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Offline electrodacusTopic starter

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Re: Force multiplier
« Reply #125 on: February 08, 2023, 07:46:51 pm »
So, after all that, do you agree or not that each of the statements 1..3 in the context shown with the HoG, etc? If not, specifically which one are you stating is false?

Just for the moment, if you can manage to control yourself, don't relate this to your problem. Just answer the question honestly, without distractions or diversions, and we can move on.

I already answered your question but likely you are not reading what I write.
All statements are correct but are for a completely different setup than the one disused here.
You statement has 3 points of contact due to HoG in contact with vehicle body while in my examples there are only two points of contact and the body is floating so no HoG involvement.
You can only do force amplification when you have 3 points of contact not possible with only two.

Offline PlainName

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Re: Force multiplier
« Reply #126 on: February 08, 2023, 08:05:23 pm »
So, after all that, do you agree or not that each of the statements 1..3 in the context shown with the HoG, etc? If not, specifically which one are you stating is false?

Just for the moment, if you can manage to control yourself, don't relate this to your problem. Just answer the question honestly, without distractions or diversions, and we can move on.

I already answered your question but likely you are not reading what I write.

OK, thank you.
Quote
All statements are correct but are for a completely different setup than the one disused here.
You statement has 3 points of contact due to HoG in contact with vehicle body

Well, let's fix that then. Let's remove the HoG from the vehicle and instead get him to turn the treadmill. Now, why won't the treadmill moving cause the wheels to turn?

Quote
while in my examples there are only two points of contact and the body is floating so no HoG involvement.

Hate to break this to you,  but your example does actually have three points of contact. The are:

1. Left wheel on block
2. Right wheel on treadmill
3. Driving force of treadmill. <--- your Hand of God

But back to the new setup. Remove the HoG from the vehicle, turn the treadmill, wheels turn and vehicle moves. I think that's the proof you were trying not to recognise.
 

Offline electrodacusTopic starter

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Re: Force multiplier
« Reply #127 on: February 08, 2023, 08:10:37 pm »
Well, let's fix that then. Let's remove the HoG from the vehicle and instead get him to turn the treadmill. Now, why won't the treadmill moving cause the wheels to turn?

The treadmill itself can not move as when the treadmill starts to apply a force F1 that will be matched by F2 equal and opposite so nothing can move unless one of the wheels is allowed to slip.


Hate to break this to you,  but your example does actually have three points of contact. The are:

1. Left wheel on block
2. Right wheel on treadmill
3. Driving force of treadmill. <--- your Hand of God

But back to the new setup. Remove the HoG from the vehicle, turn the treadmill, wheels turn and vehicle moves. I think that's the proof you were trying not to recognise.

There are only two points of contact if you understand what that means.

1. Left wheel on the red block.
2. Right wheel on the treadmill.

Offline cbutlera

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Re: Force multiplier
« Reply #128 on: February 08, 2023, 08:21:12 pm »
While you may consider the left wheel a different object from the right wheel in this particular example they are not.
The way let and right wheel are connected to each other makes them act as a single object.
This equivalent diagram to (a) may make it more visible to you

You can als imagine having this in your hands right hand applies F1 and the left hand will apply the equal and opposite force F2 else there will be no F1 so Newton's 3'rd law pair.

All this pairs in the above diagram are newton's 3'rd law pairs so F1 = F2, F3 = F4 and F5 = F6

Let’s for a moment accept your absurd mental gymnastics, and view the table with articulated legs as one rigid object, and the ground, the stationary block, and treadmill as a second rigid object. F1 and F2 would still not constitute a Newton’s third law force pair.

The total horizontal force exerted by the ground on the table would be the vector sum of F1 and F2.  The total horizontal force exerted by the table on the ground would be of equal magnitude to this vector sum, but directed in the opposite direction.  Those two combined forces do constitute a Newton’s third law force pair.

If F1 did somehow represent the total horizontal force exerted by the ground on the table, and F2 did somehow represent the total horizontal force exerted by the table on the ground, then the table would accelerate in the direction of F1.  F1 would be the only horizontal force acting on the table.  F2 would be acting on the ground (if you are going to treat it as the Newton's third law reaction force to F1) and so would not restrain this acceleration.  This is obviously wrong.
 

Offline electrodacusTopic starter

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Re: Force multiplier
« Reply #129 on: February 08, 2023, 08:36:56 pm »

Let’s for a moment accept your absurd mental gymnastics, and view the table with articulated legs as one rigid object, and the ground, the stationary block, and treadmill as a second rigid object. F1 and F2 would still not constitute a Newton’s third law force pair.

