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Hydrostatic force problem

  1. Dec 10, 2006 #1
    A gate in an irrigation canal is in the form of a trapezoid 3 feet wide at the bottom, 5 feet wide at the top, with the height equal to 2 feet. It is placed vertically in the canal with the water extending to its top. For simplicity, take the density of water to be 60 lb/ft cubed. Find the hydrostatic force in pounds on the gate.

    I am having problems setting this problem up. It looks like its really easy but I am just not sure how to start it.

    I know F = density x gravity x area x depth
     
  2. jcsd
  3. Dec 12, 2006 #2
    Because the force acts on the gate is not constant, i.e. force at the bottom is larger than that at the top , we have
    dF = density x gravity x depth x d(area)
    Do this integration over the trapezoid will yield the correct answer.
     
  4. Dec 13, 2006 #3

    HallsofIvy

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    Imagine the gate being divided into many narrow horizontal bands of width "[itex]\Delta y[/itex]". If y is the depth of a band, and [itex]\Delta y[/itex] is small enough that we can think of every point in the band as at depth y, then the force along that band is the pressure, 60(y) [NOT "times gravity"! The density of the water is weight density, not mass density!], times the area: the length of the band times [itex]\Delta y[/itex]. Of course, the length of the band depends on y: it is a linear function of y since the sides are straight lines, length(2)= 3 and length(0)= 5 so length(y)= -y+ 5. The force on that narrow band is 60y(5-y)[itex]\Delta y[/itex]. The total force on the gate is the sum of those,[itex]\Sum 60y(5-1y)\Delta y[/itex], as y goes from 0 to 2. In the limit, that becomes the integral
    [tex]60\int_0^5 y(5-y)dy[/tex]
     
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