In the figure, pulleys are smooth and strings are massless. m 1 = 1 kg and m 2 = 3 1 kg . To keep m 3 at rest, mass m 3 should be :

The string over the lower pulley has tension T = m 1 + m 2 2 m 1 m 2 g . The upper pulley supports m 3 through 2 T when m 3 is at rest: m 3 g = 2 T . m 3 = m 1 + m 2 4 m 1 m 2 = 1 + 3 1 4 × 1 × 3 1 = 1 kg .

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PhysicsNLMChallenging application of NLMMedium2 minQB

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In the figure, pulleys are smooth and strings are massless. $m_{1} = 1 kg$ and $m_{2} = \frac{1}{3} kg$ . To keep $m_{3}$ at rest, mass $m_{3}$ should be :

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Solution:

The string over the lower pulley has tension $T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$ . The upper pulley supports $m_{3}$ through $2 T$ when $m_{3}$ is at rest: $m_{3} g = 2 T$ .

$m_{3} = \frac{4 m _{1} m _{2}}{m _{1} + m _{2}} = \frac{4 \times 1 \times \frac{1}{3}}{1 + \frac{1}{3}} = 1 kg$ .

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Challenging application of NLM

⏱ 2mℹ️ Source: QB

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