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Fluids at Rest: Pressure/Force Problem

  1. Dec 4, 2011 #1
    This question has been asked several times, but I don't want the answer (I can see the stupid answer...) I need help with my solution...


    1. The problem statement, all variables and given/known data

    A large aquarium of height 5.00m is filled with fresh water to a depth of 2.00m. One wall of the aquarium consists of thick plastic 8.00m wide. By how much does the total force on that wall increase if the aquarium is next filled to a depth of 4.00m.

    2. Relevant equations

    p = ρgh

    p = F/A

    A = hw


    3. The attempt at a solution

    ρ = 0.998x103 kg/m3
    g = 9.81 m/s2
    h1 = 2.00m
    h2 = 4.00m
    w = 8.00m

    ΔF = F2 - F1
    ΔF = p2A2 - p1A1
    ΔF = ρgh2A2 - ρgh1A1
    ΔF = ρg (h2A2 - h1A1)
    ΔF = ρg (h22w - h12w)
    ΔF = ρgw (h22 - h12)

    ΔF = (.998x103 kg/m3) (9.81 m/s2) (8.00m) [ (4.00m)2 - (2.00m)2 ]

    ΔF = 9.40x105 N

    Correct Answer: 4.71x105 N


    Why am I double what the answer is suppose to be?
     
  2. jcsd
  3. Dec 5, 2011 #2

    BruceW

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    to start with, what does w=8m really mean? In the question, it says: 'One wall of the aquarium consists of thick plastic 8.00m wide.' So this isn't the width of the aquarium itself, right?

    I don't understand this problem at all. Surely you must have the horizontal area of the aquarium to know how much water is going in, but the question does not give you this...

    Edit: oh, I see, w is the horizontal length of one of the surfaces of the tank. But how are you supposed to infer the horizontal lengths of the other faces of the tank? Is it meant to be a cube, or a cuboid, or some other shape?
     
  4. Dec 5, 2011 #3
    From what I understand, you don't need to know the sizes of the other walls.

    Pressure is delta-Force over delta-Area. And for flat surfaces, Pressure ends up simply being P = F/A (no integration needed).

    Besides, the problem states that it only wants to know the pressure difference of one wall.

    At first, I thought the 8.0m was in fact the thickness of the plastic, but that is unreleastic and would mean there isn't enough information provided... Besides, it states "a thick plastic 8.0m wide" not "a wide plastic 8.0 thick." ... I could have saved myself a little headache if I actually 'read' the problem...

    Then, I figured that I should include my air pressures... But they all dropped out of the Net-Force.

    Ultimately, I got crazy and figured that "sum of the total forces" meant truly that. So, I added all my forces... But, that move was stupid... If 'subtracting' (Net-Force) was too high, then obviously adding would be high too...

    Of course, one thing I haven't done is use Integration. Though, I can see a Problem with that already... I have three things that are changing: As area increases, the magnitude of the force increases, which effects the Pressure value...

    Of course, that brings up an interesting point.
    F = ma
    A = wh; w is constant
    P = mgh; g is constant

    m = Density * V = Density * Lwh; Density, L, and w are constant

    Therefore, F = (L*density*w) * the integral of (delta-h) from [2,4].

    But, I don't know L... Well that was a useless journey... =\
     
  5. Dec 5, 2011 #4
    To find the force you integrate [itex]\displaystyle{8 \rho gh \ dh} [/itex] from 0 to the height, the reason your answer was double the actual answer was just because you left out a constant [itex]\displaystyle{\frac{1}{2}}[/itex] in your working :)
     
  6. Dec 5, 2011 #5
    (My apologies for not using the pretty font)

    Now,
    rho = D
    dh = delta-h
    dF = delta-F

    I think I see the logic behind it:

    P = F/A
    F = P A

    P = (D g dh)
    A = (w dh)

    dF = [D g dh] * [w dh]
    dF = D g w (dh)^2

    I think we can simply rewrite the end as:
    (dh)^2 = h * dh

    Therefore,
    dF = D g w h (dh)

    Integrate with respect to the limits of [2,4] to get:

    F = .5 * D g w h^2 | [2,4]
     
  7. Dec 5, 2011 #6

    BruceW

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    Woops, I missed that bit while reading. Ok, it should be possible to find the answer to the question.
     
  8. Dec 5, 2011 #7

    BruceW

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    The 'correct answer' seems to make the assumption that atmospheric pressure is equal to zero pascals. (Or that the aquarium is a perfect vacuum apart from the water contained).
     
  9. Dec 5, 2011 #8

    vela

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    The pressure is equal to [itex]p=\rho g h[/itex], not [itex]\rho g\,dh[/itex], so you should have [itex]dF = \rho g h\,dA = \rho g h w\,dh[/itex]. To find the total force on the wall, integrate from 0 to H, where H is the depth of the water.
     
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