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A little help please

  1. Jan 4, 2005 #1
    Im stuck on this one and its driving me a bit mad :confused:

    Here is what I have so far......

    Calculate the tensile stress in a nylon fishing wire of diamater 0.36mm which a fish is pulling with a force of 20 N.

    What I get:
    Stress = Force / Area

    Stress = 20 / π*(0.18x10^-3)^2

    Stress = 20 / 1.02x10^-7

    Stress = 1.96x10^8 Pa


    Area = π*(0.18)^2 mm
    Area = 1.02x10^-1 mm
    Area = 1.02x10^-4 m

    Stress = 20 / 1.02x10^-4

    Stress = 1.96x10^8 Pa

    Stress = 1.96x10^5 Pa

    WHEREAS it should be:

    Stress = 1.96x10^6 Pa

    Could somebody tell me what the answer is and PLEASE tell me how it was supposed to be worked out. This isnt really for homework per say, as its just a question I seen while working far ahead of everyone and it struck my curiosity.
  2. jcsd
  3. Jan 4, 2005 #2


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    [tex]A=\frac{\pi d^{2}}{4}\sim\frac{3.14\cdot(0.36mm)^{2}}{4}\sim 1.01\cdot 10^{-1}mm^{2}=1.01\cdot 10^{-7}m^{2} [/tex]
    [tex]Stress\sim\frac{20N}{1.01\cdot 10^{-7}m^{2}}\sim 1.97\cdot 10^{8}Pa [/tex]

    Last edited: Jan 4, 2005
  4. Jan 4, 2005 #3


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    Hmmm I get 1.96x10^8 Pa.

    Remember when converting areas that
    [tex]1mm=10^{-3}m[/tex] but

    So your second answer should also come out to 1.96x10^8 Pa.
    Last edited: Jan 4, 2005
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