Circumventing Newton's third law through Euler Inertial Forces

[swansont]

4 hours ago, John2020 said:

F_A is  the action meaning the input force that creates the torque. The N_nut is the corresponding action and the N_screw is the corresponding reaction in Newton's 3rd, these are the contact forces (action-reaction) and not F_A (this is the input force). See wikipedia: https://en.wikipedia.org/wiki/Simple_machine (mechanical advantage). I use the notation _A to show it (perpendicular t0 the axis of rotation that implies is a torque force) in contrast with where it applies in regards to Fig.1-Lower (along the axis of the non-rotating unthreaded rod).

“Input force”? That not something that goes in a free-body diagram. It doesn’t matter why the screw turns. 

You already agreed that the normal force is the action force. The latter should not appear as an item in the drawing; you already accounted for it.

Newton’s second law tells us that the acceleration of an object is the result of the forces acting on that object. Your “input force” is not a force acting on the nut. The screw pushes on the nut - the normal force. There is friction. There is no other force present.

 

 

 

[John2020]

8 hours ago, joigus said:

Don't you understand nobody who knows anything about chess can be cheated with that? It doesn't help if you say: "keep it simple," or "you overlook something."

I completely agree with you. Please @swansont, @Ghideon , @joigus and @pzkpfw , I make the mistake and fall repeatedly in the trap to explain myself about things I don't even address in my work, therefore take attention only to the device which is the main subject of this discussion. I just would like to make clear the following:

-I cannot address GR.

-I cannot address QM and especially the physics behind a quasiparticle. Ignore about what I said about Polariton and stuff (it is not addressed in my work).

-In my work, I just borrowed the definition of a theoretical quasiparticle (that may even doesn't exist) having a very special structure (presupposes that it has a mechanism to move by means of the supposed Euler internal force) in order to distinguish the supposed findings that apply just for the wider framework in Special Relativity (ignore this at the moment), in contrast to the known Einsteinian Special Relativity that applies just for bare particles  e.g. electron (no structure, therefore it cannot be propelled by means of an Euler internal force since the mechanism is absent).

-I may (as far as I can) address classical mechanics, without Hamiltonians and Lagrangians.

-Take Fig.1-Lower and Fig.1-Upper as a Physics exercise given in High School and try to solve the following question: Having as knowns what is given in the drawings (don't involve motors, there are the forces instead there present), would the system (for Lower and Upper) acquire an acceleration? Initially, I  would prefer to address the question in an ideal situation that does not involve friction and normal forces and later with friction and normal forces.

7 hours ago, swansont said:

“Input force”? That not something that goes in a free-body diagram. It doesn’t matter why the screw turns. 

You already agreed that the normal force is the action force. The latter should not appear as an item in the drawing; you already accounted for it.

I apologize @swansont. Sometimes I fall into the trap to assume what I see and describe is clear for everyone, however by mixing unintentionally definitions in the wrong place and time. When there is a conversion mechanism (leadscrew and nut), the input is the torque force (F_A) and the output is the force that appears perpendicular to the input that is responsible for the displacement of the mass m_T, in agreement with the definition of the mechanical advantage (see: https://en.wikipedia.org/wiki/Simple_machine).Additionally, we will ignore at the moment the rotation of the system due to the counter torque by assuming the torque force (input) will be entirely converted into mass m_T displacement.

Yes, regarding the drawing with the normal forces, the normal forces represent the action-reaction pair and F_A the input to the conversion mechanism (leadscrew and nut). 

 

8 hours ago, joigus said:

You cannot have a force producing an acceleration on x_com because you don't have any dependence on x_com in the potential energy. It cannot come from fictitious forces because ficticious forces are obtained by dependence on corresponding coordinates in the kinetic energy.

May I ask you something, because I am not aware in general about the Lagrangian formalism. What you write above is this because the entire system is considered as a point (in Lagrangian) as it is the case for bodies while solving problems, in general? If, yes could we use the known expression about the definition of the COM and the velocity of the COM as follow:

\[\vec{r}_{CM} = \frac{1}{m}\sum_{i=1}^{n}m_i\vec{r}_i \\
\vec{u}_{CM} = \frac{d\vec{r}_{CM}}{dt}=\frac{1}{m}\sum_{i=1}^{n}\frac{d(m\vec{r}_{i})}{dt} \\
\vec{p} = m\vec{u}_{CM} = \sum_{i=1}^{n} \vec{p}_i\]

With the above we may address all the parts of the system separately that may help identify which part could cause a potentially change in the COM of the whole system. Do you agree with this or am I wrong?

