# Equilibrium of Rigid Bodies

## Problems from IIT JEE

**Problem (IIT JEE 2005):**
A block of mass $m$ is at rest under the action of force $F$ against a wall as shown in figure. Which of the following statement is incorrect?

- $f=mg$, where $f$ is the frictional force.
- $F=N$, where $N$ is the normal reaction.
- $F$ will not produce torque about centre of mass.
- $N$ will not produce torque about centre of mass.

**Solution:**
The forces acting on the block are its weight $mg$, normal reaction $N$, frictional force $f$, and applied force $F$ (see figure). Let O be the centre of mass and $N$ acts at a distance $x$ below it. In equilibrium, $N=F$ and $f=mg$. The torque about centre of mass shall be zero i.e.,
\begin{align}
\label{bsb:eqn:1}
Nx-fa/2=Fx-mga/2=0.
\end{align}
Above equation gives $x=mga/(2F)$. The $x$ shall be less than or equal to $a/2$ for $N$ to act on the block. Thus, $F \geq mg$ for block to be in equilibrium.