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Math Help - Fluid Pressure

  1. #1
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    Exclamation Fluid Pressure

    Good Morning!
    I could use some help with this fluid pressure/force problem.

    A trough 10 ft long has a trapezoidal cross section that is 2 ft wide at the bottom, 4 feet wide at the top, and 3 feet high. If the trough is filled with oil (density 50 pounds per cubic foot) find the force exerted by the oil on one end of the trough.

    I know that F=ma, W= Int(Force Equation) and the basic physics stuff. I'm just having trouble visualizing this and getting the integral all set up. I tried drawing a picture, but I don't think it helped me much...

    Thanks in advance for any help you can provide!
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  2. #2
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    Let p(y) be the pressure exerted on the wall at a depth of y. Consider the infinitesimal strip at y with height  dy and width equal to the width w(y) of the trough at that depth. Then the force exerted on that strip is the area of the strip times the pressure exerted by the oil: p(y)w(y)dy.

    Add up the force on all these strips (integrate from y=3 to y=0 or y=0 to y=3 depeding on how you think about the problem) and you should get the total force on the wall.
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  3. #3
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    Still confused...

    I'm still really confused, I keep getting really outlandish answers and I don't think the picture I drew is helping me figure out the integral. Thanks in advance...
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  4. #4
    Behold, the power of SARDINES!
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    Fluid Pressure-capture.jpg
    Maybe this will help. Note that the two triangles are similar.

    using proportions we get \frac{3-x}{3}=\frac{y}{1} or solving for y

    y=\frac{3-x}{3}

    so the width of each strip is 2+2y=2+2 \cdot \frac{3-x}{3}

    the height of each stip is dx and the length is 10 so

    Volume=l \times w \times h

    so

    dV=10(2+2 \cdot \frac{3-x}{3})dx

    so the Volume is given by

    V=\int_0^310(2+2 \cdot \frac{3-x}{3})dx

    To finish multiply the volume by the desity of the oil per cubic ft
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