Edited October 13, 2020 by John2020

[John2020]

Correction: "the torque force (input) will be entirely converted into mass m_T change in momentum."

[Ghideon]

1 hour ago, John2020 said:

-Take Fig.1-Lower and Fig.1-Upper as a Physics exercise given in High School and try to solve the following question: Having as knowns what is given in the drawings (don't involve motors, there are the forces instead there present), would the system (for Lower and Upper) acquire an acceleration? Initially, I  would prefer to address the question in an ideal situation that does not involve friction and normal forces and later with friction and normal forces.

Ok let's do it as a High School exercise.
Answer: As we have learned from basic physics there is the law of conservation of momentum. We also know from Fig.1 that no external forces are acting on the system. This means that the system in Fig.1 will remain at rest; there will be no acceleration. 

 

1 hour ago, John2020 said:

I cannot address GR.

But you do just that all the time in your posts.

Edited October 13, 2020 by Ghideon

[swansont]

3 hours ago, John2020 said:

I apologize @swansont. Sometimes I fall into the trap to assume what I see and describe is clear for everyone, however by mixing unintentionally definitions in the wrong place and time. When there is a conversion mechanism (leadscrew and nut), the input is the torque force (F_A) and the output is the force that appears perpendicular to the input that is responsible for the displacement of the mass m_T, in agreement with the definition of the mechanical advantage (see: https://en.wikipedia.org/wiki/Simple_machine).Additionally, we will ignore at the moment the rotation of the system due to the counter torque by assuming the torque force (input) will be entirely converted into mass m_T displacement.

That’s a description of how the device works. It is not, however, a physics analysis.

Quote

Yes, regarding the drawing with the normal forces, the normal forces represent the action-reaction pair and F_A the input to the conversion mechanism (leadscrew and nut). 

F_A is not a force the screw exerts on the nut. It is exerted on the screw, so it has no part in the analysis of the motion of the nut.

There is friction, and there is the normal force, which, when accounting for the entire thread, is along the axis of the screw. It’s not reactionless, and is why the mass moves.

 

 

[pzkpfw]

@John2020 , maybe this would be a useful exercise for you?

Take a look through the threads by "LB7" over at this other forum: https://www.thenakedscientists.com/forum/index.php?topic=80651.msg615121

He also thinks he can beat conservation laws. He makes thread after thread which boil down to: straight lines are different to curves ... then with some magic (I've really never been able to see how he expects to get it in or out) he proclaims that he gets unbalanced non-conserved energy.

The exercise is: go through his threads and see if his explanations make any sense to you. Compare to your own.

Edited October 13, 2020 by pzkpfw

[John2020]

1 hour ago, swansont said:

F_A is not a force the screw exerts on the nut. It is exerted on the screw, so it has no part in the analysis of the motion of the nut.

There is friction, and there is the normal force, which, when accounting for the entire thread, is along the axis of the screw. It’s not reactionless, and is why the mass moves.

OK, however on the drawing with the thread I shared, the normal forces pair has an angle in regards to the rotation axis because of the inclination of the thread. It implies a small reactionless force will appear while the screw rotates by advancing m_T to the right.

Increasing the length of the lead will result to more reactionless force. Isn't that confirmed by the drawing of the thread?

Alternarively, when we replace the nut and the screw with their magnetic equivalents, there we have no contact forces. Wouldn't that increaee the available reactionless force?

34 minutes ago, pzkpfw said:

The exercise is: go through his threads and see if his explanations make any sense to you. Compare to your own.

My own is much clearer as concept even if we don't agree. I couldn't understand of what he is talking about.

2 hours ago, Ghideon said:

But you do just that all the time in your posts.

I didn't start the subject with GR, you did and I fall into it.

