Circumventing Newton's third law through Euler Inertial Forces

[Ghideon]

8 hours ago, John2020 said:

OK. Let us see what comes next.

As we agreed, there is no need for a rotation, the trick is a sloped shape makes the normal force push at an angle and not along the direction of travel of the nut. I have added a pusher* as an illustration, the pusher moves linear and it has the crucial angle at the point of contact with the nut. This has, as agreed, identical effect on the nut as frictionless contact force with the bolt. Ok? We have a variant of the Fig1 device with identical force to push the nut along the the guide rails. This means that whatever pushes the nut dos not have to rotate to have the required effect on the nut and allow for a reactionless drive. As the nut movement and any rotation does not need to be connected for the drive to work, can we remove rotation completely? 

 

 

 

*) As this is not (yet) and engineering topic I'll not add details about the pusher until needed. The focus now is the physical principles and explanations. 

[John2020]

2 hours ago, Ghideon said:

We have a variant of the Fig1 device with identical force to push the nut along the the guide rails. This means that whatever pushes the nut dos not have to rotate to have the required effect on the nut and allow for a reactionless drive.

I have an objection here. If we had friction this would appear along the slope in which the torque force has to overcome in order the nut to advance. This is the point of the misunderstanding, the nut is not being pushed along the guide rails but along the slope in your drawing.

Since nut motion is restricted along the axis of rotation of screw by looking directly on the nut, the contact point appears stationary and what moves is the slope itself along the inclination of the slope.

Edited October 17, 2020 by John2020

[Ghideon]

24 minutes ago, John2020 said:

I have an objection here.

Ok. Why did you not mention that earlier? Why bring this objection up once you realise there is no need for any rotation?

 

25 minutes ago, John2020 said:

If we had friction this would appear along the slope in which the torque force has to overcome in order the nut to advance.

We do not have any friction. I use Fig 1, ideal conditions and only the mechanical analyse you agree on. Does your reactionless drive require friction now? Friction is not an fictitious force by the way, it is real.

 

26 minutes ago, John2020 said:

This is the point of the misunderstanding, the nut is not being pushed along the guide rails but along the slope in your drawing.

Then where along the path of reasoning we have done regarding the detailed analysis do you now disagree? No need to start from the beginning repeating claims, just comment on the thing that you now want to take back.

28 minutes ago, John2020 said:

Since nut motion is restricted along the axis of rotation of screw by looking directly on the nut, the contact point appears stationary and what moves is the slope itself along the inclination of the slope.

Sliding along the slope has no effect, as we agreed on, since there is no friction.

If friction need to be introduced I'll fix that. 

 

[swansont]

10 hours ago, John2020 said:

But in our case the force directed to the center of the circle is not a contact force. For example the screw could have threads but empty inner volume (like a cylinder with threads imprinted on its surface).

You asked for an example and I gave it. You can’t assume force and velocity are in the same direction. Full stop.

 

10 hours ago, John2020 said:

This is also the reason the path I mentioned above is crucial because only contact forces are relevant to the action-reaction principle along with the direction of the velocity that must comply with it.

There is a force.

There are two main issues here: 1) how Newton’s laws apply to the problem, and 2) what your abomination of physics says about the problem

Your version of physics is untested and unsupported. You can’t use it to rebut case #1. 

We know your version says this is reactionless, and we’re telling you what Newton says. If you think Newton does not say that, you have to use Newton’s laws to rebut. Not your crackpot physics. 

Actual physics shows a net force along the axis, and a reaction force. I posted the free-body diagram already.

[John2020]

41 minutes ago, Ghideon said:

Sliding along the slope has no effect, as we agreed on, since there is no friction.

If friction need to be introduced I'll fix that. 

When I wrote "Let us see what comes out of this", I was waiting your analysis and not that I agree in whatever you assume from the beginning. My above reasoning about friction has to do with the following: Having friction or not, based on your drawing motion occurs along the slope that implies a torque force (project upon the slope) must be there, otherwise nut motion cannot be justified at all.

34 minutes ago, swansont said:

You asked for an example and I gave it. You can’t assume force and velocity are in the same direction. Full stop.

You misinterpret/misunderstand my reasoning. I would suggest you to re-read my post. Here is your text "An object moving in a circle at constant speed has a force directed to the center of the circle, perpendicular to the velocity. The reaction force is in the opposite direction (e.g. you pull on a rope, the rope pulls you, the mass swings in a circle.)".

