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Calculus 2: Fluid Force

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A circular "pressure release gate" is located in the vertical face of a dam. The radius of the gate is 2 metres and the gate is positioned so that its top lies 5 metres below the surface of the water held in the reservoir behind the dam.Find the total fluid force acting on the door of the gate, when it is in a fully closed position.


    2. Relevant equations
    Area= [tex]2\sqrt{4-x^2}[/tex] dy

    Pressure = pg(5+x)

    3. The attempt at a solution
    F= pg∫ [tex]2\sqrt{4-x^2}[/tex] (5+x) dy
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 18, 2009 #2

    tiny-tim

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    Hi Nyasha! :smile:

    (have a square-root: √ and a rho: ρ :wink:)

    almost right :smile:, but the area of the slice is 2√(4 - y2) dy, at depth 5+y :wink:
     
  4. Feb 18, 2009 #3

    HallsofIvy

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    I don't see your point, tiny-tim. There is no coordinate system given with the problem. It perfectly legitimate to set up a coordinate system in which x is vertical.

    [tex]pg\int_0^2 2\sqrt{4- x^2}(5+x)dx[/tex] is exactly the same as [tex]pg\int_0^2 2\sqrt{4- y^2}(5+y)dy[/tex]
     
  5. Feb 18, 2009 #4

    tiny-tim

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    erm … the point is you can't use x and y at the same time! :wink:

    (i assume that's why Nyasha was puzzled about how to integrate)
     
  6. Feb 18, 2009 #5
    If my integration is correct then how do l further simplify this integration using the area of a disk of radius "R" where x=Rsinθ and dx=Rcosθ and a simple symmetry argument
     
  7. Feb 18, 2009 #6

    tiny-tim

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    Make the substitution and see!

    What do you get? :smile:
     
  8. Feb 18, 2009 #7
    I get :


    [tex]
    pg\int_0^2 2\sqrt{4- R^2sin\theta^2}(5+Rsin\theta)Rcosd\theta
    [/tex]


    From here l don't know how to further simplify this
     
  9. Feb 18, 2009 #8

    tiny-tim

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    … and now put R = 2 … :rolleyes:

    (oh, and you haven't substituted the limits)
     
  10. Feb 18, 2009 #9

    [tex]

    pg\int_0^2 2\sqrt{4- 4sin\theta^2}(5+2sin\theta)2cosd\theta

    [/tex]


    I am not supposed to multiple this integral by two using the symmetry argument
     
  11. Feb 18, 2009 #10

    tiny-tim

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    Nyasha, you still haven't changed the limits of integration …

    and what is √(4 - 4sin2θ) ?
     
  12. Feb 18, 2009 #11

    [tex]


    pg\int_0^4 2\sqrt{4- 4sin\theta^2}(5+2sin\theta)2cosd\theta


    [/tex]


    What am l supposed to do with √(4 - 4sin2θ) ? I'm l supposed to use some trig identity
     
  13. Feb 18, 2009 #12

    tiny-tim

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    If x = 2sinθ, and x goes from 0 to 2, what does θ go from?

    And you must learn your trignonometric identities … what is 1 - sin2θ?
     
  14. Feb 18, 2009 #13

    [tex]


    pg\int_0^2 {4\cos\theta}(5+2sin\theta)2cosd\theta


    [/tex]


    The limit of the integral is from 0 to π/2. I don't know how to put it into the integral using Latex source code.


    [tex]

    pg\int_0^2 ({20\cos\theta +8\cos\theta\sin\theta})2cosd\theta



    [/tex]

    Is [tex] 8\cos\theta\sin\theta} [/tex] = [tex] sin(8\theta) ?
    [/tex]
     
  15. Feb 18, 2009 #14

    tiny-tim

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    Hi nyasha! :smile:

    (you can put π/2 into the integral with ^{\pi /2} :wink:)

    You don't like undoing those brackets, do you? :rolleyes:

    It's a combination of cos2θ and cos2θsinθ.

    No, sin2θ = 2sinθcosθ, but it doesn't work for sin8θ.

    btw, going back to the original integral, it would have been simpler to just integrate x√(4 - x2) and √(4 - x2) separately, without substituting …

    and shouldn't the limits have been -2 to +2? :smile:
     
  16. Feb 18, 2009 #15
    If we integrate x√(4 - x2) and √(4 - x2) separately then do we need x=Rsinθ. The question says in order to evaluate the integral l will need the use the formula of disk of radius "R" where x=Rsinθ.

    How did you get your limits to be -2 to +2 ?

    [tex] pg\int_0^{\pi /2} {4\cos\theta}(5+2sin\theta)2cosd\theta



    [/tex]


    Is this integral correct ?
     
    Last edited: Feb 18, 2009
  17. Feb 23, 2009 #16
    FluidForce.jpg

    According to the solutions manual when l evaluate one of my integrals l should get zero. What am l doing wrong on this question ? Please help.
     
  18. Feb 23, 2009 #17

    tiny-tim

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    Hi Nyasha! :smile:

    Your integral (from 0 to 2) is only over the top half of the gate …

    you need to integrate from -2 to 2.
     
  19. Feb 23, 2009 #18

    I used the symmetry argument so that l could integrate from 0 to 2 and then multiply it by two. Is that correct ?
     
  20. Feb 23, 2009 #19

    tiny-tim

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    Well, it would be … if it wasn't for the (5 + y).

    (in physical terms: the pressure on the bottom half of the gate is obviously more than on the top half)
     
  21. Feb 23, 2009 #20

    Thanks very much tiny-tim it now makes perfect sense.
     
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