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Thread: 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 $\displaystyle h$ above the base, and with thickness $\displaystyle \delta h$.
      .
    • Work out the area of this cross-section in terms of $\displaystyle h$, and hence its mass in terms of $\displaystyle h$ and $\displaystyle \delta h$.
      .
    • This mass of water is raised through a vertical height of $\displaystyle (2-h)$ feet. Find an expression for the work done, $\displaystyle \delta W$, when this happens.
      .
    • Write down an expression for $\displaystyle \frac{dW}{dh}$, and hence, by integrating between $\displaystyle h = 0$ and $\displaystyle h = 2$, 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 $\displaystyle b$ at height $\displaystyle y$ above the base, then, as you say:
    When $\displaystyle y = 0, b = 2$

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

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

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