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Math Help - Work problems

  1. #1
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    Work problems

    I'll get right to the point.

    A trough is 3 meters long, 2 meters wide, and 4 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 4 meters, and base, on top, of length 2 meters). The trough is full of water (density 1000 kg/m3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top. (Note: Use g=9.8 m/s2 as the acceleration due to gravity.)

    Thanks to anyone who can help. Also, if it's not too much trouble, would any would be assistors provide a suggested answer first and then explain how they got it? Rest assured, I have been through the rigamorole trying to figure out what I'm doing wrong on this one, so I don't really need step by step coaching, especially seeing as time is not on my side. Thank you.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by G-Rex View Post
    I'll get right to the point.

    A trough is 3 meters long, 2 meters wide, and 4 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 4 meters, and base, on top, of length 2 meters). The trough is full of water (density 1000 kg/m3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top. (Note: Use g=9.8 m/s2 as the acceleration due to gravity.)

    Thanks to anyone who can help. Also, if it's not too much trouble, would any would be assistors provide a suggested answer first and then explain how they got it? Rest assured, I have been through the rigamorole trying to figure out what I'm doing wrong on this one, so I don't really need step by step coaching, especially seeing as time is not on my side. Thank you.
    This type of word problem is popular here lately. Work to empty a vessel. They are all the same. Only the shape of the vessel changes.

    Okay, we will come to the answer at the end of the explanation.

    Work = weight*height, basically.

    We get an element of weight of the water in the trough, dWeight.
    dWeight = dVolume*(unit weight)

    Unit weight = mg = 1000(9.8) = 9800 Newtons per cu.m.

    The element dVolume is a horizontal plate somewhere inside the trough.
    dVolume = dV = 2x(3.0)(dy) = 6x dy cu.m --------(i)

    What is x and y, or dy?
    In the vertical cross section of the trough, let (0,0) be at the bottom point of the isosceles triangle. So the y-axis bisects the isosceles triangle. The right end of the top of the isosceles triangle is at point (1,4).
    So the equation of the inclined right side of the isosceles triangle is:
    (y -0) = [(4-0)/(1-0)](x -0)
    y = 4x
    Since we are using dy for integration,
    x = y/4
    And so dV = 6x dy = 6(y/4) dy = (3/2)y dy cu.m.
    Hence,
    dWeight = [(3/2)y dy](9800) = 14,700y dy newtons.

    To lift this dWeight to the top of the trough, we need to raise it (4-y) meters.
    So, dWork = (14,700y dy)(4 -y)
    dWork = 14,700[4y -y^2]dy joules -----(ii)

    The dWeight, or dy, goes from y=0 to y=4, so,
    Work = (14,700)INT.(0-->4)[4y -y^2]dy
    Work = (14,000)[2y^2 -(1/3)y^3]|(0-->4)
    Work = (14,700)[(2(4^2) -(1/3)(4^3)) -(0)]
    Work = (14,700)[10.667]
    Work = 156,800 joules ------------------------answer.
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