Question

Minimum value of $F$ for which the cube begins to tip about an edge is equal to $\frac{2mg}{M}$. Then, $M$ is -

• A. $2$
• B. $4$
• C. $5$
• D. $3$

Answer 1

The correct option is D $3$

In the critical condition when the block is just about to topple, the normal will shift to rightmost edge as shown in figure.


Cube will have tendency to topple about $\text{O}$.
Torque of external forces acting on cube about point $\text{O}$,
${\tau }_f=0,{\tau }_F=F\times \frac{3a}{4},$
${\tau }_N=0,{\tau }_{mg}=mg\times \frac{a}{2}$

Block will topple if ${\tau }_{\text{clockwise}}>{\tau }_{\text{anticlockwise}}$
or $F\times \frac{3a}{4}>mg\times \frac{a}{2}$
$\therefore F>\frac{2mg}{3}$
Minimum value of $F$ is $\frac{2mg}{3}$ to topple the cube, hence $M=3$.