Hydrostatic Equilibrium Problem Involving a Cube and two Liquids

In summary: Archimedes' Principle depends on the assumption that the fluid can reach all parts of the body below the upper surface of the fluid. That is not the case here.
  • #1
ViktorVask
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New poster has been reminded to show their work on schoolwork problems
Homework Statement
The hole in the bottom of a container is sealed by a cube of side 𝑎 and density 𝜌. Inside the container there are two liquids of densities 𝜌1 and 𝜌2 (𝜌1> 𝜌2), as shown in the figure. The interface between the liquids coincides with the AB line. If the upper liquid level it is at the same height as the top point of the cube, which must be the minimum density 𝜌2 of the liquid superior so that the cube remains in equilibrium?
Relevant Equations
Hi guys, I have solved the this problem but I do not have the answer for it. If you wanna share your solution so we can discuss and get a conclusion I would enjoy a lot, thank you.
figura.png
 
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  • #2
Forum rules require you to show your attempt.
 
  • #3
yes as @Orodruin set you must show your attempt preferably written with ##\LaTeX##. If you don't know latex, then post a screenshot of your written work as a last resort.
 
  • #4
I say typed non-latex is still preferable to pictures ...
 
  • #5
Orodruin said:
I say typed non-latex is still preferable to pictures ...
hmm can't fully agree to that, maybe or maybe not, it depends on the hand writing and on the quality of the picture.
 
  • #6
Delta2 said:
hmm can't fully agree to that, maybe or maybe not, it depends on the hand writing and on the quality of the picture.
In pictures, there is no way of quoting a specific part of the solution. Also, we both know that many pictures posted are hardly readable, rotated, or blurry. It is also against the homework guidelines (item 5) to just post a picture of your work.
 
  • #7
Orodruin said:
In pictures, there is no way of quoting a specific part of the solution. Also, we both know that many pictures posted are hardly readable, rotated, or blurry. It is also against the homework guidelines (item 5) to just post a picture of your work.
There is a way of manually quoting by typing
manually enter the quoted text here
but requires more effort from the person that replies. Or you can edit the picture in a program and cut and paste the quoted portion of the picture.
I agree that usually the pictures are of bad quality and discourage the reader of replying to the post.
 
  • #8
Delta2 said:
There is a way of manually quoting by typing but requires more effort from the person that replies. Or you can edit the picture in a program and cut and paste the quoted portion of the picture.
I agree that usually the pictures are of bad quality and discourage the reader of replying to the post.
I advise posters that if they must post images of work then a) make them very clear and b) number all equations.
 
  • #9
ViktorVask said:
I have solved the this problem but I do not have the answer for it.
Unusual. Do you mean merely that you do not have the official answer?
Please demonstrate that you have solved it by posting your working, with or without your answer.
 
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  • #10
haruspex said:
Unusual. Do you mean merely that you do not have the official answer?
Please demonstrate that you have solved it by posting your working, with or without your answer.
Yeah. This problem is from the Ibero-American Physics Olympiad 2017 but they did not share the solution/answer. I will attach here what I did, sorry for not doing this before.
 

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  • #11
ViktorVask said:
Yeah. This problem is from the Ibero-American Physics Olympiad 2017 but they did not share the solution/answer. I will attach here what I did, sorry for not doing this before.
I can't read it. What's your final answer?
 
  • #12
Chestermiller said:
I can't read it. What's your final answer?
𝜌2 = 4𝜌 - 𝜌1
 
  • #13
ViktorVask said:
Yeah. This problem is from the Ibero-American Physics Olympiad 2017 but they did not share the solution/answer. I will attach here what I did, sorry for not doing this before.
I need some explanation to follow your working.
You start by finding the pressure ("P") at the bottom of the tank, but what is F' exactly? It seems to be the "missing" force, i.e. the force that would be exerted up on the bottom of the cube if you were to slice off the bit the goes below the bottom of the tank, mend the hole in the tank, and allow the liquid to pass under the frustrated cube. Is that right?
But I have no idea how you are defining E.

I considered the forces due to the upper liquid. Above the boundary layer it exerts a downward force computable from the pressure at its average depth and the area it acts on there, while below the boundary layer it is an upward force based on the pressure at its full depth (depth of the boundary layer) and the area below that.
The second pressure is double the first but acts on half the area, so...?
 
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  • #14
A simpler approach is to use Archimedes principle: upthrust = weight of liquid displaced = weight of cube
 
  • #15
I solved this two different ways and came up with the same peculiar result. Did anyone else get a peculiar result with respect to ##\rho_2##?
 
  • #16
Chestermiller said:
I solved this two different ways and came up with the same peculiar result. Did anyone else get a peculiar result with respect to ##\rho_2##?
Yes. I think maybe @haruspex is hinting at this in post #13.
 
  • #17
Steve4Physics said:
A simpler approach is to use Archimedes principle: upthrust = weight of liquid displaced = weight of cube
Archimedes' Principle depends on the assumption that the fluid can reach all parts of the body below the upper surface of the fluid. That is not the case here. It is possible to correct for that by calculating what the additional force would be if the fluid could reach all parts, but in this case you have to be careful not to increase the depth of the fluid in the process.
See the first part of my post #13, relating to F'.
 
  • #18
TSny said:
Yes. I think maybe @haruspex is hinting at this in post #13.

Quite so.
It is also interesting to note how the net force due to the upper fluid varies as its depth increases from zero.
 
  • #19
haruspex said:
I need some explanation to follow your working.
You start by finding the pressure ("P") at the bottom of the tank, but what is F' exactly? It seems to be the "missing" force, i.e. the force that would be exerted up on the bottom of the cube if you were to slice off the bit the goes below the bottom of the tank, mend the hole in the tank, and allow the liquid to pass under the frustrated cube. Is that right?
But I have no idea how you are defining E.

I considered the forces due to the upper liquid. Above the boundary layer it exerts a downward force computable from the pressure at its average depth and the area it acts on there, while below the boundary layer it is an upward force based on the pressure at its full depth (depth of the boundary layer) and the area below that.
The second pressure is double the first but acts on half the area, so...?
Yeah. P is the pressure at the bottom of the tank. F' is the force that acts on that area due the pressure P. E is the buoyancy force(the famous one E = density of liquid x gravity x volume of body submersed), but here is the main problem of this kind of situation : We cannot just say that E is the total force that acts on the cube since there is a part of the cube who is not surrouded by the liquid (the bottom/ the hole) as you said there is a 'missing' force, so what I did (the famous trick for these problems) was calculating this force F' so now we can just say that the resultant upward force is Fr = E - F'.
E = 𝜌2.g.V/2 + 𝜌1.g.3V/4 (V is the total volume of the cube V = a³)
Fr must equilibrate the weight of the cube = 𝜌.g.V

I am sorry if the image was in a low quality, I think it loses quality when I upload (for me it is readable)

There is another problem very similar to this one, instead of a cube it is a sphere but the idea is (of course) the same, just different geometry. You can find this problem of the sphere in the great book ''200 puzzling physics problems''
 
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  • #20
ViktorVask said:
Yeah. P is the pressure at the bottom of the tank. F' is the force that acts on that area due the pressure P. E is the buoyancy force(the famous one E = density of liquid x gravity x volume of body submersed), but here is the main problem of this kind of situation : We cannot just say that E is the total force that acts on the cube since there is a part of the cube who is not surrouded by the liquid (the bottom/ the hole) as you said there is a 'missing' force, so what I did (the famous trick for these problems) was calculating this force F' so now we can just say that the resultant upward force is Fr = E - F'.
E = 𝜌2.g.V/2 + 𝜌1.g.3V/4 (V is the total volume of the cube V = a³)
Fr must equilibrate the weight of the cube = 𝜌.g.V

I am sorry if the image was in a low quality, I think it loses quality when I upload (for me it is readable)

There is another problem very similar to this one, instead of a cube it is a sphere but the idea is (of course) the same, just different geometry. You can find this problem of the sphere in the great book ''200 puzzling physics problems''
I confirm your value for the ##\rho_2## term, but I only get 1/3 of your ##\rho_1## term (in the equation for the bottom force). And, this supports not only the weight of the cube, but also the weights of the triangular wedges of the two fluids above.
 
  • #21
haruspex said:
Archimedes' Principle depends on the assumption that the fluid can reach all parts of the body below the upper surface of the fluid. That is not the case here. It is possible to correct for that by calculating what the additional force would be if the fluid could reach all parts, but in this case you have to be careful not to increase the depth of the fluid in the process.
See the first part of my post #13, relating to F'.
Agreed. Using Archimedes' Principle and the correction for the 'missing force' isn't hard. I tried it and got the 'peculiar' answer. But your Post ##13 explains the peculiarity (and saved me a sleepless night!).
 
