Question

Two men $A$ and $B$ are carrying a uniform bar of length $L$ on their shoulders. The bar is held horizontally such that $A$ gets $\left(\frac{1}{4}\right)th$ load. If $A$ is at one end of the bar, the distance of $B$ from that end is

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

Answer 1

Answer: (C)

$\frac{2L}{3}$

Let $F$ be the total force exerted on the men by weight. If A gets $\frac{1}{4}$ th of the load , B gets $\frac{3}{4}$ th of the load $F_B=\frac{3}{4}F$
Balancing equilibrium of moments at the center point
$\frac{F}{4}\times \frac{L}{2}=\frac{3}{4}F\times x$
$x=\frac{L}{6}$ is the distance of B from center
Distance of B from A is $\frac{L}{6}+\frac{L}{2}$
$=\frac{2L}{3}$