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Hookes law equation for all gauges of copper wire

  1. Apr 22, 2008 #1
    [SOLVED] Hookes law equation for all gauges of copper wire

    1. The problem statement, all variables and given/known data
    For my coursework i'm trying to find an equation using hookes law that works with all gauges of copper wire, i know that this means i will have to change the hookes law equation from F=ke to F=ake (a is not the area it's just a letter for the constant that i need to find)


    2. Relevant equations
    F=ke
    F=ake
    Youngs modulus


    3. The attempt at a solution
    i've tried working out the stiffness for several wire gauges and seeing if there was a pattern to them but my teacher said i should involve the youngs modulus equation.

    Any help really appreciated
     
  2. jcsd
  3. Apr 22, 2008 #2

    Hootenanny

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    Hooke's law can be derived by collecting the constants of Young's modulus. Try doing the same, but this time you want two constants, not just one.
     
  4. Apr 22, 2008 #3
    by substituting i can get E=kl/A

    A is going to be known because i'd know the wire gauge and using rho=f/a i can work out the force but i still can figure out how i'd find k unless i'd already know it when working out

    Am i getting close?
     
  5. Apr 22, 2008 #4

    Hootenanny

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    Young's modulus in it's entirety is defined thus,

    [tex]E = \frac{\sigma}{\varepsilon}= \frac{F/A_0}{\Delta \ell/\ell_0} = \frac{F \ell_0} {A_0 \Delta \ell} [/tex]

    Where [itex]F[/itex] is the applied force, [itex]A_0[/itex] is the original area, [itex]\Delta\ell[/itex] is the extension, [itex]\sigma[/itex] is the stress and [itex]\varepsilon[/itex] is the strain.

    Does that help?
     
  6. Apr 22, 2008 #5
    scratch that i can also get the equation k=EL/A but now i'm totally stuck

    i could substitute that k into the f=ke equation but i've been told that the youngs modulus was for a unit length so i would have to do something to....
     
  7. Apr 22, 2008 #6

    Hootenanny

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    What's wrong with,

    [tex]F = \frac{A_0E\Delta\ell}{\ell_0} = A_0\cdot C\Delta\ell[/tex]
     
  8. Apr 22, 2008 #7

    i think i've figured out how you get to that

    E=FL/Ae

    EA=FL/e

    EAe=FL

    EAe/L=F

    F=ACe (if C equals youngs mod/length)

    Which rearranges to e=F/A/C

    i think that's right... i hope
     
  9. Apr 22, 2008 #8

    Hootenanny

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    Yup, sounds good to me :approve:
     
  10. Apr 23, 2008 #9
    YAAAAAAAAY

    thanks for your help
     
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