Two smooth cylindrical bars weighing W 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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∴ N 1 ​ = N 2 ​ = N N sin 6 0 ∘ + N sin 6 0 ∘ = W N = W / 3 ​ ​ N cos 6 0 ∘ = F ​ 3 ​ W ​ × 2 1 ​ = F ​ ∴ F = 2 3 ​ W ​ ​

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