# How is Newton's 3rd law of action reaction forces valid when a man pulls a massless spring?

Imagine there is a massless spring of spring constant 10 newton/metre Now, a person's hand apply a pulling force of constant 100 newton on the spring Now, according to Newton's 3rd law, the spring should also exert an equal amount of pulling force on the man, but right at this MOMENT/INSTANT, the spring has 0 elongation, and according to Hooke's law, the force applied by spring is equal to spring constant multiplied by its elongation/compression, since elongation is 0, the restoring force of string is 0, therefore the restoring force of spring on the hand of the man is 0 How is Newton's 3rd law being followed here? Since the man pulled the end of the spring with 100N, the end of that spring should also pull the man with 100N, but as we calculated the spring force is 0 (according to Hooke's law)

Consider another situation

In this same scenario, because of 100N force, the spring must have elongation, let's look at the moment when elongation of spring is 1 metre As we can see, man is still pulling on the spring with a constant force of 100N, but at this moment when elongation of spring is 1 metre, the reaction force applied by spring due to action force of man's pull is 10N (according to Hooke's law, as spring restoring force is spring constant multiplied by acceleration, which is $$10\times1=10$$)

Therefore ,here also newtons 3rd law is being invalid, as man is pulling on the string with a force of 100N but the spring only pulls on the man with a force of 10N

But, when the elongation of spring is 10 metres, there is a whole different story Here, as we can see, Newton' s 3rd law is valid,as the man is pulling on the spring with a force of 100N, and the spring also pulls the man with a force of 100N (as spring force is spring constant times elongation, which is $$10\times10=100$$) at this moment

So Newton's 3rd law was invalid initially, but it became valid after spring elongated a length of 10metres

Can someone explain what's happening here? I'm pretty sure I'm doing something wrong as a law can't be wrong anytime, am I considering wrong action reaction pair of forces? Or is there some other mistake?

• I have to say, I am a bit upset at the lack of upvotes for this high quality question! Apr 25 at 22:40
• Even if the spring is massless, you have one. And this has interesting ramifications... Apr 26 at 9:07
• Could you re-phrase that, please? Could you re-name Newton's 3rd law of action reaction forces under its usual title? Could you start by explaining how your massless spring behaves differently from an ordinary, massy spring? Apr 29 at 21:47

Now, a person's hand apply a pulling force of constant 100 newton on the spring

A person’s hand cannot apply a constant 100 N force on the end of a massless spring. The assumption that this can be done is non-physical and incorrect.

Simply because a person wants or tries to apply a 100 N force doesn’t mean that they can. You cannot push with 100 N on an inextensible string that breaks at 10 N.

If Chuck Norris executes a perfect 100 N punch against a wall and then performs the biomechanically exact same punch against a fly, even for Chuck Norris it will not be 100 N.

• "A person’s hand cannot apply a constant 100 N force on the end of a massless spring" – Unless the spring happens to be elongated just right to apply exactly 100 N force on the hand. Right? Apr 25 at 1:20
• Right, but at the time of the quoted text it was at its relaxed length
– Dale
Apr 25 at 1:53

In the end, we will find that the moment where you found your contradiction doesn't exist, because as the spring mass decreases, the time it takes for the spring to elongate goes to zero. For a small but nonzero mass, the spring exerts a reaction force inertially on the person while it is in the process of elongating.

Setting something (like the mass of a spring) equal to exactly zero (or infinity) is never a useful thing to do. It makes certain things you might calculate undefined or nonsensical. There will often be logical issues with your analysis after you do this. Let me instead put a weight with a small mass $$m$$ at the end of the spring, and then we can make $$m$$ small and see what happens as the spring approaches becoming massless.

The person starts pulling on the mass at the end of the spring with a force $$F$$. It starts accelerating right with an acceleration $$F/m$$. Of course, the mass is also pulling on the person with a force $$-F$$, the person pushes on the floor with their feet and receives a force from static friction of $$F$$ so that they don't accelerate. This is kind of the answer to your question - no matter how small the mass of the spring is, it exerts a reaction force on the person because of its inertia. This is only happening while the spring is accelerating as it elongates. Setting this mass to exactly zero is what confused you (how can something have inertia without mass??).

As the spring elongates, the net force on the mass is $$F-kl$$, where $$l$$ is the length and $$k$$ is the spring constant. So when $$l=k/F$$, the mass no longer accelerates. As the mass decreases this takes less time because the spring accelerates faster throughout the elongation process. So for the "massless" spring, it is instantly elongated to the length where it's in equilibrium.

I'm ignoring the obvious issue that in this description the spring will overshoot the equilibrium and oscillate back and forth forever. Maybe the person stops pulling at the right time so that the spring becomes motionless at equilibrium and then they start pulling again to keep it there.

