Question

Two persons of equal height are carrying a long uniform wooden plank of length $l$. They are at distance $\frac{l}{4}$ and $\frac{l}{6}$ from nearest end of the rod. The ratio of normal reaction at their heads is-

• A. $2:3$
• B. $1:3$
• C. $4:3$
• D. $1:2$

Answer 1

The correct option is C $4:3$

Taking moment at point x,
$w\times (\frac{l}{2}-\frac{l}{4})=N_2\times (l-\frac{l}{4}-\frac{l}{6})$
$\therefore N_2=\frac{w\left(\frac{l}{4}\right)}{\left(\frac{7l}{12}\right)}$
$\therefore N_2=\frac{3w}{7}$
Now,
Taking moment at point y,
$w\times (\frac{l}{2}-\frac{l}{6})=N_1\times (l-\frac{l}{4}-\frac{l}{6})$
$\therefore N_1=\frac{w\left(\frac{l}{3}\right)}{\frac{7l}{12}}$
$\therefore N_1=\frac{4w}{7}$
Ratio of reaction force,
$\frac{N_1}{N_2}=\left(\frac{4w}{7}\right)÷\left(\frac{3w}{7}\right)$
$=\frac{4}{3}$