I am a little confused about springs. I just wanted to know that if I pull an ideal spring of spring constant $k$ such that the spring has been symmetrically pulled and its elongation (total) comes out to be $x$ then would the force on one side by $$F=kx$$ or $$F=kx/2$$ I am a little bit confused and hence I resorted to ask it here.
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3$\begingroup$ Possible duplicate of Extension of an unfixed spring $\endgroup$– sammy gerbilCommented Aug 11, 2016 at 17:53
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4 Answers
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An ideal spring - Three loading cases : free,compressed,stretched.
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1$\begingroup$ Cool diagrams. What software did you use? $\endgroup$ Commented Aug 12, 2016 at 14:55
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2$\begingroup$ @ja72 : GeoGebra. It's free. This Figure was drawn for other purposes (a textbook written in LaTeX) and I find it useful to insert in an answer herein. Note that the spring here is a curve in space with parameters. I added the 3D version for you. $\endgroup$– VoulkosCommented Aug 13, 2016 at 3:31
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1$\begingroup$ I've used Geogebra, but the springs threw me off. The 3D works great if the red is on the left eye and blue on the right. So the little icon on top left is wrong. $\endgroup$ Commented Aug 13, 2016 at 14:19
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1$\begingroup$ @ja72 : The previous icon did mean exactly that the red is on the left eye and blue on the right. I changed this little icon. I think that is more clear now. $\endgroup$– VoulkosCommented Aug 13, 2016 at 16:32
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$\begingroup$ @ja72 : Take a look here i.imgur.com/JtgXEHv.jpg $\endgroup$– VoulkosCommented Aug 13, 2016 at 19:35
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There is no such thing as a non-symmetrical pull of a string or some non-"total" elongation. $x$ is elongation, and that's it. $F=kx$ is the spring force, and that's it.
An object tied to one end of the spring experiences this spring force. The whole force. Something tied to the other end experiences by Newton's 3rd law the same force (in the opposite direction).
No need to half it.
Seen from each object at the ends of the spring, it would be:
You draw the free body diagram of the object at one end, and it shows a spring force. This force is the push, the spring exerts on the object because it is compressed a bit and tries to return to the uncompressed state. And it is experimentally found to be proportional to the compression as $F=kx$. Which also feels intuitive: doubling the compression doubles the tendency of it to return to the relaxed state (at least within reasonable force ranges and for usual, ideal springs where Hooke's Law applies).
You then draw the free body diagram of the object at the other end. It also feels a spring force. Same argument holds; the force from the spring is present, because the spring is compressed a bit, and the force appears to be proportional to the compression as $F=kx$.
We cannot talk about the "elongation $x$ caused at each end" and then say that "the total elongation is the sum of these". Because the elongation $x$ that happens is already caused by the two ends both being pulled - if only one was pulled (so only one end feels a force), there would be no elongation at all! (Assuming ideal mass less spring).
Force on one end does not contribute alone with one half of the total elongation; the forces on both ends cause it in collaboration.
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$\begingroup$ The same force, but in the opposite direction. $\endgroup$ Commented Aug 11, 2016 at 17:40
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$\begingroup$ "if only one was pulled (so only one end feels a force), there would be no elongation at all! (Assuming ideal mass less spring)."? What about spring attached to a wall $\endgroup$– BrianCommented Sep 26, 2020 at 15:13
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$\begingroup$ @Buraian There is a force holding the spring to the wall. Otherwise it would be be attached. $\endgroup$– SteevenCommented Sep 26, 2020 at 15:39
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$\begingroup$ So does the force which spring exerts on the wall increase as we compress the spring? $\endgroup$– BrianCommented Sep 26, 2020 at 15:40
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$\begingroup$ @Buraian Yes indeed. Just like if you place a book on the wall and push on it, then your pushing force propagates through the book to the wall. The spring does the same, just via its elastic force as an intermediate $\endgroup$– SteevenCommented Sep 26, 2020 at 17:59
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When you state $F = k x$ the variable $x$ represents the extension of the spring and not the position of one of the ends.
In a symmetric pull each end moves by $ \frac{\delta}{2} $ then the total displacement is $\delta$. The spring force is then $F = k \delta$ on each end of the spring.
answered Aug 11, 2016 at 18:34
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The other answers simply quote Hookes law, the static relationship between displacement and force. But if you consider the question more deeply, the spring has distributed mass and distributed compliance, and so a spring all by itself is dynamic and therefore does not propagate force instantaneously as Hookes law implies all by itself.
So if you push (or pull) on a spring you will displace the coils (or leafs or whatever), and the same force will be seen on the opposite side of the spring - however delayed by the dynamic response of the spring.
answered Aug 11, 2016 at 19:12
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$\begingroup$ For the purposes of simple oscillations, the recipe to fix this is to add one-third of the mass of the spring to the working mass of the system. $\endgroup$ Commented Aug 11, 2016 at 20:22
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$\begingroup$ @dmckee that's probably a good rule of thumb, but not what the OP is asking. If I'm understanding they are asking how force is transmitted through a spring (what's the opposing force). $\endgroup$ Commented Aug 11, 2016 at 20:25
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$\begingroup$ The words are a bit difficult to understand. For example, what do you mean by dynamic respose? $\endgroup$– BrianCommented Sep 26, 2020 at 15:02
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$\begingroup$ Dynamic means physical relationships as a function of time. Hookes law is static. It doesn't involve time. $\endgroup$ Commented Sep 26, 2020 at 15:17