Equilibrium of Rigid Bodies
A rigid body is said to be in static equilibrium if
- net external force on the body is zero and
- net external torque on the body is zero.
The torque may be defined about any point.
Solved Problems from IIT JEE
Problem from 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 distance $x$ 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.
More Solved Problems on Equilibrium of Rigid Bodies
Related Topics
References
- IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
- 300 Solved Problems on Rotational Mechanics by Jitender Singh and Shraddhesh Chaturvedi