Masses and Pulley: Find Accelerations of A, B, C

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SUMMARY

The discussion focuses on calculating the accelerations of three masses (A, B, and C) connected by a pulley system. The key equations derived include M_A a_A = T, where T represents the tension in the rope. The participants emphasize the importance of applying Newton's second law correctly within an inertial frame, clarifying that the only force acting on masses A and B is the tension. Misinterpretations regarding the acceleration of the frame and the role of tension are addressed, leading to a consensus on the necessity of four equations to solve the system accurately.

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  • Understanding of Newton's second law of motion
  • Familiarity with pulley systems and tension forces
  • Knowledge of kinematic constraints in multi-body systems
  • Ability to analyze forces in inertial frames
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  • Learn about tension in pulley systems and its effects on acceleration
  • Explore kinematic constraints and their applications in dynamics
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Homework Statement



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Two masses, A and B, lie on a frictionless table. They are attached to either end of a light rope of length l which passes around a pulley of negligible mass. The pulley is attached to a rope connected to a hanging mass, C.

We are supposed to find the accelerations of A, B and C.

The Attempt at a Solution



I'm sorry that I don't have the time to write out my detailed working. I didn't get the answer required. But I notied that the solution given (page 5 from http://hep.uchicago.edu/cdf/frisch/p141/ps4_solutions.pdf) simply states

M_A a_A = T

and so on. My other equations, like the kinematic constraint (relation between ac, aa, and ab) and the equation of motion for mass C is ok. We need 4 equations to solve. So 2 are ok.

My equations for MA and MB are different from the solution here. I have assumed that M_A is in a frame that is accelerating at a rate equal to a_C, in addition to the tension imparted on mass A. Same for B. Can I do it like I have done here, or is this totally wrong?
 
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bigevil said:
My equations for MA and MB are different from the solution here. I have assumed that M_A is in a frame that is accelerating at a rate equal to a_C, in addition to the tension imparted on mass A. Same for B. Can I do it like I have done here, or is this totally wrong?
It's not clear to me what you've done. The equations for Ma and Mb are just Newton's 2nd law (which applies in an inertial frame). The tension is the only horizontal force on those masses. (The fact that the pulley is accelerating will end up affecting the tension, but that will come out of solving the equations for all three masses.)
 
bigevil said:
...
My equations for MA and MB are different from the solution here. I have assumed that M_A is in a frame that is accelerating at a rate equal to a_C, in addition to the tension imparted on mass A. Same for B. Can I do it like I have done here, or is this totally wrong?

I don't understand what you mean by this. Tension T alone is the net force acting on MA, therefore its acceleration is T/MA. There is no other acceleration to be added to this. MA doesn't care how tension T is generated. All it knows is that, if tension T acts on it, its acceleration will be T/MA.
 

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