If we consider an experiment of pulling a spring with a constant force $F$, then by Newton's Third Law of Motion we should experience an equal reaction force $F$ in the opposite direction. But by Hooke's Law, a stretched spring should exert a force proportional to the amount of stretching ($x$), that is, force should be $k\times x$. How can the two possible at the same time? I am sure that I must be missing something, maybe something to do with the internal of spring, but I can't figure out what.
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1$\begingroup$ When you pull, one end of the spring by a distance $x$, you apply force of $kx$ in one direction. The spring then pulls back with a force $-kx$ (by Newton's Third Law) to keep the spring in equilibrium. What is the confusion here? $\endgroup$– SamCommented Jan 26, 2020 at 13:07
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$\begingroup$ It is not true that you can only pull with force kx, it can be anything, after all, you are the one who is pulling it. $\endgroup$– Big BrotherCommented Jan 26, 2020 at 13:09
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1$\begingroup$ To pull the spring by distance $x$, you will have to apply a force of $kx$ so long as Hooke's law is still obeyed for that distance $x$. $\endgroup$– SamCommented Jan 26, 2020 at 13:21
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$\begingroup$ Hooks low hold till elastic limit. And if you say force can be anything then they will be beyond elastic limit, so no law will be applied. $\endgroup$– आर्यभट्टCommented Jan 26, 2020 at 13:25
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1$\begingroup$ This question has nothing to do with the elastic limit. NOTHING. Can we please drop that discussion? Suggestion to the original poster: if you change the first sentence to "ideal spring" this discussion goes away on its own. $\endgroup$– garypCommented Jan 26, 2020 at 14:12
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2 Answers
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If we consider an experiment of pulling a spring with a constant force $F$, then by Newton's Third Law of Motion we should experience an equal reaction force $F$ in the opposite direction.
The spring provides a restoring force $F=kx$, as long as it is not stretched beyond capacity.
But stretched beyond capacity it will still provide a restoring force but it will no longer be proportional to $x$.
But before the response is such that $F=kx$, that is, $x $ is less than $F/k$, what is the reaction?
We need to look at this dynamically. Assume a point mass $m$ attached to the spring, where the force $F$ will act on. The spring is kept horizontal $x$-axis (so we don't need to account for gravity)
Say that at $t=0$, $x=0$ and we start applying the constant force $F$ (assume also the spring to be of $0$ mass). The spring's restorative force is also $0$ (because at that point $x=0$).
Since there is now a net force acting on the point mass, by N2L there must be acceleration:
$$F=ma$$
More generally (for $x>0$)
$$\Sigma F_i=ma$$ So: $$F-kx=m\ddot{x}$$ So for $x=\frac{F}{k}$:
$$F=kx \Rightarrow \ddot{x}=a=0$$
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$\begingroup$ I am not at all asking if force is kx till a particular capacity or not, I am just asking why Newton's Law says that force exerted by spring should be F (which is constant) and Hooke's Law says it should be kx. $\endgroup$ Commented Jan 26, 2020 at 13:20
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1$\begingroup$ I thought that was self-evident? As the force $F$ extends the spring, the spring's response gradually increases, until $F=kx$. $\endgroup$– GertCommented Jan 26, 2020 at 14:16
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$\begingroup$ But before the response is such that F=kx, that is, x is less than F/k, what is the reaction? Newton's third law says it should be F whereas Hooke's law says it should be kx(<F). $\endgroup$ Commented Jan 26, 2020 at 14:25
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$\begingroup$ So there's a mass on a spring and you are pulling it with force F. Gert's post shows that initially there is an acceleration because there is a net force on the mass because kx < F. When x is large enough so that kx = F, the acceleration is 0. Newton's third law says that the mass always exerts an equal and opposite force to F throughout this process. But that reaction force is not what is considered when finding the motion of the mass as it is not acting on the mass. $\endgroup$ Commented Jan 26, 2020 at 15:42
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The reaction force will be $F$. If the displacement of the spring is linear up to the magnitude of that force, $k = \frac{F}{x}$
answered Jan 26, 2020 at 13:20