Question

A $50kg$ homogeneous smooth sphere rests on the ${30}^{\circ }$ inclined plane $A$ and bears against the smooth vertical wall $B$ . The contact force at point $B$ is
(Take $g=10m/s^2$)

• A. $250N$
• B. Zero
• C. $\frac{500}{\sqrt{3}}N$
• D. $500N$

Answer 1

The correct option is C $\frac{500}{\sqrt{3}}N$

FBD of sphere


For equilibrium,
$\Rightarrow N_A\sin {60}^{\circ }=mg$
$\Rightarrow N_A\frac{\sqrt{3}}{2}=mg$
$N_A=\frac{2mg}{\sqrt{3}}$
$\Rightarrow N_B=N_A\cos {60}^{\circ }$
$N_B=\frac{2mg}{\sqrt{3}}\times \frac{1}{2}=\frac{50\times 10}{\sqrt{3}}$
$N_B=\frac{500}{\sqrt{3}}N$