The "table" is a rigid object due to the configuration and the forces acting on it.
Can you have F1 different from zero in a non accelerating frame of reference without the equal and opposite F2 ?
If not then that is newton's 3'rd law.
But whatever you consider or not that to be 3'rd law you can not show F2 to be twice as large as the apply F1 as you can not have a working gearbox (force multiplication) with only two points of contact (the two legs).

The total horizontal force exerted by the ground on the table would be the vector sum of F1 and F2.  The total horizontal force exerted by the table on the ground would be of equal magnitude to this vector sum, but directed in the opposite direction.  Those two combined forces do constitute a Newton’s third law force pair.

If F1 did somehow represent the total horizontal force exerted by the ground on the table, and F2 did somehow represent the total horizontal force exerted by the table on the ground, then the table would accelerate in the direction of F1.  F1 would be the only horizontal force acting on the table.  F2 would be acting on the ground (if you are going to treat it as the Newton's third law reaction force to F1) and so would not restrain this acceleration.  This is obviously wrong.

What is the net horizontal force acting on the vehicle ?

Fnet = F1 - F2 = 0 is my answer.  I'm curious to see what your answer is since whatever you wanted to say in the above comment makes no sense.

Offline cbutlera

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Re: Force multiplier
« Reply #130 on: February 08, 2023, 08:56:25 pm »
The total horizontal force exerted by the ground on the table would be the vector sum of F1 and F2.  The total horizontal force exerted by the table on the ground would be of equal magnitude to this vector sum, but directed in the opposite direction.  Those two combined forces do constitute a Newton’s third law force pair.

If F1 did somehow represent the total horizontal force exerted by the ground on the table, and F2 did somehow represent the total horizontal force exerted by the table on the ground, then the table would accelerate in the direction of F1.  F1 would be the only horizontal force acting on the table.  F2 would be acting on the ground (if you are going to treat it as the Newton's third law reaction force to F1) and so would not restrain this acceleration.  This is obviously wrong.
...
... whatever you wanted to say in the above comment makes no sense.

Do you perhaps regard F1 and F2 as bi-directional forces?  So in the case of F1 for instance, do you think of it as both pushing against the wheel in the direction shown by the arrow, and by implication also pushing back against the treadmill in the opposite direction with an equal magnitude?
« Last Edit: February 08, 2023, 09:02:28 pm by cbutlera »
 

Offline electrodacusTopic starter

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Re: Force multiplier
« Reply #131 on: February 08, 2023, 09:14:24 pm »
Do you perhaps regard F1 and F2 as bi-directional forces?  So in the case of F1 for instance, do think of it as both pushing against the wheel in the direction shown by the arrow, and by implication also pushing back against the treadmill in the opposite direction with an equal magnitude?

Yes there is of course the pair to F1 pushing against the treadmill but we are interested in the forces acting on the vehicle and not on the treadmill as we want to know if there is a net force on the "vehicle".

You can see the treadmill as a motor with the stator connected to ground and rotor connected to the vehicle right wheel but due to the way the vehicle is configured internally it will act as a solid block and so the left wheel will push against ground which is basically the same as the motor stator.
So the stator is basically rigidly connected to the rotor thus the motor can not move even if it can apply a force.
Force needs to be high enough for something to slip and depending on what slips will determine in what direction the vehicle will move.
Left wheel slips it moves in the direction of applied force.
Right wheel slips it moves less intuitively against the direction of the applied force and that can only happen with energy storage.

Offline cbutlera

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Re: Force multiplier
« Reply #132 on: February 08, 2023, 09:19:40 pm »
Do you perhaps regard F1 and F2 as bi-directional forces?  So in the case of F1 for instance, do think of it as both pushing against the wheel in the direction shown by the arrow, and by implication also pushing back against the treadmill in the opposite direction with an equal magnitude?

Yes there is of course the pair to F1 pushing against the treadmill but we are interested in the forces acting on the vehicle and not on the treadmill as we want to know if there is a net force on the "vehicle".
...

That pair to F1 is the Newton's third law reaction force to F1.
 

Offline electrodacusTopic starter

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Re: Force multiplier
« Reply #133 on: February 08, 2023, 09:45:06 pm »

That pair to F1 is the Newton's third law reaction force to F1.

