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Finding the horizontal force on the gate if the density of water is w. helpp!?

i posted this twice already but neither of the answers checked with the back of the book. i really need to understand how to do this for my test so please help me and i REALLY REALLY appreciate your time. okay heres the quesiton again.

A control gate in the form of a parabolic segment of base 12 and height 4 is submerged in water so that its base is 2 units below the surface of the water. Find the horizontal force on the gate if the density of the water is w. < thats the end of the question.

more details: the parabolic gate has the equation 9y=x^2. and there are points (-6,4) and (6,4) on the parabolic gate in the text theres these formulas. p=F/D=wh F=w h A how would i set this up? and what would be my limits ?

4 Answers

  • smci
    Lv 7
    1 decade ago
    Favorite Answer

    So first find the intersection points with the water surface:

    "height 4 is submerged in water so that its base is 2 units below the surface of the water."

    So y=1/9 x^2 shifted down by 2 becomes y=1/9 x^2 - 2

    And the intersections with the surface are at y=0:

    0=1/9 x^2 - 2

    x^2 = 18

    x = +/- 3√2

    F = Pressure x Area

    But Pressure is also a function of depth y (or d, or h, whatever):

    Pressure = w * g * y ("w" = rho, usually)

    So the incremental force at depth y, due to an infinitesimal strip dy deep and x-(-x)=2x wide, is:

    dF = dP * dA

    = (w * g * y) * (2x dy)

    It's easier to parameterize for y than x.

    So if we know y, x is given by:

    y = 1/9 x^2 - 2

    x = 3 √(y+2)


    dF = (w * g * y) * (6 √(y+2) dy)

    dF = (6wg) y sqrt(y+2) dy

    Then integrating between y=-2 and y=0 :

    F = (6wg) ∫y √(y+2) dy between y=-2 and 0

    Make a substitution z = y+2, thus dz = dy

    F = (6wg) ∫((z-2) √z dz) between z=0 and +2

    = (6wg) ∫ z^(3/2) -2z^(1/2) dz)

    = (6wg) [-2/5 z^(5/2) + 2(2/3) z^(3/2)]

    = (wg) [-12/5 z^(5/2) + 8z^(3/2)]

    = (wg) [-12/5 2^(5/2) + 8*2^(3/2)]

    = (wg2√2) [8 - 12/5 * 2]

    = (wg2√2) [8 - 4.8]

    = (wg) 6.4√2


    sahsjing, you should say your approach is the same as mine, because I posted the approach first :)

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  • 1 decade ago


    My approach might contain some errors. The thing is you didn't post the picture. I assume the gate has parabola shape, opens up and most part of the gate is above the water surface .Is this right? But I don't understand why " there are points (-6,4) and (6,4) on the parabolic gate"? These two points should be above the water surface, aren't they?

    If you can't post the picture, would you just post the answer? I am 100% sure I can figure it out.

    If p = wh, then

    F = ∫PdA

    = ∫w(2-y) 2xdy [y: 0...2], (dA = 2xdy)

    = ∫w(2-y) 6√y dy [y: 0...2], (x = 3√y)

    = 9.0501 w



    The density is a function of height. Therefore, your idea doesn't work. By the way, what do you think the parabola opens up or down?



    Your approach is the same as mine. You missed a sign only. Check it. You should get the same answer as I showed here.



    I am very glad that finally you got the same answer as I had showed here. :)

    By the way, you really don't need to modify the equation, which makes the integral more complicated.

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  • 1 decade ago

    All you need to do is integrate over the pressure of the water over the submerged area of the gate. I will do everything algebraically and put numbers in last. Things are much clearer that way and, if I misunderstood some point, it is easy for you to correct it.

    The gate has the form: y = a*x^2. Here a = 1/9

    The base (origin) is at a depth of b. Here b = 2

    the density of the water is k. k is not specified but for dimensions in feet it is 62.4 lb/ft^3 for meters it is 1000 kg/m^3

    The water pressure is proportional to the density so it is a straight line fnction with a value of b*k at y = 0. It increases with increasing depth (decreasing y) at a rate of k. These two point give the slope and intercept of the pressure function. The pressure as a function of y is:

    p = b*k - k*y = k(b - y)

    The force on a differential area is the product of the pressure and the area:

    dF = p dA = k(b - y) dy dx

    All that is left is to set up the limits and integrate. y ranges from 0 to b and x ranges the width of the gate, from -sqrt(y/a) to +sqrt(y/a). Since the x limits are a function of y, you will integrate on x first. the first integration yields: k(b - y) x dy

    Applying the limits gives: 2 k (b - y) sqrt(y/a) dy

    Integrating this on y (and simplifying) gives: 4ky(b/3 - y/5) sqrt(y/a)

    Applying the limits of 0-b:

    F = 4kb(b/3 - b/5)sqrt(b/a) = 8kb^2 sqrt(b/a)/15

    Putting in the numbers: F = 8 k 2^2 sqrt(18)/15 = 9.051 k

    The final result depends on your selection of units to choose k.

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  • 1 decade ago

    Wouldn't you just find the area of the gate and multiply that by the density?

    The area is int (4 - x^2/12) dx, from -6 to 6.

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