Solving Blocks & Pulley Homework: Find Acceleration

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Homework Statement


[PLAIN]http://img109.imageshack.us/img109/1267/picture2ae.png

[tex]m_1 = 2kg[/tex]
[tex]m_2 = 6kg[/tex]
[tex]R = .25cm[/tex]
[tex]M = 10kg[/tex]
[tex]\theta = 30º[/tex]
[tex]\mu = 0.36[/tex]

Find the acceleration of the system.

Homework Equations


The Attempt at a Solution



[tex]m_1 a = T_1 - m_1 g \mu[/tex]
[tex]m_2 a = m_2 g \sin \theta - m_2 g \mu \cos \theta - T_2[/tex]
[tex]\tau = I a/R = R T_2 - R T_1 \cos \theta[/tex]I solved this system and the result is different from the solution.
What am I missing/doing wrong?Regards
Johnny
 
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Krappy said:
[tex]\tau = I a/R = R T_2 - R T_1 \cos \theta[/tex]
Why the cosθ factor?
 
Doc Al said:
Why the cosθ factor?

Because of the cross product. |r||F| * sin(x), but in that case, sin(x) is the cos(theta).
 
Krappy said:
Because of the cross product. |r||F| * sin(x), but in that case, sin(x) is the cos(theta).
When taking the cross product, the angle is between the vectors r and F. Since the ropes are tangential to the pulley, that angle is 90 degrees.
 
I have though about this and I think that T1 must equal to T2 because, if not, the string would be extending. Am I thinking right? I ask this since before I though that it was the difference between the two tensions T1 and T2 that would accelerate the pulley, but now, I don't think that's the case.

Now I have:

[tex]m_1 a = T - m_1 g \mu[/tex]
[tex]m_2 a = m_2 g \sin \theta - m_2 g \mu \cos \theta - T - \frac{Ia}{r^2}[/tex]


Which one is correct?
 
Krappy said:
I have though about this and I think that T1 must equal to T2 because, if not, the string would be extending. Am I thinking right?
No. Without a difference in string tension there would be no net torque to accelerate the pulley.
I ask this since before I though that it was the difference between the two tensions T1 and T2 that would accelerate the pulley, but now, I don't think that's the case.
You were right the first time. (Why did you change your mind?)

Now I have:

[tex]m_1 a = T - m_1 g \mu[/tex]
[tex]m_2 a = m_2 g \sin \theta - m_2 g \mu \cos \theta - T - \frac{Ia}{r^2}[/tex]


Which one is correct?
In addition to not reflecting the difference in tensions, the second equation looks odd since it has a term relating to the pulley. But the pulley only affects m_2 via the tension in the string that attaches to m_2.

You need three equations: one for each block and one for the pulley.
 
Thank you Doc Al.

I got confused since that with the last system I got the right answer. (Notice that if I substitute the equation T2-T1 = Ia/R^2 in the original second one, I get the same thing). The thing that I was missing in the first place was that both tensions were tangential to the pulley.

Thank you very much. ;)