Question

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

Answer 1

For the box to slide
$F=\mu mg=0.4mg$

For the box not to topple, torque due to $F$ should be less than or equal to torque due to $mg$.

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

$\Rightarrow 0.4mg(\frac{a}{2}+b)\le mg\frac{a}{2}$

$\Rightarrow 0.2a+0.4b\le 0.5a$

$\Rightarrow \frac{b}{a}\le \frac{3}{4}$

$\Rightarrow b\le 0.75a$

But, here, for the box, the maximum value of $b$ can be equal to $0.5a$.

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

Hence, $50$ is the correct answer.