Question

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

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

Answer 1

Answer: (A)

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
$\therefore$ $\sum \tau =0$
$R_B\times \frac{L}{2}-R_M\times x=0$
$\Rightarrow$ $\frac{W}{4}\times \frac{L}{2}-\frac{3W}{4}\times x=0$
$\Rightarrow$ $x=\frac{L}{6}$
$\therefore$ Distance from other end, $y=\frac{L}{2}-x$
$\Rightarrow$ $y=\frac{L}{2}-\frac{L}{6}=\frac{2L}{6}=\frac{L}{3}$