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

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$\frac{W}{2}$

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

$\frac{W}{2 \sqrt{3}}$

answer is D.

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Detailed Solution

For upper cylinder

When F is minimum , action and reaction force between the lower cylinders is zero

2N cos 30 = W ....... (1)

For lower cylinder
N cos 60 = F ....... (2)
From (1) & (2)

$\frac{2 \cos 30^{0}}{\cos 60^{0}} = \frac{W}{F} \Rightarrow F = \frac{W}{2 \sqrt{3}}$

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