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Work and Fluid Force, lifting water out of a triangular prism tank

  1. Feb 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A vertical cross section of a tank is shown. Assume the tank is 16 feet long and full of water. ([tex]\delta=62.4[/tex], and that the water is to be pumped to a height of 8 feet above the top of the tank. Find the work done in emptying the tank. The tank is a triangular prisim with base=5ft and height=8ft.

    2. Relevant equations
    Not sure that there are any.

    3. The attempt at a solution
    First, I am supposed to find the force of the water. It says to find the width as a function of the height, and the book is unclear how to do this very well. From what I can gather it's the height divided by half the base equals the distance to pump the water divided by W, W being the width. So I solved and got W=[tex]\frac{5}{2}-\frac{5}{16}y[/tex]. And then since work is force times height, you just multiply that by the distance it has to be lifted, that's 16-y. And that's your integrad, but you have a 16[tex]\delta[/tex] factor too, but you just bring that outside of your integral. After I simply the integrand and anti differentiate I got [tex]40y-\frac{15}{4}y^{2}+\frac{5}{48}y^{3}[/tex] evaluated from 0 to 8, again, with the factor of 16[tex]\delta[/tex]. Fundamental theorm it and I got 132787 rounded to the nearest whole number. The answer is wrong, and I'm fairly certain I know how to do all of this except finding the width as a function of the height. The book says it has to do with similar triangles. But I don't get what they are saying. Can someone shed some light on this for me?
    Last edited: Feb 20, 2011
  2. jcsd
  3. Feb 21, 2011 #2


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    how about using conservation of energy and consider the centre of mass's
  4. Feb 21, 2011 #3
    Because this is a calculus 2 class, not a physics class, I don't know how to do it that way. I know how to find the centroid of an area that has uniform density, but that's not how we are supposed to do it.
  5. Feb 21, 2011 #4


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    ok well i would find w(h)

    then infinetsimal vol element
    dV = w(h).L.dh

    think of the work dW required to get this infinitesimal element to teh reuired hieght and then integrate over h

    its all the same thing though
    Last edited: Feb 21, 2011
  6. Feb 21, 2011 #5
    That's what I'm having trouble with. I can't find the infitesimal volume, I can't get the width as a function of the height. Could you walk me through it?
  7. Feb 21, 2011 #6


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    changed notation in last post

    ok so you know
    w(0) = 5
    w(8) = 0

    and as its a triangle, its width will vary linearly in between...

    so basically you have two points (0,5) and (8,0) find the equation of the line that connects them, and that will be w(h)
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