# Rubber bands and Hooke's Law

1. Oct 25, 2004

### Manchot

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.

2. Oct 26, 2004

### Gokul43201

Staff Emeritus
Are you sure $x = L/L_0~~and~not~~\delta L/L_0~$ ?

3. Oct 26, 2004

### Manchot

No, I'm not sure.

4. Oct 26, 2004

### Pyrrhus

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?

Last edited: Oct 26, 2004
5. Oct 26, 2004

### arildno

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..

6. Oct 26, 2004

### Manchot

7. Oct 26, 2004

### PerennialII

Which is what they give under the link. So it looks like a simple uniaxial time-independent hardening mod of sorts ... so is it just a simple made up correction or does it have any theoretical merit ?