Question

A uniform rod of length $l$ is placed symmetrically on two walls as shown in figure. The rod is in equilibrium. If $N_{1}and{N}_2$ are the normal forces exerted by the walls on the rod then:

• A. $N_1>N_2$
• B. $N_1<N_2$
• C. $N_1=N_2$
• D. $N_{1}and{N}_2$ would be in the vertical directions

Answer 1

Answer: (C)

$N_1=N_2$

As the rod is in equilibrium, thus net torque about point O is zero i.e. ${\tau }_o=0$

$\therefore$ $N_1\times \frac{l}{4}-N_2\times \frac{l}{4}=0$

$⟹N_1=N_2$