The example with the rope doesn't depict the situation with the nut. As a counter argument I wrote we could have a thin threaded cylinder with no inner volume that means there is no mass (void) along the axis of rotation. Consequently, your assertion/example does not apply (is totally wrong) in my case. The same mistake is made by Ghideon analysis that means the torque force pushes the nut along the slope.

34 minutes ago, swansont said:

There is a force.

There are two main issues here: 1) how Newton’s laws apply to the problem, and 2) what your abomination of physics says about the problem

Your version of physics is untested and unsupported. You can’t use it to rebut case #1. 

We know your version says this is reactionless, and we’re telling you what Newton says. If you think Newton does not say that, you have to use Newton’s laws to rebut. Not your crackpot physics. 

Actual physics shows a net force along the axis, and a reaction force. I posted the free-body diagram already

If you cannot address what I suggested by insisting only on what is familiar to you then please do not label people as crackpots. First think and then apply what you know. 

What is wrong in your approach is the notion that the action-reaction principle is independent of the direction of the velocity of the objects. In others words, you break/misuse Newton's 3rd law by ignorance.

Edited October 17, 2020 by John2020

[swansont]

20 minutes ago, John2020 said:

When I wrote "Let us see what comes out of this", I was waiting your analysis and not that I agree in whatever you assume from the beginning. My above reasoning about friction has to do with the following: Having friction or not, based on your drawing motion occurs along the slope that implies a torque force (project upon the slope) must be there, otherwise nut motion cannot be justified at all.

You misinterpret/misunderstand my reasoning. I would suggest you to re-read my post. Here is your text "An object moving in a circle at constant speed has a force directed to the center of the circle, perpendicular to the velocity. The reaction force is in the opposite direction (e.g. you pull on a rope, the rope pulls you, the mass swings in a circle.)".

The example with the rope doesn't depict the situation with the nut.

I never claimed it did. I was pointing out that you can’t assume that they are, because you’ve made several incorrect assumptions elsewhere.

 

20 minutes ago, John2020 said:

 

As a counter argument I wrote we could have a thin threaded cylinder with no inner volume that means there is no mass (void) along the axis of rotation. Consequently, your assertion/example does not apply (is totally wrong) in my case. The same mistake is made by Ghideon analysis.

What is meant by “mass along the axis of rotation”? Mass is not a vector.

A nut already has no mass on the axis, but that’s where it’s center of mass is located

 

20 minutes ago, John2020 said:

If you cannot address what I suggested by insisting only on what is familiar to you

What I’m familiar with is Newtonian physics. The only way I can address your claims is to tell you what will actually happen, in accordance with physics

I can’t check your model because there is none. I can’t address your experimental results because there are none.

20 minutes ago, John2020 said:

then please do not label people as crackpots.

I did not call you a crackpot. I called your “model” (there’s no actual model) crackpot physics. By this I mean that it’s contrary to accepted physics, and like most crackpot claims, there is no evidence to support it. No model, no experiment. Just repeated assertion.

 

20 minutes ago, John2020 said:

First think and then apply what you know. 

I am. Newtonian physics. What I posted is an accurate account. If you disagree, you have to also use Newtonian physics to show it. Not your crackpot physics.

20 minutes ago, John2020 said:

What is wrong in your approach is the notion that the action-reaction principle is independent of the direction of the velocity of the objects. 

When you say “wrong in my approach”, are you saying it’s inconsistent with your physics, or Newtonian physics?

Since you haven’t properly formulated your physics, I don’t see how you can hold anybody to this. To be wrong, you have to have given us your version of the laws of motion. You haven’t.

[John2020]

42 minutes ago, Ghideon said:

Sliding along the slope has no effect, as we agreed on, since there is no friction.

If friction need to be introduced I'll fix that.

Again, the problem is not the friction but that motion occurs across the slope that presupposes a torque force is at play.

3 minutes ago, swansont said:

What is meant by “mass along the axis of rotation”? Mass is not a vector.

A nut already has no mass on the axis, but that’s where it’s center of mass is located

In a previous post of yours if I remember correctly you claimed the nut is pushed to the right and the reaction pushes the rest of the system over the screw and along the axis of rotation. There are two misconceptions in this approach:

a) Checking the diagram of Ghideon, motion occurs along the thread slope where the thread contact is relevant to forces analysis.

b) I used as a counter argument a thin threaded cylinder having no inner volume (no inner mass). Your assertion of having a reaction force along the axis of rotation, obviously does not hold because assuming there would be a force, it would have no mass to push (in the opposite direction). Consequently, the momentum of the nut is not counteracted by an opposing momentum (since there is no mass to push along the axis of rotation).

