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Math Help - Joint variation + percent question

  1. #1
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    Joint variation + percent question

    The weight (W) that can be supported by a wooden beam varies directly as the square of its diameter (d) and inversely as its length (l).
    a What percentage increase in the diameter would be necessary for a beam twice as long to support an equivalent weight?

    What percentage change in the weight would be capable of being supported by a beam
    three times as long with twice the diameter?

    I know how to do this one:

     W \alpha \frac {d}{l}

     W = \frac {kd}{l}

     W = \frac {2}{3}

    So the weight needs a 33% increase.


    But for this question:
    What percentage increase in the diameter would be necessary for a beam twice as long to support an equivalent weight?

    How do I do this?
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  2. #2
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    Quote Originally Posted by jgv115 View Post
    The weight (W) that can be supported by a wooden beam varies directly as the square of its diameter (d) and inversely as its length (l).
    a What percentage increase in the diameter would be necessary for a beam twice as long to support an equivalent weight?

    What percentage change in the weight would be capable of being supported by a beam
    three times as long with twice the diameter?

    I know how to do this one:

     W \alpha \frac {d}{l}

     W = \frac {kd}{l}

     W = \frac {2}{3}

    So the weight needs a 33% increase.


    But for this question:
    What percentage increase in the diameter would be necessary for a beam twice as long to support an equivalent weight?

    How do I do this?
    Dear jgv115,

    W=\frac{kd}{l}---------(1)

    When the beam is twice as long, let the diameter be D that supports the weight W,

    W=\frac{kD}{2l}---------(2)

    By (1) and (2),

    D=2d

    So the percentage increase in the diameter would be 100%.
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