Question

A steel wire can withstand a load up to2940N. A load of150kgis suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is (in degrees)

Answer 1

$$\begin{gathered} \mathrm{Required}\mathrm{condition}: \\ \frac{{\mathrm{mv}}^2}{\mathrm{L}}+\mathrm{mg}=\mathrm{F} \\ \frac{\mathrm{m}}{\mathrm{L}}\left[2\mathrm{gL}\left(1-\mathrm{cos\theta }\right)\right]+\mathrm{mg}=\mathrm{F} \\ 2\mathrm{mg}\left(1-\mathrm{cos\theta }\right)+\mathrm{mg}=\mathrm{F} \\ 3\mathrm{mg}-2\mathrm{mgcos\theta }=\mathrm{F} \\ 2\mathrm{mgcos\theta }=3\mathrm{mg}-\mathrm{F} \\ \mathrm{cos\theta }=\frac{3\mathrm{mg}-\mathrm{F}}{2\mathrm{mg}} \\ =\frac{\left(3\times 150\times 9.8\right)-2940}{\left(2\times 150\times 9.8\right)} \\ =0.5 \\ \mathbf{\theta }\mathbf{=}{\mathbf{60}}^{\mathbf{0}} \end{gathered}$$