# Do two springs compressed springs placed against each other obey Newton's third law?

I've been mulling over the following scenario.

Say I take two springs whose force is given by some well-defined function of position (for example, Hooke's law F=kx will do). Using this law, I choose to compress spring A by a distance x that will guarantee me that the spring will exert a force of 10N.

I then choose to compress spring B by a distance such that it will exert a force of 7N.

I put the two springs next to each other, release them and allow each to exert its force against the other. It's easier to imagine along the horizontal axis, and we can ignore gravity and friction and any other potential meddlers.

Spring A should exert a force of 10N on spring B according to its force law and compression, but spring B should exert force of 7N back on spring A according to ITS force law and compression.

Thus the force that spring A exerts on spring B is NOT equal and opposite to the force spring B exerts on spring A.

This is a violation of Newton's Third law. What am I missing here?

Suppose I take a spring at equilibrium and push on the end of it with a force of ten Newtons. After a very short time, the spring has not compressed appreciably, so the $F=-kx$ law says the force it exerts on my hand is nearly zero, but Newton's third law says it must be ten Newtons. Which is right?

The second answer, that the spring exerts a ten-Newton force on my hand, is right. The spring is a dynamical object. Try thinking of it as 100 springs each with .01 the mass and 100 times the spring constant all connected in series. What you'll see is that the first spring, the one I'm touching, can be more compressed than the second, third, fourth ones etc. The spring doesn't have uniform tension in it any more, which is the assumption behind the $F = -kx$ law.

In other words, Hooke's law is built on assumptions, and when your spring is massive and accelerating differently in different parts, as real springs do, those assumptions fail. Your two springs will exert equal and opposite forces on one another as Newton's third law dictates.

"Spring A will exert 10 N on Spring B and Spring B will exert 7N on Spring A " - This approximation is wrong.

Spring A will exert x N on Spring B and Spring B will also exert x N on Spring A. The value of x can be calculated if their mass is given and some constraint such as the point of contact never changes position or something like that is given. (Although I think its complex to calculate). As a result of this x, the whole individual springs will move in opposite directions and will expand (uncoil) in the gap created by such movement.

• Although I think this answer can be better written, the underlying gist is correct. I don't understand why someone downvoted it and didn't give helpful comments too. Jan 11, 2017 at 8:05
• Ok.. What's the problem with it ?.. I think "a down voter must comment reason for down voting" -> this feature must be included in Stack Exchange sites Jan 12, 2017 at 11:31
• Better format and flow, don't worry it will happen if you continue writing. Jan 12, 2017 at 12:47

What you're missing is the following:

Let length and compression of string 1 be $l_1$ and $x_1$ respectively. Similarly, for the second string $l_2$ and $x_2$.

When you put the two strings adjacently, what is fixed is the distance between their fixed ends, which is $l_1+l_2-x_1-x_2$. The individual compressions $x_1$ and $x_2$ are not fixed.

Just like when you pull a string and leave it, the string goes back to its initial position, similarly in this case, as soon as you put them together and leave the strings, the individual compressions will change so that ${x_1}^{new}={x_2}^{new}$, so that magnitude of forces is the same$=kx$.

In a nutshell, there is nothing holding the strings to be exactly at compressions equivalent to 100 N and 70 N. They will adjust to make forces equal and opposite. Clear?