Rubber bands and Hooke's Law

by Manchot
Tags: bands, hooke, rubber
 P: 725 I have found a website which claims that rubber bands obey a force law $$F=-kT(x-\frac{1}{x^2})$$ $$x=\frac{L}{L_0}$$ 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.
 PF Patron Sci Advisor Emeritus P: 11,137 Are you sure $x = L/L_0~~and~not~~\delta L/L_0~$ ?
 P: 725 No, I'm not sure.
HW Helper
P: 2,274

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

$$\sigma = Y \delta L$$

Since Stress = Force/Area

$$\frac{F}{A} = Y \delta L$$

$$F = YA \delta L$$

You know

$$\delta L = \frac{\Delta L}{L_{o}}$$

$$F = YA \frac{\Delta L}{L_{o}}$$

Rearranging

$$F = \frac{YA}{L_{o}} \Delta L$$

we have

$$F = \frac{YA}{L_{o}} \Delta L$$

Hooke's Law

$$F = k \Delta x$$

where k in our equation is (x = L)

$$k = \frac{YA}{L_{o}}$$

The people from that page probably tried something similar, can you give us the website?
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 The given formula, in order to be meaningful must have $$x=\frac{L}{L_{0}}$$ Rewritten slightly, it simply says: $$F=-kT\delta{L}({1+\frac{1}{x}+\frac{1}{x^{2}}})$$ Hence, it predicts a hardening for compression of the rubber. I don't know if it actually is good, though..
 P: 725 This is the website that I got the information from: http://www.newton.dep.anl.gov/askasc...0/phy00525.htm . It's about two-thirds down the page.
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