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Q.

A uniform cubical box of side a is placed on a rough floor and it is to be moved by applying minimum possible force F at a point which is at a distance b above its centre of mass. If the coefficient of friction is $μ = 0 .4$ , then what is the maximum possible value of $100 \times \frac{b}{a}$ for which the box will not topple before moving?

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answer is 50.

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Detailed Solution

For no toppling
$\begin{array}{l} F (\frac{a}{2} + b) \leq mg \frac{a}{2} \\ \Rightarrow μ \frac{a}{2} + μb \leq \frac{a}{2} \\ \Rightarrow 0 .2 a + 0 .4 b \leq 0 .5 a \\ \Rightarrow 0 .4 b \leq 0 .3 a \\ \Rightarrow b \leq 0 .75 a \end{array}$
(in limiting case)
But it is not possible as maximum value of b can be equal to 0.5a only
${(100 \frac{b}{a})}_{max} = 50 .00$

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