Hookes law equation for all gauges of copper wire

  • Thread starter groom03
  • Start date
  • #1
groom03
27
0
[SOLVED] Hookes law equation for all gauges of copper wire

Homework Statement


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)


Homework Equations


F=ke
F=ake
Youngs modulus


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
 

Answers and Replies

  • #2
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,624
8
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.
 
  • #3
groom03
27
0
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?
 
  • #4
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,624
8
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?
 
  • #5
groom03
27
0
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...
 
  • #6
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,624
8
What's wrong with,

[tex]F = \frac{A_0E\Delta\ell}{\ell_0} = A_0\cdot C\Delta\ell[/tex]
 
  • #7
groom03
27
0
What's wrong with,

[tex]F = \frac{A_0E\Delta\ell}{\ell_0} = A_0\cdot C\Delta\ell[/tex]


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
 
  • #8
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,624
8
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
Yup, sounds good to me :approve:
 
  • #9
groom03
27
0
YAAAAAAAAY

thanks for your help
 

Suggested for: Hookes law equation for all gauges of copper wire

Replies
1
Views
267
Replies
5
Views
1K
Replies
4
Views
496
Replies
5
Views
373
Replies
1
Views
801
Replies
5
Views
600
Replies
4
Views
356
Top