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Pressure exerted by water on Sphere partially exposed at end

  1. Jul 12, 2015 #1
    1. The problem statement, all variables and given/known data

    There is a cylindrical container that has a small hole of radius ##a ( < R)## at the bottom. A sphere of radius ##R## and density ##\rho_{s} > \rho_{w}## is placed in the cylinder such that it completely covers the hole (a part of it sticks out as in attached figure). The container is now filled with water (##\rho_{w}##) to a height ##h##.

    The objective is to determine the force exerted by the water on the sphere. (specifically on its dependence with the height of water) Buoyancy.png

    2. Relevant equations

    I have two methods in mind, but not sure which is correct.
    - Using Archimedes Principle
    - Using pressure integration

    3. The attempt at a solution

    Method 1:

    If we assume that the hole along with the portion that remains outside the container did not exist .. and the resulting contact surface between the flat portion of the sphere and the bottom of the cylinder was not perfectly smooth, then the force by the liquid would be equal to the weight of the liquid displaced (##V \rho_{w}g##).

    This force would be comprised of the force on the bottom surface and the force on the remaining cylindrical surface. Now, the force on the bottom surface is ##\rho_{w}hg \pi a^{2}##.

    So the force on the cylindrical section is ## V \rho_{w} g - \rho_{w} h g \pi a^{2}##.

    Since this is the force we want for the original question, this should be the answer after calculating ## V ## appropriately.

    Method 2.

    This involves integration.

    Buoyancy1.png

    Consider the diagram.

    The pressure at a circular (shell) element that makes an angle ##\theta## is given by ##P = (h - R \cos \alpha - R \cos \theta) \rho_{w}g ##. The elemental area ## A = 2 \pi R \sin \theta R d \theta##.

    So the net vertical force on the element (downward) is ## F_{v} = PA \cos \theta##

    Integrating this from ## \theta = 0 ## to ## \theta = \pi - \alpha ## should give the answer.

    However, evaluating this integral does not seem to match the answer for the above method.

    Is one of the two methods conceptually wrong?
    Any help would be highly appreciated.
     
  2. jcsd
  3. Jul 12, 2015 #2

    SteamKing

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    I'm not sure I follow your reasoning here. It seems much too complicated since you are applying Archimedes' Principle, i.e. the force applied by the water = the weight of water actually displaced by the sphere. It seems to me, the trick is to calculate the volume of the spherical cap which emerges from the hole in the bottom of the cylinder and subtract this volume from that of the whole sphere. This gives you the volume of displaced water.

    Fortunately, the formula for calculating the volume of a spherical cap can be found quite readily:

    https://en.wikipedia.org/wiki/Spherical_cap

    I think, looking at your diagram for the integration of hydrostatic pressure over the wetted surface of the sphere, it would be better to take h as the distance from the surface of the water in the cylinder measured to the center of the sphere. Then, you can use regular spherical coordinates to set up the integration.

    It's not clear if you are using vector calculus in your integration, because, while the hydrostatic pressure acts normal to the surface of the sphere, you are interested only in the net vertical force acting on the sphere due to this pressure.
     
  4. Jul 12, 2015 #3

    TSny

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    I believe your two methods are correct. I get the same result using both methods.
     
  5. Jul 12, 2015 #4
    @SteamKing

    For the first method, I can calculate the volume. What I intend to find out is the force exerted by the liquid on the curved section of the hypothetical ball (which will be the answer as needed for the original question). Since the buoyant force force for the hypothetical case is equal to the force acting on the bottom + the force on the curved surface, it should be easy to find out

    For the second method, I ignored the horizontal component of the forces, since they would negate out to zero. The change of measurement of h seems like good advice, probably can reduce calculations a bit.

    @TSny

    It's a bit late here. I'll post my complete calculations in roughly around 8 hours. I was getting conflicting results. Hope you can review my calculations and help me out if needed.
     
  6. Jul 12, 2015 #5

    haruspex

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    That seems to be addressing how to find V in litjee's post, so it is complementary to litjee's work, but does not replace it. Litjee is correct that the pressure the water would exert over the area of the hole (were the sphere replaced by water) has to be subtracted.
    Note in particular that the weight of water displaced does not depend on depth, but the subtracted term does. This explains why a sufficient depth of water will hold the sphere in place, while shallow water might allow it to float up.
     
    Last edited: Jul 12, 2015
  7. Jul 13, 2015 #6
    Thanks everyone.

    I went through the calculations again .. checking each and every line and voila the culprit was that I missed a factor of 2.

    Just for completeness, I get the force by the liquid (curved area) to be ##\displaystyle \pi \rho _{w} g \left [ a^{2} h - \frac{1}{3} \left ( a^{2} \sqrt{R^{2} - a^{2}} + 2R^{3} + 2R^{2} \sqrt{R^{2} - a^{2}}\right ) \right ]##

    Did I get it right?
     
    Last edited: Jul 13, 2015
  8. Jul 13, 2015 #7

    haruspex

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    Try a couple of sanity checks: a=R, a=0. Do those answers look right?
     
  9. Jul 13, 2015 #8
    Woopsie, there was a typo in the earlier post.

    ##a = 0## gives ##- \pi \rho_{w} g \cdot \frac{4}{3}R^{3}## which seems right.

    ## a = R ## gives ## \pi \rho_{w} g R^{2} h - \frac{2}{3} \pi R^{3} \rho_{w} g ## which also seems right
    Positive indicates a downward force.

    Yay! sanity checks are holding.. Thanks! I'm reasonably certain these are correct now.
    It's strange how you forget doing basic things like sanity checks, checking for miscalculations when you've been out of touch from actually going to paper pencil calculations and not relying on a piece of software.
     
    Last edited: Jul 13, 2015
  10. Jul 13, 2015 #9

    haruspex

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    Yes, except that you are being inconsistent wrt sign. You ought to state which direction is positive for your answer.
     
  11. Jul 13, 2015 #10
    True, will edit the relevant post.

    Thanks! :
     
  12. Jul 13, 2015 #11

    TSny

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    For what it's worth, I get the same result.
     
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