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 $= \mu ).$ The pulley is frictionless and of negligible mass. The downward acceleration of mass $m_1$ is $(Assume\, m = m_2 = m_3 = m)$
AIPMT 2014, Difficult
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Force of friction
on mass $m_2=\mu m_2 g$
Force of friction on mass
$ m_3=\mu m_3 g$
Let a be common acceleration of the system.
$\therefore a=\frac{m_1 g-\mu m_2 g-\mu m_3 g}{m_1+m_2+m_3}$
Here, $m_1=m_2=m_3=m$
$\therefore a =\frac{m g-\mu m g-\mu m g}{m+m+m}=\frac{m g-2 \mu m g}{3 m}$
$ =\frac{g(1-2 \mu)}{3}$
Hence, the downward acceleration of mass $m_1$ is $\frac{g(1-2 \mu)}{3}$.
on mass $m_2=\mu m_2 g$
Force of friction on mass
$ m_3=\mu m_3 g$
Let a be common acceleration of the system.
$\therefore a=\frac{m_1 g-\mu m_2 g-\mu m_3 g}{m_1+m_2+m_3}$
Here, $m_1=m_2=m_3=m$
$\therefore a =\frac{m g-\mu m g-\mu m g}{m+m+m}=\frac{m g-2 \mu m g}{3 m}$
$ =\frac{g(1-2 \mu)}{3}$
Hence, the downward acceleration of mass $m_1$ is $\frac{g(1-2 \mu)}{3}$.
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