Question

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

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

Answer 1

Answer: (A)

Weight of the rod= $W$
Reaction of boy $R_{B}=\frac{w}{4}$
Reaction of man $R_{M}=\frac{3 w}{4}$


As the rod is in rotational equilibrium
$\therefore $ $\displaystyle \sum \tau = 0$
$R_{B}\times \frac{L}{2}-R_{M}\times x=0$
$\Rightarrow $ $\frac{W}{4}\times \frac{L}{2}-\frac{3 W}{4}\times x=0$
$\Rightarrow $ $x=\frac{L}{6}$
$\therefore $ Distance from other end, $y=\frac{L}{2}-x$
$\Rightarrow $ $y=\frac{L}{2}-\frac{L}{6}=\frac{2 L}{6}=\frac{L}{3}$