Physics - NLM Question with Solution | TestHub

FBD in frame of the car:

$$\begin{array}{rlrlrlrlrl}& \mathrm{T}\cos \alpha +\mathrm{ma}/2=\mathrm{mg}\dots (1)& \mathrm{ma}\frac{\sqrt{3}}{2}=\mathrm{T}\sin \alpha & \mathrm{T}=\mathrm{ma}\frac{\sqrt{3}}{2\sin \alpha }& \mathrm{ma}\frac{\sqrt{3}}{2}\cot \alpha +\frac{\mathrm{ma}}{2}=\mathrm{mg}& \frac{1}{2}(\sqrt{3}\cot \alpha +1)\mathrm{a}=\mathrm{g}& \sqrt{3}\cot \alpha +1=\frac{2\times 10}{5}& \cot \alpha =\sqrt{3}& \alpha =30^{\circ }& \mathrm{T}=\mathrm{ma}\frac{\sqrt{3}}{2}2=\mathrm{m}5\sqrt{3}\end{array}$$