Question 230: Laws of Motion

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 = $\mu $). The pulley is frictionless and of negligible mass. The downward acceleration of mass m
1
is (Assume m
1
= m
2
= m
3
= m)
![](https://neetcare-image.cdn.examgoal.net/7khtfk4gew5pr/81ff39ba-6930-4a85-8b46-2cea914fed8e/7a56d7f0-2465-11ea-9e2f-733b46d80788/file-7khtfk4gew5ps-320w.jpg)

Accepted Answer

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.
$ 	herefore $ $a = {{{m_1}g - \mu {m_2}g - \mu {m_3}g} \over {{m_1} + {m_2} + {m_3}}}$
Here, m
1
= m
2
= m
3
= m
$ 	herefore $ $a = {{mg - \mu mg - \mu mg} \over {m + m + m}} = {{mg - 2\mu mg} \over {3m}} = {{g\left( {1 - 2\mu } ight)} \over 3}$
Hence, the downward acceleration of mass m
1
is ${{g\left( {1 - 2\mu } ight)} \over 3}$

Answer

${{g\left( {1 - g\mu } \right)} \over 9}$

Answer

${{2g\mu } \over 3}$

Answer

${{g\left( {1 - 2\mu } \right)} \over 3}$

Answer

${{g\left( {1 - 2\mu } \right)} \over 2}$

Laws of Motion

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