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Homework Help: Hydrostatic force, half circle

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the hydrostatic force on any side of a bottom half-circle with a 6 m diameter with the top 1 m above water level.

    http://img377.imageshack.us/my.php?image=45095885io3.png"

    2. Relevant equations

    P = 1000 * 9.8 (mass density of water) * displacement
    A = l * w
    F = P * A


    3. The attempt at a solution

    Pressure = density * gravity * displacement
    displacement = (2 - y)

    x = [tex]2\sqrt{9 - y^2}[/tex]
    A = x * delta y (for any rectangular strip)

    I'm concerned with my displacement; would it be 2 - y?
    Would I be integrating from -1 to -3?

    EDIT: I looked over the problem again, I think what we really have is a semi circle with a radius of 2 at sea level.
    The area would be 2 * sqrt(4 - y^2) and the displacement just y, I think :)
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Dec 3, 2008 #2

    HallsofIvy

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    Science Advisor

    No, you do not have a semi-circle with a radius of 2: the problem specifically says a "half-circle with a 6 m diameter" so you have a semi-circle with a radius of 3 m, not two. The fact that the water only comes up to 1 m below the top means that the depth for any y (with [itex]x= \sqrt{9- y^2}[/itex]) is y-1. You will want to integrate from y= 1 to y= 3.
    (Note: I've taken my positive y axis downward to avoid negatives.)
     
  4. Dec 4, 2008 #3
    Thanks, but I'm having trouble with the second integral in the problem:

    [tex]2pg \int (y-1) * \sqrt{9-y^2}[/tex]

    For: [tex]2pg\int \sqrt{9-y^2}[/tex] from 1 - 3 (when integrals are seperated)

    doesn't that evaluate to [tex]2pg \int9*cos^2\theta[/tex] ?

    I'm not getting the correct answer when I evaluate this integral.
     
    Last edited: Dec 4, 2008
  5. Feb 13, 2011 #4
    I'm having a problem with the exact same problem. The answer should be 6.7x104N, but I have yet to get this answer.

    I evaluate [tex]2pg\int(y-1)\sqrt{9-y^{2}}dy[/tex] from 1 to 3

    And get: [tex]2pg\inty\sqrt{9-y^{2}dy - 2pg\int\sqrt{9-y^{2}}dy[/tex] from 1 to 3.

    I then do u substitution for the first integral using [tex]u=9-y^{2}, du=-2y, -\frac{1}{2}du=ydy[/tex]

    For the second integral, since the are is a semicircle, I use [tex]2pg\pi3^{2}[/tex]

    So I have: [tex]-pg\inty\sqrt{u}du - 2pg\pi3^{2}[/tex] from 1 to 3.

    Then I take the antiderive of the first integral to get

    [tex]-pg\frac{2}{3}(9-y^{2})^{\frac{3}{2}} - 2pg\pi3^{2}[/tex]

    From here I plug in the numbers and get:

    [tex]-147832-277088[/tex] which give me [tex]-424920[/tex] I then divide it by 2, since it says to find the Hydrostatic Force on one side of the semicircle to get [tex]2.1x10^{5}[/tex]

    I think I'm making a little mistake somewhere, but I can't seem to figure out where. Any help would be appreciated.
     
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