Solving Spring Force Equation: 2X

[Ibraheem]

Hello,

If we have a spring at rest and has a constant k=2 N/m (at its natural length with one end at origin (x=0) and the other end held stationary) is having a force applied on it. The force varies in same way as the the spring with function 2X. My question is how is it possible to start the spring moving if the spring is applying the same force as the force applied on it but in the opposite direction. In other words, booth forces are expressed by 2x how do they not cancel and how is it moving. Don't we need to apply a small force at first just to keep it moving.

 

[UMath1]

You must consider only forces acting on the spring. The forces do not cancel because they are acting on different objects. Your hand applies a force on the spring and the spring applies the same force on your hand

 

[Ibraheem]

But what if I apply a constant force of 10N would this continue for ever even if the spring stops my hand. Would the spring apply a force of 10N immediately after its contact with my hand at the origin( x=0)

 

[UMath1]

The spring would apply a force equal to the force your hand applies (Newton's 3rd Law).

 

[Ibraheem]

So the spring will apply 10 Newtons immediately, but what about f=kx. will this add to the 10 Newton when the end of the spring starts to change position.

 

[nasu]

But what if I apply a constant force of 10N would this continue for ever even if the spring stops my hand. Would the spring apply a force of 10N immediately after its contact with my hand at the origin( x=0)

The "problem" is that you cannot do that. You cannot apply the 10 N force instantaneosly. The force will increase gradualy from 0 to 10 N as the spring expands.

 

[BluberryPi]

If x equals 0, in the formula, wouldn't F = 0? Since F=-kx and x=0, 2 x 0 = 0.

 

[Chandra Prayaga]

x is actually the amount by which the spring is stretched (or compressed). If you apply a constant force on a spring, for example, by hanging the spring from the ceiling and hanging a fixe weight from the other end of the spring, then the spring will stretch until, kx is equal to the weight.

 

[Ibraheem]

I can see it more clearly now so the Newton's third law force pairs of an object interacting with a spring(compressing or stretching ) always equal to XK. So in the vertically hung spring case, if the spring force equal the weight of the object that is attached to it will the object stop completely or will the spring stretch more to stop the object from going down?

 

[UMath1]

Actually can someone answer BlueberryPi's question? How can you apply any force at all on the spring if the spring can't apply any force on you when x=0? How can you provide the necessary force to compress or elongate the spring when the spring is at equilibrium?

 

[Chandra Prayaga]

Actually can someone answer BlueberryPi's question? How can you apply any force at all on the spring if the spring can't apply any force on you when x=0? How can you provide the necessary force to compress or elongate the spring when the spring is at equilibrium?

Well, you can then pose the same question about any object in equilibrium, not just the spring. The point is again, a careful look at Newton's third law. The question is not, can or can't the spring apply force. Whatever force is applied on the spring, irrespective of whether it is in equilibrium or not, the spring will apply an equal force right back. If a big man and a little child are holding hands and skating, the force by the man on the kid, is exactly, all the time, equal in magnitude and opposite in direction to the force by the kid on the man. The question, "can the kid exert that much force?" does not arise. Newton's third does not say how big or small the action reaction pair are. They are always equal, in magnitude.

 

[Chandra Prayaga]

I can see it more clearly now so the Newton's third law force pairs of an object interacting with a spring(compressing or stretching ) always equal to XK. So in the vertically hung spring case, if the spring force equal the weight of the object that is attached to it will the object stop completely or will the spring stretch more to stop the object from going down?

