A boy and a man carry a uniform rod of length L horizontally in such a way that the boy gets 1 4 t h of the load. If the boy is at one end of the rod, the distance of the man from the other end is

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TestHub · Accepted Answer

Let normal reaction on the rod due to boy is $N_{1}$ and due to man is $N_{2}$
Given, $N_{1} = \frac{W}{4}$
$∴$ $N_{2} = W - \frac{W}{4} = \frac{3 W}{4}$

Let $x$ be the distance of the man from the other end.

For rotational equilibrium net torque must be zero.
$\Rightarrow N_{1} \times \frac{L}{2} = N_{2} \times (\frac{L}{2} - x)$
$\Rightarrow$ $\frac{W}{4} \times \frac{L}{2} = \frac{3 W}{4} \times (\frac{L}{2} - x)$
$\Rightarrow$ $x = \frac{L}{3}$

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