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

Answer: 50.00

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