The pulleys A and C are fixed while the pulley B is movable . A mass `M_(2)` is attached to pulley B , while the string has masses `M_(1) "and " M_(3)` at the two ends . Find the acceleration of each mass .

Explanation

The force acting on masses `M_(1),M_(2)"and " M_(3)` are shown in figure .
Let `a_(1) "and" a_(3)` be upward acceleration of masses `M_(1)"and" M_(3) "and" a_(2)`
the downward acceleration of mass `M_(2)`.
For motion of mass M_(1)
`T-M_(1) g=M_(1)a_(1)` ....(1)
For motion of mass `M_(3)`
`T-M_(3)g=M_(3) a_(3)` ....(2)
For motion of mass `M_(2)`
`M_(2)g-2T=M_(2)a_(2)` ....(3)
Let `l_(1),l_(2),l_(3)` be the lengths of vertical portions of the string between pulley at any instant . `M_(1) ` goes up through `x_(2) M_(2)` goes down through y and `M_(3)` goes up through z.
Then `l_(1)+2l_(2)+l_(3)=(l_(1)-x)+2(l_(2)+y)+l_(3)-z`
`rArr x+z =2y rArr (d^(2)x)/(dt^(2))+(d^(2)z)/(dt^(2))=2(d^(2)y)/(dt^(2))`
`a_(1)+a_(3)=2a_(2) rArr a_(2)=(a_(1)+a_(3))/2` ....(4)
From (1), a_(1)=T/M_(3)-g` ....(5)
From (2),` a_(3)=T/M_(3)-g` ....(6)
substituting `a_(1) "and" a_(3)` in (4) , we get
`a_(2)=T/2(1/M_(1)+1/M_(3))-g` ....(7)
From (3), `T=(m_(2)(g-a_(2))/2` ....(8)
substituting this value in (7), we get
`a_(2)=M_(2)/4 (g-a_(2))(1/M_(1)+1/M_(3))-g "or" a_(2)=(M_(2)(g-a_(2))(M_(3)+M_(1)))/(4M_(1)M_(3))-g`
or `4M_(1)M_(3)a_(2)-M_(2)(M_(1)+M_(3))g-M_(2)(M_(1)+M_(3)a_(2)-4M_(1)M_(3)g`
or `[4M_(1)M_(3)+M_(2)(M_(1)+M_(3))]a_(2)=(M_(1)+M_(3))M_(2)g-4M_(1)M_(3)g`
`a_(2)=((M_(1)+M_(3))M_(2)g-4M_(1)M_(3)g)/(4M_(1)M_(3)+M_(2)(M_(1)+M_(3))) a_(2)=(M_(1)M_(2)+M_(2)M_(3)-4M_(1)M_(3))/(M_(1)M_(2)+M_(2)M_(3)+4M_(1)M_(3)) ....(9)
Substituting value of `a_(2)` in above equation , we get
`a_(1)=(-M_(1)M_(2)+3M_(2)M_(3)-4M_(1)M_(3))/(M_(1)M_(2)+M_(2)M_(3)+4M_(1)M_(3))g` ....(10)
similarly from (6)and (9)
`a_(3)=(3M_(1)M_(2)-M_(2)M_(3)-4M_(1)M_(3))/(M_(1)M_(2)+M_(2)M_(3)-4M_(1)M_(3))` ....(11)