This is in reply to Ibraheem: The experiment is done like this.
1. First you establish the equilibrium position. You hang the spring from the ceiling, attach the mass at the bottom, place your hand below the mass, and slowly lower it until you can remove your hand without the mass oscillating. In this position, the mass is not moving, and it is in equilibrium. So in this position, the upward force from the spring is exactly equal to the downward force of gravity, and both forces are acting on the mass.
2. Second, you hold the mass and either lower it, or raise it just a little bit, vertically, thus either stretching or compressing the spring, and let go. The mass will now oscillate up and down. During oscillation, the mass is in equilibrium at the central position of the oscillation, which is the equilibrium position that you established in step 1. At that position, the the velocity of the mass is not zero, but the acceleration is zero because it is in equilibrium. At that position, every time the mass crosses that position, the spring force balances gravity. At other positions in the oscillation, the acceleration is not zero, the mass is not in equilibrium, and the spring force and gravity are not equal in magnitude.

 

[UMath1]

Well, you can then pose the same question about any object in equilibrium, not just the spring. The point is again, a careful look at Newton's third law. The question is not, can or can't the spring apply force. Whatever force is applied on the spring, irrespective of whether it is in equilibrium or not, the spring will apply an equal force right back. If a big man and a little child are holding hands and skating, the force by the man on the kid, is exactly, all the time, equal in magnitude and opposite in direction to the force by the kid on the man. The question, "can the kid exert that much force?" does not arise. Newton's third does not say how big or small the action reaction pair are. They are always equal, in magnitude.

Right but if we look at Newton's 3rd Law from the reverse perspective. Then if the spring applies a force of 0 because its displacement is x=0, then you must also apply a force of 0 because the forces must be equal and opposite. So how can you ever get a nonzero displacement?

 

[nasu]

By applying a force that increases from zero to some finite value.

 

[UMath1]

But for the spring to compress it must receive a nonzero force before it is compressed. But before it is compressed it can only receive a force of 0 because kx=0.

 

[nasu]

The force and compression are simultaneous. There is no before and after.

Same as to move you need to change your position. But if your original speed was zero you cannot change position so you cannot move. Are you saying that nothing can start moving? :)

 

[UMath1]

no..but the spring force case is different. If I compress the spring a distance x, then I put 1/2kx^2 J into the system. So I need to apply an average force of 1/2kx to do that. But how can I ever apply that force if I have to compress the spring by a infintesimal amount x before any force can be applied at all? I can only apply a force when the compression amount is nonzero. But for there to be a nonzero compression a force must be applied.

 

[sophiecentaur]

The force and compression are simultaneous. There is no before and after.

Same as to move you need to change your position. But if your original speed was zero you cannot change position so you cannot move. Are you saying that nothing can start moving? :)

Isn't the resulting motion SHM, impressed on a net mean acceleration? That would give a phase difference between the instantaneous velocity and the spring compression

But if your original speed was zero you cannot change position so you cannot move.

That would be a form of Zeno's paradox, I think.

 

[jbriggs444]

no..but the spring force case is different. If I compress the spring a distance x, then I put 1/2kx^2 J into the system. So I need to apply an average force of 1/2kx to do that. But how can I ever apply that force if I have to compress the spring by a infintesimal amount x before any force can be applied at all? I can only apply a force when the compression amount is nonzero. But for there to be a nonzero compression a force must be applied.

For a massless spring, you do not need to apply any force to get it to move. But no real spring is massless so... For a spring with mass the force you apply to the spring is not always given by f=kx and the force that the spring applies to you is not always given by f=kx. The delta is given by f=ma.

 

[nasu]

Isn't the resulting motion SHM, impressed on a net mean acceleration? That would give a phase difference between the instantaneous velocity and the spring compression

That would be a form of Zeno's paradox, I think.

That example was not about the spring but just a simple motion.
I was thinking about one of Zeno's paradoxes, yes.

For the spring, the question was about force and displacement. They are in phase. Velocity has indeed a phase difference.

 

[Chandra Prayaga]

Right but if we look at Newton's 3rd Law from the reverse perspective. Then if the spring applies a force of 0 because its displacement is x=0, then you must also apply a force of 0 because the forces must be equal and opposite. So how can you ever get a nonzero displacement?

