"Shifting" non-inertial frame in Accelerating Atwood Machine

[ln(]

Hello,

I was referring to this thread: https://www.physicsforums.com/threads/accelerating-atwoods-machine-problem.44305/ to solve a problem on accelerating atwood machines in which, naturally, one attempts to find the acceleration of each of the individual masses.

Although I now understand how to solve the problem thanks to that thread, I have difficulty identifying the issue with my first approach, where I solve for the accelerations as if the atwood machine were NOT accelerating and then adding the acceleration of the entire system to these accelerations. Doing this results in a different answer than the correct approach mentioned in the thread.

What is wrong with this approach? Thanks.

PS: My apologies if this should be in the Homework section. I was unsure where to put it since technically I know how to do the problem; I am just unsure of the general nature of changing our reference frame in this instance.

 

[Orodruin]

Here is some food for thought. Imagine the Atwood machine was in free fall. How would the masses move? How does that change when you stop the free fall?

 

[ln(]

Here is some food for thought. Imagine the Atwood machine was in free fall. How would the masses move? How does that change when you stop the free fall?

I'm not sure, since my intuition is clearly wrong. Without freefall, solving the accelerations is easy. With freefall, I clearly cannot just subtract g from the accelerations in the system without freefall.

 

[Orodruin]

Will there be any tension in the string in free fall? (Imagine placing an Atwood machine on the International Space Station, which is in free fall)

 

[ln(]

Will there be any tension in the string in free fall? (Imagine placing an Atwood machine on the International Space Station, which is in free fall)

I believe not.

 

[Orodruin]

So therefeore the acceleration is?

 

[ln(]

So therefeore the acceleration is?

Just g downwards.

 

[Orodruin]

Just g downwards.

Yes, for all masses. This is clearly different from just adding the gravitational acceleration compared to the case when the system is not in free fall. When the system is in free fall, all masses are falling with the same acceleration.

Does that help?

 

[ln(]

Yes, for all masses. This is clearly different from just adding the gravitational acceleration compared to the case when the system is not in free fall. When the system is in free fall, all masses are falling with the same acceleration.

Does that help?

Yes, thank you!

The other issue is that my "logic" here is, to my understanding, essentially what ehild applied in his approach with the accelerations relative to the pulley., with the difference being that he said that $a_r = -a_r$ and $a_1 = a-a_r$ and $a_2 = a+a_r$ during the derivation while I attempt to apply this after the fact.
Although what I did is wrong, and thank you for giving me an intuitive sense of that, I still don't know intuitively WHEN it is appropriate to actually apply this transformation (as in adding the $a$ to the accelerations relative to the pulley itself).

Another possibility is that such a transformation (adding $a$ to the accelerations relative to the pulley) is always valid. If this is indeed the case, that means what I find by pretending there is zero acceleration on the system is not actually the accelerations of the masses with respect to the pulley. So then, how can you find them while accelerating with the pulley itself, i.e. while in the frame of reference of the pulley?

 

[Orodruin]

In an accelerated frame there is an additional inertial force on any massive object. This force is equal to the object’s mass multiplied by the acceleration and opposite in direction to the acceleration. Thus, it acts exactly as a gravitational force and the accelerating system is equivalent to the non-accelerating one, just with what appears to be a different gravitational acceleration.

 

[ln(]

In an accelerated frame there is an additional inertial force on any massive object. This force is equal to the object’s mass multiplied by the acceleration and opposite in direction to the acceleration. Thus, it acts exactly as a gravitational force and the accelerating system is equivalent to the non-accelerating one, just with what appears to be a different gravitational acceleration.

Ah, fantastic.

So does this mean that, given a mass on a table in an elevator, the table will have to support the mass with more force if the elevator is accelerating upwards than if it were on ground?

And say we have a mass of mass $m_1$ connected to one side of a simple atwood machine and $m_2$ to the other and $m_2 > m_1$. Let this system be accelerating with acceleration $a'$. I will use the convention of $a$, the acceleration of the masses relative to the pulley, going in the direction of the overall motion. If I want to write down expression of the forces acting on them with respect to the accelerated frame, is it correct to say that $$m_1a = F_\textrm{tension} - m_1g - m_1a'$$ and $$m_2a = m_2g + m_2a' - F_\textrm{tension}$$ This seems to be correct, at least.