[swansont]

20 minutes ago, John2020 said:

 

In a previous post of yours if I remember correctly you claimed the nut is pushed to the right and the reaction pushes the rest of the system over the screw and along the axis of rotation. There are two misconceptions in this approach:

By claiming “misconception” you are implying that it’s not in keeping with Newtonian physics. But you fail to show any discrepancy. You only discuss how it’s not consistent with your version of physics.

 

Quote

a) Checking the diagram of Ghideon, motion occurs along the thread slope where the thread contact is relevant to forces analysis.

I’m talking about my diagram. But “motion” is ambiguous. Linear motion refers to the center of mass of the object, not the surface. Rotational motion can reference the surface.

 

Quote

b) I used as a counter argument a thin threaded cylinder having no inner volume (no inner mass). Your assertion of having a reaction force along the axis of rotation, obviously does not hold because assuming there would be a force, it would have no mass to push (in the opposite direction). Consequently, the momentum of the nut is not counteracted by an opposing momentum (since there is no mass to push along the axis of rotation).

A massless nut is unphysical. You apply a torque and it would have infinite angular acceleration. Apply a force and it has infinite linear acceleration. It has no momentum. So this shows nothing. It’s also changing the example. Please stop doing that. 

To apply the laws of motion, there has to be some mass.

[John2020]

12 minutes ago, swansont said:

I’m talking about my diagram. But “motion” is ambiguous. Linear motion refers to the center of mass of the object, not the surface. Rotational motion can reference the surface.

Your diagram with the N1 is the same as that of Ghideon with N. This means the contact of the thread of the nut moves along the slope (actually the contact is motionless and slope of the screw is the one that moves). Is it correct or not?

Edited October 17, 2020 by John2020

[swansont]

8 minutes ago, John2020 said:

Your diagram with the N1 is the same as that of Ghideon with N. This means the contact of the thread of the nut moves along the slope. Is it correct or not?

Yes. The threads move with respect to the other threads. The diagram will look the same after some time t, except that it will have translated along the x axis (i.e. the rotational axis) The force is normal to the thread surface, meaning there is a component along the axis and a component perpendicular to it (radial)

[John2020]

Just now, swansont said:

Yes. The threads move with respect to the other threads. The diagram will look the same after some time t, except that it will have translated along the x axis (i.e. the rotational axis) The force is normal to the surface, meaning there is a component along the axis and a component perpendicular to it (radial)

I don't understand this point. What force are you addressing now? Isn't that the action-reaction principle applies to contact forces and in our case those contact forces are the normal forces being perpendicular to the slope?

[swansont]

2 minutes ago, John2020 said:

I don't understand this point. What force are you addressing now? Isn't that the action-reaction principle applies to contact forces and in our case those contact forces are the normal forces being perpendicular to the slope?

The normal force, same as before. (I’m just anticipating you claiming that there is no force along the axis, when the x-component of the normal force clearly is)

Do you understand you can express a vector as the sum of its components?

The normal force is perpendicular to the surface. Since it’s a FBD, no reaction forces are shown (only forces acting on the nut are depicted in my drawing, or any  FBD)

 

[John2020]

13 minutes ago, swansont said:

The normal force, same as before. (I’m just anticipating you claiming that there is no force along the axis, when the x-component of the normal force clearly is)

Do you understand you can express a vector as the sum of its components?

Of course, I understand a vector is the sum of its components. OK, regarding what I said about having no normal force on the x-axis (the x-component) was when one does not make the analysis at the slope level. Let's continue, there is an x-component of the normal force. What do you claim as next?

 

edit: removed "negligible due to the helix angle". Wrong assumption.

17 minutes ago, swansont said:

The normal force, same as before. (I’m just anticipating you claiming that there is no force along the axis, when the x-component of the normal force clearly is)

Somewhere has the torque force to be introduced, otherwise we speak nonsense. The torque force is perpendicular to the x-axis.

Edited October 17, 2020 by John2020

[swansont]

1 hour ago, John2020 said:

Of course, I understand a vector is the sum of its components. OK, regarding what I said about having no normal force on the x-axis (the x-component) was when one does not make the analysis at the slope level. Let's continue, there is an x-component of the normal force. What do you claim as next?

This x-component is responsible for the acceleration of the center of mass. F=ma

 

1 hour ago, John2020 said:

Somewhere has the torque force to be introduced, otherwise we speak nonsense. The torque force is perpendicular to the x-axis.

If you wish to obtain a numerical solution, yes, absolutely. The geometry of the screw will relate the torque to the force. But qualitatively speaking, the reason for the force is irrelevant. What we know is there is a net force exerted on the nut, along the x-axis.

We also know the nut exerts a force on the bolt. That’s the reaction force.

[John2020]

11 minutes ago, swansont said:

What we know is there is a net force exerted on the nut, along the x-axis.

We also know the nut exerts a force on the bolt. That’s the reaction force.

But all those forces you are speaking about address a stationary situation (in absence of a torque force) as also apply to a dynamic (with a torque force) one.

Consequently, they cannot affect the motion of the system. What is relevant for the motion of the nut, is just the net torque force (as I did in my analysis in the first post).

[swansont]

Just now, John2020 said:

But all those forces you are speaking about address a stationary situation (in absence of a torque force) as also apply to a dynamic (with a torque force) one.

Consequently, they cannot affect the motion of the system. What is relevant for the motion of the nut, is just the net torque force (as I did in my analysis in the first post).

Excuse me? How can they not affect the motion of the system? If you exert a net force, you get an acceleration. Newton’s first and second laws.

[John2020]

6 minutes ago, swansont said:

Excuse me? How can they not affect the motion of the system? If you exert a net force, you get an acceleration. Newton’s first and second laws

The analysis you made in absence of a torque force does not imply acceleration, thus no net force on x-axis.

Edited October 17, 2020 by John2020

[Ghideon]

 

3 hours ago, John2020 said:

When I wrote "Let us see what comes out of this", I was waiting your analysis and not that I agree in whatever you assume from the beginning.

We started from your device. The initial assumption was that you would understand the analysis and the involved physical principles. But I think I get it, you dodged the question (the bold part) below.

15 hours ago, Ghideon said:

Here is an illustration @John2020, I have reused fig 1 to make it more clear which case we discuss. I have made part of the nut transparent so we see the exact shape inside. The nut is not threaded inside but has one point touching the bolt's thread; the triangular element is touching the bolt. Material fatigue would probably be issue in a real case but should not affect the analyse at this stage.

When the bolt is turning the nut is pushed along the guide rails. Nut is not rotating and pushed by this one point of contact. This setup should behave according to your claims?

The rotating thread and the point of contact on the nut is frictionless; we may now look at action/ reaction pairs etc required to get an explanation? 

No more dodging. We start with your fig1 again and simplify it, step by step and check where the device stops being a reactionless device. 

With the internal of the nut constructed as above, will or will not the device behave as predicted by you; circumventing Newton's third law?

[John2020]

8 minutes ago, Ghideon said:

With the internal of the nut constructed as above, will or will not the device behave as predicted by you; circumventing Newton's third law?

If the above is the original drawing you posted with a single contact point of the nut thread, then yes it is expected as I claim.

As I also mentioned to swansont, there is not net force along the x-axis if we don't t introduce the torque force. What swansont claims is flawed becaused it is based on static analysis (no torque force) that implies no net force on x-axis.

[swansont]

47 minutes ago, John2020 said:

The analysis you made in absence of a torque force does not imply acceleration, thus no net force on x-axis.

There’s a normal force. You agreed there’s a component of this force along the axis.  How does that not imply acceleration? F=ma  If there’s a force, there’s an acceleration. There is no getting around this

I also did NOT say there is no torque.

 

28 minutes ago, John2020 said:

If the above is the original drawing you posted with a single contact point of the nut thread, then yes it is expected as I claim.

As I also mentioned to swansont, there is not net force along the x-axis if we don't t introduce the torque force. What swansont claims is flawed becaused it is based on static analysis (no torque force) that implies no net force on x-axis.

The torque is the source of the force. They will be proportional to each other. We don’t need to know the details to qualitatively analyze the problem.

28 minutes ago, John2020 said:

If the above is the original drawing you posted with a single contact point of the nut thread, then yes it is expected as I claim.

As I also mentioned to swansont, there is not net force along the x-axis if we don't t introduce the torque force. What swansont claims is flawed becaused it is based on static analysis (no torque force) that implies no net force on x-axis.

It’s not based on a static analysis. I conclude there’s an acceleration. That’s not static.

[John2020]

14 minutes ago, swansont said:

It’s not based on a static analysis. I conclude there’s an acceleration. That’s not static.

Addressing just the normal forces along the x-axis without a torque force, there is no net force according to Newton's laws.

A net force due to the torque force is required to overcome the normal force along the x-axis (since there is no additionally a friction force to overcome). In this case we have a non zero net force that accelerates the nut. This is how Newton's laws work.

30 minutes ago, swansont said:

There’s a normal force. You agreed there’s a component of this force along the axis.  How does that not imply acceleration?

Because there is a normal and a counter normal force along the x-axis, otherwise you break Newton's 3rd law between the screw and the nut. One has to introduce a torque force to overcome this balancing of forces on two bodies.

If you may confirm this then we may contnue.

Edited October 17, 2020 by John2020

[Ghideon]

1 hour ago, John2020 said:

If the above is the original drawing you posted with a single contact point of the nut thread, then yes it is expected as I claim.

As I also mentioned to swansont, there is not net force along the x-axis if we don't t introduce the torque force. What swansont claims is flawed becaused it is based on static analysis (no torque force) that implies no net force on x-axis.

Thanks, yes it is the original drawing, based on your design. I think this "No net force along the x-axis" is a crucial point of misunderstanding. Once that is resolved the rest will fall into place. I may have proceeded too fast for you to follow earlier, I'll try a slower pace.  Do you agree that the thread in my drawing of a device, operations according to your claims, is in principle a wedge? The nut is affected, at the point of contact, by only a normal force from the sloped shape.

 

 

Side note: neglect the part below if it disturbs the dialogue with @swansont,

51 minutes ago, John2020 said:

One has to introduce a torque force to overcome this balancing of forces on two bodies.

If that is true then a guillotine is impossible to construct (linear motion, no torque). 

(I have no specific experience from guillotines but a fair share of experience from vintage wood splitting machinery. There are those requiring torque and there are those using linear force. )

 

 

Edited October 17, 2020 by Ghideon
clarified

[John2020]

45 minutes ago, Ghideon said:

If that is true a guillotine is impossible to construct (linear motion, no torque).

It is wrong what I wrote about balancing. I just wanted to stress the fact a torque force is required to overcome the normal force on x-axis (thw same applies from the side of the screw).

45 minutes ago, Ghideon said:

The nut is affected, at the point of contact, by only a normal force from the sloped shape.

I assume you still speak about the case in absence of a torque force, like being idle (no motion is taking place).

45 minutes ago, Ghideon said:

The nut is affected, at the point of contact....

I just would like to clear out something for my own understanding: Are there contact forces (normal mechanical forces) in the construction (screw is contacting the nut) in absence of external forces e.g. gravity and torque forces?

Edited October 17, 2020 by John2020

[swansont]

2 hours ago, John2020 said:

Addressing just the normal forces along the x-axis without a torque force, there is no net force according to Newton's laws.

There are forces and there are torques. There are not “torque forces”

The torque on the bolt is the source of the normal force. If we ever get that far, you can solve for the mathematical relationship between the two.

But: F=ma

Forces determine linear acceleration 

Torques determine angular acceleration 

They are separate calculations. Different degrees of freedom.

Quote

A net force due to the torque force is required to overcome the normal force along the x-axis (since there is no additionally a friction force to overcome). In this case we have a non zero net force that accelerates the nut. This is how Newton's laws work.

“overcome”?

 

Quote

Because there is a normal and a counter normal force along the x-axis, otherwise you break Newton's 3rd law between the screw and the nut. One has to introduce a torque force to overcome this balancing of forces on two bodies.

What is exerting this “counter normal force”? There’s bolt in contact with the nut. What else is there? Invisible unicorns?

( reminder that we are applying Newton’s laws here. Not anything you’ve made up)

 

[John2020]

14 minutes ago, swansont said:

The torque on the bolt is the source of the normal force.

Thank you for clarifying that. I thought since two mechanical parts contact each other then it should be there besides friction (if we had), a mechanical force. 

Let's continue. The force that creates the torque (therefore I called it torque force), equals to the normal force along the x-axis that pushes the nut to the right, correct?

Now acording to action-reaction principle a reaction should appear upon the screw, right?