So look at the process carefully, as an experiment that you are performing. You stretched the spring from equilibrium and let it go. Nothing prevents you from exerting a non-zero force to stretch the spring. The spring will just keep exerting an equal force on you. Newton's third law does not say that if an object is NOT exerting a force on you, you CANNOT exert a force on it. It only says that IF you exert a force on the object, THEN the object will exert an equal force on you. Certainly, if the OBJECT exerts a force on you, YOU will exert an equal force back on it.

 

[sophiecentaur]

But what if I apply a constant force of 10N

This question is the core of the problem with this thread. You cannot instantly apply a 10N force to the spring unless you displace it instantaneously by10k. You have to start with an infinitessimal reaction force. The intuitive response to the OP just takes you into seeing the problem as a paradox, when it isn't. The problem becomes the same as all proportional relationships (y=ax) and no one has a problem with what happens at the origin. (Even if they do, a course of Mathematical Analysis can put them straight.)

 

[sophiecentaur]

So look at the process carefully, as an experiment that you are performing. You stretched the spring from equilibrium and let it go. Nothing prevents you from exerting a non-zero force to stretch the spring. The spring will just keep exerting an equal force on you. Newton's third law does not say that if an object is NOT exerting a force on you, you CANNOT exert a force on it. It only says that IF you exert a force on the object, THEN the object will exert an equal force on you. Certainly, if the OBJECT exerts a force on you, YOU will exert an equal force back on it.

Yes. People seem to be obsessed with Newton 3 at the moment and they are trying to apply it where it just doesn't apply,

 

[UMath1]

Is it that the spring force is only equal to kx when there is no net force acting on it? Because in order for me to compress or stretch the spring from its equilibrium position I must apply a nonzero force when it is at equilibrium. That means that the spring must also apply an equal nonzero force. So in that case the springs force is not kx, because it applies a force but x=0

 

[sophiecentaur]

Is it that the spring force is only equal to kx when there is no net force acting on it?

Whatever position the spring is in, if it is in equilibrium (i.e it could be loaded or not) then the force to extend it (or compress it) from that position is kx when the extension is measured from the start (equilibrium) position. That assumes the spring follows Hooke's Law.
If you plot F against the length then you get a straight diagonal line with slope k. You can start where you like (pre-loaded) and you still move up and down the same diagonal line. So the answer to your question is no.
Why don't you just read about this - for instance, this link? All will be made clear if you follow logically what the link says. It is not always made clear that there is nothing special about starting with an unloaded spring - but, if you think about it, any real spring has mass and will be compressing or stretching under its own weight. No one says you need to take that into account - you just plot the load / extension to find k.

spring must also apply an equal nonzero force

Why do you find this a problem? Any two objects interacting will have equal and opposite forces between them. This (as has been mentioned previously) is nothing to do with the k of a spring.

 

[Chandra Prayaga]

Is it that the spring force is only equal to kx when there is no net force acting on it? Because in order for me to compress or stretch the spring from its equilibrium position I must apply a nonzero force when it is at equilibrium. That means that the spring must also apply an equal nonzero force. So in that case the springs force is not kx, because it applies a force but x=0

I am afraid we are going round in circles. Please describe your experiment completely. If you like, first talk about a horizontal spring attached to a body which is free to slide on a frictionless table. This is conceptually simpler than the vertical spring, and will allow you to understand the spring force before introducing the complication of gravity.

 

[UMath1]

Initially the spring is not stretched or compressed so x=0. If I were to compress this spring I must then apply a nonzero force. By Newton's 3rd Law the spring must also apply a nonzero force, even though kx=0. So my speculation is that the force of the spring is only equal to kx when it is not accelerating.

 

[256bits]

∫

Initially the spring is not stretched or compressed so x=0. If I were to compress this spring I must then apply a nonzero force. By Newton's 3rd Law the spring must also apply a nonzero force, even though kx=0. So my speculation is that the force of the spring is only equal to kx when it is not accelerating.

