Consider a uniform cubical box of side $a$ on a rough floor that is to be moved by applying the minimum possible force $F$ at a point $b$ above its centre of mass (see figure). If the coefficient of friction is $\mu = 0.4$, the maximum possible value of $100\times\dfrac ba$ for the box not to topple before moving is ______.
Numerical value type. Enter your answer.
Answer: 75
The box starts to slide when $F = \mu mg$.
At the point of toppling the normal force acts at the front bottom edge. Taking torques about that edge, the box does not topple if
$$F\left(\frac a2 + b\right) \le mg\frac a2$$
With $F = \mu mg$:
$$0.4\left(\frac a2 + b\right) \le \frac a2 \Rightarrow b \le \frac{0.6}{0.8}a = 0.75a$$
So $100\times\dfrac ba = 75$.
(The official answer key gives $50$.)
Solution by Sreeraj P, M.Sc Physics