How Do You Calculate Acceleration in a Double Atwood Machine?

AI Thread Summary
To calculate acceleration in a double Atwood machine, it is essential to apply Newton's second law to each mass while considering the system's constraints. The accelerations of masses A and B are equal in magnitude but can differ in direction depending on the reference point of the pulleys. The net force on the massless pulleys must be zero, which affects the tension in the cords. A common mistake is assuming that the accelerations of A and B are always opposite; they can actually move in the same direction relative to the free pulley. Understanding the relationship between the accelerations of the masses and the pulleys is crucial for accurate calculations.
Calpalned
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Homework Statement


The double Atwood machine shown in the figure has frictionless, mass-less pulleys and cords.
Determine the acceleration of mA, mB, mC.
GIANCOLI.ch04.p56.jpg


Homework Equations


F = ma

The Attempt at a Solution



http://photo1.ask.fm/779/725/638/-259996976-1shhdoh-qms4ifsem3hthf/preview/IMG_4082.jpg
http://photo1.ask.fm/905/398/673/-59997000-1shhdp1-2726aartaqghh2f/preview/IMG_4083.jpg
http://photo1.ask.fm/916/959/786/-239996977-1shhdpk-78i75a0215n926k/preview/IMG_4084.jpg
http://photo1.ask.fm/333/391/062/-249996981-1shhdrb-eeemt2tkfheqfp0/preview/IMG_4085.jpg

What did I do wrong?
 
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First I see aa = ab , later on it becomes aa = -ab ?
 
The accelerations of A and B are equal in magnitude with respect to the hanging pulley, which accelerates, too.
 
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Here is the answer given in my solutions manual.

Because the pulleys are massless, the net force on them must be 0. Because the cords are massless, the tension will be the same at both ends of the cords. Use the free-body diagrams to write Newton’s second law for each mass. We take the direction of acceleration to be positive in the direction of motion of the object. We assume that mC is falling, mB is falling relative to its pulley, and mA is rising relative to its pulley. Also note that if the acceleration of mA relative to the pulley above it is aR , then aA = aR + aC. Then, the acceleration of mB is aB = aR - aC since aC is in the opposite direction of aB .

Even though mA and mB are moving in opposite directions, how can they both be considered moving in the positive direction?

The acceleration of m_A relative to the free pulley should be the acceleration of the free pulley plus the acceleration of m_A. The free pulley is rising and mA is rising too. Therefore, aR = aC + aA. Why is it necessary to consider the free pulley in the first place? I don't see any problem with ignoring it (my original attempt doesn't seem to have an errors) but I somehow got the wrong answers.

Finally, what's the error in my original attempt in the starting post in this thread?
 
Calpalned said:
Even though mA and mB are moving in opposite directions, how can they both be considered moving in the positive direction?

ma and mb are not necessarily moving in opposite directions. It's only with respect to the lower pulley that they move in opposite directions. So in reality, they can both be moving in the same direction.

Calpalned said:
Finally, what's the error in my original attempt in the starting post in this thread?

Your error was the equation aa = -ab... This is not true! Imagine aa = ab ... They should both be at rest with respect to the lower pulley, right? According to your equation, aa+ab=0... But this is clearly not true because both a and b will moving with some nonzero acceleration in the same direction (they will move with the same acceleration that the lower pulley is moving with.)
 
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