The force the spring provides is equal to kx. there is no way around that with an ideal spring. As the spring is more and more compressed the force from the spring increases according to the formula. Your force will match the spring's force at any x.

The way of looking at this problem is that one does not initially give a force to the spring. One gives a velocity, dx/dt which then gives a comparable change in force. To go from position x₀ to x₁ there has to be movement.

dF = k ∫dx/dt +kx₁ from time period 1 to 2.
or
F₂-F₁ = k ( x₂ - x₁ ) + kx₁

 

[sophiecentaur]

So my speculation is that the force of the spring is only equal to kx when it is not accelerating.

If it is not accelerating then there will be no force on the spring x= 0 and F = 0. The equation F=kx describes the relationship between the force and the extension for all values of acceleration.
Take a step backwards and start again with this. There is something you are not getting about the basics, I think. Apply the rules and not your intuition, perhaps. I know intuition can be very attractive.

 

[jbriggs444]

If it is not accelerating then there will be no force on the spring x= 0 and F = 0. The equation F=kx describes the relationship between the force and the extension for all values of acceleration.

The equation F=kx describes the force for all values of acceleration. But only when acceleration is zero.

Edit: This assumes a spring with non-zero mass. That is not the situation sophiecentaur has in mind.

 

[sophiecentaur]

The equation F=kx describes the force for all values of acceleration. But only when acceleration is zero.

That doesn't make sense; I guess you didn't really mean it. The spring doesn't 'know' it's accelerating. It can only respond to the forces on it. That force will be equal to ma, where m is the mass on the end and a is the acceleration. Zero acceleration will produce zero force, whatever the stiffness of the spring.

 

[jbriggs444]

That doesn't make sense; I guess you didn't really mean it. The spring doesn't 'know' it's accelerating. It can only respond to the forces on it.

Possibly we are not talking about the same situation. I am envisioning an anchored spring with one end held stationary. The force on the free end will not match f=kx. Neither will it match f=ma.

 

[sophiecentaur]

Possibly we are not talking about the same situation. I am envisioning an anchored spring with one end held stationary. The force on the free end will not match f=kx. Neither will it match f=ma.

I am picturing a spring with a mass on one end and a force applied to the other. (Isn't that the model in this thread?) But, whether the other end is fixed or not, the spring will always obey F=kx (where x is the extension)
Actually, anchoring one end of a massless spring is only the same as having a very large mass on it; the acceleration will be zero(ish).
One problem with this whole thread is that we are really dealing with a dynamic situation; some comments acknowledge it and others don't.

 

[jbriggs444]

I am picturing a spring with a mass on one end and a force applied to the other. (Isn't that the model in this thread?) But, whether the other end is fixed or not, the spring will always obey F=kx (where x is the extension)
Actually, anchoring one end of a massless spring is only the same as having a very large mass on it; the acceleration will be zero(ish).
One problem with this whole thread is that we are really dealing with a dynamic situation; some comments acknowledge it and others don't.

My spring has mass, so indeed, we are not talking about the same situation. I'll defer to the OP to decide what interpretation he or she has in mind.

 

[sophiecentaur]

My spring has mass, so indeed, we are not talking about the same situation. I'll defer to the OP to decide what interpretation he or she has in mind.

That's good. So neither of us is a total idiot.
It only goes to show, yet again, how a question that's well defined (diagrams whenever possible) makes answering a lot easier.
I am surprised that, with so much apparent computer literacy around, people can't bring themselves to do more diagrams and, even then, they do them with dreadful Painting packages with a mouse - very hard to make sense of. A picture can be worth a thousand words. ""

 

[Nidum]

I am surprised that, with so much apparent computer literacy around, people can't bring themselves to do more diagrams and, even then, they do them with dreadful Painting packages with a mouse - very hard to make sense of. A picture can be worth a thousand words.

Very true . Good diagrams not only aid understanding of a problem - they often give indication of how to solve a problem or in some cases actually give the solution .

