Question

A rod of length 100 m is pivoted at a distance of 60 m from one end. A block of mass 10 kg is suspended at the other end of the rod. If the rod is in equilibrium, find the mass of the rod. Take acceleration due to gravity $g$ = $10m{s}^{-}2$

• A. 20 kg
• B. 30 kg
• C. 40 kg
• D. 60 kg

Answer 1

The correct option is C

40 kg

Let the mass of rod be $m$.
Let us draw all the forces acting on the rod.


For a body to be in equillibrium, net force = 0 and net torque = 0

Acceleration due to gravity $g$ = $10m{s}^{-}2$

$\therefore$ Normal reaction $N$ = $mg+10g$
Torque due force 'mg' = $mg\times 10m=100mNm$ (anti-clockwise)
Torque due to force '100 N'= $100N\times 40m=4000Nm$ (clockwise)

Torque due to $N$ = $0$ ( bcz force is passing through axis of rotation)
For equillibrium, net Torque = 0
Clockwise Torque = Anti-clockwise Torque
$\Rightarrow$ $4000=100m$
$\Rightarrow$$m=40kg$