Q. 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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A

$\frac{W}{2}$

B

$E W$

C

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

D

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

Solution:

For upper cylinder

$2 N \cos 30^{\circ}=W$ ...(i)
For lower block
$N \cos 60^{\circ}=F$ ...(ii)
From (i) & (ii)
$\frac{2 \cos 30^{\circ}}{\cos 60^{\circ}}=\frac{W}{F}$
$\Rightarrow F=\frac{W}{2 \sqrt{3}}$

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