# B Spring Attached to Wall

1. Jul 25, 2015

### andyrk

When we compress a spring attached to a wall, the wall also exerts the same force as we do, otherwise the spring would accelerate (which it doesn't). So does that mean the wall compresses the spring? If not, why? If yes, then why do we have compression of x (done by us) only?

2. Jul 25, 2015

### Staff: Mentor

Both the wall and yourself have to exert a force on the spring for it to compress. The difference between is that the wall doesn't move while you yourself do.

3. Jul 25, 2015

### andyrk

How can the spring not compress even though the wall is applying force on it?

4. Jul 25, 2015

### Staff: Mentor

From the point of view of whichever hand is doing the compressing, it is in fact the wall that is compressing the spring, while it itself isn't moving.

5. Jul 25, 2015

### andyrk

You mean to say even though I am compressing the spring, it is actually the wall that is compressing it? Then the question again arises..why is my hand not compressing the spring? Both wall and hand should compress because of Hooke's Law.

6. Jul 25, 2015

### Staff: Mentor

I'm saying that it doesn't matter who you say does the compressing. Both you and the wall are exerting a force on the spring, so it's probably most accurate to say both of you are compressing it. However, we live in a world where we like to make things easier by saying that it is you that does the compressing, and not the wall. That way you can simplify the equations, making things like Hooke's law much easier to work with. Only having to worry about the distance one end of the spring moves is much easier than worrying about both ends.

7. Jul 25, 2015

### Nathanael

If you're asking why the compression is x=F/k and not x=2F/k, then it's simply a matter of the way the spring constant, k, is defined.

8. Jul 25, 2015

### Orodruin

Staff Emeritus
String compression is a property of the string as a whole and needs two forces, one on each side. The spring constant relates the force to the compression/extension of the string, i.e., the difference between the spring rest length and the spring's actual length.

9. Jul 25, 2015

### andyrk

What if I say that I don't want to make things easier and look at things just the way they are? Would it then mean that I compressed the spring by $x/2$ even though I clearly compressed it by $x$?

10. Jul 25, 2015

### Staff: Mentor

No. You'd need to modify the spring equation somehow.

11. Jul 25, 2015

### andyrk

It would always remain $F = -kx$.

12. Jul 25, 2015

### Orodruin

Staff Emeritus
Look, it is very very simple, $x$ is the difference between the spring length and the spring rest length, $F$ is the force (on either side) and $k$ the spring constant. Do not overcomplicate things!

13. Jul 25, 2015

### andyrk

If I take a spring in my hand and stretch it a length x more than its original length, with both my hands. Now, do both my hands experience an inward force $F = -kx$?

14. Jul 25, 2015

### Staff: Mentor

Ah, so you wouldn't have to change the spring equation. Good to know.

15. Jul 25, 2015

### andyrk

But no one replied to post #13.

16. Jul 25, 2015

### Qwertywerty

Yes .

17. Jul 25, 2015

### Orodruin

Staff Emeritus

18. Jul 25, 2015

Staff Emeritus
PF is a collection of volunteers who are selflessly devoting their time to helping you understand. They are not your servants to be berated when they didn't answer your question after a mere twenty minutes.

As it happens, post #12 has the exact answer to your post #13.

19. Jul 25, 2015

### Staff: Mentor

I'm going back to your original post, because I think I can help relate the kinematics to the force in a little different way that might work for you. The tension in the spring is always going to be T = kΔL, where ΔL is the change in length, relative to the original unextended length. If ΔL is negative, then the spring is in compression, and the tension is negative.

Let xL represent the x coordinate of the left end of the spring at time t, and let xR represent the x coordinate of the right end of the spring at time t. Let the original unextended length of the spring be L0. So we are going to allow the possibility that both ends of the spring can move, and not just one end. So the left end can move by moving your left hand, and the right end can move by moving your right hand. With this description, the change in length of the spring ΔL relative to its original unextended length is given by:
$$ΔL=x_R-x_L-L_0$$
It doesn't matter which hand is moving and in which direction, the only thing that matters is the distance between your left hand and your right hand, xR-xL. The tension that each hand feels is then given by:
$$T=k(x_R-x_L-L_0)$$
If the term in parenthesis is negative, then the spring is in compression.

Hope this helps.

Chet

20. Jul 25, 2015

### andyrk

So just applying a force on one end doesn't mean that the spring would extend or compress?

21. Jul 25, 2015

### Staff: Mentor

Yes, that's right. It isn't possible to apply a force to one end of a mass-less spring to try to get it to extend or compress, without in some way anchoring the other end. In one case, you've anchored it by attaching it to the wall, and, in the other case, you've anchored it by holding it with your other hand. Or, you could move both ends simultaneously, but the forces on your two hands would always be of the same magnitude.

The characteristic of a mass-less spring is that the forces at the two ends are always equal in magnitude and opposite in direction.

Chet

22. Jul 25, 2015

### andyrk

Very interesting and fascinating. But isn't that true for springs that have mass as well?

23. Jul 25, 2015

### Staff: Mentor

No. For springs that have mass and involve acceleration, the force on one end is not equal to the force on the other end, and the tension varies along the spring. You saw that kind of deal last week when you analyzed the tension variations along a rotating rod that has mass.

Chet

24. Jul 25, 2015

### andyrk

But I don't think the spring can compress or extend that way then. Also, if you compress a spring with both hands..so that its length changes by x, both the hands experience a force $F = -kx$ in opposite direction. But how do we know how much did each hand compress the spring individually?

25. Jul 25, 2015

### Staff: Mentor

It is not possible to identify the individual amount that each hand compresses the spring. There is no such thing. The only thing that matters is the distance between the two hands, and not their individual locations. For example, if you hold your left hand fixed and move your right hand to the left x inches, you might think that your right hand has compressed the spring. But now, what if you move both hands to the right x inches, while holding the length of the spring fixed. The spring still has the same force, but now it looks like the right hand was fixed, and the left hand compressed the spring. If someone had left the room before any compression was done, and then came back afterwards, they would not be able to tell.

Chet