A boy and a man carry a uniform rod of length L horizontally in such a way that the boy gets `(1/4)^(th)` of the load. If the boy is at one end of the rod, the distance of the man from the other end is:

Options

A. `L/3`
B. `L/4`
C. `(2L)/3`
D. `(3L)/4`

Answer

A

Explanation

Let normal reaction on the rod due to boy is `N_(1)` and due to man is `N_2` Given, `N_(1) = W/4`
`therefore N_(2) = W -W/4 = (3W)/4`

Let x be the distance of the man from the other end. For rotational equilibrium net torque must be zero.
`rArr N_(1) xx L/2 = N_(2) xx (L/2 -x)`
`rArr W/4 xx L/2 = (3W)/4 xx (L/2 -x)`
`rArr x = L/3`