Question

Force $F$ is applied horizontally at the vertex of equilateral triangular wedge of side $l$ and mass $m$ placed on the ground. Friction is considered to be absent everywhere. The minimum value of $F$ required to topple the wedge is

• A. $\frac{mg}{2}$
• B. $\frac{\sqrt{3}mg}{2}$
• C. $\frac{mg}{\sqrt{3}}$
• D. $mg$

Answer 1

The correct option is B $\frac{\sqrt{3}mg}{2}$

By force balance on the wedge$N=mg--\left(i\right)$Before it topples net torque at centre of mass (point P) is zero.
When it is about to topple the torque due to force F is balanced by the torque due to Normal reaction acting at point C.
$\therefore N\times \frac{l}{2}=F\times \frac{l}{\sqrt{3}}--\left(ii\right)$
By using equ (i) and (ii) we get
$F=\frac{\sqrt{3}mg}{2}$