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Find the minimum value of $F$ for the block to topple about an edge.

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$F_{m i n} = \frac{M g}{2 b}$

$F_{m i n} = M g a b$

$F_{m i n} = \frac{M g a}{2 b}$

$F_{m i n} = \frac{M g a}{4 b}$

answer is C.

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Detailed Solution

When the block is about to topple the normal reaction $N$ shifts to the edge through $O$ . FBD during toppling taking torque about $O$

$F (b) = M g (\frac{a}{2}) \Rightarrow F_{m i n} = \frac{M g a}{2 b}$ .

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