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Fluid Mechanics

  1. Jul 18, 2015 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations


    3. The attempt at a solution

    On integrating the force due to fluid pressure on the cap , I have arrived at the correct result ##ρgπR^2(H-\frac{2}{3}R)## . This in turn would be the force with which the cap presses the bottom.

    But I would like to solve this problem with less maths i.e may be by using Archimedes principle .

    The weight of the volume of fluid displaced would be ##ρgπ\frac{2}{3}R^3## . Shouldn't this be the force exerted by fluid on the cap ?

    I would be grateful if somebody could help me with the problem .
     

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    Last edited: Jul 18, 2015
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  3. Jul 18, 2015 #2

    ShayanJ

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    You can't use Archimedes principle here because the force is not due to buoyancy but pressure.
    Anyway, I get a different result than yours. Are you sure you didn't make any mistake?

    EDIT: Bad wording!!! Of course buoyancy is due to pressure too, I just meant Archimedes principle is for the upward force.
     
  4. Jul 18, 2015 #3

    TSny

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    The weight of the fluid displaced by a submerged hemisphere would be equal to the upward buoyant force acting on the hemisphere. Note that this upward force is due to fluid pressure on the curved surface as well as the flat surface of the hemisphere. You only want the force on the curved surface. So, you have little more work to do.
     
  5. Jul 18, 2015 #4
    Hello TSny
    But force on the flat surface is not due to the fluid . It is due to the floor on which the vessel is resting .
     
  6. Jul 18, 2015 #5

    TSny

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    In this problem, the force on the flat surface is due to atmospheric pressure since the flat surface is exposed to the atmosphere because of the hole.

    Archimedes principle assumes that the object is completely surrounded by fluid. [EDIT: Or the object is partially submerged. Here, the bottom of the hemisphere is not exposed to the fluid since it is covering a hole. So, the weight of fluid displaced by the hemisphere does not equal the net force of the fluid on the hemisphere. So, in this sense, Archimedes' principle does not apply.]
     
    Last edited: Jul 19, 2015
  7. Jul 18, 2015 #6

    ShayanJ

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    Using Pascal's law, the pressure at a depth z under the water is ## P=\rho g z ##. We want the pressure on the surface of the hemisphere, so z should be calculated accordingly. If we consider a spherical coordinate system with the origin at the centre of the hemisphere, we can write ## z=H-R\sin\theta##. So the downward force at each surface element of the hemisphere is ## \rho g (H-R\sin\theta)R^2\sin^2\theta d\theta d\varphi ##. Integrating it over the surface of the hemisphere gives ## \pi \rho g R^2 (\pi H-\frac{8R}{3}) ##. I don't see how Archimedes principle can help.
     
  8. Jul 18, 2015 #7
    I think I am misunderstanding some things .

    Isn't the vessel resting on a horizontal surface ? Hasn't the cap plugged the hole ? How is atmospheric pressure exerting force on the flat surface of the hemisphere ?
     
  9. Jul 18, 2015 #8
    You may be right . I am quite prone to making mistakes . May be TSny could verify the result.

    I am just looking at a more intuitive solution . By no means I am saying that this problem could be solved by Archimedes Principal . But it might very well be :smile: .
     
  10. Jul 18, 2015 #9
    I get the same answer that Tanya got. I don't see an easier way of doing the problem.

    Chet
     
  11. Jul 18, 2015 #10
    Hi Chet

    Thanks for the confirmation . Do you agree that Archimedes Principle is not applicable in this situation ? If not , what is the reason ? Is it because the cap rests on a surface i.e does not have fluid under it .
     
  12. Jul 18, 2015 #11
    Yes. See TSny post #5.

    Chet
     
  13. Jul 18, 2015 #12

    TSny

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    You can use Archimedes' principle as a way to get to the answer. You just have to modify the weight of the fluid displaced to take into account that you only want the force due to pressure on the curved surface alone. The weight of the fluid displaced by the volume of hemisphere will give you the net upward force due to fluid pressure acting on all surfaces of the hemisphere if the hemisphere were completed surrounded by fluid. Since your hemisphere is not completely surrounded by fluid, you will need to modify what Archimedes' principle gives you with a simple calculation.
     
  14. Jul 18, 2015 #13
    No headway :smile: . How to do what you are suggesting ?
     
  15. Jul 18, 2015 #14

    TSny

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    Suppose the hemisphere is completely submerged in the fluid so that the hemisphere is completely surrounded by fluid. So, right now there is no hole in the container and the hemisphere is not at the bottom of the container.

    Archimedes' principle gives you the buoyant force. But the buoyant force is just the net upward force due to pressure of the fluid acting on the curved and flat surfaces of the hemisphere. What would you have to do this buoyant force to get only the net force acting on the curved surface alone? Assume the hemisphere is oriented as in your problem with the flat surface horizontal.
     
  16. Jul 18, 2015 #15
    Force due to liquid on curved surface + Force due to liquid on flat surface = ##ρgπ\frac{2}{3}R^3## .

    Force due to liquid on curved surface = ##ρgπ\frac{2}{3}R^3## - Force due to liquid on flat surface .

    Is this what you are suggesting ?
     
    Last edited: Jul 18, 2015
  17. Jul 18, 2015 #16

    TSny

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    Yes, just be careful with signs due to force on curved surface being downward while force on flat surface is upward.
     
  18. Jul 18, 2015 #17
    Now if liquid was present , force due to liquid on flat surface would have been = ##ρgπR^2H## .

    So, Force due to liquid on curved surface = ##ρgπ\frac{2}{3}R^3 - ρgπR^2H##

    But then ,this expression has a negative sign as opposed to what I got in the OP .
     
    Last edited: Jul 18, 2015
  19. Jul 18, 2015 #18

    TSny

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    In OP you calculated the downward force on the curved surface. In #17 you calculated the upward force on the curved surface.
     
    Last edited: Jul 18, 2015
  20. Jul 18, 2015 #19

    TSny

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    Assuming that the system is surrounded by the atmosphere, you should make sure that you have taken atmospheric pressure into account. For example, in your OP the force that you calculated by integrating the pressure over the curved surface should include an additional term due to atmospheric pressure. But would this affect the answer for the force that the hemispherical cap presses on the bottom of the container? :wideeyed:
     
    Last edited: Jul 18, 2015
  21. Jul 18, 2015 #20
    Ok . So the force should be ##P_0πR^2+ρgπR^2(H-\frac{2}{3}R)## . Right ??
     
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