[swansont]

2 minutes ago, John2020 said:

OK, however on the drawing with the thread I shared, the normal forces pair has an angle in regards to the rotation axis because of the inclination of the thread. It implies a small reactionless force will appear while the screw rotates by advancing m_T to the right.

I addressed this already. You have shown only a single point of contact. When you integrate over 360 degrees (one full traverse of the circle), the radial components cancel. Also it’s not reactionless.

2 minutes ago, John2020 said:

Increasing the length of the lead will result to more reactionless force. Isn't that confirmed by the drawing of the thread?

There is no reactionless force. The nut exerts a reaction force on the screw.

2 minutes ago, John2020 said:

Alternarively, when we replace the nut and the screw with their magnetic equivalents, there we have no contact forces. Wouldn't that increaee the available reactionless force?

There is no reactionless force. The magnetic materials will be exerting the action and reaction forces. They would be magnetic forces in this case.

[John2020]

3 minutes ago, swansont said:

I addressed this already. You have shown only a single point of contact. When you integrate over 360 degrees (one full traverse of the circle), the radial components cancel. Also it’s not reactionless.

How can they be cancelled from the moment the normal pair appears a constant angle even when you integrate 360° in regards to the axis of rotation?

[swansont]

2 minutes ago, John2020 said:

How can they be cancelled from the moment the normal pair appears a constant angle even when you integrate 360° in regards to the axis of rotation?

You can spin a nut freely on a bolt. there is no net radial force

[John2020]

Just now, swansont said:

You can spin a nut freely on a bolt. there is no net radial force

I cannot follow you. Do you speak about the F_A or about the normal forces pair,?

[swansont]

4 minutes ago, John2020 said:

I cannot follow you. Do you speak about the F_A or about the normal forces pair,?

Since F_A is not a force the bolt exerts on the nut, I am certainly not talking about it. And you should stop, too. It’s wrong to include it.

 

[John2020]

4 minutes ago, swansont said:

Since F_A is not a force the bolt exerts on the nut, I am certainly not talking about it. And you should stop, too. It’s wrong to include it.

 

Then you ignore the normal forces pair is all the time at a constant angle with the rotation axis that means they are not along the rotation axis that implies a net force that pushes the nut appears reactionless.

Edited October 13, 2020 by John2020

[Ghideon]

51 minutes ago, John2020 said:

I didn't start the subject with GR, you did and I fall into it.

I'll try to clarify. Newtonian physics and General Relativity (GR) have common parts. Newtonian mechanics make reasonable predictions within the theory's applicability such as at velocities v << c or invariant mass m >> 0. You claim that your reactionless drive possesses abilities* that are impossible in Newton's physics and you try to support those claims by (incorrectly) applying Newtonian physics. Unfortunately for your device the claims are equally impossible in GR and leads to more contradictions. Newtons is a useful approximation of GR under the circumstances presented which means that whether you intend it or not you keep rejecting GR as well as Newton. I thought GR is a good way to think outside the box for while to see from a fresh angle how your idea is doomed within physics as we know it; sorry that I proposed an analyse and discussion is outside your level of knowledge. 

I'll leave GR for a while and move back to the internals of the device again.

 

*) For instance an earlier post about gravity comment=1155341

 

Edited October 13, 2020 by Ghideon

[swansont]

11 minutes ago, John2020 said:

Then you ignore the normal forces pair is all the time at a constant angle with the rotation axis that means they are not along the rotation axis that implies a net force that pushes the nut appears reactionless.

The net force is along the axis. I will work on a drawing to show why (I’ve realized my earlier explanation isn’t quite right)

The fact that there is a component along the axis is sufficient to prove it’s not reactionless. The normal forces of the nut and bolt are an action-reaction force pair. 

 

[John2020]

2 minutes ago, swansont said:

The normal forces of the nut and bolt are an action-reaction force pair. 

I am saying this from the moment I shared the drawing with the thread. Furthermore, the pair appears a constant angle with the rotation axis (because of the thread inclination) while the translation screw rotates.

10 minutes ago, Ghideon said:

sorry that I proposed an analyse and discussion is outside your level of knowledge. 

