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Math Help - A Deck Arch Bridge

  1. #1
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    A Deck Arch Bridge

    OKay i have done the first parts of this question, but now i am stuck
    It goes like this:
    a bridge has a deck arch in the shape of a parabola. The road deck runs
    horizontally along the top of the arch. The structure is further strengthened
    by 14 vertical steel cables, equally spaced, between the road and the arch.
    In addition, you know that if the base of one of the pylons is taken as the origin,
    then the equation, y, of the arch is given by: y= -1/100x sqaured + 27/10x (x being the unknown)

    a) solve algebraically for the width of the bridge

    I calculated it to be 270metres long

    b) What is the horizontal distance between each steel cable?

    I calculated it to be 18metres between each

    c) solve algebraically for the greatest height of the arch.

    well its undoubtedly the middle of the parabola. therefore the middle of the width. the way it is drawn shows that is you find the height of the bridge, the answer will be the same, as the parabola touches the top of the bridge.... if im not explaining right plz tell me what u dont understand. i may be able to scan the picture..??
    Last edited by CaptainBlack; May 7th 2007 at 03:32 AM.
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  2. #2
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    visual attachment

    okay, maybe a picture will help. (attached)
    Attached Thumbnails Attached Thumbnails A Deck Arch Bridge-gash.jpg  
    Last edited by CaptainBlack; May 7th 2007 at 03:33 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by el Naiad View Post
    OKay i have done the first parts of this question, but now i am stuck
    It goes like this:
    a bridge has a deck arch in the shape of a parabola. The road deck runs
    horizontally along the top of the arch. The structure is further strengthened
    by 14 vertical steel cables, equally spaced, between the road and the arch.
    In addition, you know that if the base of one of the pylons is taken as the origin,
    then the equation, y, of the arch is given by: y= -1/100x sqaured + 27/10x (x being the unknown)

    a) solve algebraically for the width of the bridge

    I calculated it to be 270metres long

    b) What is the horizontal distance between each steel cable?

    I calculated it to be 18metres between each

    c) solve algebraically for the greatest height of the arch.

    well its undoubtedly the middle of the parabola. therefore the middle of the width. the way it is drawn shows that is you find the height of the bridge, the answer will be the same, as the parabola touches the top of the bridge.... if im not explaining right plz tell me what u dont understand. i may be able to scan the picture..??
    That all seems correct, for the middle of the bridge x=135, so:

    y=-135^2/100 + 27*135/10 = 182.25 m.

    RonL
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  4. #4
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    Hello, el Naiad!

    A bridge has a deck arch in the shape of a parabola.
    The road deck runs horizontally along the top of the arch.
    The structure is further strengthened by 14 vertical steel cables,
    equally spaced, between the road and the arch.
    In addition, you know that if the base of one of the pylons is taken as the origin,
    then the equation of the arch is given by: .y .= .-0.01x + 2.7x

    a) Solve algebraically for the width of the bridge
    . . I calculated it to be 270m long . Right!

    b) What is the horizontal distance between each steel cable?
    . . I calculated it to be 18m between each . Yes!

    c) Solve algebraically for the greatest height of the arch.

    You are right . . . It occurs at the exact middle of the bridge.
    Since it is 270m long, the peak must occur at x = 135.
    Hence: .y .= .(-0.01)(135) + (2.7)(135) .= .182.25m


    If we must use a more algebraic approach . . .
    . . the vertex of a parabola occurs at: .x = -b/2a

    Our parabola has: .a = -1/100, b = 27/10

    Hence, the vertex is: .x .= .-(27/10) 2(-1/100) .= .135

    Got it?

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