How Does Gravity Affect the Stretching of a Heavy Spring?

[Saulius]

I was watching a curled up electric cord hanging lose in my bathroom and started wondering
about the way it stretches - the beginning is stretched the most, since it is being stretched by the whole cord below, as it goes down it is stretched less and less until at the end it is essentially not streched at all.

And so I started wondering, how could one find it's density distribution, assuming it is not a curled spring, but say, a rubber band, heavy enough to be streched by gravity of it's own weight, but not as much so that Hooke's law would be innapplicable.

Now when I try writting equations, I get into trouble. I assumed, that each infinitessimal lengh interval could be increased by some extra amount due to stretching of the spring below it, but according to Hooke's law, that is a finite amount, which is proportional to the mass below. Thus if I integrate that, I'll get an infinity. Either way I attack the problem, I get a mismatch of infinitesimal vs. finite quantities.

So the question is, how should I attack this problem and get consistent equations ? Are any of my assumptions wrong (Hooke's law e.g.) ?

 

[Dadface]

A nice problem.May I advise you to try it for a wire whose cross sectional area is initially uniform?Springs are rather difficult to anayse because for low loads they tend to unwind and straighten out rather than stretch,hence the reason why some electric cords are coiled up.Rubber is also difficult because of its polymer structure-it stretches easily to begin with but becomes more difficult to stretch when the polymer straightens out(rather like a spring).

 

[Saulius]

I don't really care about what exactly is being stretched, as long as it adheres to Hooke's law. My problem is actually to get consistent equations when something is being stretched by it's own weight - I can't seem to write them down, since, as I've stated, each infinitesimal length of the cord can be though of being stretched by a finite amount (the mass of the cord below the interval), hence making the total stretch infinite.

 

[Doc Al]

Now when I try writting equations, I get into trouble. I assumed, that each infinitessimal lengh interval could be increased by some extra amount due to stretching of the spring below it, but according to Hooke's law, that is a finite amount, which is proportional to the mass below. Thus if I integrate that, I'll get an infinity. Either way I attack the problem, I get a mismatch of infinitesimal vs. finite quantities.

How could an infinitesimal length stretch by a finite amount? Clearly something's wrong in your analysis.

So the question is, how should I attack this problem and get consistent equations ? Are any of my assumptions wrong (Hooke's law e.g.) ?

Let x be the position along the unstretched spring and dx be an infinitesimal element of that unstretched spring. The first thing to do is find the spring constant of that spring element dx. (Assume the overall spring constant of the spring is k if you ignored its weight. Imagine it horizontal instead of vertical.) Then you can find the stretched length of dx (another infinitesimal) and integrate to find the overall increase of length of the spring under its own weight.

 

[Saulius]

The first thing to do is find the spring constant of that spring element dx. (Assume the overall spring constant of the spring is k if you ignored its weight. Imagine it horizontal instead of vertical.)

Um, well I always thought that the spring constant is well .. a constat, and it doesn't matter if I chop up the spring and take only a dx peace of it. And my problem was that when I took an element dx, k dx, according to Hooke's law, was mass of the spring below * g, which is finite and doesn't make sense.

I guess the question now is, how do I find the k of an element dx ? But even if I know that, still, the force acted on dx is finite, which is the part that I'm confused about.

 

[davieddy]

Label each layer of the spring (wire or whatever) by x, its unstretched
distance from the bottom.
Let y(x) be the displacement of layer x from its unstretched position.
The strain is dy/dx, which is proportional to the weight of the spring beneath
the layer, which is in turn proportional to x.

David

 

[Doc Al]

Um, well I always thought that the spring constant is well .. a constat, and it doesn't matter if I chop up the spring and take only a dx peace of it.

That's incorrect. The spring constant is a property of the spring as a whole. (The parameter that characterizes the stiffness of the spring material, independent of length, is Young's modulus, not the spring constant.) Imagine two identical springs with spring constant k. If you attached them in series, will the double length spring have the same spring constant? Think about it. Imagine pulling the single spring and the double spring with the same force. Which stretches more?

Similarly, imagine cutting a spring in half. What happens to the spring constant?

And my problem was that when I took an element dx, k dx, according to Hooke's law, was mass of the spring below * g, which is finite and doesn't make sense.

Once you figure out the spring constant for a length of spring dx, it will all make sense.

Follow my reasoning above, then take it to the limit as you divide the spring into smaller pieces. Hint: Call the original length of the spring L and its spring constant k.

