Question

Cylinders $\text{B}$ and $\text{C}$ are fixed and cylinder $\text{A}$ is kept above them as shown in figure.


If the system is at rest, find the normal reaction at the contact point of $\text{A}$&$\text{B}$.
$($all cylinders are identical and have mass $m)$

• A. $\frac{mg}{2\sqrt{2}}$
• B. $\frac{mg}{\sqrt{3}}$
• C. $\frac{\sqrt{3}}{2}mg$
• D. $\frac{2}{3}mg$

Answer 1

The correct option is B $\frac{mg}{\sqrt{3}}$

The contact force due to cylinder $\text{B}$ on $\text{A}$, will be directed normally and passes through the center of cylinder $\text{A}$.

Similarly, the normal reaction on $\text{A}$ due to cylinder $\text{C}$ will pass through the center of $\text{A}$.


As shown in figure, the $\Delta A^{\prime }{B}^{\prime }{C}^{\prime }$ formed by the line joining center of the touching cylinders will be equilateral triangle with side $2r$, where $r$ is radius of each cylinder.

FBD of cylinder $\text{A}:$


Due to symmetry, the normal reaction exerted by $\text{C}$ and $\text{B}$ on $\text{A}$ will be the same. Let it be $N$.

Applying equilibrium conditions on $\text{A}$,

$\sum F_y=0$

$\Rightarrow N\cos {30}^{\circ }+N\cos {30}^{\circ }=mg$

$\Rightarrow 2N\cos {30}^{\circ }=mg\Rightarrow N=\frac{mg}{\sqrt{3}}$

$\therefore N=\frac{mg}{\sqrt{3}}$

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: In this problem the arrangement of}\\ \text{cylinders gives rise to a symmetrical}\\ \text{structure. Thus,}{N}_{CA}=N_{BA}\text{and normal}\\ \text{reaction at any contact point will be}\\ \text{along line joining center of cylinders.}\end{array}$