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Homework Help: Minimum force required to form sphere

  1. May 27, 2015 #1
    1. The problem statement, all variables and given/known data
    ?temp_hash=7ef78ca52a47c03b0eac6a90ab6716fc.png


    2. Relevant equations


    3. The attempt at a solution

    Doing a vertical force balance 2Fcosθ=mg ,where m is the mass of water .

    Not sure how to proceed .

    What role does the pin hole at the top play ?

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

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  2. jcsd
  3. May 27, 2015 #2

    Simon Bridge

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    ... that is the key question.
    Think about it - what does putting a pinhole in the top do that not having a hole there not do?
    i.e. imagine there is also a pinhole at the bottom - what difference would a hole at the top make?
     
  4. May 27, 2015 #3

    haruspex

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    Pace Simon, but the pinhole at the top is only to get around a complication that the setter did not want to get tangled up with. It looks like you didn't notice the complication anyway, so what you have done so far is ok. But you need another equation. The answer to the question should be an actual number, not dependent on any variables.
     
  5. May 27, 2015 #4
    Here's a hint Tanya: Think hydrostatic.

    Chet
     
  6. May 27, 2015 #5
    I have thought about it . Please elaborate .
     
  7. May 27, 2015 #6
    The pressure (force per unit area) on each hemisphere is going to increase with depth. To get the horizontal force on each hemisphere, you need to integrate. I haven't figured out yet whether you also need to do a moments balance.

    Chet
     
  8. May 27, 2015 #7
    Chet ,

    I somehow feel there is a simpler , less mathematically rigorous solution to this problem .

    Is the force across the interface between the spheres given by $$2 ρg\int_0^R \sqrt{R^2 - (R-h)^2}h dh$$ ? But then ρ comes in picture and the integral also doesn't look nice .
     
  9. May 27, 2015 #8

    TSny

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    The integrand looks good. Check the limits.

    If you want to avoid the integration, consider a vertically oriented circular area submerged in a static fluid. Consider two, symmetrically placed horizontal strips of area as shown in blue. How do the pressures at each of these strips compare to the pressure at z = 0? Does the sum of the forces on the two strips simplify?
     

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    Last edited: May 27, 2015
  10. May 27, 2015 #9
    It should be from 0 to 2R . Right?

    Yes .

    If we denote the pressure at z=0 to be P ,pressure at the upper blue strip is P-ρgz . Pressure at the lower blue strip is P+ρgz

    2P . Is that so ?

    Edit : 2P is sum of pressures .
     
  11. May 27, 2015 #10

    TSny

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    Yes.
    Yes.
    This is the total force on the two strips?
     
  12. May 27, 2015 #11
    Total force on two strips is 4P##\sqrt{R^2-z^2}dz## .
     
  13. May 27, 2015 #12

    TSny

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    Yikes. (Correct, but messy.) Let the area of both strips together be ΔA. Express the sum of the forces on the two strips in terms of P and ΔA.
     
  14. May 27, 2015 #13
  15. May 27, 2015 #14

    TSny

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    Well, I was taking ΔA as representing the sum of the areas of the two strips. If so, then your expression is not quite corrrect.
     
  16. May 27, 2015 #15
    Total force on vertically oriented circular area should be ## \pi ρgR^3 ## . Is that correct ?
     
  17. May 27, 2015 #16

    TSny

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    Should atmospheric pressure be included?
     
  18. May 27, 2015 #17
    It should be ##(P_{atm}+ρgR)πR^2## ?
     
  19. May 27, 2015 #18

    TSny

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    Yes, the total force on the circle is the pressure at the center times the total area of the circle. Do you see why they put the little hole at the top?
     
  20. May 27, 2015 #19
    What should be the next step ?
     
  21. May 27, 2015 #20

    TSny

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    The next step should probably be to take a break and sleep on it. (Isn't it late there?)

    Anyway, one thing to think about is how your result for the circle can be used to get the net horizontal force of the fluid on the inner surface of the hemisphere. Also think about the effect of external atmospheric pressure acting on the outer curved surface of the hemisphere.
     
  22. May 27, 2015 #21
    Excellent thinking TSny !!!

    Got the correct answer :oldsmile: .

    And , yes ,its quite late here . But it is always a pleasure interacting with you .
     
  23. May 27, 2015 #22

    TSny

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    OK. I guess we'll let Chet worry about the balance of torques. :smile:
     
  24. May 27, 2015 #23
  25. May 27, 2015 #24
    Yes. To find out where the force has to be applied, you first need to get the moment of the pressure distribtuion around z = 0. You might be able to determine this from knowing the moment of inertia of a semicircle, without actually doing the integration. If not, get the moment the same way that TSny indicated to get the force.

    Chet
     
  26. Sep 28, 2015 #25
    Hello,
    Could you please explain as to why the force exerted by external atmospheric pressure acting on the outer curved surface of the hemisphere would be PatmπR2 ?
     
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