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Buoyancy Situation Doubt - Archimedes Principle

  1. Jun 5, 2013 #1
    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 penetrate


    Thanks to all of you for your help and opinion!
     
  2. jcsd
  3. Jun 5, 2013 #2

    russ_watters

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    Staff: Mentor

    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.
     
  4. Jun 5, 2013 #3
    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!
     
  5. Jun 7, 2013 #4
    may I ask- How are the balls connected?
     
  6. Jun 7, 2013 #5
    with a wire (volume negligible)
     
  7. Jun 7, 2013 #6
    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:

    image.jpg

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

    image.jpg

    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!
     
  8. Jun 7, 2013 #7
    Is the image give problems...

    30ic4s6.jpg

    mwuopj.jpg
     
  9. Jun 7, 2013 #8
  10. Jun 7, 2013 #9

    rcgldr

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    Homework Helper

    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.
     
  11. Jun 8, 2013 #10
    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.
     
  12. Jun 8, 2013 #11

    rcgldr

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    Homework Helper

    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.
     
  13. Jun 8, 2013 #12

    jbriggs444

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    Science Advisor

    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.
     
  14. Jun 9, 2013 #13
    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 explaination!
     
  15. Jun 10, 2013 #14

    jbriggs444

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    Science Advisor

    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]
     
    Last edited: Jun 10, 2013
  16. Jun 10, 2013 #15
    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:

    2uorm78.jpg

    Isn't it?
     
  17. Jun 10, 2013 #16

    jbriggs444

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    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.
     
  18. Jun 10, 2013 #17
    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.
     
  19. Jun 10, 2013 #18

    jbriggs444

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    Do the computation. Report the result.
     
  20. Jun 10, 2013 #19
    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.
     
  21. Jun 10, 2013 #20

    jbriggs444

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    Science Advisor

    Then that is the computation that you should perform.

    What is the weight of water on that hemisphere?
     
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