An equilateral prism of mass m rests an 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 $F = \frac{mg}{\sqrt{x}}$ Then the value of x is____________________

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answer is 3.

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Detailed Solution

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

$F (\frac{\sqrt{3}}{2} a) = mg (\frac{a}{2}) \Rightarrow F = \frac{mg}{\sqrt{3}}$

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