Consider a uniform rod of mass $40kg$ and length $8m$, pivoted about a point $P$ $3m$ from one end as shown in the figure. Few external forces are acting on the rod as shown in figure.mark out the correct statement (s). [Take $g=m/s^2$]

 

The rod is in translation equilibrium in any case as it is privoted, now let us check for its rotational equilibrium.

For rotational equilibrium, the net torque acting about any point must be zero

Let us take the torque about $P$.

$\underset{ }{\overset{ }{\sum \zeta =400\times 1-20\times 5-100\times 3=0}}$

[Taking clockwise as $+ve$and anticlock wise as $-ve$]

So rod is in rotational equilibrium, also. If a body is in rotational equilibrium then

$\sum _{ }^{ }{\zeta }_{ext}=0$ about any point.

The force exterted by a pivot on rod maintains the translational equilibrium in horizontal and vertical directions. Let $R_x$ and $R_y$ be the horizontal and vertical components of force exetred by pivot on rod, respectively.

$R_x=200-50=150N$

$R_y=400+100-20=480N$

Then, $R=\sqrt{{\left(105\right)}^2+{\left(480\right)}^2}\simeq 503N$

The rod will exert equal and opposite force on the pivot.