Frame of reference (non inertial/inertial) in an Atwood Mach

[fontseeker]

Homework Statement

The pulley in an Atwood's machine is given an upward acceleration *a*, as shown below. Find the acceleration of each mass and the tension in the string that connects them. Hint: a constant upward acceleration has the same effect as an increase in the acceleration due to gravity

Find the acceleration of blocks 1 and 2 in both a non-inertial frame of reference (an observer accelerating with the machine) and an inertial frame of reference (observer at earth)

Homework Equations

F = ma

The Attempt at a Solution

I solved the question successfully for the observer at earth. I got the following for my acceleration equations:

Those equations are right for the inertial frame of reference portion. However, now that I am trying to solve for the acceleration of block 1 and 2 in a non-inertial frame of reference, I am unsure on how to consider acceleration.

I though of the following (for the non-inertial portion):

1) Since the observer is moving with the machine, could I just cancel out the added acceleration and use g (9.8 m/s^2) for the acceleration of the system

2) I was told that acceleration was absolute. So then the acceleration of a non-inertial frame and an inertial frame would be the same, meaning from both perspectives the acceleration would be (a+g)

I am not sure on how to consider the acceleration of the machine when the observe is moving with the machine. (accelerating with it).

 

[kuruman]

In the non-inertial frame, the acceleration $a$ up becomes a fictitious force $ma$ in the down direction on a body of mass $m$.

 

[fontseeker]

In the non-inertial frame, the acceleration $a$ up becomes a fictitious force $ma$ in the down direction on a body of mass $m$.

I don't understand why it would go in the down direction, since it is said that the acceleration is going up. What confuses me is that, in my inertial frame portion, I treated this acceleration a as an addition on gravity. So, if I treated the acceleration as an addition to gravity as well in the non-inertial frame, they would both give the same equations.

 

[kuruman]

I don't understand why it would go in the down direction, since it is said that the acceleration is going up.

When you are in a car accelerating forward, the fictitious force points towards the rear and pins you to your seat, no?

So, if I treated the acceleration as an addition to gravity as well in the non-inertial frame, they would both give the same equations.

The same form of equations where "g" in the inertial frame is replaced with "g+a" in the non-inertial frame.

 

[fontseeker]

When you are in a car accelerating forward, the fictitious force points towards the rear and pins you to your seat, no?

The same form of equations where "g" in the inertial frame is replaced with "g+a" in the non-inertial frame.

I did that for my equation in the inertial frame of reference. I replaced g with (a + g) because I have to represent the addition on acceleration. However, do I replace g with (a+g) for both inertial and non-inertial?

 

[kuruman]

However, do I replace g with (a+g) for both inertial and non-inertial?

Let's do this from the top, starting with the non-inertial calculation. What are the two equations (one for each mass) in that frame?

 

[fontseeker]

Let's do this from the top, starting with the non-inertial calculation. What are the two equations (one for each mass) in that frame?

I haven't done the calculations for non-inertial because I am confused with that frame of reference. I did the calculations for an inertial frame.

So I got the tension, and then I replaced with (a+g). @kuruman do you recommend doing the same for both?

 

[kuruman]

I recommend that you derive the expression as I indicated in post #6. You have to understand the justification for the replacement. For the non-inertial frame draw free body diagrams, one for each mass, with additional fictitious forces $m_1a$ and $m_2a$ acting on the appropriate masses.

Write down Newton's Second Law twice using $a_{com.}$ as the common acceleration of the masses. As usual, you should get a system of two equations and two unknowns, the tension and the common acceleration. Solve for the unknowns.

 

[fontseeker]

I recommend that you derive the expression as I indicated in post #6. You have to understand the justification for the replacement. For the non-inertial frame draw free body diagrams, one for each mass, with additional fictitious forces $m_1a$ and $m_2a$ acting on the appropriate masses.

Write down Newton's Second Law twice using $a_{com.}$ as the common acceleration of the masses. As usual, you should get a system of two equations and two unknowns, the tension and the common acceleration. Solve for the unknowns.

I did that. I solved using a (which I knew it wouldn't matter since it would cancel out). I solved for acceleration in mass 1 and mass 2, and then set them equal to each other. Then I got tension (as displayed in post #7). And then I replaced g by (a + g). However, my question is if there are fictitious forces m1a and m2a on the non-inertial frame of reference, then how would you solve it in an inertial frame of reference?

Also would it be accepted to draw fictitious forces in my free body diagram? Or is that just for my reference?

 

[kuruman]

I did that. I solved using a (which I knew it wouldn't matter since it would cancel out). I solved for acceleration in mass 1 and mass 2, and then set them equal to each other. Then I got tension (as displayed in post #7).

Please show me the process and the equations. If you did what I asked you in post #8, you would not have to replace g with g+a. I suspect what you did is find the tension with a = 0 and then replaced the common acceleration of the masses with g + a. Perhaps you don't understand that there are two accelerations in the inertial frame and the non-inertial frame. These are $a$, the given acceleration of the Atwood machine, and $a_{com.}$ the common acceleration of the two masses. These enter in the equations for both inertial and non-inertial frames.

Sorry, but for your sake, you have to convince me you understand what you are doing and why.

 

[fontseeker]

Please show me the process and the equations. If you did what I asked you in post #8, you would not have to replace g with g+a. I suspect what you did is find the tension with a = 0 and then replaced the common acceleration of the masses with g + a. Perhaps you don't understand that there are two accelerations in the inertial frame and the non-inertial frame. These are $a$, the given acceleration of the Atwood machine, and $a_{com.}$ the common acceleration of the two masses. These enter in the equations for both inertial and non-inertial frames.

Sorry, but for your sake, you have to convince me you understand what you are doing and why.

I am sorry, I forgot to post the equations of my work:

 

[kuruman]

Is this calculation that you show in the inertial or non-inertial frame? How are accelerations a₁, a₂ and a₃ defined? What is their relation to the given acceleration a?

 

[Chestermiller]

Let $a_{2r}$ represent the upward acceleration of mass m2 relative to the pulley. Since the string has constant length, in terms of $a_{2r}$, what is the upward acceleration of mass m1 relative to the pulley? In terms of the pulley acceleration a and the relative acceleration $a_{2r}$, what is the absolute upward acceleration of mass m2? In terms of the pulley acceleration a and the acceleration of mass m1 relative to the pulley, what is the absolute upward acceleration of the mass m1? In terms of these absolute accelerations, what are the force balances on masses m1 and m2?

 

[kuruman]

Also would it be accepted to draw fictitious forces in my free body diagram? Or is that just for my reference?

It would be acceptable to draw fictitious forces as long as you specify in your free body diagram that it is in the non-inertial frame.