 

[Saulius]

Thanks, the spring constant vs. Young's modulus explanation fixed my problem! :>

 

[Saulius]

OK, I'm having one more problem here. Using the definition of Young's modulus, I was able to get the amount of stretching dy on each infinitesimal interval of the spring dx, which is

$$dy = \frac{mg}{SEL} x \, dx$$

If I integrate that, I'll get the total stretch, which is

$$\Delta L = \int_0^L \! \frac{mg}{SEL} x \, dx = \frac{mgL}{2SE}$$

All of that is fine, but now I want to find the density distrubution of the stretched spring $$\rho(z)$$ such that

$$\int_0^{L+\Delta L} \rho(z) \, dz = m$$

My approach is taking an interval dx, and basically saying that

$$\rho(x + \Delta L(x)) = \frac{dm}{dz} = \rho_0 \frac{dx}{dx + dy}$$

But the density I get from this reasoning doesn't yield the mass of the spring when integrated along the whole stretched spring. What am I doing wrong ?

 

[Doc Al]

Since

$$dz = dx + dy$$

Then

$$\rho(z) dz = \rho_0 \frac{dx}{dx + dy}(dx + dy) = \rho_0 dx$$

 

[davieddy]

It is a pity you had to revert to a wire and Young's modulus.
k the "spring constant" may refer to the entire spring, but it
doesn't take Einstein to guess that a length x of that spring
will extend by kx/l.

 

[Dadface]

Daviedy for a spring Mg= ke where k is the spring constant and the equation assumes that the same weight acts on all parts of the spring .In the exercise carried out here the thread starter is taking into account the weight of the spring itself this weight increasing as you go from top of the spring to the bottom.

There is another problem with wires which perhaps hasnt been noticed this being that as we move to the infinitessimal elements at the top of the wire the increasing stresses and strains result in the wire stretching more and becoming increasingly thinner as a result.The cross sectional area therefore varies along the wire length.Of course to a good approximation this thinning effect may be considered as negligible.

 

[davieddy]

Spare me Dadface.

 

[Dadface]

I knew what you was getting at davieddy.Nice to meet you.

 

[Saulius]

Thanks everyone, I finally found the correct density distribution! :)

 

[interference]

OK, I'm having one more problem here. Using the definition of Young's modulus, I was able to get the amount of stretching dy on each infinitesimal interval of the spring dx, which is

$$dy = \frac{mg}{SEL} x \, dx$$

If I integrate that, I'll get the total stretch, which is

$$\Delta L = \int_0^L \! \frac{mg}{SEL} x \, dx = \frac{mgL}{2SE}$$

Apologies for bringing this up after so long. I too was looking at a similar problem to Saulius and I merely want to find the total stretch of a heavy spring under its own weight. I have tried my best to read Doc Al's advice but I cannot figure out how to find the stretching dy on each infinitesimal interval of the spring dx, which Saulius figured to be:
$$dy = \frac{mg}{SEL} x \, dx$$

>>>

Now I know that the elastic modulus E of the spring is

$$E = \frac{F}{S} \frac{dx}{dy}$$ where S is the cross-area of the spring

Hence I can also write that
$$F = \frac{ES}{dx} dy$$

However, how does one find this tension the infinitesimal interval of spring is experiencing? I understand it is the weight of the spring below that point $$\left( \frac{l - x}{l}\right) mg$$ but I do not understand how one arrives at Saulius' solution.

Did I do something wrong or have a misguided understanding of something?

I appreciate the advice!

 

[Doc Al]

Now I know that the elastic modulus E of the spring is

$$E = \frac{F}{S} \frac{dx}{dy}$$ where S is the cross-area of the spring

Hence I can also write that
$$F = \frac{ES}{dx} dy$$

However, how does one find this tension the infinitesimal interval of spring is experiencing? I understand it is the weight of the spring below that point $$\left( \frac{l - x}{l}\right) mg$$ but I do not understand how one arrives at Saulius' solution.

Did I do something wrong or have a misguided understanding of something?

So far, so good. Keep going:

$$\left( \frac{l - x}{l}\right) mg = \frac{ES}{dx} dy$$

$$\left( \frac{l - x}{l}\right) mg \:dx = ES \:dy$$

Now just integrate both sides. Note that the integral of dy will equal ΔL.

 

[interference]

Thanks Doc Al. I must have confused myself! :) Problem solved.