Friction between the wedge shown and the horizontal floor is sufficient to prevent the wedge from sliding. The mass of the wedge is M and its angle of inclination is $θ$ . A small block of mass m is just placed near the top of the wedge and released. Friction coefficient between the wedge and the block is $μ$

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The wedge will topple for any value of m irrespective of $μ$ .

If $μ > \tan θ$ , the wedge will not topple for any value of m.

If $μ < \tan θ$ , the wedge will topple for $m > \frac{M}{3 \sin θ (\sin θ - μ \cos θ)}$

If $μ < \tan θ$ , the wedge will not topple for $m > \frac{M}{3 \sin θ (\sin θ - μ \cos θ)}$

answer is A, B.

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Detailed Solution

$N \sin θ h > M g \frac{b}{3} + μ m g \cos θ h \cos θ$

$m g \cos θ . \sin θ h > M g \frac{b}{3} + μ m g h \cos^{2} θ$

$3 m \sin θ (\sin θ - μ \cos θ) > M$

$m > \frac{M}{3 \sin θ (\sin θ - μ \cos θ)}$

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