Question

A rod of mass $m$ is placed on two supports as shown. Calculate the normal reaction of each support.

• A. $N_1=\frac{1}{3}mg,N_2=\frac{2}{3}mg$
• B. $N_1=\frac{2}{3}mg,N_2=\frac{1}{3}mg$
• C. $N_1=\frac{mg}{2},N_2=\frac{mg}{2}$
• D. $N_1=\frac{mg}{4},N_2=\frac{3mg}{4}$

Answer 1

The correct option is A $N_1=\frac{1}{3}mg,N_2=\frac{2}{3}mg$

As th rod is in equilibrium,
$\because {\tau }_A=0$
$\therefore (N_1\times 0)+(N_2\times \frac{3L}{4})-(mg\times \frac{L}{2})=0$
$N_2=\frac{2}{3}mg$
As, $N_1+N_2=mg$
and $N_1=\frac{1}{3}mg$