Relation between masses in a three-mass pulley system at rest

[IhasSwag]

Homework Statement

Find the relation between the mass of block 1 and the mass of block 2. Assume that the pullies are massless and frictionless and that the rope is inelastic. All masses are unknown.
Possible answers: m_block2 = m_block1[/B]

m_block2 > m_block1
m_block2 < m_block1
The system below is at rest.​

Homework Equations

F = mg = T

The Attempt at a Solution

I'm quite clueless on how to approach this problem, but this is how far I came.
m₁*g - T₁ = m₁*a
m₂*g - 0.5*T₂ = m₂*a

∑F = m₁*a + m₂*a = m₁*g - T₁ +m₂*g - 0.5*T₂

I'm not sure how to go any further.

Thanks in advance for the help!

 

[Delta2]

Focus on Pulley with block 2. Since the pulley is at rest, the sum of forces on that pulley equals zero.

 

[IhasSwag]

So T₁ + T₃ - m₂*g = 0 ?
m₁*g + m₃*g = m₂*g
m₁ + m₃ = m₂ ?
Am I approaching this problem the right way now?

I found out that I made a mistake in the possible answers. It should be:
mblock2 = 2*mblock1
mblock2 > 2*mblock1
mblock2 < 2*mblock1

 

[TSny]

The rope has the same tension throughout.

 

[Delta2]

Yes correct. Now notice that the tension should be the same so $T_1=T_3$.

 

[IhasSwag]

If T₁ = T₃ then m₁ = m₃
This means that m₂ = 2*m₁ ?

But now I face another problem because according to my teacher, the answer should have been: m₂ < 2*m₁

 

[Delta2]

Is the rope considered to be massless?

 

[IhasSwag]

It doesn't state that the rope is massless, so I guess not.

 

[IhasSwag]

I meant to say that the rope has mass

 

[Delta2]

If the rope is massless then what we did here is correct and your teach is wrong.

However, if the rope is NOT massless, then the tension forces $T_1+T_3$ have to lift the weight of the segment of rope that is attached to pulley 2, so it will be $T_1+T_3=m_2g+m_rg$ where m_r the mass of the rope along the circumference of pulley 2.

 

[IhasSwag]

The rope is indeed not massless.
Thank you very much for your help.