A weightless rod is acted upon by upward parallel forces of 2N 4N at ends A - Physics - System Of Particles And Rotational Motion

Question

A weightless rod is acted upon by upward parallel forces of 2N
& 4N at ends A & B respectively Total length of rod AB is
3m To keep the rod in equilibrium, a force of 6N should act on the
rod in the following manner:
a) downwards at end A
b) downwards at mid point of AB
c) downwards at at point C such that AC=1m
d) downwards at at point D such that BD=1m

Ujjval Chauhan · Accepted Answer

Dear student

The rod will be in equilibrium if the net force and torque on the
rod will be zero
The total force in the upward direction=4N+2N=6N
so 6N force will be downward direction
Let x be the distance of the action of force on the rod at point D from the midpoint of the rod
and let this position is on the side of point B
Now the net torque due to these forces will be zero
The torques due to forces 2N and 6N will be in clockwise and torque due to 4N will be anticlockwise soNet torque about the mid point of the rod2×1
5+6×x=4×1 53+6x=66x=3x=0
5 cmso distance of the position of 6N force from A and B will be respectively2cm and 1cm
hence option d is correct