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  1. #1
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    Question Urgent Math-Physics question

    Could you please urgently help me to solve the following question. I have to turn it in until tomorrow (monday) night.
    Thank you.

    A mass of 1019Kg is supported by two wires A and B each made of a different material. Before loading the wires are of equal length. After loading they are still of equal length. Find the load carried by each wire.


    Wire Diameter E

    Wire A 1 mm 206,000 Nmm 2

    Wire B 2 mm 116,000 Nmm 2
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by salvavida
    Could you please urgently help me to solve the following question. I have to turn it in until tomorrow (monday) night.
    Thank you.

    A mass of 1019Kg is supported by two wires A and B each made of a different material. Before loading the wires are of equal length. After loading they are still of equal length. Find the load carried by each wire.


    Wire Diameter E

    Wire A 1 mm 206,000 Nmm 2

    Wire B 2 mm 116,000 Nmm 2


    You know that the sum of the loads/tensions on/in the two wires sums to
    1019g Newtons:

    F_A+F_B=1019 g\ \ \ \dots(1)

    and that the original lengths and the extensions of the wires are equal.

    If Y_A and Y_B are the Youngs moduli of the two wires (Presumably your E's)
    and A_A and A_B their crossectional areas then if F_A and F_B are the
    tensions in the two wires we have:

    <br />
Y_A=\frac{F_A/A_A}{\Delta l_A/l_A}<br />

    and:

    <br />
Y_B=\frac{F_B/A_B}{\Delta l_B/l_B}<br />

    but the original length and the extension of each wire are
    equal so:

    \frac{Y_A}{Y_B}=\frac{F_A/A_A}{F_B/A_B}\ \ \ \dots(2).

    Equations (1) and (2) should allow you to solve
    for the two loads F_A and F_B

    RonL
    Last edited by CaptainBlack; June 18th 2006 at 12:08 PM.
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  3. #3
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    Red face Urgent help for Math-Physics question

    Dear RonL,

    thanks a lot for your reply, but could you please give me a bit more informatin on the solution, as I have not figured out the path to the answer yet.
    Thanks,
    Salvavida
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  4. #4
    Grand Panjandrum
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    We found that:

    \frac{Y_A}{Y_B}=\frac{F_A/A_A}{F_B/A_B}\ \ \ \dots(2).

    plugging the values of Youngs modulus for the two materials, and the
    crossectional areas gives:

    \frac{206}{116}=\frac{F_A/(\pi\ 0.5^2)}{F_B/(\pi\ (1^2)},

    so:

    F_A=\frac{206}{116}\frac{F_B}{4}\approx0.444\ F_B\ \ \ \dots(3).

    But we also had:

    <br />
F_A+F_B=1019 g\ \ \ \dots(1)<br />
,

    so substituting from (3):

    1.444\ F_B=1019g=1019\times 9.81=9996.39,

    solving:

    F_B=6920.79\ N,

    and so:

    F_A=9996.39-F_B=3075.6.

    RonL
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