In the pulley system shown the movable pulleys `A,B` and `C` have mass m each, `D` and `E` are fixed pulleys. The strings are vertical light and inextensible Then
.

Explanation

Constraint relations,
Let `y_(1), y_(2) and y_(3)` are the positions of pulleys A, B and C respectively at any instant with respect to a dotted line shown in The total length of string
`2y_(1)+y_(2)+(y_(2)-y_(1))+y_(3)+(y_(3)-y_(2))+l_(0)=l`
or `y_(1)+y_(2)+2y_(3)+l_(0)=l` ...(i)
Where `l_(0)` is the length of part of string over the pulleys, which is constant
Differentiating equation (i) w.r.t. time, we get
`dy_(1)/(dt)+dy_(2)/(dt)+2(dy_(3))/(dt)=0`

or `v_(1)+v_(2)+2v_(3)=0` ...(ii)
also `a_(1)+a_(2)+2a_(3)=0`
Let `a_(1)=a` upward
and `a_(2)=a` upward
then `a=((a_(1)+a_(2))/2)=a` downward
Since string is same throughout and uniform the tension in it will be same every where, thus
From pulley A,
`2T-(T+1g)=1a,` ...(i)
Form pulley B,
`2T-(T+1g)=1a` ...(ii)
From pulley C,
`1g - 2T = 1a` ...(iii)
Solving above equations , we get
`a_(1)=-g/3, a_(2)=(-g)/3, a_(3)=g/3`
and `T=(2g)/3=6.53N`