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If we have an Atwood machine where masses $m_A$ and $m_B$ rest on the ground, then we apply an upwards force $F$ to the Atwoods machine.

What is the acceleration of the blocks when $F=124N$, $m_A = 20$kg, $m_B = 10kg$?

I am confused to about this problem,because I am not sure how this upward force $F$ on the Atwoods machine effects the two masses.

For example, for each mass I would normally do this:

$$F_{netA} = m_A*g - T = m_A*a$$ $$F_{netB} = T - m_B*g = m_B*a$$

But now I have another force directed upwards, of $124N$. Where does this come into play? I have thought about adding it to each of the equations, but I don't think this is right, because the force is acting on the Atwood's machine, not the mass. I believe this effects the Tension, but I'm not exactly sure how.

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As you lift the pulley, the masses stay on the ground until the string is taut. As the pulley is massless, you accelerate upward infinitely fast. It is better to start the simulation with the string taut. Now you have twice the tension in the string pulling down on the pulley and $F$ pulling up on the pulley.

Physically, one of three things can happen, depending on $F$. If $F$ is small, you won't lift either mass off the ground, the tension in the string will be $\frac F2$ and masses will be motionless. If $F$ is somewhat larger, the heavy mass will stay on the ground, the pulley will accelerate upward, and the light mass will accelerate upward twice as fast as the pulley. If $F$ is larger yet, both masses will lift off the ground. The pulley will accelerate upward at $\frac F{m_A+m_B}$. The lighter mass will accelerate upward faster than the heavy one.

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  • $\begingroup$ For the second case, why does the lighter mass accelerate faster than the pulley? Shouldn't they move at the same acceleration? $\endgroup$ – Jason Jun 12 '14 at 23:29
  • $\begingroup$ Because the heavy mass is stationary on the ground and the string is fixed in length. Let the string be 5m long. When you start, the pulley is 2.5m high. When the pully is 3m high (so has risen 0.5m), the light mass is 2m below it, so is 1m high. You are running a 2:1 block and tackle in reverse. $\endgroup$ – Ross Millikan Jun 12 '14 at 23:31

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