Question

A boy and a man carry a uniform rod of length $L$ horizontally in such a way that the boy gets ${\left(\frac{1}{4}\right)}^{th}$ of the load. If the boy is at one end of the rod, the distance of the man from the other end is

• A. $\frac{L}{3}$
• B. $\frac{L}{4}$
• C. $\frac{2L}{3}$
• D. $\frac{3L}{4}$

Answer 1

Answer: (A)

Let normal reaction on the rod due to boy is $N_1$ and due to man is $N_2$
Given, $N_1=\frac{W}{4}$
$\therefore$ $N_2=W-\frac{W}{4}=\frac{3W}{4}$

Let $x$ be the distance of the man from the other end.
For rotational equilibrium net torque must be zero.
$\Rightarrow N_1\times \frac{L}{2}=N_2\times \left(\frac{L}{2}-x\right)$
$\Rightarrow$ $\frac{W}{4}\times \frac{L}{2}=\frac{3W}{4}\times \left(\frac{L}{2}-x\right)$
$\Rightarrow$ $x=\frac{L}{3}$