Question

Two smooth cylindrical bars weighing W each lie next to each other in contact. A similar third bar is over the two bars as shown in figure. Neglecting friction, the minimum horizontal force on each lower bar necessary to keep them together is

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

Answer 1

Answer: (C)

$\frac{W}{\sqrt{3}}$

For the system to be in equilibrium, the normal force between the lower cylinders should be zero which can be achieved by applying and external force $F$ as mentioned in the figure.
Need to find:
$F=?$

On the basis of free body diagram of left lower cylindrical bar, we can write:

$F=N\sin {30}^{\circ }$ ----------[1]

and, On the basis of free body diagram of left lower cylindrical bar, we can write:

$W=2N\cos {30}^{\circ }$ ----------[2]

From eq $\left[1\right]and\left[2\right]$: $\left[1\right]/\left[2\right]$

$\Rightarrow \frac{F}{W}=\frac{N\sin {30}^{\circ }}{2N\cos {30}^{\circ }}$

$\Rightarrow F=\frac{W\sin {30}^{\circ }}{2\cos {30}^{\circ }}=\frac{W\tan {30}^{\circ }}{2}$

On putting, $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$

$\Rightarrow F=\frac{W\left(\frac{1}{\sqrt{3}}\right)}{2}$
$\Rightarrow F=\frac{W}{2\sqrt{3}}$

Hence, the minimum horizontal force on each lower bar necessary to keep them together is, $F=\frac{W}{2\sqrt{3}}$
So, the correct option is $\left(D\right)$