Question

Assuming frictionless contact everywhere, determine the magnitude of external horizontal force $P$ applied at the lower end for equilibrium of the rod as shown in figure. The rod is uniform and its mass is ${}^{\prime }{m}^{\prime }$.

• A. $\frac{mg}{2}\cot \theta$
• B. $mg\tan \theta$
• C. $\frac{mg}{2}\sin \theta$
• D. $mg\cos \theta$

Answer 1

The correct option is A $\frac{mg}{2}\cot \theta$

Let the length of the rod $=l$. Let $N_1$ and $N_2$ be the normal reaction forces acting on the end $A$ and $B$ in a direction $\perp$ to the contact surface.


For rod to be in complete equilibrium, $F_{\text{net}}=0\&{\tau }_{net}=0$,
For translational equilibrium along horizontal $\&$ vertical direction
$\sum F_x=0$
$\Rightarrow N_2=P...\left(i\right)$ for horizontal direction.
Similarly $\sum F_y=0$ along vertical direction
$\Rightarrow N_1=mg...\left(ii\right)$

For rotational equilibrium the net torque be zero about point $B$, ${\tau }_{\text{net}}=0...\left(iii\right)$
Taking anticlockwise sense of rotation for torque as $+ve$
$\&\tau =F\times r_{\perp }$
$\because$Force $N_1$ passes through reference point $B$, hence its torque about point $B$ is zero.


Substituting the torquex with proper signs in Eq $\left(iii\right)$,
$\because \sum \tau =0$
$\Rightarrow -N_2\left(l\sin \theta \right)+mg\left(\frac{l}{2}\cos \theta \right)+0=0$
$\therefore N_2=\frac{mg}{2}\cot \theta$

Substituting $N_2=P$ from equation $\left(i\right)$
$\Rightarrow P=\frac{mg}{2}\cot \theta$
Option $\left(a\right)$ is correct.