Solving for Tension of Inextensible String System

[markosheehan]

a light inextensible string passes over a fixed pulley A, under a movable pulley B of mass M and then over a second fixed pulley C. A mass m is attached to one end of the string and a mass 3m is attached to the other end. the system is released from rest. (i) prove that the tension t in the string is given by the equation t(1/m + 1/3m)=g (g stands for gravity) i can't prove this even though i feel like i am doing all the the right equations. the 3 equations i get by using f=ma and looking at all the particles separately are t-mg=ma t-3mg=3mb Mg-2t=M(a/2 + b/2) a stands for the acceleration the particle of mass m. b stands for the acceleration of the particle of mass 3m. a/2 + b/2 stands for the acceleration . when i get a and b on there own from the first 2 equations and put this in for a and b in the last equation i do not get the answer required. i can post a picture of the diagram in the question if anyone needs it. any help much appreciated

 

[IanCg]

a light inextensible string passes over a fixed pulley A, under a movable pulley B of mass M and then over a second fixed pulley C. A mass m is attached to one end of the string and a mass 3m is attached to the other end. the system is released from rest. (i) prove that the tension t in the string is given by the equation t(1/m + 1/3m)=g (g stands for gravity) i can't prove this even though i feel like i am doing all the the right equations. the 3 equations i get by using f=ma and looking at all the particles separately are t-mg=ma t-3mg=3mb Mg-2t=M(a/2 + b/2) a stands for the acceleration the particle of mass m. b stands for the acceleration of the particle of mass 3m. a/2 + b/2 stands for the acceleration . when i get a and b on there own from the first 2 equations and put this in for a and b in the last equation i do not get the answer required. i can post a picture of the diagram in the question if anyone needs it. any help much appreciated

I am new to these forums and noticed your problem in the unanswered list so I thought I would have a look at it.
The good news is that your mechanics was good and I agree with your equations, you must have made a slip in their solution.
Here is my working.

(1) t - mg = ma
(2) t - 3mg = 3mb
3times equation (1) minus (2) gives (3) 2t = 3ma - 3mb

(4) Mg - 2t = M(a/2 + b/2) mutiply this by 6m (5) 6Mmg - 12mt = 3Mma +3Mmb
(5) + M times (3) gives (6) 6Mmg - 12mt + 2Mt = 6Mma

6M times equation (1) gives (7) 6Mt - 6Mmg = 6Mma
putting (6) and (7) together 6Mmg -12mt +2Mt = 6Mt - 6Mmg
rearranging 12Mmg = 12mt + 4Mt dividing this by 12Mm gives the result required.