# Why is more difficult to pull a short spring than a long spring? [closed]

Why is more difficult to pull a short spring than a long spring?

I guess it has to do with the law of the levers but I am not sure

• This is unclear. Are you talking about two springs that are identical except for their length? More difficult to pull how far? Mar 2, 2021 at 21:29
• Every inverse relationship you'd encounter in mechanics will not be due to levers ;) Where is the lever in a spring?
– user87745
Mar 3, 2021 at 6:23

Imagine two truly identical and ideal springs: same $$k$$ and same equilibrium length.

Now we cut one of them exactly in half, discarding one of the halves.

It's obvious that to make the halved one the same length as the one that hadn't been cut we need to exert an extending force.

This shows that in the given and described conditions, the halved, thus shorter one, has now a higher spring constant, say $$k'$$.

• "This shows that in the given and described conditions, the halved, thus shorter one, has now a higher spring constant, say 𝑘′." -- How? The argument only says that you need to apply force to create strain, not that you need to apply more force per unit elongation. E.g., nothing in the thought experiment says that it takes more force to elongate the new spring to its previous length than it would take the original spring to be elongated to 3/2 times its original length, i.e., it doesn't show that the same amount of elongation from the unstrained position needs more force for the shorter one.
– user87745
Mar 3, 2021 at 5:56

Suppose you have one spring of equilibrium length $$l$$ and constant $$k$$. A thought experiment you may conduct is to consider it made of two springs of length $$\frac{l}{2}$$ and constant $$k'$$. Let's try to determine $$k'$$.

Well, by the equation which describes series associations of springs, you should have $$\frac{1}{k}=\frac{1}{k'}+\frac{1}{k'} \Rightarrow k'=2k$$.

Thus, it is is more difficult to pull a short spring than a long spring (assuming they are similar).

@Lemoine's answer is basically correct -- also, intuitive and quite minimal. +1. All the other answers are not answering the question for the reasons I point out in the respective comments.

I would like add two points. I would assume springs with constant area of cross-section along one direction so that its geometry can be fully characterized by the shape of the constant cross-section, its area, and the length of the spring. Really, the last two factors are the only ones that matter, the shape of the constant cross-section is irrelevant.

• The basic reason as to why @Lemoine's answer works is because it is true that you can treat a single spring of natural length $$l$$ as comprising of two springs of the same material, same constant cross-section but of lengths $$l/2$$. Whenever this happens in physics, the name for the situation is that the system is linear, i.e., you add together two things and the composite thing is nothing more or less than the sum of its parts. In other words, we can do this because the springs we are talking about are, by assumption, linear springs. Once you accept that, the argument of deriving the spring constant for the smaller spring by the force-balancing trick that has been described works splendidly.
• Now, why do we consider linear springs? Well, it's because we do the experiments and it turns out that most of the materials out of which the springs are made out do behave linearly as far as the strain is not too high. In particular, you would hear (rightly so) that the spring constant of the halved spring is double the spring constant of the original spring because of something called Hooke's law which follows from the more broader Young's law. As I said, these laws are valid for the materials out of which springs are made in cases where the strain in the spring is not too high. In particular, Young's law states that the tension in a bar of constant cross-section $$A$$ and length $$l$$ when you elongate it by $$\Delta l$$ is given by $$F=-\frac{YA}{l}\Delta l$$where $$Y$$ is a constant called Young's modulus which depends on the material, and the minus sign suggests that the tension produced is of such nature that it tries to re-establish the bar's original length. Such bars are what the springs can be modelled as. Thus, the famous relation between the force produced in a spring and the elongation, namely, $$F=-k\Delta l$$ can be re-understood in light of Young's law and one can recognize that the spring constant $$k$$ is given by $$k = \frac{YA}{l}$$One can now clearly see that the spring constant is inversely proportional to the natural length of the spring. This inverse relationship is what is called Hooke's law. I presume that the reason is that Hooke independently observed this about springs before Young gave his more general law about the stress--strain relationship in (elastic) materials.

If the springs are identical (in terms of the material and the structure, e.g., the degree of coiling) but just have different lengths (fewer/more elements), then it will be more difficult to extend the shorter one by the same distance as you have to extend it by a larger percentage (you have to stretch each element more). In other words, the spring constant for a shorter spring is larger. If you extend the springs by the same percentage instead (rather than the same absolute length) you will feel the same force, so in this case your observation does not apply.

• They cannot be identical but of different length. That is an oxymoron.
– Gert
Mar 2, 2021 at 21:44
• You are just rewording the things that the OP already knows via just introducing the word spring constant. How does it answer OP's question?
– user87745
Mar 3, 2021 at 5:57
• @Dvij D.C.I did not just reword the OP's question. I made clear that his observation implicitly assumes that the springs are extended by the same absolute distance. It does not apply if they are extended by the same length percentage instead, Mar 3, 2021 at 8:23