Yes that pair will also be Newton's third law reaction force. So will the F2 and the counterpart acting on the red box.
But to know if vehicle moves you will only look at the balance of forces acting on the vehicle and that will be F1 and F2

Reducing everything to just the electric motor powering the treadmill you can sat the the rotor is connected to the right wheel and the stator to the left wheel.
Since the way the belt connects the two wheels only allows the wheel to move in the same direction the motor will stall so no movement just a force proportional with the current applied to the motor.  All energy will be dissipated as heat and there will be no movement.

If you push on a wall with F1 using your hands the wall pushes back on you with an equal and opposite force but then at your feet there is also another pair of forces you pushing against ground and ground pusing on you and at the base of the wall there is also an say F2 both wall pushing against ground and ground against the wall.
In any case nothing will move the same as in the example (a) unless the applied force is large enough for the wheel to slip at the connection to ground.

I'm no longer sure what you are confused about. Is only that vehicle is a "locked gearbox" so thus can be treated as a single solid object or even if that was the case you still do not agree with the fact that F2 is equal and opposite to F1.

Offline cbutlera

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Re: Force multiplier
« Reply #134 on: February 08, 2023, 10:05:52 pm »

That pair to F1 is the Newton's third law reaction force to F1.

Yes that pair will also be Newton's third law reaction force. So will the F2 and the counterpart acting on the red box.
But to know if vehicle moves you will only look at the balance of forces acting on the vehicle and that will be F1 and F2
...
This is true, F1 and its pair constitute a Newton's third law force pair, as do F2 and its pair.  Newton's third law has now been applied to both F1 and F2 independently.  It tells you nothing about the relationship between F1 and F2.

Quote
If you push on a wall with F1 using your hands the wall pushes back on you with an equal and opposite force but then at your feet there is also another pair of forces you pushing against ground and ground pusing on you and at the base of the wall there is also an say F2 both wall pushing against ground and ground against the wall.
This is also true and again each of those pairs constitute a Newton's third law force pair.

Quote
In any case nothing will move the same as in the example (a) unless the applied force is large enough for the wheel to slip at the connection to ground.
...
Whether or not anything moves has nothing further to do (beyond its application to these force pairs) with Newton's third law.  So you cannot keep using it as the reason why you think that everyone else is wrong.

Edit: added text in brackets for clarity.
« Last Edit: February 08, 2023, 10:11:21 pm by cbutlera »
 

Offline electrodacusTopic starter

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Re: Force multiplier
« Reply #135 on: February 08, 2023, 10:26:04 pm »

Whether or not anything moves has nothing further to do (beyond its application to these force pairs) with Newton's third law.  So you cannot keep using it as the reason why you think that everyone else is wrong.


Can F2 be different than F1 in a non accelerating reference frame and if so what will be the relation between F2 and F1 ?

If you are not able to provide that relation and insist vehicle moves then you did not understood why it moves.

F1 will not accelerate the vehicle thus F2 equal and opposite to F1 else F1 can not exist according to Newton's 3'rd law.

In non accelerating reference frame in case (a) F2 = F1 and the real tests confirm that is the case.
« Last Edit: February 08, 2023, 10:28:17 pm by electrodacus »
 

Offline cbutlera

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Re: Force multiplier
« Reply #136 on: February 08, 2023, 10:27:59 pm »
...
I'm no longer sure what you are confused about. Is only that vehicle is a "locked gearbox" so thus can be treated as a single solid object or even if that was the case you still do not agree with the fact that F2 is equal and opposite to F1.

F1 may or may not be equal to F2, but they do not constitute a Newton's third law force pair.

If you assume that the vehicle is not accelerating, then Newton's second law will tell you that F1 is equal and opposite to F2 (if you ignore any frictional effects). And if you assume that F1 is equal and opposite to F2 then Newton's second law will tell you that the vehicle will not accelerate.  But you cannot prove that the vehicle will not accelerate this way.
 

Offline electrodacusTopic starter

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Re: Force multiplier
« Reply #137 on: February 08, 2023, 10:40:38 pm »

F1 may or may not be equal to F2, but they do not constitute a Newton's third law force pair.

If you assume that the vehicle is not accelerating, then Newton's second law will tell you that F1 is equal and opposite to F2 (if you ignore any frictional effects). And if you assume that F1 is equal and opposite to F2 then Newton's second law will tell you that the vehicle will not accelerate.  But you cannot prove that the vehicle will not accelerate this way.

Newton's second law has no application in a non accelerating reference frame.

Videos I showed you multiple times show that F2 = F1 and there is no acceleration.
In order for vehicle to move relative to ground one of the wheels needs to slip.

