Can anybody check if I solved this correctly?(tension forces + pulley)

[PhyIsOhSoHard]

Homework Statement

The pulley can rotate and has no friction.
Block m_2 is twice as big as m_1
Find the tensions, T_1, T_2, T_3

Homework Equations

Newton's second law

The Attempt at a Solution

Tension 1:
By creating a free body diagram only for the block, I got with Newton's second law:
T_1-mg=ma
T_1=m(a+g)

Tension 2:
T_2-2mg=ma
T_2=m(a+2g)

Tension 3:
T_3-T_1-T_2=ma

Inserting above values for the other tensions:
T_3=ma+m(a+g)+m(a+2g)
T_3=3m(a+g)

Did I solve this correctly? What about the acceleration? Can I do anything about that?

 

[Enigman]

T_3-T_1-T_2=ma

the pulley has no net force acting on it.

 

[PhyIsOhSoHard]

the pulley has no net force acting on it.

What exactly do you mean?
Are the 2 first tensions correct?

 

[PhyIsOhSoHard]

So we have:

T_1-mg=ma
T_1=m(a+g)

2mg-T_2=ma
T_2=2m(g-a)

Since the tension is uniform on the string:
m(a+g)=2m(g-a)
a=\frac{1}{3}g

Substituting this into the first equation:
T_1-mg=\frac{1}{3}mg
T=\frac{4}{3}mgThe tension of the pulley since it has no net force acting:
T_3-T_1-T_2=0

Since T_1=T_2=T
T_3=2T

Inserting the tensions:
T_3=2\frac{4}{3}mg

 

[haruspex]

So we have:

T_1-mg=ma
T_1=m(a+g)

2mg-T_2=ma
T_2=2m(g-a)

Since the tension is uniform on the string:
m(a+g)=2m(g-a)
a=\frac{1}{3}g

Substituting this into the first equation:
T_1-mg=\frac{1}{3}mg
T=\frac{4}{3}mg


The tension of the pulley since it has no net force acting:
T_3-T_1-T_2=0

Since T_1=T_2=T
T_3=2T

Inserting the tensions:
T_3=2\frac{4}{3}mg

Assuming the pulley has no moment of inertia, that all looks correct. I wouldn't write the final answer as T_3=2\frac{4}{3}mg though. It looks like you mean "two and four thirds".