No worries. It is perfectly understood for someone knowledgeable in Physics to use modern argumentation (see GR) in contrast to old fashioned non fascinating, boring and a closed chapter like classical mechanics. Well, from my side I explore just the possibility whether Newton's laws are incomplete (not discarded, just incomplete) or not. I would be totally careless to dismiss proven science. 

[Ghideon]

17 minutes ago, John2020 said:

Then you ignore the normal forces pair is all the time at a constant angle with the rotation axis that means they are not along the rotation axis that implies a net force that pushes the nut appears reactionless.

Attempting an analogy: Put a frictionless screw with a in a vertical postion. Put a nut on the top end. Gravity will make the nut start rotating down the screw. There is no "reactionless" forces involved even if nothing mechanical is acting on the nut.

 

9 minutes ago, John2020 said:

I would be totally careless to dismiss proven science. 

Yes, you dismiss GR while saying that you do not understand GR, "careless" may be a suitable word to describe that. 

(Personally I try another approach when I happen to post something careless and/or dismissive about mainstream science. Other members steps in and posts corrections and I get curios about the thing i obviously did not understand as good as I thought I did. Result: More knowledge)

[John2020]

2 minutes ago, Ghideon said:

There is no "reactionless" forces involved even if nothing mechanical is acting on the nut.

I am aware about this analogy. Well this is not exactly the same as in my apparatus. The nut is being affected by a constant external gravitational force that implies constant evolution of nut around the screw that means constant displayment speed.

In my case is accelerated displacement. The displacement force is not rectilinear (as we see with real contact forces) and that was the reason I called it fictitious and more accurately an artificial fictitious force because of the translation screw construction that implies an helix trajectory is at play (not rectilinear motion).

However, the question remains: Where appear the normal forces pair (action-reaction) in your analogy and what is their angle in regards to the rotation axis or the evolution axis (your case).

[Ghideon]

4 minutes ago, John2020 said:

The nut is being affected by a constant external gravitational force that implies constant evolution of nut around the screw that means constant displayment speed.

The gravitational force is constant*. The velocity is not constant, the nut will accelerate down the screw. The acceleration will depend on the nut shape.

 

 

*) the difference in height is neglected

[John2020]

26 minutes ago, Ghideon said:

The gravitational force is constant*. The velocity is not constant, the nut will accelerate down the screw. The acceleration will depend on the nut shape.

No it's not. Please check the mechanical advantage definition: https://en.wikipedia.org/wiki/Simple_machine

This is the reason I used a varying angular velocity (accelerating) which is a requirement for the accelerating mass transfer that would imply in an accelerating change of the COM and eventually acceleration of the system as caused by the assumed reactionless force (fictitious in nature, it is not a rectilinear real force since it emerges through the evolution of the nut around the axis of rotation of the translation screw (when one uses your analog or my own where here the screw rotates).

Edited October 13, 2020 by John2020

[joigus]

15 hours ago, John2020 said:

[...]

May I ask you something, because I am not aware in general about the Lagrangian formalism. What you write above is this because the entire system is considered as a point (in Lagrangian) as it is the case for bodies while solving problems, in general? If, yes could we use the known expression about the definition of the COM and the velocity of the COM as follow:

[...]

With the above we may address all the parts of the system separately that may help identify which part could cause a potentially change in the COM of the whole system. Do you agree with this or am I wrong?

The key to the Lagrangian formalism is not to consider the composite system as a point. The key is to think about variables in an analytic manifold rather than vectors and directions. Positions of all parts of the system become points in an n-dimensional surface. They are called "configurations."

Generally there are two ways to proceed. One is to identify all the independent variables to specify a configuration and write down the evolution equations, which involve so-called "generalised momenta" and "generalised forces." The other is to ignore dependences (constraints) and solve the constraints later by a method called Lagrange multipliers.

In your case you would need 3 variables to specify where your system as a whole is. Then, internal variables to specify where every internal part is with respect to this centre (typically the COM.) Because no internal interaction energy depends on where the system is, no energy exchange depends on where you place it, there can be no "generalised force" on the COM, and thereby no second derivative of it involved in the dynamics.

I don't have time to be more explicit now. Maybe tonight, or tomorrow.

 

Edited October 14, 2020 by joigus

[Ghideon]

11 hours ago, John2020 said:

No it's not. Please check the mechanical advantage definition: https://en.wikipedia.org/wiki/Simple_machine

What is not? I am discussing the following comment regarding a nut rotating down a vertical screw due to gravitational force:

11 hours ago, John2020 said:

The nut is being affected by a constant external gravitational force that implies constant evolution of nut around the screw that means constant displayment speed.

Please explain in detail what you mean by "constant evolution of nut around the screw that means constant displayment speed" in the example I gave. When we place the nut on top of the vertical screw the initial velocity is zero. A constant velocity means that the velocity remains zero and the nut will not move. That can't be what you mean, the nut will start to rotate down the screw (=acceleration) and the vertical velocity as well as the rate of rotation will increase.

 

Side note 1: We still discuss an ideal situation with no friction and where force of gravity is constant. In reality there will be friction, vibrations, air resistance etc that limits the maximum velocity. In reality, for a long enough screw, there will me a maximum velocity reached and the nut may continue to move at that constant maximum velocity until it reaches the end of the screw. But again, that is not part of the analysis and adds unnecessary details 

Side note 2: This is a more interesting part of the discussion; locating a specific point where we disagree that allows for a detailed analysis within the realms of ordinary physics.)

 

[swansont]

15 hours ago, John2020 said:

Then you ignore the normal forces pair is all the time at a constant angle with the rotation axis that means they are not along the rotation axis that implies a net force that pushes the nut appears reactionless.

If there was a net radial force that pushes, the nut would accelerate in that direction. Why doesn't it? Because it pushes up against the bolt, and the bolt pushes back with another normal force.

(edit: bolt is blue, nut is orange)

 

The radial component of N₁ cancels with N₂, leaving no radial force. There is a net force along the axis.

(I am ignoring friction, which will balance this out if there is no acceleration)

N₁ and N₂ each have a reaction force from the nut acting on the bolt. If there are reaction forces (and there always are) then there is nothing "reactionless" about this system.

 

14 hours ago, John2020 said:

No it's not. Please check the mechanical advantage definition: https://en.wikipedia.org/wiki/Simple_machine

This is the reason I used a varying angular velocity (accelerating) which is a requirement for the accelerating mass transfer that would imply in an accelerating change of the COM and eventually acceleration of the system as caused by the assumed reactionless force (fictitious in nature, it is not a rectilinear real force since it emerges through the evolution of the nut around the axis of rotation of the translation screw (when one uses your analog or my own where here the screw rotates).

Mechanical advantage has nothing to do with this. If there is a net force, there will be acceleration (Newton's 1st and/or 2nd law. Take your pick) Gravity in this case is basically doing the job of your motor in your machine, except it's the nut that's rotating.

[John2020]

1 hour ago, Ghideon said:

Please explain in detail what you mean by "constant evolution of nut around the screw that means constant displayment speed" in the example I gave. When we place the nut on top of the vertical screw the initial velocity is zero.

You are right, my mistake.

27 minutes ago, swansont said:

Why doesn't it? Because it pushes up against the bolt, and the bolt pushes back with another normal force.

I will come on this later. In the meantime your new drawing doesn't seem to be correct. The opposing Normal forces must be collinear (N1 and N2 are not).

Later in the evening. I go back to work now.

[swansont]

34 minutes ago, John2020 said:

You are right, my mistake.

I will come on this later. In the meantime your new drawing doesn't seem to be correct. The opposing Normal forces must be collinear (N1 and N2 are not).

Later in the evening. I go back to work now.

They are not opposing normal forces, and they are not collinear because the surfaces are at different orientations.

They are normal forces from different contact points of the nut and bolt. One is from the flat part of thread-on-thread contact (N₁) and the other is on the edge of the thread, contacting the housing of the nut (N₂)_. They are both forces exerted by the bolt on the nut, and each has a reaction force exerted by the nut (these are not shown, because I am describing the forces exerted on the nut)

The radial component of N₁ is balanced by N₂. We know this has to be the case because the nut does not accelerate in the vertical direction. The axial component of N₁ remains, and is the reason the nut accelerates.