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Math Help - calculating work from to pump water over the top of a trough

  1. #1
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    calculating work from to pump water over the top of a trough

    A water trough is 6 feet long, its vertical cross section is as isosceles trapezoid with lower base 2 feet, upper base 3 feet and altitude 2 feet. if the trough is full, how much work is done in pumping all of the water over the top of the trough? Use 62.5 lb/ft^3 as the weight density factor for water.
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  2. #2
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    Hello victorxz90044
    Quote Originally Posted by victorxz90044 View Post
    A water trough is 6 feet long, its vertical cross section is as isosceles trapezoid with lower base 2 feet, upper base 3 feet and altitude 2 feet. if the trough is full, how much work is done in pumping all of the water over the top of the trough? Use 62.5 lb/ft^3 as the weight density factor for water.
    Here are the steps you need to solve the problem:

    • Consider a horizontal cross-section through the water at a height h above the base, and with thickness \delta h.
      .
    • Work out the area of this cross-section in terms of h, and hence its mass in terms of h and \delta h.
      .
    • This mass of water is raised through a vertical height of (2-h) feet. Find an expression for the work done, \delta W, when this happens.
      .
    • Write down an expression for \frac{dW}{dh}, and hence, by integrating between h = 0 and h = 2, evaluate the total work done.

    Can you do this now? Let us know where you got stuck, if not.

    Grandad

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  3. #3
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    as far as i have gone...

    (b1+b2)/2 y = A

    pump distance= 2-y

    y =0 , b=2
    y=2, b=3

    -b2 = (b1+(1/2)y)/2


    6*b2*dy
    6(b1+(1/2)y)dy

    dw = int 62.5 [6(2+2y)dy] (2-y)


    now what
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  4. #4
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    Hello victorxz90044
    Quote Originally Posted by victorxz90044 View Post
    (b1+b2)/2 y = A

    pump distance= 2-y

    y =0 , b=2
    y=2, b=3

    -b2 = (b1+(1/2)y)/2


    6*b2*dy
    6(b1+(1/2)y)dy

    dw = int 62.5 [6(2+2y)dy] (2-y)


    now what
    Each cross-section is a rectangle whose length is 6 feet. If the breadth is b at height y above the base, then, as you say:
    When y = 0, b = 2

    When y = 2, b = 3
    and the relation is clearly linear, so:
    b = 2+\tfrac12y
    So the area of cross-section is 6b = 12 + 3y square feet, and the volume of this thin slice is (12+3y)\delta y cubic feet. Its mass is therefore 62.5(12+3y)\delta y pounds. It is raised through a distance (2-y) feet. So the work done is:
    \delta W = 62.5(12+3y)\delta y(2-y) foot pounds. (I don't know whether that's the correct unit; I usually work in SI.)

    \Rightarrow \frac{dW}{dy}=62.5(12+3y)(2-y)

    \Rightarrow W = \int_0^262.5(12+3y)(2-y)\;dy
    Grandad
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