In fact this is also part of the problem in your question. More realistically, the way one would elongate a spring is start by applying a tiny force $$\delta_f$$ for a small time $$\delta_t$$. So the spring is moving to the right with velocity $$v=\delta_f\delta_t/m$$. Thereafter just apply a force $$vkt$$, so that you counteract the slowly increasing spring force and the spring end keeps moving. Then when you get to the desired location, apply a force $$vkt-\delta_f$$ again for a time $$\delta_t$$ and the spring stops in equilibrium with $$vkt=100\text{ N}$$, assuming you wanted to be applying 100N in the end. It's quite weird to suddenly apply the final force. That's not how one would naturally elongate a spring.

After a discussion in the comments, I see that your main confusion was that you thought that the reaction force had to come from the spring. So I'll clarify that in equilibrium, the way you should see it is this: The person pulls on the mass with a force of 100N. The mass pulls back on the man with a force of 100N because it just does. Because newton's third law tells you it does. No other reason whatsoever. Then the spring pulls on the mass with a force of 100N, and that's why the mass isn't accelerating - all forces are balanced. Indeed also the mass pulls on the spring with a force of 100N because for every force there is an equal and opposite reaction force. The spring doesn't accelerate because the wall pulls on it (and it pulls on the wall). The wall doesn't move because it pulls on the earth (and the earth pulls on it). The man doesn't move because he gets some force from the earth from static friction. The earth doesn't move because the force it experiences from the man is the same as the force it experiences from the wall. All good-everything in equilibrium, nothing is moving.

When the spring isn't elongated far enough to pull with a force of 100N, the mass is accelerating. It's still pulling on the man with 100N, but the force on the mass isn't balanced by a force from the spring, so it accelerates.

• Apr 24 at 19:39
• re: receives a force from static friction of F so that they don't accelerate - but they do accelerate - at least the parts of the body that are in contact with the spring, so the force exerted on the ground and the force exerted on the spring will certainly be different. The mass of the man's hand and arm (and the acceleration required to move them) seems far more significant than the mass of the spring. If the extension is enough that the man needs to move their whole body, even more so. Apr 26 at 6:29

Massless springs do not exist but an approximation is made by assuming the mass of a spring is very much less then any other masses within the system.
The other thing to note is that in the real world there is no such thing as an instantaneous change, a change must happen over a period of time which could be much less than any other time interval in relation to the system under consideration.
However assume that the spring is massless and what that means is that the net force on the spring must always be zero.
In your example how does that work?
In the final (steady) state, at your end of the spring you exert a force of $$100\,\rm N$$ on the spring and the spring exerts a force of $$100\,\rm N$$ on you in the opposite direction - Newton's third law.
At the other end of the spring the wall exert a force of $$100\,\rm N$$ on the spring and the spring exerts a force of $$100\,\rm N$$ on the wall in the opposite direction - Newton's third law.
Thus, in the "ideal" world of massless springs and instantaneous changes all that should concern you is the initial state and the final state and not the detail as to how the system moved from one to the other.

As I have pointed out that in the real world the spring (which has mass) will have to be accelerated and it will take time to move to a new position, and hence have a net force acting on it, but at all times Newton's third law will be obeyed.

• You exert only one force on the spring. The spring exerts two forces, one on you and one on the wall, and the wall exerts a force on the spring. Apr 24 at 14:53
• Is the 100N force exerted on me by the spring the spring constant/restoring force (kx)??or are spring constant/restoring force(kx) and the 100N reaction force exerted on me by the spring different forces???? Apr 24 at 18:50
• @Aakash for a real (massive) spring, the force you exert on the spring must account for both the acceleration and extension of the spring: $F = ma + kx$, with a being the second derivative of the position, $a = \ddot x = d^2x/dt^2$ Apr 24 at 20:03
• In the limit of zero spring mass I suppose the man accelerates instantly to the final position (applying an infinite force to move himself, but a finite force on the spring). If the man can't apply an infinite force to move himself then he can't apply a 100N force instantly on the spring. Apr 24 at 21:32

Hooke's Law cannot be used in all cases. It only applies for Hookean springs. This means that when a force is applied, it will deform. If/When (at some point all springs act like non-hookean) it doesn't get deformed, then it is not Hookean. A non-hookean spring is mathematically equivalent to a rigid body. So, for a non-hookean spring, if the force applied is increased, the reaction force increases too even if the deformation is the same.

To summarize, Newton's Third Law is a fundamental law of the universe and is valid in all cases. However, Hooke's Law gives correct answers only in case of a Hookean spring.

If the spring is massless, the entire inertia is because the fact that the wall pulls the spring. In this case, the force will be exerted due to wall and not because of spring. The action force is actually applied on wall and the reaction force is applied by wall. The spring just acts like a medium.