Pistons placed on top of communicating vessels

[highschoboy004]

Homework Statement

A U-shaped pipe is sealed by two pistons of mass m. Initially, both pistons are held such that the heights of the air columns in both branches are h. Cross sectional area of left branch is 2S, right branch's is S. The initial air pressure in the pipe is equal to atmospheric pressure p0. Then both pistons are released, find the heights of both pistons when they reach equilibrium. Supposed that the temperature of the air inside the pipe remains constant.

Relevant Equations

F/f=S/s;...

I have tried using Pascal's law for the pipe but it turned out that the law only works for incompressible fluids, whereas this pipe contains air, which is compressible. I thought then this problem might be the same as one with a cylinder containing air sealed with a piston, but there might be some things I've missed out.

 

[kuruman]

According to our rules, to receive help, you need to show some credible effort towards answering the question. How about telling us what you do know and how you would approach this problem?

Please read, understand and follow our homework guidelines, especially item 4, here
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

 

[Lnewqban]

Welcome, @highschoboy004 !

Any progress in your work that we could see?
Do you know how to complete and show to us a free body diagram of each piston?

This may help you to keep moving in the right direction:
https://en.wikipedia.org/wiki/Pascal's_law#Applications

 

[highschoboy004]

Welcome, @highschoboy004 !

Any progress in your work that we could see?
Do you know how to complete and show to us a free body diagram of each piston?

This may help you to keep moving in the right direction:
https://en.wikipedia.org/wiki/Pascal's_law#Applications

Thank you for asking! I figured out that the air pressure inside the pipe must be constant such that its volume remains constant in order for Pascal's law to work.

 

[Lnewqban]

Thank you for asking! I figured out that the air pressure inside the pipe must be constant such that its volume remains constant in order for Pascal's law to work.

That is a good first step.

But each combination of mass-piston induces a different pressure in the gas below it.
Therefore, after both pistons are released, each will naturally slide to a new height in order to reach equilibrium.

Could you calculate each of those heights?
What value that equilibrium pressure will be?

 

[highschoboy004]

I can't help thinking that since both pistons have the same mass m, their weight should cancel each other when equilibrium is achieved, keeping both pistons at the initial height h.

 

[haruspex]

I can't help thinking that since both pistons have the same mass m, their weight should cancel each other when equilibrium is achieved, keeping both pistons at the initial height h.

But are they subject to the same forces?
(Btw, the question makes no sense to me.)

 

[highschoboy004]

But are they subject to the same forces?

This should be where I'm stuck at, I can't picture the force distribution between the two pistons.

 

[haruspex]

This should be where I'm stuck at, I can't picture the force distribution between the two pistons.

What forces act on each?

 

[highschoboy004]

What forces act on each?

I think there are two forces acting on each piston: their weight and the fluid's force. I'm asking about the latter force.

 

[jbriggs444]

I think there are two forces acting on each piston: their weight and the fluid's force. I'm asking about the latter force.

You are given the cross-sectional area of each piston. What is the formula for the fluid's force on each?

 

[jbriggs444]

(Btw, the question makes no sense to me.)

If I am interpreting it right, the scale of the apparatus would have to be quite large.

 

[haruspex]

If I am interpreting it right, the scale of the apparatus would have to be quite large.

So large that we have to find the weight of a column of air the density of which varies because of its own weight?

 

[highschoboy004]

If I am interpreting it right, the scale of the apparatus would have to be quite large.

I did not expect this, how insightful!

 

[highschoboy004]

So large that we have to find the weight of a column of air the density of which varies because of its own weight?

Thank you all for all of your help <3, I've managed to express the respective height of each piston, substituting all that air with water. This is just a random physics problem I picked up online so it can be quite "broken" sometimes.

 

[jbriggs444]

So large that we have to find the weight of a column of air the density of which varies because of its own weight?

Yes. That is the "easy" part. That gives you the relative heights -- about 5.5 km per engineering toolbox.

Then you have to utilize the information about the total mass of contained air by making an assumption about the cross-sectional area of the tubing on the bottom and about the reference location where $p_0$ is measured. That would allow you to deduce the absolute heights. The diagram makes the cross-section of the bottom tubing appear to be $S$.

Add the assumption of an isothermal ideal gas which means that real world tables like the one I used will be wrong.

I've not actually worked it. Just planned out an approach.

 

[Lnewqban]

I can't help thinking that since both pistons have the same mass m, their weight should cancel each other when equilibrium is achieved, keeping both pistons at the initial height h.

That is the reason to draw a free body diagram, whenever possible.
We have symmetry of masses and their weights, but not of piston's areas.

$P_{right}=F/A=mg/S$

$P_{left}=F/A=mg/2S$

Now combine that concept with your correct statement in post #4:
"I figured out that the air pressure inside the pipe must be constant such that its volume remains constant in order for Pascal's law to work."

Then, we conclude that one of the above equations must be incorrect, since we can only have one unique air pressure underneath both pistons.
Which one would it be?
Which mass would not be in equilibrium with the upward pressure-induced-force of its piston?

Thank you all for all of your help <3, I've managed to express the respective height of each piston, substituting all that air with water.

Could you show us your calculations and results?

I believe that it is important that you understand the physical principles that are involved in this problem.
Let's go back to try solving it with air, please.

 

[haruspex]

Let's go back to try solving it with air, please.

With air it is a very different problem. To a first approximation, the extra weight of air under the piston in the taller column is balanced by the extra weight of air above the piston in the shorter column. Why should it ever reach a balance?

 

[haruspex]

That gives you the relative heights -- about 5.5 km per engineering toolbox.

I would guess that is a heuristic result, so includes variation in temperature. We are told to assume constant temperature inside the tube. For that, I get that $p_{bot}=p_{top}e^{gh\rho_{top}/p_{top}}$.

 

[Lnewqban]

With air it is a very different problem. To a first approximation, the extra weight of air under the piston in the taller column is balanced by the extra weight of air above the piston in the shorter column. Why should it ever reach a balance?

I am not sure that either balance or equilibrium are proper words in this case.
Nevertheless, I can see a geometrical limit to the downward movement of either piston.

"The initial air pressure in the pipe is equal to atmospheric pressure p0.
Then both pistons are released"
... which makes me also see changes in the internal pressure, as well as assume that the value of h is within the centimeters-meters range rather than kilometers.

"... find the heights of both pistons when they reach equilibrium."
... which makes me see a second and final change in the value of that internal pressure and in the height of each air column.

Perhaps my reasoning is incorrect, please comment.
I only hope @highschoboy004 is following.

 

[haruspex]

I am not sure that either balance or equilibrium are proper words in this case.
Nevertheless, I can see a geometrical limit to the downward movement of either piston.

"The initial air pressure in the pipe is equal to atmospheric pressure p0.
Then both pistons are released"
... which makes me also see changes in the internal pressure, as well as assume that the value of h is within the centimeters-meters range rather than kilometers.

"... find the heights of both pistons when they reach equilibrium."
... which makes me see a second and final change in the value of that internal pressure and in the height of each air column.

Perhaps my reasoning is incorrect, please comment.
I only hope @highschoboy004 is following.

I'll reply by PM.