1.  No wheel slip F2 = F1 according to Newton's 3'rd law.

2. Left wheel (output wheel) slip during acceleration period F1 = F2 + (m * a) thus both 3'rd and 2'nd law are added and F1 is larger so vehicle moves to the left (it is dragged to the left as there is no wheel rotation).
After acceleration stops and vehicle is dragged at constant speed F2 = F1

3. Right wheel slip  F2 = F1 while energy storage is charged so right wheel rotates but vehicle is not accelerated then stored energy is converted to vehicle kinetic energy F2 = F1 + (m * a) again both 3'rd and 2'nd law applies this time the acceleration is due to stored energy and so vehicle moves to the right.

All 3 cases where demonstrated in my videos showing the real world experiment.
Notice that in none of the cases the F2 = 2 * F1 that will be the case for a 2:1 gearbox. But there is no working gearbox as that requires 3 points of contact and will not work with just 2.   

Offline james_s

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Re: Force multiplier
« Reply #138 on: February 08, 2023, 10:55:07 pm »
Seriously? Another one of these threads? Why do you guys waste time arguing with a troll? He never posts anything except for nonsense like this trying to stir people up.
 
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Offline electrodacusTopic starter

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Re: Force multiplier
« Reply #139 on: February 08, 2023, 11:01:56 pm »
Seriously? Another one of these threads? Why do you guys waste time arguing with a troll? He never posts anything except for nonsense like this trying to stir people up.

The troll seems to be you. You have nothing to say but decide to comment anyway.

Offline cbutlera

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Re: Force multiplier
« Reply #140 on: February 08, 2023, 11:04:39 pm »
Newton's second law has no application in a non accelerating reference frame.
That is complete nonsense.  You can have accelerating objects in a non accelerating frame of reference.

Quote
Videos I showed you multiple times show that F2 = F1 and there is no acceleration.
In order for vehicle to move relative to ground one of the wheels needs to slip.

1.  No wheel slip F2 = F1 according to Newton's 3'rd law.
...

Way back in this discussion, Nominal Animal posted a very detailed kinematic analysis of this system.  You dismissed his analysis out of hand by falsely claiming that it was contrary to Newton's third law, but this claim would only be true if Newton's third law did require that F1 and F2 were equal and opposite.  Then a few messages ago you identified the correct Newton's third law reaction forces to both F1 and F2. Now you have gone straight back to claiming that F2 is the Newton's third law reaction force to F1.  It isn't.  You are going around in circles.

« Last Edit: February 08, 2023, 11:07:17 pm by cbutlera »
 

Offline IanB

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Re: Force multiplier
« Reply #141 on: February 08, 2023, 11:11:58 pm »
You are going around in circles.
Not so much going around in circles, as carefully avoiding and dodging any deduction that might lead to a rational conclusion, artfully changing the story and backtracking when it looks like things are getting too close to the correct answer. In all the previous threads it has been the same.

Notice how the behavior is always to goad other people into responding and doing work in order to prolong the game?
 

Offline electrodacusTopic starter

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Re: Force multiplier
« Reply #142 on: February 08, 2023, 11:17:52 pm »

That is complete nonsense.  You can have accelerating objects in a non accelerating frame of reference.

There is only one object of interest here the vehicle and it is either accelerating in which case Newton's second law is involved or it is not accelerating in which case only Newton's 3'rd law is involved.


Way back in this discussion, Nominal Animal posted a very detailed kinematic analysis of this system.  You dismissed his analysis out of hand by falsely claiming that it was contrary to Newton's third law, but this claim would only be true if Newton's third law did require that F1 and F2 were equal and opposite.  Then a few messages ago you identified the correct Newton's third law reaction forces to both F1 and F2. Now you have gone straight back to claiming that F2 is the Newton's third law reaction force to F1.  It isn't.  You are going around in circles.

Kinematics is not useful if you want to predict how the real world object moves as forces are not considered at all. Thus real world test show that veicle does not move the way it is shown by that Kinematics model.

Because F1 cannot exist without the equal and opposite F2 when no wheel slip is allowed.

Correct me if I'm wrong but you (same as others) claim that F2 = 2 * F1 for a 2:1 gear ratio.  No such thing is observed in any of the real world tests because you can not have force multiplication done by a device with only two points of contact.

Offline electrodacusTopic starter

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Re: Force multiplier
« Reply #143 on: February 08, 2023, 11:20:24 pm »
You are going around in circles.
Not so much going around in circles, as carefully avoiding and dodging any deduction that might lead to a rational conclusion, artfully changing the story and backtracking when it looks like things are getting too close to the correct answer. In all the previous threads it has been the same.

Notice how the behavior is always to goad other people into responding and doing work in order to prolong the game?

Do you agree that a theory is incorrect if it can not predict what happens in a real world experiment ?

If so show me where F2 = 2 * F1 for a 2:1 gear ratio since that is what you seem to claim.
I claim F2 = F1 and that is observed in my tests. Vehicle is either stationary or moving at constant speed dragged in the direction of applied force.

Offline cbutlera

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Re: Force multiplier
« Reply #144 on: February 08, 2023, 11:27:41 pm »
You are going around in circles.
Not so much going around in circles, as carefully avoiding and dodging any deduction that might lead to a rational conclusion, artfully changing the story and backtracking when it looks like things are getting too close to the correct answer. In all the previous threads it has been the same.

Notice how the behavior is always to goad other people into responding and doing work in order to prolong the game?

Yes, I think that you are right and I have been wasting my time.  After finding that paper describing the surprisingly common "Newton’s Second Law – Net Force” (N2-NF) misconception, I thought that maybe I could help.  It looks like I was wrong about that.
 

Offline electrodacusTopic starter

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Re: Force multiplier
« Reply #145 on: February 08, 2023, 11:39:52 pm »
Problem with real numbers.

Wheels in diagram (a) are not allowed to slip but the red box can slip relative to ground gear ratio 2:1.
coefficient of static friction us = 0.5
coefficient of kinetic friction uk = 0.3
Vehicle mass 5kg with a 50% distribution so 2.5kg on the right and 2.5kg on the left wheel.
Red box is 1kg

A) What is the minimum force F1 needed for the red box to start moving ?
B) What is F2 in relation to F1 if F1 is less than needed for the red box.
C) What is F2 in relation to F1 if F1 is larger than needed for the box to start accelerating and in what direction will the red box move.
D) What is F2 in relation to F1 when box speed is constant and what is that speed of the red box relative to treadmill speed.

Consider the treadmill infinitely long so no worry about the wheel falling of the treadmill same for the red box.

« Last Edit: February 08, 2023, 11:46:09 pm by electrodacus »
 

Offline PlainName

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Re: Force multiplier
« Reply #146 on: February 08, 2023, 11:49:58 pm »
Quote
There are only two points of contact if you understand what that means.

1. Left wheel on the red block.
2. Right wheel on the treadmill.

Well something is moving the treadmill, and to do that it has to be pushing (or pulling) against some part of your setup, otherwise nothing will move1. Perhaps missing that is the source of your mistake.

---
[1] Actually, if something drives the treadmill and it's not attached to anything else, the treadmill will just disappear off to the right. But it can't, because it is attached to the ground to which the left block is also attached. Thus it's another point of contact, as you call it, and completes the circuit.
 

Offline electrodacusTopic starter

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Re: Force multiplier
« Reply #147 on: February 08, 2023, 11:54:38 pm »
Quote
There are only two points of contact if you understand what that means.

1. Left wheel on the red block.
2. Right wheel on the treadmill.

Well something is moving the treadmill, and to do that it has to be pushing (or pulling) against some part of your setup, otherwise nothing will move1. Perhaps missing that is the source of your mistake.

---
[1] Actually, if something drives the treadmill and it's not attached to anything else, the treadmill will just disappear off to the right. But it can't, because it is attached to the ground to which the left block is also attached. Thus it's another point of contact, as you call it, and completes the circuit.

Hey this is very good.  Maybe this is where the confusion comes from.
Treadmill is attached to ground so the force it applies to the right wheel is relative to ground.
There is an electric motor inside the treadmill with the stator connected to ground and rotor connected to the treadmill belt.
Is this helpful ?

Offline PlainName

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Re: Force multiplier
« Reply #148 on: February 08, 2023, 11:56:39 pm »
Seriously? Another one of these threads? Why do you guys waste time arguing with a troll?

It's like trying to bang a nail in with a hammer, but whatever you do and however you hold the hammer, the nail always manages to be slightly somewhere else at the critical moment. Keep coming back because it's one of those things you can't quite believe unless you see it  ::)
« Last Edit: February 09, 2023, 12:00:34 am by PlainName »
 
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Offline PlainName

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Re: Force multiplier
« Reply #149 on: February 08, 2023, 11:59:09 pm »
Quote
Is this helpful ?

You tell me. I reckon I know what you'll say, but say it anyway.
 


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