How Do You Calculate Acceleration and Tension in a Double Atwood Machine?

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SUMMARY

The discussion focuses on calculating acceleration and tension in a double Atwood machine, which consists of frictionless, massless pulleys and cords. The equations derived include F=ma, leading to the acceleration formula for masses A and B as aA,B = g * (mB - mA) / (mB + mA). The tension in the cords is represented as FTC = 2FTA, where FTA is the tension affecting masses A and B. The complexity arises from determining the directions of the masses' movements, particularly when considering mass C's influence on the system.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of tension in strings and pulleys
  • Familiarity with the concept of acceleration in mechanical systems
  • Basic algebra for solving equations
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  • Study the dynamics of multi-mass pulley systems
  • Learn about the effects of friction in pulley systems
  • Explore advanced topics in classical mechanics, such as Lagrangian mechanics
  • Practice solving problems involving multiple connected masses
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Homework Statement


[/B]
GIANCOLI.ch04.p56.jpg


The double Atwood machine has frictionless, massless pulleys and cords. Determine the acceleration of masses mA,mB,mC, and the tensions in the cords.

Homework Equations


[/B]
F=ma

The Attempt at a Solution


[/B]
First I looked down the bottom section. Assuming mB> mA

mB*g - FTA = mB*a
FTA - mA*g = mA*a

I got the following from the equations above, so I think this must be the acceleration for the masses A and B.

aA,B = g* (mB - mA)/(mB + mA)

Now Assuming mC*g>FTC

mC*g - FTC = mC*a

This is where I choked. What should be the second equation for this pulley? FTC = 2FTA?
If I were to find FTA in terms of mA, mB, and aA,B, would that be enough for aC's equation?
 
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hitemup said:

Homework Statement


[/B]
GIANCOLI.ch04.p56.jpg


The double Atwood machine has frictionless, massless pulleys and cords. Determine the acceleration of masses mA,mB,mC, and the tensions in the cords.

Homework Equations


[/B]
F=ma

The Attempt at a Solution


[/B]
First I looked down the bottom section. Assuming mB> mA

mB*g - FTA = mB*a
FTA - mA*g = mA*a

I got the following from the equations above, so I think this must be the acceleration for the masses A and B.

aA,B = g* (mB - mA)/(mB + mA)

A and B hang from the rope around the hanging pulley. Their accelerations are equal with respect to the pulley, but not with respect to the ground.
I suggest to write the acceleration of C first. It will be the same (with opposite sign) as the acceleration of the hanging pulley. It has no mass, so 2FTA=FTC.
 
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ehild said:
A and B hang from the rope around the hanging pulley. Their accelerations are equal with respect to the pulley, but not with respect to the ground.
I suggest to write the acceleration of C first. It will be the same (with opposite sign) as the acceleration of the hanging pulley. It has no mass, so 2FTA=FTC.

What confuses me here is that there are too many possibilities, especially for A and B.

To write correct equations with respect to ground, I think I should consider the effect of FTC on A and B like you said. But what are their directions?
I mean if we assume C is going down, no problem. But either A or B could be going down or up in this situation. Should I just assume their directions too?
 
Assume the directions, and solve.
 

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