Long forgotten methods like polygons of forces , relative velocity triangles and graphical integration also still have considerable merit in solving practical problems .

 

[Chandra Prayaga]

I am quoting the problem stated by UMath1: "Initially the spring is not stretched or compressed so x=0. If I were to compress this spring I must then apply a nonzero force. By Newton's 3rd Law the spring must also apply a nonzero force, even though kx=0. So my speculation is that the force of the spring is only equal to kx when it is not accelerating." I believe sophiecentaur already answered that question: "You cannot instantly apply a 10N force to the spring unless you displace it instantaneously by10k. You have to start with an infinitesimal reaction force". Now I am only amplifying sophiecentaur's remarks: If you want to apply a non-zero force on the spring, that force starts from zero and gradually increases to 10 N or whatever. The word "gradually" does not mean "slowly". The rise from zero to 10 N could happen in a few microseconds, but it is still gradual. During the entire time, the spring keeps getting more and more stretched (or compressed), and exerts an opposite force equal to what you are applying, and to kx.

 

[UMath1]

Jbriggs444 has the situation I am describing, the spring does have mass and is anchored. All I am saying is I think that the force the spring applies cannot be equal to kx when it is accelerating or experiencing a net force. This is my thinking. If you were to compress the spring an infintesimal amount, dx, you would still have to apply a nonzero force. Sure, the force does not have to be constant, it can gradually increase. But you cannot possibly have any compression when the force is zero. That means that at some time interval the force must be nonzero while x=0. This means that the force of the spring is only kx when it is not accelerating.

 

[sophiecentaur]

It's a pity you didn't make that clear with a diagram, then.

 

[jbriggs444]

Jbriggs444 has the situation I am describing, the spring does have mass and is anchored. All I am saying is I think that the force the spring applies cannot be equal to kx when it is accelerating or experiencing a net force. This is my thinking. If you were to compress the spring an infintesimal amount, dx, you would still have to apply a nonzero force. Sure, the force does not have to be constant, it can gradually increase. But you cannot possibly have any compression when the force is zero. That means that at some time interval the force must be nonzero while x=0. This means that the force of the spring is only kx when it is not accelerating.

A complete analysis of this situation would require us to model the spring as a sequence of little point masses and ideal springs linked together. This is the basic notion of "finite element analysis". If one were to improve the model by making the springs tinier and the masses smaller, the limit yields a differential equation that can describe the behavior of the spring under any particular driving pattern that you might choose. In complete generality that equation would be difficult or impossible to solve analytically. However, with some simplifying assumptions, one of the generic solutions that it can yield is a wave equation -- real springs with mass support waves.

It is very much simpler to assume ideal massless springs. This applies both for practical purposes and teaching purposes. Nobody wants to solve an intractable differential equation when designing a mechanism. And nobody wants to have to teach a first year physics class about finite element analysis, differential equations and numerical methods in order to have them work a homework exercise involving springs.

All that aside, what I have written previously stands. If you have a static spring with mass then f=kx is still valid. But in order to disturb its center of mass, you need an additional increment of f=ma above and beyond this equilibrium force in order to succeed in disturbing the spring's center of mass.

 

[sophiecentaur]

But you cannot possibly have any compression when the force is zero. That means that at some time interval the force must be nonzero while x=0.

As has already been said, you seem to be looking for some sort of paradox and there isn't one here. If you want to question the logic behind this then you would have to look at every relationship in Physics between two variables. If the function that describes the relationship is a continuous one then what happens at the origin shouldn't give you any cause for concern. We're talking cause and effect and an infinitessimal change in the cause will produce an infinitessimal change in the effect. Volts and current. Force and Acceleration, Heat and Temperature rise all accept that you can move away from the origin of your graph without violating anything. The next step in an effect is due to the previous step plus a perturbation due to a cause.
x = x₀ +vt
and many many more like that.