How will a spring stretch in the following cases?

[Lotto]

TL;DR Summary

I am currently thinking about springs and have three scenarios with a spring of a spring constant $k$.

1) I have a spring on the ground with no friction and the spring is not attached from one end. If I apply a force $F$ and the spring is massless, will it stretch? I think that it won't. But if it has a mass $m$, will it stretch now? Will it be $x=\frac{F}{k}$? I don't know, but I imagine the spring to stretch and then to start oscillating. But with what period? Because this differs from the case when the spring is massless and there is a body at one end.

2) If I apply a force $F_1$ on one side of the spring and a force $F_2 < F_1$ on the other side, how will it stretch? The spring isn't massless (if it was, would it stretch with an infinite acceleration?) The ground is frictionless. Will it be $x=\frac{F_1-F_2}{k}$? Or $x=\frac{F_1}{k}$?

3) The spring of a mass $m$ is on the ground with a coefficient of dynamic friction $f$. If I apply a force $F$, how will it stretch? Will it be $x=\frac{F-mgf}{k}$?

 

[kuruman]

"On the ground" does not provide complete information. Is the axis of the spring parallel or perpendicular to the ground?

 

[Lotto]

"On the ground" does not provide complete information. Is the axis of the spring parallel or perpendicular to the ground?

It lies on the ground, so the axis is parallel to the ground.

 

[berkeman]

I have a spring on the ground

Meanwhile, in the northern hemisphere...

 

[Baluncore]

Meanwhile, in the northern hemisphere...

Spring has sprung.

Is it a flat spring or a helical wire spring.
You need a "Free Body Diagram" showing:
Case 1. The connection, distributed mass of the spring, and friction with the ground.
Case 2. The attachments to both ends.
Case 3. What shape is the spring, and how is it held in place?

 

[Lnewqban]

TL;DR Summary: I am currently thinking about springs...

Weren't you sinking boats recently...?

It seems to me that "will it stretch?" implies that the springs are being pulled by one or two forces.

For the case of not anchored at one end massive spring on frictionless horizontal surface, inertia would be the only thing opposing the movement of the center of mass of that stretching spring.

I believe that the acceleration of that CM should be half the value of the acceleration of the end receiving the pulling force.

 

[A.T.]

I am currently thinking about springs and have three scenarios with a spring of a spring constant $k$.

You cannot have a non-zero net force on a mass-less spring. You can approximate a massive spring as masses connected by mass-less springs.

 

[Lotto]

Spring has sprung.

Is it a flat spring or a helical wire spring.
You need a "Free Body Diagram" showing:
Case 1. The connection, distributed mass of the spring, and friction with the ground.
Case 2. The attachments to both ends.
Case 3. What shape is the spring, and how is it held in place?

My spring looks like this

and let's say that I just grab it in my hand at its end. But I think that we can suppose that the force is acting on the axis of the spring.

I am most curious about the second case - let's say for the beginning (or isn't it basically the same situation?) that the spring is massless and there are two bodies of the same masses at its ends (the mass of each body is $m/2$ so it has in total the same mass as the spring in my original question). How will the spring stretch? If the two forces had equal magnitudes, the centre of mass of the system would be stationary. So I suppose that when the forces aren't equal, the net force acting in it is $F_1-F_2$. But can I somehow use this information to calculate $x$?

 

[Lotto]

Well, I think that when we have a massless spring with two bodies of masses $m_1$ and $m_2$ and apply forces $F_1$ and $F_2 <F_1$, I think I already know how to calculate $x$.

The whole spring has a spring constant $k$. From the reference frame connected with the centre of mass of the system, a force acting of the first body is $F_2$ and a force acting on the second body is $F_1$. If we divide the spring into two springs which are connected in the centre of mass of the system, their spring constants will be $k_1=k\frac{m_1+m_2}{m_2}$ and $k_2=k\frac{m_1+m_2}{m_1}$. So then $x_1=\frac{F_2 m_2}{k(m_1+m_2)}$ and $x_2=\frac{F_1 m_1}{k(m_1+m_2)}$. So then $x=\frac{F_1m_1+F_2m_2}{k(m_1+m_2)}$.

Is it correct?

 

[Lnewqban]

In this last case, what would happen for a huge value of k?
The CM of the F1-F2 system would be accelerated in the direction of F1.
Would your equation reflect that?

 

[Lotto]

In this last case, what would happen for a huge value of k?
The CM of the F1-F2 system would be accelerated in the direction of F1.
Would your equation reflect that?

Well, if $k$ is huge, $x$ is small. If $k$ goes to infinity, $x$ goes to zero. So I see no problem here. But maybe I miss something?

 

[sophiecentaur]

The whole spring has a spring constant k.

The value k assumes that the load on the spring and the incremental expension are considered to be constant over the length of the spring and ignores the mass of the spring. If you hold just one end and accelerate that end at a then you can write and solve a differential equation, starting with an elemental length of the spring. You could start with the problem of the shape of a spring when supported under a stationary point. (acceleration g) There's an old PF post which discusses this. You could then move on to a suspension point that's under a general acceleration a.

But maybe I miss something?

Yes; you miss all the fun!

 

[Lotto]

The value k assumes that the load on the spring and the incremental expension are considered to be constant over the length of the spring and ignores the mass of the spring.

Now I don't know if I understand it. Do you mean by it that if we change the length of the spring by stretching, the value $k$ changes as well? So if we double the length, the value $k$ is twice smaller?

 

[sophiecentaur]

Now I don't know if I understand it. Do you mean by it that if we change the length of the spring by stretching, the value $k$ changes as well? So if we double the length, the value $k$ is twice smaller?

You are right, of course, for an ideal spring. k is the change in extension per unit force, which applies to every part of the spring, even an elemental length. In a 'real' suspended or pulled spring, the force difference over each section is proportional to the extension. It's the force that changes and not the k, over the spring length.

However, k is an approximation. The Young Modulus, which is Stress/Strain, is an inherent property of the spring material but, as a wire is stretched, its cross sectional area will change, meaning that the stress (force / cross sectional area) is actually modified by the length (the volume of steel is constant). A coil spring uses the torsion in the coils which still affects the effective CSA but the helical design greatly reduces the 'k' factor compared with a straight wire. But that's academic. If you look at the link I provided you will get a proper idea of how to work out the forces involved on a self - loaded spring. Something to start your investigation with.

 

[Lotto]

You are right, of course, for an ideal spring. k is the change in extension per unit force, which applies to every part of the spring, even an elemental length. In a 'real' suspended or pulled spring, the force difference over each section is proportional to the extension. It's the force that changes and not the k, over the spring length.

However, k is an approximation. The Young Modulus, which is Stress/Strain, is an inherent property of the spring material but, as a wire is stretched, its cross sectional area will change, meaning that the stress (force / cross sectional area) is actually modified by the length (the volume of steel is constant). A coil spring uses the torsion in the coils which still affects the effective CSA but the helical design greatly reduces the 'k' factor compared with a straight wire. But that's academic. If you look at the link I provided you will get a proper idea of how to work out the forces involved on a self - loaded spring. Something to start your investigation with.

The link you provided leads to this post, so could you send it again, please?

 

[sophiecentaur]

The link you provided leads to this post, so could you send it again, please?

This works for me. Try again

 

[Frabjous]

This works for me. Try again

Refers to this thread for me also

 

[sophiecentaur]

Refers to this thread for me also

Oh what a plonker I am. Sorry
Try this link.