Buoyancy Situation Doubt - Archimedes Principle

[Kamira]

Hi to all, i will appreciate your help in this.

This is the situation:

I have a tank, with a water column in it. This tank at the bottom has an "ideal" seal (a seal that permit the passing of object from bottom to upwards, but not the water to fall down.)

In this system i have sphere with density < water density, connected with cord.Click the link for images:

https://www.dropbox.com/s/0k93koeqbmtt9ot/global1.jpg

Bottom detail:

https://www.dropbox.com/s/ls3go84h0oop3og/detail1.jpg

I would like some explanation about forces and formulas. I have already written some, tell me your opinion.

I have moreover the situations case, i would hear your opinion on these too:

-CASE 1 (the second balance case is impossible because the opposing force to penetration for the ball in the bottom is always bigger than the buoyancy force of the other balls.

Click for image:

https://www.dropbox.com/s/8z2pc43tj7kj1yc/case1.jpg
-CASE 2 (i have more balls out the water in the bottom side)

https://www.dropbox.com/s/upyetwatlioagci/case2.jpg

Note: LIMIT CASE: if i would have in place of all the submerged spheres a cylinder floater, with dimension: A:section, H hight (same hight of the tank), compleatly submerged, the buoyancy force of the cylinder will be the same of the opposing force at the bottom for the penetration of the ball ( density water*g*H*A). So neither in that limit case the ball will penetrateThanks to all of you for your help and opinion!

 

[russ_watters]

Welcome to PF!

If you are just asking if there are any problems with what is posted, I don't see any, but beyond that it isn't very clear what you are asking.

Also if you are looking to turn this into a perpetual motion machine, please note that hat that violates the laws of physics and therefore the rules of this forum.

 

[Kamira]

Hi Russ!

Thanks for the welcome.

I am not at all a believer of perpetual motion (as you can see and read in my first post!) (my objective is to prove it doesn't and never works, without using Termodinamic principles) only with basic Phisic, so as a Galilean i want to prove with formulas and cases.

I asked for the cases, if my interpretations are correct, and formulas are correct, especially for case 2.
Just to know your opinion and, if you want, to talk about that.

Thank you guys!

 

[Grombely]

may I ask- How are the balls connected?

 

[Kamira]

with a wire (volume negligible)

 

[Kamira]

Hi guys, i will propose you the problem in another way, More clear, and images.

I have a tank with water (10 m hight) , with an ideal seal at the bottom (water can't fall down, but can enter bodies).

I have a system of 6 cubes ( of plystynere) with dimension 1x1x1 m.
These cubes are connected with a rope (volume negligible).

As you can see in this image:

In the next image i have a situation (1) which is unstable.

My question is:

in a stable situation it will be as in the balance situation (2) (see upper figure),

and...if the answer is yes... how much will be de measure b? (and how you reach that result...).

I will post a video too, just the clear you the scene.

Thank you for support and helping!

 

[Kamira]

Is the image give problems...

 

[Kamira]

The video:

https://www.youtube.com/watch?v=JrhurV3pp1I

 

[rcgldr]

Assuming you have that magical seal that allows a mass to extend below the bottom water line, the result will be unstable. The further downwards that the extended mass is, the greater the amount of pressure from the water since the pressure increases with depth.

To create a stable situation, you'd need a lower density mass floating at the upper surface of the water, supporting any higher density masses below it.

 

[Kamira]

Yes in the first image is unstable, i ask for the stable situation in the second image.

The buoyancy force of the cubes pulls up, but the pressure at the bottom will resist for the bottom cube. The resultant of these forces will permit the cube to penetrate a little?
If yes...how much?

I need the b measure in the stable situation and the steps to get it.

 

[rcgldr]

I ask for the stable situation in the second image.

It's not stable, assuming that the second image is a boundary state, if the mass moves up, it continues to move up, if the mass moves down, it continues to move down. There's no self correcting tendency for the mass to remain at the position in the second image.

 

[jbriggs444]

Due to the use of cubical shapes, there is a discontinuity in the system. If the top surface of the bottom cube is just below the seal there is no pressure on it. If the top surface of the bottom cube is just above the seal there is full pressure of water-at-depth on it.

That means that the point of stability (if there is one) is almost certainly with the top surface of the bottom cube exactly at the seal. This is the point of discontinuity. The pressure on the top surface of the bottom cube is undefined at that point.

