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Hydrostatic fluids question

  1. Sep 30, 2015 #1
    MODERATOR'S NOTE: THIS THREAD HAS BEEN MOVED FROM ANOTHER FORUM, SO THERE IS NO TEMPLATE.


    BlXNu3u.png
    This is the problem here. I am having trouble making progress on this problem after getting the wrong answer for the vertical hydrostatic force.

    The equations I am using are: fx= (density)(gravity)(Zcg)(Projected area)

    For this I used
    p= 1000 kg/m^3
    g = 9.81 m/s^2
    Zcg = 1.5m + (0.75/2)m
    A = (0.75(1.2) m^2

    this gave me 16,554.4 N, which is correct (for the horizontal component).

    Now, trying to solve for the vertical hydrostatic force I used the same equation, but with Zcg being 1.5+0.75 meters as it is at the bottom of the quarter-circle. I am not sure what else to do, but this doesn't give me the right answer, it is short by about 3,000 N.

    I have seen other solutions that solve for the vertical force by finding the "weight of the missing water" above the quarter-circle. I don't understand this method, and it was never explained in class or in my book, so I am weary to use it.

    If anyone could help that would be great!!

    Thank you.
     
    Last edited by a moderator: Sep 30, 2015
  2. jcsd
  3. Sep 30, 2015 #2
    Argh, I just realized I posted this in the wrong section. My apologies!
     
  4. Sep 30, 2015 #3

    SteamKing

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    For the hydrostatic force in the vertical direction, have you accounted for the variation in pressure due to depth, otherwise known as Pascals's Law:

    https://en.wikipedia.org/wiki/Pascal's_law

    Because pressure changes with depth, the c.g. and the center of pressure don't coincide in the vertical direction.
     
  5. Sep 30, 2015 #4
    Well, I changed the Zcg to be the entire radius of the quarter-circle, as this would put it at the depth of the bottom of the circle, whereas for the horizontal component was the radius/2.
     
  6. Sep 30, 2015 #5

    SteamKing

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    Yes, but as you found with your calculations, that didn't provide the correct answer.
     
  7. Sep 30, 2015 #6
    This is true, but I'm not sure what else to do with your suggestion, as I feel I already tried that. The vertical component of the force would act on the horizontal plane of the bottom of the quarter-circle, so wouldn't the cp and cg coincide for that case?
     
  8. Sep 30, 2015 #7
    "The vertical component of pressure force on a curved surface equals in magnitude
    and direction the weight of the entire column of fluid, both liquid and atmosphere,
    above the curved surface."

    This is from my textbook, and is adding to my confusion, because in this problem the surface technically has no water above it, but I know the pressure at the surface from the inside is affected by the column of water above it, which I thought my equation would handle, but it's not :/
     
  9. Sep 30, 2015 #8
    With regard to the vertical force, if the "missing water" region were actually filled with water, it's weight would just balance the upward force exerted on the arc from below. In other words, in this filled situation, if the arc weren't there, all the water would still be in equilibrium. It wouldn't know that the arc is there.

    Chet
     
  10. Sep 30, 2015 #9
    I thought about that too, Chet. I tried to calculate the force by finding the volume of the water above the arc (as if there was water there), finding it's weight, and the force. This also gives me the wrong answer. I am starting to wonder if the answer in the back of the book is wrong. They get 22,600 N.
     
  11. Sep 30, 2015 #10
    You can get the volume much more easily by just subtracting the area of the quarter circle.
     
  12. Sep 30, 2015 #11
    That's what I did. Acted as if it were a rectangular cube, and then subtracted the area of the cylinder. This gives me 14,664.5N when converted to a force.
     
  13. Sep 30, 2015 #12
    I am at my wits end with this problem. I will probably just go with this answer and hope the back of the book is wrong. If anyone has some suggestions otherwise, that would be great. Thanks again.
     
  14. Sep 30, 2015 #13
    My answer matches yours, plus I derived the equation by integrating the pressure from below over the surface, and got the same answer. So the book must be wrong.

    Chet
     
  15. Sep 30, 2015 #14
    Thank you for being so thorough! I think I can finally move on to the next problem. FEWF! 1/5 complete, and only 3 hours later :D
     
  16. Sep 30, 2015 #15
    Actually, I integrated it first because I couldn't figure out a simpler method of doing the problem. Then I realized that this was the same as the weight of the missing water.
     
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