How Do You Solve a Double Atwood's Machine Problem?

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Homework Statement


Masses m1 and m2 are connected by a light during A over a light frictionless pulley B. The axel of pulley B is connected by a light string C over a second light frictionless pulley D to a mass m3. Pulley D is attached o the ceiling. The system is released from rest.
In terms of m1, m2, m3 and g what are
a) the acceleration of the block m3
b) the acceleration of pulley B
c) the acceleration of block m1 and m2
d) The tension in the string A
e) The tension in the string C

Homework Equations


F = ma

The Attempt at a Solution


a) As the strings are weightless the tension either side of the pulley will be the same.
I came up with equations
TA - m1g = m1a1
TA - m2g = -m2a1
TB - m3g = m3a2
TB - (m1 + m2)g = -(m1 + m2)a2
TB = 2TA

Rearranging the first 2 equations i got
TA = (2m1m2g)/(m1 + m2)

I then substituted this into equation 3 to get
(4m1m2g)/(m1 + m2) - m3g = m3a
which when i rearrange goes
(4m1m2g - m1m3g - m2m3g)/(m1m3 + m2m3) = a

However this is not the answer stated in my textbook and I'm not sure where I've gone wrong.
Any help would be much appreciated!
 
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Your first equations would be right only if the pulley B were fixed.
 
Jilang said:
Your first equations would be right only if the pulley B were fixed.
So would it be TA - m1g = m1(a1 + a2) as it is accelerating not only on its own pulley system but also has the acceleration of pulley B?
This would also change equation 2 to T1 - m2g = -m2(a1 + a2)
 
If a1, a2, and a3 are accelerations measured relative to the earth, then equation 1 is OK. The next three need modification. You are right that you are going to need to think about relative accelerations.