cos\theta_0$ $2\cos\theta_0 = 1 \implies \cos\theta_0 = \frac{1}{2}$ $\theta_0 = 60^\circ$