**Problem:**
A cubical block of side L rests on a rough horizontal surface with coefficient of friction $\mu$. A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high, so that the block does not slide before toppling, the minimum force required to topple the block is
(IIT JEE 2000)

- infinitesimal
- mg/4
- mg/2
- $mg(1-\mu)$

**Solution:**
Let C be the centre of mass (geometrical centre) of the cubical block of mass $m$. The forces acting on the block are its weight $mg$, frictional force $f$, applied force $F$, and reaction forces $R_1$ and $R_2$.

In equilibrium, \begin{align} &R_1+R_2=mg,\\ &f=F. \end{align} Also, torque about the centre of mass C is zero i.e., \begin{align} FL/2-R_2L/2+fL/2+R_1L/2=0. \end{align} The block starts toppling when its left end is just above the ground i.e., $R_1=0$. Substitute in above equations and solve to get \begin{align} F=R_2/2=mg/2. \end{align}

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