1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    [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


    User Avatar
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook