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Limiting Friction

Three equal w...

Question

Three equal weights A, B and C of mass $2 k g$ each are hanging on a string passing over a fixed frictionless pulley as shown in the figure. The tension in the string connecting weights B and C is:

z e r o

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13 N

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3.3 N

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19.6 N

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Solution

The correct option is B $13 N$
The Free body diagram of the system is shown.
For block A: $T - m g = m a$
$T - 2 g = 2 a$
We get $T = 2 a + 2 g$
For block B + C system : $m g + m g - T = (m + m) a$
$2 g + 2 g - T = (2 + 2) a$
$⟹ 4 g - T = 4 a$
Or $4 g - 2 a - 2 g = 4 a$
Or $2 g = 6 a$
This gives $a = g / 3$
For block C: $m g - T^{'} = m a$
$2 g - T^{'} = 2 a$
Or $2 g - T^{'} = \frac{2 g}{3}$
Thus we get $T^{'} = 4 g / 3 = \frac{4 \times 10}{3} = 13 N$

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Three equal weights A, B and C of mass 2kg each are hanging on a string passing over a fixed frictionless pulley as shown in the figure. The tension in the string connecting weights B and C is:

Three equal weights A, B and C of mass $2 k g$ each are hanging on a string passing over a fixed frictionless pulley as shown in the figure. The tension in the string connecting weights B and C is: