Jump to content
John2020

Circumventing Newton's third law through Euler Inertial Forces

Recommended Posts

4 minutes ago, John2020 said:

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?

No. A force along the x-axis can’t create a torque. The torque will come from the perpendicular component. 

 

4 minutes ago, John2020 said:

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

Yes. 

Share this post


Link to post
Share on other sites
1 hour ago, swansont said:

No. A force along the x-axis can’t create a torque. The torque will come from the perpendicular component. 

I am not saying this. We have a translation mechanism that implies the perpendicular component is converted to an induced force along the x-axis that is equals to the Normal force along the x-axis (the concept of the simple machine). If not then, what is the cause behind the Normal force on x-axis? Isn't the Normal force on the x-axis pushing the nut or not?

Edited by John2020

Share this post


Link to post
Share on other sites
3 hours ago, John2020 said:

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).

A torque or a force? On the bolt or the nut? Torque can rotate the bolt. Torque does not move the nut
-The nut can't rotate in fig1
-The nut is restricted by two guiding rods (that I assume is frictionless).
-Torque, for the nut, is useful when analysing the force applied to the guiding rods. Torque has nothing to do with the nut moving (sliding along the guiding rods)

 

3 hours ago, John2020 said:

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

I speak of the picture based directly on your Fig 1 in opening post; bolt is rotating due to an applied torque. 

image.png

 

3 hours ago, John2020 said:

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?

Assuming absence of external forces and assuming the bolt does not accelerate then in an ideal system there is no contact force. I assume we speak of macroscopic forces, not any attraction on molecular level or similar.  

Share this post


Link to post
Share on other sites
8 minutes ago, Ghideon said:

A torque or a force? On the bolt or the nut? Torque can rotate the bolt. Torque does not move the nut
-The nut can't rotate in fig1
-The nut is restricted by two guiding rods (that I assume is frictionless).
-Torque, for the nut, is useful when analysing the force applied to the guiding rods. Torque has nothing to do with the nut moving (sliding along the guiding rods)

Yes, this is all clear. We don't have to repeat ourselves. The torque on the bolt induces a counter torque on on the nut although this does not rotate, however as we have already said the contact of the thread ascribes a helix in opposite direction of that of the bolt.

11 minutes ago, Ghideon said:

Torque has nothing to do with the nut moving (sliding along the guiding rods)

Here the cause is the induced counter torque and due to the translation screw mechanism, it makes sense the perpendicular component a.k.a counter torque to be converted to the Normal force on the x-axis. I asked swansont the same (see above). If the converted counter torque to Normal force on the x-axis is not the cause of the nut motion then what is?

 

Share this post


Link to post
Share on other sites
1 hour ago, John2020 said:

I am not saying this.

You did say this. “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”

They are not equal. One is N sin(pitch angle) while the other is N cos(pitch angle).

(they will be equal only if your threads are at 45 degrees)

 

Quote

We have a translation mechanism that implies the perpendicular component is converted to an induced force along the x-axis that is equals to the Normal force along the x-axis (the concept of the simple machine). If not then, what is the cause behind the Normal force on x-axis? Isn't the Normal force on the x-axis pushing the nut or not?

Yes, the normal force component along the x-axis is what gives the linear acceleration. The force that creates the torque comes from the normal force component acting along the y-axis.

 

 

Share this post


Link to post
Share on other sites
2 hours ago, swansont said:

No. A force along the x-axis can’t create a torque. The torque will come from the perpendicular component. 

Correction on the previous post:

I am not saying this. We have a translation mechanism that implies the perpendicular component (in our case the counter torque and not that applies on the bolt) is converted to an induced force along the x-axis that is equals to the Normal force along the x-axis (the concept of the simple machine). If not then, what is the cause behind the Normal force on x-axis? Isn't the Normal force on the x-axis pushing the nut or not?

Note: We know the nut does not rotate, however the counter torque is responsible for ascribing the thread contact of the nut a counter clockwise evolution in contrast with the applied torque on the bolt (clockwise).

6 minutes ago, swansont said:

The force that creates the torque comes from the normal force component acting along the y-axis.

I think it is backwards, otherwise it does not make sense. In absence of torque there is no Normal y-component. So, on applied torque implies Normal y-component.

 

Edited by John2020

Share this post


Link to post
Share on other sites
16 minutes ago, John2020 said:

Yes, this is all clear. We don't have to repeat ourselves. The torque on the bolt induces a counter torque on on the nut although this does not rotate, however as we have already said the contact of the thread ascribes a helix in opposite direction of that of the bolt.

Yes, but we agreed the helix shape of the threads of the nut does not matter since it works just as well with one point of contact between the nut and the bolt. I assume that "ascribes a helix" is a shape, not a movement.

16 minutes ago, John2020 said:

Here the cause is the induced counter torque and due to the translation screw mechanism, it makes sense the perpendicular component a.k.a counter torque to be converted to the Normal force on the x-axis. I asked swansont the same (see above). If the converted counter torque to Normal force on the x-axis is not the cause of the nut motion then what is?

The nut can't rotate. You have guiding rods in fig 1. Unless guiding rod(s) break there is balance between forces so no rotation takes place. It the nut was only affected by torque around the axis of rotation of the bolt the nut would remain stationary. Torque has zero effect on the nut movement and cant move the nut due to the construction in fig 1. Would you like an analogy to help you see? 

 

(I now realise how you failed to follow my previous analysis that went from fig1 via a sloped shape pushing at the nut, we need to explore torque vs force some more)

Edited by Ghideon

Share this post


Link to post
Share on other sites
Just now, John2020 said:

Correction on the previous post:

I am not saying this. We have a translation mechanism that implies the perpendicular component (in our case the counter torque and not that applies on the bolt) is converted to an induced force along the x-axis that is equals to the Normal force along the x-axis (the concept of the simple machine). If not then, what is the cause behind the Normal force on x-axis? Isn't the Normal force on the x-axis pushing the nut or not?

The motion of the bolt exerts both a force and a torque on the nut. It exerts a normal force, because there is a surface in contact. The force has has x and y components.

The force in the x direction causes an acceleration. There is no other axial force on the nut.

Just now, John2020 said:

Note: We know the nut does not rotate, however the counter torque is responsible for ascribing the thread contact of the nut a counter clockwise evolution in contrast with the applied torque on the bolt (clockwise).

The counter-torque on the nut ensures no rotation, but has no effect on the motion in the x direction. 

Share this post


Link to post
Share on other sites
21 minutes ago, swansont said:

Yes, the normal force component along the x-axis is what gives the linear acceleration. The force that creates the torque comes from the normal force component acting along the y-axis.

Then from the action-reaction principle:

a) an opposing normal force component on x-axis is exerted on the screw

b) an opposing normal force component on y-axis is exerted on the screw due to the applied torque upon the nut

c) Since (a) and (b) are manifested because of the torque on the nut then, the reaction should be of the same type, namely a counter torque upon the screw

Are the above statements correct?

Note: The counter torque (as  above has nothing to do with the torque of the system but that of the screw. We assume there is no system rotation (this is a mistake I have to correct in my paper. So we assume ideally the system will not rotate and the torque will be entirely converted to nut displacement as I declared this in my first posts) while the torque applies due to the conservation of angular momentum.

11 minutes ago, Ghideon said:

Yes, but we agreed the helix shape of the threads of the nut does not matter since it works just as well with one point of contact between the nut and the bolt. I assume that "ascribes a helix" is a shape, not a movement.

We have gone all through these. You understand very well what I mean.

11 minutes ago, Ghideon said:

The nut can't rotate. You have guiding rods in fig 1. Unless guiding rod(s) break there is balance between forces so no rotation takes place. It the nut was only affected by torque around the axis of rotation of the bolt the nut would remain stationary. Torque has zero effect on the nut movement and cant move the nut due to the construction in fig 1. Would you like an analogy to help you see? 

This is all perfectly clear but you still miss the point. The mistake from my side (that someone from you pointed out) is when on applies a torque, the entire system will rotate. I have to fix this on my paper (I will explain in another occasion). Then, I said, let us for a moment assume the system will not rotate when we apply the torque on the nut and this torque will be totally converted to nut displacement. 

All other you mention are perfectly clear and I understand how the mechanism works.

Edited by John2020

Share this post


Link to post
Share on other sites
15 minutes ago, John2020 said:

Then, I said, let us for a moment assume the system will not rotate when we apply the torque on the nut and this torque will be totally converted to nut displacement. 

Torque around the x axis on the nut can't be converted to nut displacement, that's obvious. You are confusing torque and force. But you are on the right track separating into x and y components.

 

 

15 minutes ago, John2020 said:

All other you mention are perfectly clear and I understand how the mechanism works.

If that were the case this thread would not exist. At least it would have no claims abut circumventing Newton. 

 

(I will let swansont answer the questions that were directed at swansont )

Edited by Ghideon

Share this post


Link to post
Share on other sites
16 minutes ago, Ghideon said:

If that were the case this thread would not exist. At least it would have no claims abut circumventing Newton. 

I admit I didn't communicate the entire idea as most would expect as also I made several mistakes while defining some things.

However, I would be interested to have your view on the above three points a,b,c, too.

Edited by John2020

Share this post


Link to post
Share on other sites
29 minutes ago, John2020 said:

Then from the action-reaction principle:

a) an opposing normal force component on x-axis is exerted on the screw

b) an opposing normal force component on y-axis is exerted on the screw due to the applied torque upon the nut

c) Since (a) and (b) are manifested because of the torque on the nut then, the reaction should be of the same type, namely a counter torque upon the screw

Are the above statements correct?

Looks OK

29 minutes ago, John2020 said:

Note: The counter torque (as  above has nothing to do with the torque of the system but that of the screw. We assume there is no system rotation (this is a mistake I have to correct in my paper. So we assume ideally the system will not rotate and the torque will be entirely converted to nut displacement as I declared this in my first posts) while the torque applies due to the conservation of angular momentum.

You can’t “entirely convert” the torque to nut displacement. Nut displacement is from the axial force.

 

Share this post


Link to post
Share on other sites
1 hour ago, John2020 said:

a) an opposing normal force component on x-axis is exerted on the screw

b) an opposing normal force component on y-axis is exerted on the screw due to the applied torque upon the nut

c) Since (a) and (b) are manifested because of the torque on the nut then, the reaction should be of the same type, namely a counter torque upon the screw

a) seems true; a nut being accelerated along the positive x-axis exerts a reaction force on the screw in negative x-direction

b) Seems ok, guide rails prevents rotation so the screw is affected by an equal and opposite force in the y direction.

c) false, a and b are not manifest because of torque in the nut. They manifest because of torque on the screw resulting in a normal force affecting the nut. 

 

edit: I notice my comment regarding c) differs from @swansont

It is possible that I misinterpreted c), I will edit or follow up

 

 

Side note, primarily at Swansont: 

(Disclaimer: Section below may lack rigour, it's kind of "garage talk" than science & physics)

37 minutes ago, swansont said:

You can’t “entirely convert” the torque to nut displacement.

I think that is a good way to put it in mechanical terms, I'll reuse that when needed! (I sometimes discuss mechanical stuff with individuals that have no formal education but extensive practical knowledge). I think the other way around holds as well, you can't "entirely convert" all nut displacement to torque. I think both these "impossible conversions" would require an impossibly steep thread angle; 90 degrees (grooves parallel to centre of the bolt, no spiral) 

 

Edited by Ghideon
x-post w swansont

Share this post


Link to post
Share on other sites
25 minutes ago, swansont said:

You can’t “entirely convert” the torque to nut displacement. Nut displacement is from the axial force.

Ok  we have already clarified this with the x and y components of Normal forces. 

25 minutes ago, swansont said:

Looks OK

1 So we confirm there is a translation mechanism where a torque (over the x,y normal forces components) is converted to nut displacement.

2.On the other hand, the reaction is a counter torque (again over the analysis with the x,y components of the normal force) exerted upon the screw, that does not result in a mass being displaced (as we have with the nut) since nothing is being displaced in that direction (opposite to nut displacement).

Is there any objection on the above?

 

 

17 minutes ago, Ghideon said:

It is possible that I misinterpreted c), I will edit or follow up

We speak now about the reaction of the screw while the nut is being displaced.

Edited by John2020

Share this post


Link to post
Share on other sites
43 minutes ago, John2020 said:

Ok  we have already clarified this with the x and y components of Normal forces. 

1 So we confirm there is a translation mechanism where a torque (over the x,y normal forces components) is converted to nut displacement.

2.On the other hand, the reaction is a counter torque (again over the analysis with the x,y components of the normal force) exerted upon the screw, that does not result in a mass being displaced (as we have with the nut) since nothing is being displaced in that direction (opposite to nut displacement).

Is there any objection on the above?

 

Yes. If the nut exerts a force on the bolt, how does the bolt not get displaced? 

 

 

59 minutes ago, Ghideon said:

 

c) false, a and b are not manifest because of torque in the nut. They manifest because of torque on the screw resulting in a normal force affecting the nut. 

There are multiple torques. One causes the bolt to turn, one exerted on the nut as you show. And the torque from the guide bars on the nut, as well as the nut on the bolt.

So I think you and I are looking at different ones

 

 

Share this post


Link to post
Share on other sites
7 minutes ago, swansont said:

Yes. If the nut exerts a force on the bolt, how does the bolt not get displaced?

Because the normal forces that are exerted on the bolt result just in its rotation. Furthermore, there is no mass coupled (as we have with the nut) in the screw that will result in an opposite direction to nut displacement.

Share this post


Link to post
Share on other sites
38 minutes ago, John2020 said:

Because the normal forces that are exerted on the bolt result just in its rotation.

The force along its axis can’t exert a torque.

Quote

Furthermore, there is no mass coupled (as we have with the nut) in the screw that will result in an opposite direction to nut displacement.

The bolt is massless? See my earlier comments about that. Otherwise, there’s mass, which must accelerate in accordance with F=ma

 

 

Here’s a little thought experiment. The mass of the nut is very, very, very large - we affix it to something with the mass of the earth. Or a person, holding the nut. We exert a torque on the bolt.

You really think the bolt won’t move? 

Share this post


Link to post
Share on other sites
31 minutes ago, swansont said:

So I think you and I are looking at different ones

It is all those you mention without the torque from the guiding lines.

2 minutes ago, swansont said:

You really think the bolt won’t move? 

I don't understand where you are pointing with the thought experiment.

10 minutes ago, swansont said:

The bolt is massless? See my earlier comments about that. Otherwise, there’s mass, which must accelerate in accordance with F=ma

When the mass of the bolt is very small compared to the nut and the system as a whole, what will the system do while nut is being displaced?

Share this post


Link to post
Share on other sites
52 minutes ago, swansont said:

There are multiple torques. One causes the bolt to turn, one exerted on the nut as you show. And the torque from the guide bars on the nut, as well as the nut on the bolt.

So I think you and I are looking at different ones

Yes I think we do but I hope that will not hinder further progress. I will follow along on your line of thought in next post(s).

But I'll dare to try to clarify my reasoning. Due to my engineering background and the stated "torque, counter torque" I began at the asymmetry for how the torques comes into play (order of events). The torque of the screw (applied my motor) will impose torque on the nut resulting in torque on the guide rails and displacement of the nut. The complete device was not allowed to rotate so we can't impose torque by spinning the machine with accelerating external force. So torque applied to the nut by something else than the engine would be kind of a wrench, held by someone riding along on the device,  trying to twist the nut. The nut is not turning due to the guide rails, no additional torque is translated to the engine. In other words, additional torque from the engine (for instance by increased power) will add torque to the nut and hence to the guide bars. But the opposite does not apply, trying to twist the nut against the rotation of the engine would not increase the load on the engine, it affects only the force on the guide rails. 

On the other hand, pressing the nut in the x direction, given zero friction and not to steep thread angle, would impose torque on the screw and hence on the engine. An engine incapable of any torque could spin backwards in this case. This asymmetry of the setup allows, in my opinion, for different answers and interpretations regarding torque in the questions a-c depending on what one consider to be cause and action when saying torque and counter torque. 

Note 1: The above is not a critique against Swansont answers, It's Ghideon's mechanical experiences ramblings only. It does not state I'm right, it tries to clarify how I had a different opinion at that point. It should not get in the way of ongoing analysis of device in Fig 1.

Note 2: @John2020 any contradictions resulting from the above is due to different interpretation or misinterpretation of points  a) - c) on my part. It has nothing to do with opinions regarding the device or regarding Newton. It's a side note from practical mechanics / engineering, not so much physics.

 

Edited by Ghideon

Share this post


Link to post
Share on other sites
1 hour ago, Ghideon said:

any contradictions resulting from the above is due to different interpretation or misinterpretation of points  a) - c) on my part.

Thank you for your practical view on this subject. No worries we just make a theoretical discussion by removing all those technical matters that make the entire undertaking more complex and more challenging.

The main focus is the action reaction principle and momentum conservation on a pure ideal device on a theoretical level.

1 hour ago, swansont said:

The force along its axis can’t exert a torque.

I would like to add my view on the above:

If the normal force on x-axis presupposes the presence of nut then,from the reaction side (upon the screw) we shouldn't have a reaction on x-axis.

An additional argument that may support the above is, in absence of nut there cannot be a normal force component on x-axis. Consequently, in absence of nut we have just the rotation of screw as result and nothing else.

Note: I understand the above breaks the action reaction symmetry, however if the above arguments are true then, we have to see what is wrong in our analysis.

Edited by John2020

Share this post


Link to post
Share on other sites
1 hour ago, John2020 said:

 

If the normal force on x-axis presupposes the presence of nut then,from the reaction side (upon the screw) we shouldn't have a reaction on x-axis.

Based on what physics?

Quote

An additional argument that may support the above is, in absence of nut there cannot be a normal force component on x-axis. Consequently, in absence of nut we have just the rotation of screw as result and nothing else.

If there’s no nut, there’s no action force, so of course there’s no reaction. That doesn’t support your assertion.

 

 

2 hours ago, John2020 said:

It is all those you mention without the torque from the guiding lines.

I don't understand where you are pointing with the thought experiment.

Disproving your thesis. Your claim is there is no force on the bolt, and only the nut moves. 

Quote

When the mass of the bolt is very small compared to the nut and the system as a whole, what will the system do while nut is being displaced?

Newton tells you the bolt will move a lot more than the nut. Your claim is the bolt won’t move.

Try it.

How do you loosen or remove a bolt/screw if there’s no force on it?

Share this post


Link to post
Share on other sites
7 hours ago, John2020 said:

Thank you for your practical view on this subject. No worries we just make a theoretical discussion by removing all those technical matters that make the entire undertaking more complex and more challenging.

The main focus is the action reaction principle and momentum conservation on a pure ideal device on a theoretical level.

Thanks. Side note: applying some practical thinking to your claims is a quick way to do an initial analyse that intuitively tells how the claims cant be correct. That intuition is of course not very useful as evidence, it's just useful to quickly find a reasonable starting point. Hence we try more rigorous approaches based on mathematical models supported by evidence.


Have you tried to vary the steepness of the thread angle and look at the theoretical results in the ideal situation described in your fig 1? Looking at the extreme values may help you realise how Swansont is correct regarding x and y components.

Share this post


Link to post
Share on other sites
10 hours ago, swansont said:

Based on what physics?

Based on the fact there is no conversion of rotational energy to kinetic through the same mechanism, thus the reaction part will have just rotational energy that does not violate the Energy conservation. In general this can be expressed as follow:

\[U_{rotation_{action-on-nut}} + U_{rotation_{reaction-on-screw}} = 0, \\
U_{rotation_{action-on-nut}} = U_{kinetic_{nut}}, \\
U_{kinetic_{screw}} = 0 \Rightarrow
U_{rotation_{reaction-on-screw}} = -U_{kinetic_{nut}}.\]

 

10 minutes ago, Ghideon said:

Have you tried to vary the steepness of the thread angle and look at the theoretical results in the ideal situation described in your fig 1? Looking at the extreme values may help you realise how Swansont is correct regarding x and y components.

I understand the reasoning of swansont but something does not feel right (I give much attention to intuition and observation than the maths (maths are by nature non-intuitive)). With your comment (a big thanks) I just realized my idea will work when the steepness (decreasing the cross section of the screw and increasing the lead) goes to a very large finite value (not infinity), meaning the inclination of the slope will be almost 90°.

Note: Swansont and maybe others have mentioned something similar (as the massless bolt) but back then I didn't realize what was the purpose of such suggestion. Now I see.

That way the Fn*cos(θ) (action on x-axis) will be almost zero that would imply an almost zero reaction upon the screw. Theoretically, for an infinite steepness the device becomes an ideal reactionless drive. As you see my intuition was right from the beginning, the problem was to justify the idea with maths (where I admit I struggle a lot since it is not my daily business (testing software, test scripts etc). The derivation I provided in my first post (not so nice actually), it actually depicts the ideal case.

A big thanks to all @Ghideon, @swansont, @joigus, @pzkpfw , @Phi for All and all all readers who like or dislike my comments.

Consequently, when we could have a carbon fiber screw with a relative high steepness and a small mass relative to the nut (screw 1 unit and nut 10) then it would be possible to practically demonstrate the principle of the reactionless drive in classical mechanics.

Edited by John2020

Share this post


Link to post
Share on other sites
1 hour ago, John2020 said:

Based on the fact there is no conversion of rotational energy to kinetic through the same mechanism, thus the reaction part will have just rotational energy that does not violate the Energy conservation. In general this can be expressed as follow:

Work is being done in your original example (the motor), so this analysis assumes just a nut and a bolt with no work, which is fine, but keep this in mind.

The next step is to find the relationship between rotation speed and linear speed of the nut and the various energies. It’s probably easiest if you assume the nut and bolt have the same mass (it has to work for any mass, so if it fails for the simple case, it just fails). See if you can conserve energy. The math will be easier because you have discarded the constraint of momentum conservation 

1 hour ago, John2020 said:

 

 

Urotationactiononnut+Urotationreactiononscrew=0,Urotationactiononnut=Ukineticnut,Ukineticscrew=0Urotationreactiononscrew=Ukineticnut.

 

 

If these are supposed to be kinetic energies, how can they sum to zero, unless they are identically zero?

The screw is turning, so its rotational energy is not zero.

You can’t have energies being the negative of each other. KE is always a positive value.

 

Share this post


Link to post
Share on other sites
55 minutes ago, John2020 said:

Consequently, when we could have a carbon fiber screw with a relative high steepness and a small mass relative to the nut (screw 1 unit and nut 10) then it would be possible to practically demonstrate the principle of the reactionless drive in classical mechanics.

And at 90º "steepness" the drive would go to zero, because cos(90º) = 0.

Ergo, no drive.

Maths saves you a lot of thinking work. Maths are by nature non-intuitive perhaps, but once you learn them, they overcome most all flaws coming from our intuition. You can autopilot most of the time.

But you've been one of the nicest, most civil speculators in these forums. For that, I thank you.

The sad thing about this is, IMHO, that had you taken the criticism, maybe there would be an idea behind worth considering (by using frictionless superconductors, high "steepness" combined, who knows.) In the way of an efficient drive mechanism, rather that spontaneous drive, which is a physical impossibility.

5 minutes ago, swansont said:

If these are supposed to be kinetic energies, how can they sum to zero, unless they are identically zero?

Ditto. I can't add more reactions today. I would need some action. ;)

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.