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

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Homework Statement

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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

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F=ma

The Attempt at a Solution

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First I looked down the bottom section. Assuming m_B> m_A

m_B*g - F_TA = m_B*a
F_TA - m_A*g = m_A*a

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

a_A,B = g* (m_B - m_A)/(m_B + m_A)

Now Assuming m_C*g>F_TC

m_C*g - F_TC = m_C*a

This is where I choked. What should be the second equation for this pulley? F_TC = 2F_TA?
If I were to find F_TA in terms of m_A, m_B, and a_A,B, would that be enough for a_C's equation?

 

[ehild]

Homework Statement

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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 m_B> m_A

m_B*g - F_TA = m_B*a
F_TA - m_A*g = m_A*a

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

a_A,B = g* (m_B - m_A)/(m_B + m_A)

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 2F_TA=F_TC.

 

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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 2F_TA=F_TC.

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 F_TC 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?

 

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Assume the directions, and solve.

 

[HalfLostAndSearching]

To solve for the acceleration of mass C, we need to consider the forces acting on it. The mass C is connected to two ropes, one going over the pulley attached to mass A and one going over the pulley attached to mass B. This means that the tension in the rope attached to mass C is equal to the sum of the tensions in the two ropes attached to masses A and B.

Therefore, we can write the following equation:

mC*g - (FTA + FTC) = mC*a

Since we already have an equation for FTA in terms of mA, mB, and aA,B, we can substitute that into this equation:

mC*g - (g* (mB - mA)/(mB + mA) + FTC) = mC*a

Now we need to find the value of FTC. To do this, we can use the fact that the tension in the rope attached to mass B is equal to the tension in the rope attached to mass A:

FTA = FTC

Substituting this into the equation above, we get:

mC*g - g* (mB - mA)/(mB + mA) - FTA = mC*a

Finally, we can solve for the acceleration of mass C:

aC = g* (mC - (mB - mA)/(mB + mA))/(mC + (mB - mA)/(mB + mA))

To find the tensions in the ropes, we can use the equations we already found for FTA and FTC:

FTA = g* (mB - mA)/(mB + mA)

FTC = g* (mB - mA)/(mB + mA)

Note that the tension in the rope attached to mass C will be equal to the sum of these two tensions.