Question

Consider three identical smooth balls placed between two vertical walls as shown in figure. Mass of each ball is $m$ and radius $r=\frac{5R}{9}$, where $2R$ is separation between the walls.

• A. Force between middle ball and wall is highest
• B. Force between top ball and wall is highest
• C. Force between top ball and wall is least
• D. Force between bottommost ball and wall is least

Answer 1

Answer: (A), (C)

The correct options are
A Force between middle ball and wall is highest
C Force between top ball and wall is least

$r+r+2r\cos \theta =2R$
$\Rightarrow 1+\cos \theta =\frac{R}{r}$
$\cos \theta =\frac{9}{5}-1=\frac{4}{5}\Rightarrow \sin \theta =\frac{3}{5}$
Various contact forces are as shown
Considering all 3 sphere together,
$N_1=3mg\to \left(1\right)$
and $N_2+N_4=N_3\to \left(2\right)$
For vertical equilibrium of lower ball,
$N_5\sin \theta +mg=N_1\to N_5\left[\frac{3}{5}\right]=2mg\Rightarrow N_5=\frac{10}{3}mg$
For vertical equilibrium of topmost ball,
$N_6\sin \theta =mg\Rightarrow N_6=\frac{5mg}{3}$
For horizontal equilibrium of topmost ball,
$N_4=N_6\cos \theta =\frac{5}{3}mg\times \frac{4}{5}=\frac{4}{3}mg$
For horizontal equilibrium of middle ball,
$N_3=(N_6+N_5)\cos \theta =[\frac{5}{3}+\frac{10}{3}]\frac{4}{5}mg=4mg$
From (2) , $N_2=4\text{mg}-\frac{4}{3}\text{mg}=\frac{8}{3}\text{mg}$
Hence $N_3$ is largest and its value is $4\text{mg}$
$N_4$ is smallest and its value is $\frac{4}{3}\text{mg}$.