Question

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

Answer 1

Step 1. Given Data,

Side of a uniform cubical box $=a$

Distance between center of mass and force $=b$

Force above its center of mass $=F$

Coefficient of friction $\mu =0.4$,

Step 2. We have to calculate the maximum value of $100\times \frac{b}{a}$,

Force balances kinetic friction so that the block can move.

So, Force $F=\mu mg$(where, $\mu$is frictional constant, $m$is mass, $g$is gravity)

For no toppling, the net torque about the bottom right edge should be zero.

$$\begin{gathered} F\left[\left(\frac{a}{2}\right)+b\right]\le mg\left(\frac{a}{2}\right) \\ \Rightarrow \mu mg\left[\left(\frac{a}{2}\right)+b\right]\le mg\left(\frac{a}{2}\right) \\ \Rightarrow \mu \left[\left(\frac{a}{2}\right)+b\right]\le \left(\frac{a}{2}\right) \\ \Rightarrow 0.2a+0.4b\le 0.5a \\ \Rightarrow 0.4b\le 0.3a \\ \Rightarrow b\le \left(\frac{3}{4}\right)a \end{gathered}$$

But, the maximum value of $b$ can only be $0.5$ .

Hence the maximum value of $100\times \frac{b}{a}$ is $50.$