First slide

Rotational motion

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 b above its center of mass (see figure). If the coefficient of friction is $μ = 0.4$ , the maximum possible value of $100 \times \frac{b}{a}$ for box not to topple before moving is _____

Moderate

Solution

$\begin{array}{l} S l i d i n g s h o u l d n o t o c c u r, F = μ m g ................ (1) \\ T o p p l i n g s h o u l d n o t o c c u r, F (b + \frac{a}{2}) = m g \frac{a}{2} ............... (2) \\ \Rightarrow μ m g (b + \frac{a}{2}) = m g \times \frac{a}{2} \\ \Rightarrow (b + \frac{a}{2}) μ = \frac{a}{2} \\ \Rightarrow 0.4 = μ = \frac{a}{2 b + a} \\ \Rightarrow 0.8 b + 0.4 a = a \\ \Rightarrow 0.8 b = 0.6 a \\ \Rightarrow \frac{b}{a} = \frac{3}{4} \\ \Rightarrow \frac{b}{a} \times 100 = \frac{3}{4} \times 100 = 75 \end{array}$

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