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Rubber bands and Hooke's Law 
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#1
Oct2504, 09:51 PM

P: 728

I have found a website which claims that rubber bands obey a force law
[tex]F=kT(x\frac{1}{x^2})[/tex] [tex]x=\frac{L}{L_0}[/tex] While this is similar to Hooke's Law in the sense that it *almost* approaches it for large values of x, it is also quite different. Can anyone confirm or deny the formula's reliability? Thanks. 


#2
Oct2604, 11:12 AM

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PF Gold
P: 11,155

Are you sure [itex]x = L/L_0~~and~not~~\delta L/L_0~[/itex] ?



#3
Oct2604, 11:26 AM

P: 728

No, I'm not sure.



#4
Oct2604, 11:34 AM

HW Helper
P: 2,277

Rubber bands and Hooke's Law
Well if you're familiar with elasticity you can formulate Hooke's Law in its terms,
Stress = Modulus of Elasticity x Relative Deformation For a longitudinal deformation, the modulus is called Young's modulus [tex] \sigma = Y \delta L [/tex] Since Stress = Force/Area [tex] \frac{F}{A} = Y \delta L [/tex] [tex] F = YA \delta L [/tex] You know [tex] \delta L = \frac{\Delta L}{L_{o}} [/tex] [tex] F = YA \frac{\Delta L}{L_{o}} [/tex] Rearranging [tex] F = \frac{YA}{L_{o}} \Delta L [/tex] we have [tex] F = \frac{YA}{L_{o}} \Delta L [/tex] Hooke's Law [tex] F = k \Delta x [/tex] where k in our equation is (x = L) [tex] k = \frac{YA}{L_{o}} [/tex] The people from that page probably tried something similar, can you give us the website? 


#5
Oct2604, 12:01 PM

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PF Gold
P: 12,016

The given formula, in order to be meaningful must have [tex]x=\frac{L}{L_{0}}[/tex]
Rewritten slightly, it simply says: [tex]F=kT\delta{L}({1+\frac{1}{x}+\frac{1}{x^{2}}})[/tex] Hence, it predicts a hardening for compression of the rubber. I don't know if it actually is good, though.. 


#6
Oct2604, 01:54 PM

P: 728

This is the website that I got the information from: http://www.newton.dep.anl.gov/askasc...0/phy00525.htm . It's about twothirds down the page.



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