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An equilateral prism of mass m rests on a rough horizontal surface with coefficient of friction $μ$ . A horizontal force F is applied on the prism as shown in figure. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, the minimum force required to topple the prism is

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$\frac{mg}{\sqrt{3}}$

$\frac{mg}{4}$

$\frac{2 mg}{\sqrt{3}}$

$\frac{7 mg}{4}$

answer is A.

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Detailed Solution

The tendency of rotating will be about point C. For the minimum force, the torque of F about C has to be equal to the torque of mg about C.

$\begin{array}{l} F (a \frac{\sqrt{3}}{2}) = mg (\frac{a}{2}) \\ \Rightarrow F = \frac{mg}{\sqrt{3}} \end{array}$

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