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Math Help - I need help with this problem about work done a trapezoidal trough

  1. #1
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    I need help with this problem about work done a trapezoidal trough

    I've never understood how to do these problems, so step-by-step help on how to find the solution would be very much appreciated. Thank you.

    A trough has a trapezoidal cross section with a height of 3 m and horizontal sides of width 1.5 m and 3 m. Assume the length of the trough is 10 m. How much work is required to pump the water out of the trough (to the level of the top of the trough) when it is full. Use 1000 kg/m^3 for the density of water and 9.8 m/s^2 for the acceleration due to gravity.
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  2. #2
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    Re: I need help with this problem about work done a trapezoidal trough

    Here's an overview of the process. Post again if you're still having trouble.

    Take a horizontal slice through the trough. Let x be the distance from the top of the trough.

    The width of this slice is calculated by fitting a line to the two points w=3m when x=0, w=1.5m when x=3m.
    The volume of this slice is length times width times height: dV=(10m)(w)(dx)
    The weight of this slice is density times volume times g: dF=(1000)(dV)(9.8)
    The work done to pump this slice out of the trough is weight times distance: dW=(dF)(x)
    And then you add up the work for all the slices: \int_{x=0}^{x=3}dW

    Once you substitute everything, you should have the integral of some function of x (the dx is buried in all the substitutions), and in this case it is pretty easy to solve.

    If you like to keep track of units (I always do), remember the conversions N=\frac{kg\cdot{m}}{sec^2} and J=N\cdot{m}.

    - Hollywood
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  3. #3
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    Re: I need help with this problem about work done a trapezoidal trough

    So then.. it would be something along the lines of Integral of 9800*10m*w*dx from 0 to 3? I'm confused as to what w is though, since you listed two values, 3m and 1.5m. What do you do with those?

    I think all the substitutions confused me.
    Last edited by bwoinski; November 8th 2012 at 02:34 PM.
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  4. #4
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    Re: I need help with this problem about work done a trapezoidal trough

    Quote Originally Posted by bwoinski View Post
    A trough has a trapezoidal cross section with a height of 3 m and horizontal sides of width 1.5 m and 3 m. Assume the length of the trough is 10 m. How much work is required to pump the water out of the trough (to the level of the top of the trough) when it is full. Use 1000 kg/m^3 for the density of water and 9.8 m/s^2 for the acceleration due to gravity.
    work = \int WALT

    W = weight-density = 9800 \, N/m^3

    A = cross-sectional area of a representative horizontal slice of water (a rectangle in this case) = 10\left(\frac{y+3}{2}\right) = 5(y+3)

    L = lift distance for a representative slice of water = (3-y)

    T = slice thickness = dy

    W = \int_0^3 9800 \cdot 5(y+3) \cdot (3-y) \, dy
    Attached Thumbnails Attached Thumbnails I need help with this problem about work done a trapezoidal trough-troughwork.png  
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  5. #5
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    Re: I need help with this problem about work done a trapezoidal trough

    Here's what I was aiming for:

    The width of this slice is calculated by fitting a line to the two points w=3m when x=0, w=1.5m when x=3m.
    w=3+\frac{x}{3}(-1.5)=(3-\frac{x}{2})

    The volume of this slice is length times width times height: dV=(10m)(w)(dx)
    dV=(10)(3-\frac{x}{2})dx=(30-5x)dx

    The weight of this slice is density times volume times g: dF=(1000)(dV)(9.8)
    dF=9800(30-5x)dx

    The work done to pump this slice out of the trough is weight times distance: dW=(dF)(x)
    dW=9800(30-5x)(x)dx

    And then you add up the work for all the slices: \int_{x=0}^{x=3}dW
    \int_0^39800(30-5x)(x)dx

    Skeeter's y is from the bottom of the trough, and my x is from the top of the trough, if you substitute x=3-y, you get skeeter's integral:
    \int_0^39800(30-5(3-y))(3-y)d(3-y)
    =\int_3^09800(5)(y+3)(3-y)d(3-y)
    =\int_0^39800(5)(y+3)(3-y)dy

    - Hollywood
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