Question

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

• A. L/3
• B. L/4
• C. 2L/3
• D. 3L/4.

Answer 1

Answer: (A)

L/3

Answer 2

Weight of the rod = W

Reaction of boy $R_{B} = \frac{W}{4}$

Reaction of man $R_{M} = \frac{3W}{4}$

As the rod is in rotational equilibrium

∴ $\sum\overrightarrow{\tau} = 0$

$R_{B} \times \frac{L}{2} - R_{M} \times x = 0$⇒ $\frac{W}{4} \times \frac{L}{2} - \frac{3W}{4} \times x = 0$⇒$x = \frac{L}{6}$

∴ Distance from the second side, $y = \frac{L}{2} - x$

⇒ y$= \frac{L}{2} - \frac{L}{6} = \frac{2L}{6} = \frac{L}{3}$