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

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Homework Help Overview

The discussion revolves around calculating the acceleration and tensions in a double Atwood machine, which consists of frictionless, massless pulleys and cords. Participants are exploring the relationships between the masses involved and the forces acting on them.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to derive equations based on the forces acting on the masses. There is a focus on the relationship between the tensions and accelerations of the different masses, particularly mA, mB, and mC. Questions arise regarding the assumptions made about the directions of the accelerations and the implications of these assumptions on the equations.

Discussion Status

Some participants have provided equations based on their assumptions about the masses and their accelerations. There is ongoing exploration of how to relate the tensions and accelerations, particularly in the context of the hanging pulley. Multiple interpretations of the directions of the masses are being discussed, indicating a productive exchange of ideas without a clear consensus yet.

Contextual Notes

Participants are considering the effects of different mass configurations and the resulting tensions, as well as the implications of assuming certain directions for the accelerations. There is an acknowledgment of the complexity introduced by these assumptions.

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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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