Rubber bands and Hooke's Law

In summary, the website claims that rubber bands obey a force law which is similar to Hooke's Law, but it is also quite different. Can anyone confirm or deny the formula's reliability? Thanks.
  • #1
Manchot
473
4
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.
 
Physics news on Phys.org
  • #2
Are you sure [itex]x = L/L_0~~and~not~~\delta L/L_0~[/itex] ?
 
  • #3
No, I'm not sure.
 
  • #4
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?
 
Last edited:
  • #5
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..
 
  • #7
The given formula, in order to be meaningful must have x=L/L0 ...

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 ?
 

1. What is Hooke's Law and how does it apply to rubber bands?

Hooke's Law states that the force required to stretch or compress an object is directly proportional to the displacement of the object. In the case of rubber bands, this means that the force needed to stretch a rubber band is directly related to how far it is stretched.

2. Why do rubber bands stretch and then return to their original shape?

Rubber bands are made of a material called polyisoprene, which has a unique molecular structure that allows it to stretch and then return to its original shape. When a rubber band is stretched, the polymer chains within the material are straightened out. However, once the stretching force is removed, these chains will return to their original coiled state, causing the rubber band to return to its original shape.

3. Can rubber bands break if stretched too far?

Yes, rubber bands can break if stretched too far. This is because the stretching force can cause the polymer chains to break, leading to a permanent change in the rubber band's shape. Additionally, repeated stretching and releasing of a rubber band can weaken it over time, making it more likely to break.

4. How does temperature affect the elasticity of rubber bands?

Temperature can affect the elasticity of rubber bands. When rubber bands are exposed to high temperatures, the polymer chains within the material become more mobile, making it easier for the rubber band to stretch. Conversely, low temperatures can cause the polymer chains to become less mobile, making the rubber band less stretchy.

5. What other factors can affect the behavior of rubber bands according to Hooke's Law?

Apart from temperature, other factors that can affect the behavior of rubber bands according to Hooke's Law include the thickness and size of the rubber band, the type of material used to make the rubber band, and the amount of force applied to the rubber band. Additionally, the age and condition of the rubber band can also impact its elasticity and how it follows Hooke's Law.

Similar threads

  • Introductory Physics Homework Help
2
Replies
35
Views
2K
  • Classical Physics
Replies
6
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
3K
Replies
25
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
5K
  • Thermodynamics
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
6K
  • Other Physics Topics
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
7K
Back
Top