Understanding the Relationship Between a Compressed Spring and Attached Wall

[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?

 

[Drakkith]

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.

 

[andyrk]

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

 

[Drakkith]

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.

 

[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.

 

[Drakkith]

You mean to say even though I am compressing the spring, it is actually the wall that is compressing it?

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.

 

[Nathanael]

why do we have compression of x (done by us) only?

Both wall and hand should compress because of Hooke's Law.

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.

 

[Orodruin]

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.

 

[andyrk]

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.

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 $x$?

 

[Drakkith]

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 $x$?

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

 

[andyrk]

It would always remain $F = -kx$.

 

[Orodruin]

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!

 

[andyrk]

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!

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 $F = -kx$?

 

[Drakkith]

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!

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

 

[andyrk]

But no one replied to post #13.

 

[Qwertywerty]

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=−kxF = -kx?

Yes .

 

[Orodruin]

But no one replied to post #13.

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!

Please read the posts better.

 

[Vanadium 50]

But no one replied to post #13.

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.

 

[Chestermiller]

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 x_L represent the x coordinate of the left end of the spring at time t, and let x_R 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, x_R-x_L. 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

 

[andyrk]

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 x_L represent the x coordinate of the left end of the spring at time t, and let x_R 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, x_R-x_L. 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

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

 

[Chestermiller]

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

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

 

[andyrk]

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

 

[Chestermiller]

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

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

 

[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?

 

[Chestermiller]

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?

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

 

[andyrk]

But now, what if you move both hands to the right x inches, while holding the length of the spring fixed.

I can't visualize this situation.

 

[Chestermiller]

I can't visualize this situation.

Sorry, then all I can say is that it is not possible to identify the individual amount that each hand compresses the spring.

With regard to the rod example, the rod is just the same as a spring with mass that has a very large spring constant (i.e., very stiff).

Chet

 

[olivermsun]

I can't visualize this situation.

If you push harder with your left hand than you push back with you right, then your left hand will just end up pushing the spring and all (including your right hand) to the right.

 

[A.T.]

But how do we know how much did each hand compress the spring individually?

The same way you know how much noise each hand makes individually, when you clap them.

 

[Drakkith]

But how do we know how much did each hand compress the spring individually?

The problem is that 'how much each hand compresses the spring' is not a well-define statement. The force exerted by each hand is identical and as the spring compresses each hand moves inwards relative to its center of mass by the same amount. So according to the spring, each hand compresses it by the same amount.

However, if I keep my right hand hand stationary with respect to my body, while my left hand moves inward and pushes the spring, from my point of view I might say that it is my left hand that does the compressing since my right hand does not move.

Both of these are potentially correct descriptions.

 

[Orodruin]

Compare to hanging a mass in the end of a massless spring in a gravitational field. If you hold the other end, the spring extends because there are forces on both ends. If you let it go, it simply accelerates with the mass, so the force of gravity on the mas is not enough to extend the spring.

 

[Chestermiller]

To expand on Drakkith's post #30, the spring condition situation can also be expressed in terms of the following kind of terminology:

1. As reckoned by an observer from a frame of reference that is at rest with respect to your left hand, it is your right hand that compressed the spring, and there is nothing that could convince him otherwise.
2. As reckoned by an observer from a frame of reference that is at rest with respect to the COM of the spring, both your hands compressed the spring equally, and there is nothing that could convince him otherwise.
3. As reckoned by an observer from a frame of reference that is at rest with respect to your right hand, it is your left hand that compressed the spring, and there is nothing that could convince him otherwise.

Chet

 

[andyrk]

Can someone provide a link which describes in detail everything about a spring in physics terms? Because the book I have says nothing about the fact that spring exerts the same force in opposite direction at both ends, but it uses this fact to solve many problems that follow it.

 

[russ_watters]

Can someone provide a link which describes in detail everything about a spring in physics terms? Because the book I have says nothing about the fact that spring exerts the same force in opposite direction at both ends, but it uses this fact to solve many problems that follow it.

My suspicion is that you aren't reading your source correctly, but try this: https://en.wikipedia.org/wiki/Spring_(device)

 

[andyrk]

My suspicion is that you aren't reading your source correctly, but try this: https://en.wikipedia.org/wiki/Spring_(device)

Even this doesn't say anything about the spring applying equal and opposite forces at both ends.