First slide

Equilibrium of a particle

Question

Two massless rings slide on a smooth circular loop of the wire whose axis lies in a horizontal plane. A smooth massless inextensible string passes through the rings, which carries masses m₁ and m₂ at the two ends and mass m₃ between the rings. If there is equilibrium when the line connecting each ring with centre subtends an angle 30^o with vertical as shown in figure. Then the ratio of masses are

Moderate

A

$m_{1} = 2 m_{2} = m_{3}$
B

$2 m_{1} = m_{2} = 2 m_{3}$
C

$m_{1} = m_{2} = m_{3}$
D

None of these

Solution

As no friction is involved, the tensions in the segments AC and AE of the string must be the same. Let its magnitude be T. For the ring A to be at rest on the smooth loop, the resultant force on it must be along AO, O being the center of the loop; otherwise there would be a component tangential to the loop.
Hence
$∠ O A E = ∠ O A C = ∠ A O E = 30^{\circ}$
The same argument applies to the segments BD and BE. Then by symmetry the point E at which the string carries the third weight must be on the radius HO, H being the highest point of the loop, and the tensions in the segments BD and BE are also T. Now consider the point E. Each of the three forces acting on it, which are in equilibrium, are at an angle of 120^o to the adjacent one. As two of the forces have magnitude T, the third force must also have magnitude T. Therefore, the three weights carried by the string are equal.

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