Problem 1.161:A system consists of three blocks of 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. Find the downward acceleration of mass $m_1$. Assume $m_1\:=\:m_2\:=\:m_3\:=m$.

Solution:

When released from rest, the hanging block will speed up downward, pulling the other to with it. This is possible due to the force exerted by the string which is called tension.

There will be frictional force from the rough surface, opposing the sliding motion of the other two blocks.Different forces acting on the system are as shown in the diagram below.

Since the blocks are connected by strings, they will have acceleration of  same magnitude.

Unbalanced force in any block will accelerate it.

For first block

Normal force from the floor, $N\:=\:mg$.

Force of friction, $f\:=\:\mu\:mg$.

The block is moving to the right side. Therefore force $T_2$ must be greater than force of friction.

So $T_2\:-\:f\:=\:ma$.

Or, $T_2\:-\:\mu mg\:=\:ma$.....$(1)$

For second  block

Normal force from the floor, $N\:=\:mg$.

Force of friction, $f\:=\:\mu\:mg$.

The block is moving to the right side. Therefore force $T_1$ must be greater than the sum of  force of friction and tension $T_2$.

So $T_1\:-\:(T_2\:+\:f)\:=\:ma$.

Or, $T_1\:-\:T_2\:-\:\mu mg\:=\:ma$......$(2)$

For third block

The block is moving downward. Therefore weight $mg$ must be greater than tension $T_1$.

So $mg\:-\:T_1\:=\:ma$.......$(3)$

Adding equations $(1)$, $(2)$ and $(3)$ we get,

$T_2\:-\:\mu mg\:+\:T_1\:-\:T_2\:-\:\mu mg\:+\:\:mg-\:T_1\:=\:ma\:+\:ma\:+\:ma$.

Or, $mg\:-\:2\mu mg =\:3ma $

From this acceleration, $a\:=\:\frac{mg\:-\:2\mu mg}{3m}$.

Or, $a\:=\:\frac{(1\:-\:2\mu)mg}{3m}$.

Then, $a\:=\:\frac{(1\:-\:2\mu)g}{3}$.

Tension $T_2$.

We have, from equation $(1)$

$T_2\:-\:\mu mg\:=\:ma$

Or, $T_2\:=\:ma\:+\:\mu mg$

Using this value of acceleration, $a\:=\:\frac{(1\:-\:2\mu)g}{3}$,

We get, $T_2\:=\:m(\frac{(1\:-\:2\mu)g}{3})\:+\:\mu mg$

Or, $T_2\:=\:mg(\frac{(1\:-\:2\mu)}{3}\:+\:\mu)$

Or, $T_2\:=\:mg\frac{(1\:-\:2\mu+\:3\mu)}{3}$

Then Tension, $T_2\:=\:mg\frac{(1\:+\mu)}{3}\:$.

Tension $T_1$

We have, from equation $(3)$

$mg\:-\:T_1\:=\:ma$

From this, $mg\:-\:ma\:=\:T_1$

Or $T_1\:=\:mg\:-\:ma$.

Using this value of acceleration, $a\:=\:\frac{(1\:-\:2\mu)g}{3}$,

We get, $T_1\:=\:mg\:-\:m(\frac{(1\:-\:2\mu)g}{3})$.

Or, $T_1\:=\:mg(1\:-\:\frac{(1\:-\:2\mu)}{3})$

Then, $T_1\:=\:mg\frac{(3\:-\:1\:+\:2\mu)}{3}$

or, $T_1\:=\:mg\frac{(2\:+\:2\mu)}{3}$

Then Tension $T_1\:=\:2mg\frac{(1\:+\:\mu)}{3}$

Tension in the vertical string is twice that in the horizontal  string.