A particle of small mass $m$ is joined to a very heavy body of mass $M$ by a light string passing over a light pulley. Both bodies are free to move. The total downward force on the pulley is nearly

m g

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2 m g

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4 m g

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Solution

The correct option is C $4 m g$
From the help of free body diagram, the equations of motion are:
$M a = M g - T$
$m a = T - m g$
By solving the two equations we can find T to be
$T = \frac{2 m M g}{M - m}$
Now, $M >> m$ so $M - m \approx M$
$T = 2 m g$
The net downward force on the pulley $= 2 T = 4 m g$

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A particle of small mass m is joined to a very heavy body of mass M by a light string passing over a light pulley. Both bodies are free to move. The total downward force on the pulley is nearly

The correct option is C 4mgFrom the help of free body diagram, the equations of motion are: Ma=Mg−Tma=T−mgBy solving the two equations we can find T to beT=2mMgM−m Now, M>>m so M−m≈MT=2mgThe net downward force on the pulley =2T=4mg

A particle of small mass $m$ is joined to a very heavy body of mass $M$ by a light string passing over a light pulley. Both bodies are free to move. The total downward force on the pulley is nearly