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Q.

Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of friction is $μ =$ 0.4 , the maximum possible value of $100 \times \frac{b}{a}$ for box not to topple before moving is

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answer is 50.

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Detailed Solution

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To prevent toppling

$F (\frac{a}{2} + b) \leq mg \frac{a}{2}$

$μ \frac{a}{2} + μb \leq \frac{a}{2} [∵ f = μmg]$
$0 .2 a + 0 .4 b \leq 0 .5 a \Rightarrow 0 .4 b \leq 0 .3 a$
$b \leq 0 .75 a$ (in limiting case)
But it is not possible as maximum value of b can be equal to 0.5a only(distance of centre of mass from edge )

$\Rightarrow \frac{b}{a} = 0.5 \Rightarrow \frac{b}{a} \times 100 = 0.5 \times 100 = 50$

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