Question

Two persons $P$ and $Q$ of the same height are carrying a uniform beam of length $3m$. If $Q$ is at one end, then the distance of $P$ from the other end such that $P$, $Q$ receive loads in the ratio $5:3$ is:

• A. $0.5m$
• B. $0.6m$
• C. $0.75m$
• D. $1m$

Answer 1

The correct option is B $0.6m$

$N_P+N_Q=Mg$
also $\frac{N_P}{N_Q}=\frac{5}{3}$
$\Rightarrow N_P=\frac{5}{8}Mg;N_Q=\frac{3}{8}Mg$
Calculating torque about the end opposite to $Q$
$N_P{d}_1+N_Q(3-d_2)=Mg\frac{3}{2}$
$\frac{5}{8}Mg{d}_1+\frac{3}{8}Mg\times 3=Mg\frac{3}{2}$
$\therefore \frac{5}{8}Mg{d}_1+\frac{3}{8}Mg\times 3=Mg\frac{3}{2}$
$\frac{5}{8}Mg{d}_1=\left(\frac{12-9}{8}\right)Mg$
$d_1=\frac{3}{5}=0.6m$