Question

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 $\mu =0.4,$ then what is the maximum possible value of $100\times \frac{b}{a}$ for which the box will not topple before moving?

Answer 1

For no toppling
$F\left(\frac{a}{2} + b\right)\leq mg\frac{a}{2}$
$\Rightarrow \mu \frac{a}{2}+\mu b\leq \frac{a}{2}$
$\Rightarrow 0.2a+0.4b\leq 0.5a$
$\Rightarrow \, 0.4b\leq 0.3a$
$\Rightarrow \, b\leq \frac{3 a}{4}$
$\Rightarrow \, b\leq 0.75a$
(in limiting case )
But it is not possible as maximum value of $b$ can be equal to $0.5a$ only
$\therefore \, \, \left(100 \frac{b}{a}\right)_{max}=50.00$