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Standard XII

Physics

The Complete Equilibrium

A system cons...

Question

A system consists of three masses $m_{1}, m_{2}$ and $m_{3}$ connected by a string passing over a pulley $P$ . The mass $m_{1}$ hangs freely and $m_{2}$ and $m_{3}$ are on a rough horizontal table (the coefficient of friction $= μ$ ). The pulley is frictionless and of negligible mass. The downward acceleration of mass $m_{1}$ is (Assume $m_{1} = m_{2} = m_{3} = m$ )

\frac{g (1 - 2 μ)}{9}

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\frac{2 g μ}{3}

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\frac{g (1 - 2 μ)}{3}

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\frac{g (1 - 2 μ)}{2}

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Solution

The correct option is C $\frac{g (1 - 2 μ)}{3}$
Tension in the string attached to mass $m_{1}$ is $T_{1}$
Tension in the string attached to mass $m_{3}$ is $T_{2}$

Apply newtons Second law, for each mass we get
$m_{1} g - T_{1} = m_{1} a . . . (1)$
$T_{1} - f_{2} - T_{2} = m_{2} a . . . (2)$
$T_{3} - f_{3} = m_{3} a . . . (3)$

Adding above equations we get
$m_{1} g - f_{2} - f_{3} = (m_{1} + m_{2} + m_{3}) a$
As $m_{1} = m_{2} = m_{3} = m$ and $f_{2} = f_{3} = μ m g$
On substituting in above equation, we get

$m g - 2 μ m g = 3 m a$
$a = \frac{g (1 - 2 μ)}{3}$

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