Solving Blocks & Pulley Homework: Find Acceleration

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Homework Help Overview

The discussion revolves around a physics problem involving a system of blocks and a pulley. The original poster presents a scenario with two masses, a pulley, and various forces acting on the system, seeking to find the acceleration. The problem includes parameters such as mass, radius, angle, and friction coefficient.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationships between tensions in the string, the forces acting on the blocks, and the torque on the pulley. Questions arise regarding the role of the cosine factor in torque calculations and the equality of tensions T1 and T2.

Discussion Status

There is an ongoing examination of the equations governing the system. Some participants suggest that the tensions should be equal, while others argue that a difference in tension is necessary for the pulley to accelerate. The discussion reflects a mix of interpretations and attempts to clarify the relationships between the forces and torques involved.

Contextual Notes

Participants note confusion regarding the application of torque and the implications of tension differences on the system's acceleration. There is acknowledgment of the need for multiple equations to fully describe the dynamics of the system.

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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. ;)
 

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