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Math Help - application

  1. #1
    Junior Member
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    application

    Theorem: The force on a submerged vertical plane area equals the product of the weight per unit volume, the submerged area, and the depth of the centroid of the area below the surface
    F=wAy
    1. What force must be withstood by a vertical dam 80ft long and 20ft deep? Ans. 500 tons
    2. What force must be withstood by a trapezoidal dam 100ft long at the top, 80ft long at the bottom, and 20ft deep? Ans. 17,333w
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  2. #2
    MHF Contributor
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    What force must be withstood by a vertical dam 80ft long and 20ft deep? Ans. 500 tons

    Is the dam for water? The dam is withstanding a body of water?

    w for water is 62.4 lbs/cu.ft
    1 metric ton = 2,204 lbs
    1 short ton = 2,000 lbs
    1 long ton = 2,240 lbs

    Submerged 20 ft deep?
    then centroid is at (20/2) = 10 ft below water surface.
    So,
    F = (62.4)(80*20)(10) = 998,400 lbs.

    If your ton is metric ton,
    F = 998,400 /2204 = 453 tons

    If short ton,
    F = 998,400 /2000 = 499 tons

    If long ton,
    F = 998,400 /2240 = 456 tons.

    So your ton is short ton.
    [It is easier to solve questions if they are defined very well, and are with complete information. ]

    What force must be withstood by a trapezoidal dam 100ft long at the top, 80ft long at the bottom, and 20ft deep? Ans. 17,333w

    Divide the trapezoid into
    ---a rectangle 80ft wide by 20ft high
    ---and an "inverted" triangle that is 20ft at at the top, zero at the bottom and 20ft high also.

    Force on the rectangle is
    F1 = w(80*20)(20/2) = 16,000w weight units.

    Force on the "inverted" triangle is
    [the centroid is (1/3)(20) from the top]
    F2 = w*[(1/2)(20)(20)]*[(1/3)(20)] = 1,333w weight units

    So, total force on the trapezoid = F1 +F2
    = 16,000w +1,333w
    = 17,333w weight units
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