  • #22
Chestermiller said:
I confirm your value for the ##\rho_2## term, but I only get 1/3 of your ##\rho_1## term (in the equation for the bottom force).
I don't see a problem with the equation for F', but in the equation for E I get 1/2 of the ##\rho_1## term: ##E=Vg(\frac 12\rho_2+\frac 38\rho_1)##.
And where is the post you quote in your post #19? @ViktorVask , did you delete it?
 
  • #23
haruspex said:
I don't see a problem with the equation for F', but in the equation for E I get 1/2 of the ##\rho_1## term: ##E=Vg(\frac 12\rho_2+\frac 38\rho_1)##.
And where is the post you quote in your post #19? @ViktorVask , did you delete it?
For the pressure at the base, I get $$p=\rho_2g\frac{a}{\sqrt{2}}+\rho_1 g \frac{a}{2\sqrt{2}}$$ Is that not what others get?
 
  • #24
Chestermiller said:
For the pressure at the base, I get $$p=\rho_2g\frac{a}{\sqrt{2}}+\rho_1 g \frac{a}{2\sqrt{2}}$$
Yes, but doesn't that match the expression for F' in Photo1? In post #19 I thought you were saying you disagreed with it.
 
  • #25
ViktorVask said:
Yeah. P is the pressure at the bottom of the tank. F' is the force that acts on that area due the pressure P. E is the buoyancy force(the famous one E = density of liquid x gravity x volume of body submersed), but here is the main problem of this kind of situation : We cannot just say that E is the total force that acts on the cube since there is a part of the cube who is not surrouded by the liquid (the bottom/ the hole) as you said there is a 'missing' force, so what I did (the famous trick for these problems) was calculating this force F' so now we can just say that the resultant upward force is Fr = E - F'.
E = 𝜌2.g.V/2 + 𝜌1.g.3V/4 (V is the total volume of the cube V = a³)
Fr must equilibrate the weight of the cube = 𝜌.g.V

I am sorry if the image was in a low quality, I think it loses quality when I upload (for me it is readable)

There is another problem very similar to this one, instead of a cube it is a sphere but the idea is (of course) the same, just different geometry. You can find this problem of the sphere in the great book ''200 puzzling physics problems''
This is weird. When I replied to Chet's post #20 earlier it was numbered #19 and your post #19 was not shown to me.
As I wrote in post #22, I think the 3/4 in there should be 3/8.
I had no problem reading the image (I find it works better if I select to open it in a new window), my only difficulty was knowing how your variables were defined.
 
  • #26
@ViktorVask has made a simple algebra mistake. Post 10, Figure 2 states:
$$F_r =\frac { \rho_2 g a^3 } {2} + \frac {3 \rho_1 g a^3 }{8} - \frac {a^3 g} {2} \left( \rho_2 + \frac {\rho_1}{2}\right)$$The right hand side is then incorrectly simplified to:$$\frac { a^3 g } {2} \left( \rho_2 + \frac {3\rho_1}{ 4} - \frac{\rho_2}{2}- \frac{\rho_1}{4} \right)$$But this should be:
$$\frac {a^3 g } {2} \left( \rho_2 + \frac {3\rho_1}{ 4} - \rho_2- \frac{\rho_1}{2} \right)$$This then leads to the (counterintuitive) answer explained by @haruspex in Post #13.
 
  • #27
Correct me if I am wrong. I integrated the vertical component of the pressure forces over the surfaces of the cube. On the portion immersed in fluid 2, I got a downward force of ##\frac{\rho_2 ga^3}{2}## and, on the portion immersed in fluid 1, I got an upward force of ##\frac{\rho_2 ga^3}{2}+\frac{\rho_1ga^3}{8}##, so the net fluid pressure force on the cube was upward, and equal to ##\frac{\rho_1ga^3}{8}##, independent of ##\rho_2##. I got the exact same result by calculating the upward force on the base of the stack, and subtracting the weights of the 4 triangular fluid wedges above.
 
  • #28
Chestermiller said:
Correct me if I am wrong. I integrated the vertical component of the pressure forces over the surfaces of the cube. On the portion immersed in fluid 2, I got a downward force of ##\frac{\rho_2 ga^3}{2}## and, on the portion immersed in fluid 1, I got an upward force of ##\frac{\rho_2 ga^3}{2}+\frac{\rho_1ga^3}{8}##, so the net fluid pressure force on the cube was upward, and equal to ##\frac{\rho_1ga^3}{8}##, independent of ##\rho_2##.
That's what I get also.
 
