0
$\begingroup$

This is taken from "Physics a general course, volume I" by I.V.Savelyev.

enter image description here

  1. How is it that "Any part of the spring acts on another part..."? I understand that at each "point" we can assign $F_{el}$, but I can't see a connection.

  2. How to prove ".. if we cut the spring in half, an identical elastic force will appear in each half with the elongation being half the original value"? And does he mean that the force will be equal to the current elastic force?

  3. I am failing to notice why "Hence ... the magnitude of the elastic force is determined not by $\Delta l$, but by its relative elongation $\Delta l/l_0$ (which the author calls strain further ahead when talking about bars), could someone please explain?

  4. What is exactly meant by elastic strains? Elastic forces?

$\endgroup$
1
$\begingroup$

Boolean's answer was pretty good, I just have a slightly different way to explain it. This actually directly relates to what you were asking in chat earlier today too.

  1. By "any part of a spring acts on another part with a force determined by eq 2.25" what they are essentially saying is that if you look at any section of the spring, both sides will be pulling away from or pushing against each other with the elastic force. It is like how a massless rope has uniform tension (except springs can go both ways).
  2. So your question 1 establishes that the $F_{el}$ is felt through each part of the spring. If you looked at the one spring as two half springs connected together (i.e. cut it in half), each half-spring would still have the same $F_{el}$ acting on it, it's basically repeating what 1. said. We know that each one must stretch $\frac {\Delta l}2$. If not, the simple act of thinking of it as two separate but connected springs would somehow change the elongation, which doesn't make sense.
  3. This directly follows from what was just shown, but wasn't necessarily apparent. In the case of spring of length $l_0$ cut in half, the each half spring still had the same force (from 1), and had half the elongation (from 2), each spring was also half the length. If you cut it into 3 springs, each spring would still have the same force $F_{el}$, and each would only have 1/3 of the elongation, while being 1/3 of the length. What this shows is that the force on the spring $F_{el}$ gives the same relative elongation, regardless of spring length. A force that compresses a $1 \ m$ long spring $0.1 \ m$ would compress a $0.1 \ m$ long spring $0.01 \ m$, if the springs were the same coil dimensions and material.
  4. Elastic strains are different than elastic forces, but they are related. Strains are relative deformations in material. This is why it keeps referencing $\frac {\Delta l}{l_0}$, because in one direction, that shows a strain. Strain is directly proportional to stress for elastic materials, which obey Hooke's Law; which also ties it in nicely with springs.
$\endgroup$
  • $\begingroup$ I haven't concentrated on this matter for some days, sorry if I forgot something from the last chat discussion. "So your question 1 establishes that the $F_{el}$ is felt through each part of the spring." How is that? Won't the middle point, for example, since it has been displaced by half the displacement, be feeling $F_{el}/2$?. $\endgroup$ – Luyw Jul 16 '19 at 18:39
  • 1
    $\begingroup$ @Luyw No, and that's basically the crux of the matter. The $F_{el}$ on the spring is like a tension in a massless string. It's uniform throughout the entire length; it doesn't only exist on the ends. Any one part of the spring is pushing or pulling on the part of the spring directly beside it with $F_{el}$. Think of a free body diagram of a section of the spring. Obviously parts of the spring don't accelerate relative to each other, because the net forces of the parts of the spring around it cancel out. By pushing or pulling on both ends you create a uniform internal force in the spring. $\endgroup$ – JMac Jul 16 '19 at 19:14
2
$\begingroup$

Honestly, this text is rather confusing, but my interpretation is this: 1. In saying "Any part of the spring acts on another part" the author is essentially saying that elastic deformation on either side of the "part" contributes to the force on that part. The author is probably trying to suggest the intuition that each adjacent section of the spring stretches those parts adjacent to itself.

  1. What the author is trying to say here is that one could imagine a single spring as two springs each displaced half as much as the previous, and each conforming to Hooke's Law. For that matter, it could be as many springs as you might want, essentially if the original spring stretches to $x_o$ as the equilibrium, and is deformed to $x_o + l$, then it is equivalent to two springs in "series", essentially one spring with eq. at $\frac{x_o}{2}$ stretched to $\frac{l}{2}$ and another doing the same thing. By Hooke's Law, $F_{spring 1} = k_s(\frac{l}{2}) = F_{spring2}$, so the net force is still the same as with only one spring. This is true even if an infinite number of springs are created in "series", such that $\vec{F_{n}} = k_s(\frac{\Delta l}{n})$ is the elastic force of each individual spring in a system of n springs, and so if n approaches infinity, each individual force becomes essentially 0, but $\Sigma\vec{F} = k_s \Delta l.$

  2. This is the weirdest part of the text, essentially the author is probably trying to introduce the idea of strain on the bar, but the idea of force being proportional to the relative displacement is rather foreign to me. Essentially, it seems that the idea here is that the "strain" on the material is essentially the force on each part of the spring, which if we conclude that the force is evenly distributed throughout the length of the spring (part 2), then clearly the strain must be proportional to the amount of "pull" ($\Delta l$) and to the amount of spring that can take that strain ($l_o$).

  3. It seems that in this context the author means "elastic force" to be the internal forces felt within the spring, however, this is rather confusing because elastic force could also refer directly to the force described in Hooke's Law.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.