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Thread: 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 $\displaystyle p(y)$ be the pressure exerted on the wall at a depth of $\displaystyle y$. Consider the infinitesimal strip at $\displaystyle y$ with height $\displaystyle dy$ and width equal to the width $\displaystyle w(y)$ 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: $\displaystyle p(y)w(y)dy$.

    Add up the force on all these strips (integrate from $\displaystyle y=3$ to $\displaystyle y=0$ or $\displaystyle y=0$ to $\displaystyle y=3$ 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 $\displaystyle \frac{3-x}{3}=\frac{y}{1}$ or solving for y

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

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

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

    $\displaystyle Volume=l \times w \times h$

    so

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

    so the Volume is given by

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

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