MHB Solving for Tension of Inextensible String System

AI Thread Summary
The discussion focuses on proving the tension in a system involving a light inextensible string over pulleys with different masses attached. The key equation to prove is t(1/m + 1/3m) = g, where t is the tension and g is gravity. Participants agree on the initial equations derived from applying Newton's second law to the masses involved, but one user struggles with the algebraic manipulation to reach the desired result. Another user provides a detailed step-by-step solution, confirming the correctness of the initial equations and guiding the original poster through the algebraic process. The conversation emphasizes the importance of careful algebraic handling in mechanics problems.
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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 seperately 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
 
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markosheehan said:
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 seperately 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.
 
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