Question

A solid cube of side ${}^{\prime }{a}^{\prime }$ is shown in the figure. Find maximum value of $100\frac{b}{a}$ for which the block does not topple before sliding.

Answer 1

For just sliding condition.
$F=\mu mg$ ....(i)


Just before toppling, normal reaction will act at the edge as shown.
For no toppling (taking torque balance about point $A$):
$F(\frac{a}{2}+b)\le mg\frac{a}{2}$
$\Rightarrow \mu \frac{a}{2}+\mu b\le \frac{a}{2}$ [from eqn (i)]
$\Rightarrow 0.2a+0.4b\le 0.5a$ [given $\mu =0.4$]
$\Rightarrow 0.4b\le 0.3a$
$\Rightarrow b\le \frac{3a}{4}=0.75a$ (in limiting case)
But it is not possible, as $b$ can have a maximum value equal to $0.5a$ (from figure)
Hence, $b_{max}=0.5a$
and ${\left(100\frac{b}{a}\right)}_{\text{max}}={\left(100\frac{0.5a}{a}\right)}_{\text{max}}=50$