Question

A uniform of rod of length $l$ is placed symmetrically on two walls as shown in the 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 $N_1$ $N_2$

Answer 1

Answer: (C)

At equilibrium, $N_1+N_2=mgcos\theta$

Also ${\tau }_A=0⟹N_1\left(0\right)+mg\frac{l}{4}cos\theta -N_2\frac{l}{2}$

$\frac{mg}{2}cos\theta =N_2$

Thus $N_1=\frac{mg}{2}cos\theta$

Hence $N_1=N_2$