Equilibrium of Rigid Bodies

Problems from IIT JEE

Problem (IIT JEE 2005): Equilibrium of Rigid Bodies 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?

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

Solution: Equilibrium of Rigid Bodies 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.