Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible…

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

Solution

For no toppling
$\mathrm{F}\left(\frac{\mathrm{a}}{2}+\mathrm{b}\right)\le \mathrm{m}\mathrm{g}\frac{\mathrm{a}}{2}$
$\mathrm{\mu }\frac{\mathrm{a}}{2}+\mathrm{\mu }\mathrm{b}\le \frac{\mathrm{a}}{2}$
$0.2\mathrm{a}+0.4\mathrm{b}\le 0.5\mathrm{a}$
$0.4\mathrm{b}\le 0.3\mathrm{a}$
$\mathrm{b}\le \frac{3\mathrm{a}}{2}$
$\mathrm{b}\le 0.75\mathrm{a}$ (in limiting case)
But it is not possible as $\mathrm{b}$ can maximum be equal to $0.5\mathrm{a}$
$\therefore {\left(100\frac{\mathrm{b}}{\mathrm{a}}\right)}_{\mathrm{m}\mathrm{a}\mathrm{x}}=50.00$

Asked in: JEE Main 2020 (07 Jan Shift 2)

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