If instead of cubes one used spheres, no such discontinuity would exist.

 

[Kamira]

Interesting jbriggs444 !

In case of sphere or cylinders so...which do you think will be the balanced situation?

Thanks you! It was a very good explanation!

 

[jbriggs444]

Assume that the spheres are one meter in diameter and that five of them completely inside the chamber. Assume that they are of negligible density and that the water and spheres are incompressible.

By inspection, the acceleration of gravity (9.8 meters/sec²) is irrelevant. If there is an equilibrium at one gee, then the same equilibrium would exist at any other gee force as well.

By inspection, the density of water is irrelevant. If an equilibrium exists at a density of 1000 kg per cubic meter, the same equilibrium would exist at any other density as well.

So let us choose to work in a system of units where there is a buoyancy of one unit of force per unit of volume. In this system of units, the pressure on the bottom will be measured in units of depth.

The pressure at the bottom is 10 per square meter.
The buoyancy of each sphere is 4/3 pi r³ = pi/6.

With 5 spheres fully contained in the chamber, that's a total buoyancy of 5pi/6... This is approximately 2.6

Now add the sixth sphere.

If its top is just touching the bottom seal then we've just calculated a total upward buoyancy of 2.6.

If it makes it half-way in then it contributes pi/12 to the buoyancy for a total of 11pi/12 (about 2.9). The pressure of water-at-depth on its cross section would be 10 pi r² = 25 pi / 4(about 7.9). That's a net downward force of about 5.3.

[edit -- corrected factor of four error on the calculated downforce in the half-inserted case above]

The key point is that with the sixth ball just touching the seal there is a net upward force and with the sixth ball halfway in there is a net downward force. The equilibrium is somewhere in between.

If you had a formula for the volume of a truncated piece of a sphere, you could write down an equation and solve for the equilibrium point. Such a formula is not difficult to derive, but in the interest of simplicity, I'm going to skip that and go with an approximation.

Ignore the contribution of buoyancy from the sixth sphere. The equilibrium will be attained when the force of water-at-depth on its cross-section at the seal is equal to the total buoyancy of the other five balls.

cross-section ~= 2.6 / 10 = 0.26

0.26 = pi r²

So r = √(0.26/pi) ~= 0.28

That r is the radius of the circle that is the cross-section.

Apply a little trig. If you measure from the center of the sphere, that circle is a cone at 34 degrees from the vertical. sin(theta) = 0.28/0.5

That means that the top of the sphere is intruding into the chamber by 0.5 * (1 - cos(theta) ) ~= 0.09 meters.

Call it 9 centimeters. [Bearing in mind that it's even odds whether I screwed up the calculation]

 

[Kamira]

Thanks Jb for your answer and time!
You are very clear.

I will analyze it very carefully.

Right now just a question.

You write: "If it makes it half-way in then it contributes pi/12 to the buoyancy for a total of 11pi/12 (about 2.9)"

But an half introduced sphere in the bottom doesn't experiment buoyancy. The pressure only will act as in the image:

Isn't it?

 

[jbriggs444]

You can calculate the force on a submerged object in a number of equivalent ways.

For instance, you can add up the pressure on all of its surfaces due to the fluid in which it is immersed. Call this the "pressure approach". Or you can calculate the volume it displaces, multiply that by the density of the fluid and the acceleration of gravity. Call this the "buoyancy approach".

The diagram that you show above lends itself to the pressure approach. But that's difficult to calculate. The pressure is not uniform -- it varies with depth. And the pressure is not all in the same direction.

So I chose to use a hybrid approach. Compute the supporting force that would be present due to buoyancy and then subtract the pressure on the cross-sectional area where no supporting fluid is present.

So the net force on the partially-inserted sphere is equal to the buoyancy that the portion inside the chamber would have experienced minus the pressure of fluid-at-depth that would have been present on the bottom surface if you had chopped the sphere in two right at the seal.

 

[Kamira]

We can call the approch as we want, but an half sphere in the bottom of a tank doesn't experiment any buoyancy, no force thant pull upwards.
The only force due the water is pushing the ball down.

 

[jbriggs444]

We can call the approch as we want, but an half sphere in the bottom of a tank doesn't experiment any buoyancy, no force thant pull upwards.
The only force due the water is pushing the ball down.

Do the computation. Report the result.

 

[Kamira]

I am sure of that without computation. No buoyancy for a semisphere posed on the bottom of a tank of water. Only the weight of the water on it.

