A pulley-rope-mass arrangement is shown in figure. Find the acceleration of block $m_{1}$ when the masses are set free to move. Assume that the pulley and the ropes are ideal.

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Solution

Constraint relations. Method 1 :

For the upper string, the length of string $l_{1}$ not to change and for this string not to slacken, acceleration of $m_{1}$ w.r.t. the fixed pulley=acceleration of the movable pulley w.r.t. the fixed pulley
or $| a_{1} | = | a_{p} |$ .......(i)
Constraint relations for string 2
Let us assume that block $m_{1}$ is moving up with acceleration $a_{1}$ and blocks $m_{2}$ and $m_{3}$ are moving down with acceleration $a_{2}$ and $a_{3}$ , respectively w.r.t. ground.
Total length of string 2 is constant.
Therefore,
$l_{2} = (x_{2} - x_{P_{2}}) + (x_{3} - x_{P_{2}}) + + {l^{'}}_{0}$
Where $l_{0}$ is the length of string on pulley 2 passing through the pulley $P_{2}$
$l_{2} = x_{2} + x_{3} - 2 x_{P_{2}} + {l^{'}}_{0}$ ........(ii)
Differentiating Eq. (ii) w.r.t. time.
$\frac{{d l}_{2}}{d t} = 0$ (as total length of string is constant)
$\frac{{d l}_{0}^{'}}{d t} = 0$
(as length of string over pulley is constant)
Differentiating Eq. (ii) twice w.r.t. time
$2 a_{1} = a_{3} + a_{2}$ ..........(iii)

Method 2 :

The acceleration of block $m_{1}$ and pulley 2 will be same in magnitude. Now considering pulley 2 and block 2 and 3.
Change in the length of segment (1)
$Δ l_{1} = ({+ y}_{2}) + ({- y}_{P_{2}}) = y_{2} - y_{P_{2}}$
change in the length of segment (2),
$Δ l_{2} = ({+ y}_{3}) + ({- y}_{P_{2}}) = y_{3} - y_{P_{2}}$
Total sum of change in the segment length of string should be zero.
$Δ l = 0 = [y_{2} - y_{P_{2}}] + [y_{2} - y_{P_{2}}]$
$\Rightarrow$ $Δ l = 0 = y_{2} + y_{3} - 2 y_{{P 2}}$
$\Rightarrow$ $2 y_{P_{2}} = y_{3} + y_{2} \Rightarrow 2 a_{P_{2}} = a_{3} + a_{2}$
or $2 a_{1} = a_{3} + a_{2}$
This equation is same as (iii) calculated in method 2.

Method 3 :

For the rope length $l_{2}$ not to change and for this rope not to slacken acceleration of $m_{2}$ w.r.t. movable pulley = - acceleration of $m_{3}$ w.r.t.the movable pulley
or $(a_{2} - a_{p}) = - (a_{3} - a_{p})$
$a_{2} + a_{3} - 2 a_{p} = 0 \Rightarrow 2 a_{0} = a_{2} + a_{3}$
This equation is same as Eq. (iii) as calculated in method 1 and 2.
Applying Newton's laws of motion equations, Free-body diagram $F i g . 6.140$
Equations of motion
For $m_{1}$ : $2 T - m_{1} g = m_{1} a_{1}$ ......(iv)
For $m_{2}$ : $m_{2} g - T = m_{2} a_{2}$ ......(v)
For $m_{3}$ : $m_{3} g - T = m_{3} a_{3}$ .......(vi)
After solving (iii), (iv), (v) and (vi), we get
$a_{1} = \frac{4 m_{2} m_{3} g - m_{1} (m_{2} + m_{3}) g}{4 m_{2} m_{3} + m_{1} (m_{2} + m_{3})}$

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