Question

Find the normal reaction at the contact points of sphere with walls.


Consider $N_1$ and $N_2$ to be the normal reactions by first and second wall respectively. [ Take $g=10{\text{m/s}}^2$ ]

• A. $N_1=25\sqrt{3}\text{N},N_2=50\sqrt{3}\text{N}$
• B. $N_1=N_2=\frac{100}{\sqrt{3}}\text{N}$
• C. $N_1=N_2=\frac{100}{\sqrt{2}}\text{N}$
• D. $N_1=\frac{25}{\sqrt{3}}\text{N},N_2=\frac{100}{\sqrt{3}}\text{N}$

Answer 1

The correct option is B $N_1=N_2=\frac{100}{\sqrt{3}}\text{N}$

Both the inclined walls are making angle ${30}^{\circ }$ from horizontal, thus due to symmetry the normal reaction exerted by them will be equal.

The normal reaction from wall will pass through the center of sphere.

From geometry the angle made by normal reaction from horizontal is,

$\theta ={90}^{\circ }-{30}^{\circ }={60}^{\circ }$

$\text{FBD of sphere}:$



For the equilibrium of sphere,

$N_1\sin {60}^{\circ }+N_2\sin {60}^{\circ }=mg$

$\Rightarrow 2N_1\sin {60}^{\circ }=mg[\because N_1=N_2]$

$\Rightarrow N_1=\frac{mg}{2\sin {60}^{\circ }}$

$\Rightarrow N_1=\frac{100}{2\times \left(\frac{\sqrt{3}}{2}\right)}$

$\therefore N_1=N_2=\frac{100}{\sqrt{3}}\text{N}$

Hence, option (b) is the correct answer.

Why this question? Whenever a spherical body comes into picture, be careful on statement that ''normal reaction always acts perpendicular at a contact point''. Thus normal reaction will always pass through center of sphere.