 

[jbriggs444]

Then that is the computation that you should perform.

What is the weight of water on that hemisphere?

 

[Kamira]

In every point of the sphere there is a pressure ρ*g*h
with ρ water and h the high of water in the point.

To obtain a righ computation , it's necessary consider those pressure and the surface on which are applicated (integrating in the case of sphere).

No buyancy force anyway that push upwards a semishere posed on the bottom .

 

[jbriggs444]

Where is your computed result?

 

[Kamira]

In your explanation has committed an inaccuracy.
No need to compute to demostrate that, pure physics.

There is no approach that can allow an half sphere like that to receive an upwards force of buoyancy.
I am sorry. Maybe you noted too that mistake and so the "there are many approaches.."

The approach is: in physics what happen to a semisphere perfectly posed in the bottom of a tank.

Answer: no buyancy, don't move upwards.

You can compute or approach as you want. That's it.

So your anterior statement: "If it makes it half-way in then it contributes pi/12 to the buoyancy for a total of 11pi/12 (about 2.9)" is quite an error.

 

[jbriggs444]

If had you actually tried to compute it, you might have realized that there is a short-cut that could be taken to simplify the computation. That is why I had asked you to "Do the computation. Report the result".

You would be trying to compute the weight of a cylindrical column of water with a hemispherical hole cut out at the bottom. The weight of such a column would be equal to the weight of a cylindrical column of water with no hole minus the weight of the hole. The weight of the hole is equal to the weight of the water displaced by the hemisphere. The weight of the water displaced by a hemisphere is equal to the buoyancy that the hemisphere would have if it were surrounded by water.

I chose to regard the situation as five and a half buoyant balls minus a correction for the bottom surface of that bottom hemisphere that is not exposed to water. It was and is obvious to me that this treatment produces the correct calculated net force.

I did not make this intuitive leap clear in that first posting and used terminology that, as you point out, ascribes "buoyancy" to a situation where the concept does not apply.

Is that what you wanted to hear me say?

Now then...

If we discard the labels on the forces, do you agree that the net force on five fully immersed spheres plus the one half-immersed sphere at the bottom seal is equal to the buoyancy that five and a half spheres would have minus water pressure at the depth of the bottom seal times the cross-sectional area of the bottom sphere at the seal?

 

[rcgldr]

We can call the approch as we want, but an half sphere in the bottom of a tank doesn't experiment any buoyancy, no force thant pull upwards. The only force due the water is pushing the ball down.

You have the pressure from the atmosphere pushing up on the ball from below, but the pressure from the atmosphere is less than the pressure of the water except at the upper surface of the water.

 

[Kamira]

Thanks Rc and Jbriggs for explaining.

Trying to do calculation (and using cylinders and real case)...

I think (hope) that's correct. :yuck:

 

[Kamira]

jdbriggs?

 

[jbriggs444]

Your computation for the upward buoyancy force from the five fully immersed cylinders appears correct. 38,500 Newtons is what I get as well.

Your computation for the downward weight from the six cylinders appears correct. 924 Newtons is what I get as well.

I do not understand your wording for "seal opposing force downwards". The forces that I see being applied on the partially immersed cylinder are:

1. Water pressure over the curved upper surface
2. Air pressure over the bottom surface -- this cancels out and we can consider it to be zero.
3. Force from the seal -- but the seal is ideal, so this is zero.
4. Upward force from the 5 tethered cylinders above -- but that's already being accounted for.
5. Weight -- but that's already being accounted for.

So the figure you are trying to quantify is water pressure over that curved upper surface.

As an approximation, you could consider that the pressure over that surface is constant and is equal to 10 meters of water head -- 98100 Pa. I will assume that this is what you are doing.

With this understanding, your equation for the water pressure on the top of the partially immersed cylinder is what I get as well.

Solving for the value of b for which the upward and downward forces balance, I also get 38.3 cm.

Applying the pythagorean theorem, I agree that you get c = 3.8 cm.

With that value in hand, you can put a bound on the error that was introduced by approximating the pressure on the upper surface of the cylinder with a constant. [This is the "buoyancy" factor that we had vigorously discussed]

I get that b will only go up to 38.4 cm at most with that factor accounted for.

 

[Kamira]

Thank you JD!

I have tried to solve it with a more precise way (integrals) but with no success.
Calculating that volume in function of angle teta.

(