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  • #29
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  • #30
haruspex said:
This is weird. When I replied to Chet's post #20 earlier it was numbered #19 and your post #19 was not shown to me.
As I wrote in post #22, I think the 3/4 in there should be 3/8.
I had no problem reading the image (I find it works better if I select to open it in a new window), my only difficulty was knowing how your variables were defined.
Oh yeah,sorry. It should be 3/8 (as in the photo) . I will edit this.
 
  • #31
Chestermiller said:
Correct me if I am wrong. I integrated the vertical component of the pressure forces over the surfaces of the cube. On the portion immersed in fluid 2, I got a downward force of ##\frac{\rho_2 ga^3}{2}## and, on the portion immersed in fluid 1, I got an upward force of ##\frac{\rho_2 ga^3}{2}+\frac{\rho_1ga^3}{8}##, so the net fluid pressure force on the cube was upward, and equal to ##\frac{\rho_1ga^3}{8}##, independent of ##\rho_2##. I got the exact same result by calculating the upward force on the base of the stack, and subtracting the weights of the 4 triangular fluid wedges above.
I will try to do it by integrating too. I realized now that I really made a simple mistake and the resultant force is what you found. But now I am a little bit confused, the total force does not depend of 𝜌2 . Isnt that weird ? What should the answer for 𝜌2 be ?
 
  • #32
ViktorVask said:
I will try to do it by integrating too. I realized now that I really made a simple mistake and the resultant force is what you found. But now I am a little bit confused, the total force does not depend of 𝜌2 . Isnt that weird ? What should the answer for 𝜌2 be ?
Yes, it certainly sounds counterintuitive, but can't be argued with (since two completely different methods give exactly the same answer).

the answer for rho2 should be that it can be as low as zero.
 
  • #33
Chestermiller said:
Yes, it certainly sounds counterintuitive, but can't be argued with (since two completely different methods give exactly the same answer).

the answer for rho2 should be that it can be as low as zero.
I just did the another solution. I will attach here. Got the same as you again.

In the first photo I calculated the force due the liquid 2
In the second one the force due the liquid 1 (taking into account the pressure that liquid 2 keeps exerting)
The last one I found the resultant
I think now it is okay
 

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  • #34
ViktorVask said:
But now I am a little bit confused, the total force does not depend of 𝜌2 . Isnt that weird ?
At first, it does take you by surprise. But the simple explanation is given in the second paragraph of post #13 by @haruspex.
 
  • #35
TSny said:
At first, it does take you by surprise. But the simple explanation is given in the second paragraph of post #13 by @haruspex.
Taking it further, if we consider starting with only the denser fluid and gradually add the lighter one, the buoyant force increases but with a negative quadratic term. It reaches a max when the fluid is half way up the face of the cube, then declines to its original value in the diagram position.
Beyond that, the buoyant forces declines linearly, eventually becoming negative overall, so a net downward force.
 
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1. What is hydrostatic equilibrium?

Hydrostatic equilibrium is a state in which the pressure within a fluid is balanced, meaning that the forces acting on any small volume of the fluid are equal in all directions.

2. How does a cube and two liquids relate to hydrostatic equilibrium?

In this problem, the cube represents a solid object submerged in two different liquids with different densities. The pressure exerted on the cube by each liquid creates a hydrostatic equilibrium, causing the cube to remain stationary.

3. What factors affect the hydrostatic equilibrium in this problem?

The densities of the two liquids and the depth at which the cube is submerged are the main factors that affect the hydrostatic equilibrium in this problem.

4. How can the hydrostatic equilibrium equation be used to solve this problem?

The hydrostatic equilibrium equation states that the pressure at a certain depth in a fluid is equal to the product of the density of the fluid, the acceleration due to gravity, and the depth. This equation can be used to calculate the pressure at different depths in each liquid and determine the equilibrium point.

5. What are some real-world applications of hydrostatic equilibrium problems?

Hydrostatic equilibrium problems are commonly used in engineering and physics to analyze the stability of structures such as dams, ships, and submarines. They are also used in meteorology to understand atmospheric pressure and in geology to study the movement of fluids in underground reservoirs.

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