Force multiplier

[PlainName]

Not quite sure you read what I wrote at point b)

I am very sure you didn't read (or understand) my point about slip. Have you never wondered how a clutch works? You can have slip and yet still transfer power - it is not all or nothing. Thus your statement that " If slip is allowed acceleration will be zero if the right wheel slips" is one extreme. It could slip and still allow acceleration.

[PlainName]

[...]

Apologies for butting in. Don't know what came over me, and you were doing a far better job. I shall leave you to it

[electrodacus]

You have agreed that the two horizontal forces that act on the vehicle are not Newton's third law force pairs, and so they do not have to be equal and opposite in all circumstances.  When they are not equal and opposite, the vehicle will be subject to a net force.  This net force must cause the vehicle to accelerate, this is Newton's second law.  Are you claiming that this law should include the proviso that it is only true if something slips?

Sorry if I was not clear. In my diagrams F1 = F2 and is Newton's 3'rd law that applies there.
There can not be motion without slip in any of my diagrams.
If you look closely at how the belt is connected you will understand that this is a "locked box". Meaning it can only move if the left wheel slips. If the right wheel slips then only the treadmill surface moves the "vehicle/gearbox" will not move relative to ground.

Te examples given for third law almost always involve gravity and that is not a requirement for Newton's 3'rd law thus the failing in understanding what the 3'rd law is about.

Say gear ratio is 2:1 in case A you can consider any type of reference frame you want. Will  F2 = 2 * F1 ?
The answer is clear no and F1 = F2 in a non accelerating frame or F1 = F2 + m*a in an accelerating frame. Thus F1 is always equal or larger than F2 and never the other way around.
When both energy storage and stick slip hysteresis are involved the discussion is very different but those two are not provided in the simple example A.

[cbutlera]

You have agreed that the two horizontal forces that act on the vehicle are not Newton's third law force pairs, and so they do not have to be equal and opposite in all circumstances.  When they are not equal and opposite, the vehicle will be subject to a net force.  This net force must cause the vehicle to accelerate, this is Newton's second law.  Are you claiming that this law should include the proviso that it is only true if something slips?

Sorry if I was not clear. In my diagrams F1 = F2 and is Newton's 3'rd law that applies there.
...

But F1 and F2 are not a Newton’s third law force pair.  Carefully read again the paragraph from page 5 of the paper regarding the N2-NF misconception.  It addresses precisely this issue.  Have you now decided that you do not agree with the arguments presented in this paper?

“Consider an object in static equilibrium, under the influence of just two forces. In the Newton’s Second Law – Net Force” (N2-NF) misconception, students note that the two forces must sum to zero (a correct application of the Second Law) and hence the two forces must be equal in magnitude and opposite in direction (also correct, as a direct mathematical result of the first statement), and thus are a Third Law force-pair (incorrect). This reasoning is similar to the (il)logical sequence, “All cats have four legs; my dog has four legs; therefore, my dog is a cat”. While a Third Law force-pair are equal in magnitude and opposite in direction, not all forces which are equal in magnitude and opposite in direction are a Third Law pair. This subtlety is often lost on novice students; but the mental model which arises from the flawed sequence of reasoning is strong and resistant to instruction (Wilson & Low, 2015)."

[electrodacus]

But F1 and F2 are not a Newton’s third law force pair.  Carefully read again the paragraph from page 5 of the paper regarding the N2-NF misconception.  It addresses precisely this issue.  Have you now decided that you do not agree with the arguments presented in this paper?

What will you say that F1 = F2 are a consequence of if not Netwonds 3'rd law ?
Or are you saying F1 is not equal with F2 in my examples ?

If so please explain how F1 can be different from F2. What is the mechanism to allow that ?


And to be clear what I claim is the same as what is observed in a real world test.

If F1 is less than what is needed for any of the wheels to slip then F1 = F2
If F1 is large enough that a wheel can slip it depends on which one slips.
    -if is the one on the right F1 = F2  (vehicle is not moving)
    -if is the one on the left F1 = F2 + (m * a)  (vehicle moves to the left in the direction the applied F1 is pointing) video proof:  https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0

In order for the vehicle to move in the opposite direction of applied force as seen in this other video both energy storage and stick slip hysteresis needs to be involved  https://odysee.com/@dacustemp:8/wheel-cart-energy-storage-slow:8

In this last case forces are not constant over time. Initial phase F1 = F2 and increasing in value as only right wheel spins stretching the belt (storing potential elastic energy). At some point force is large enough for the right wheel to slip and in that moment the stored elastic energy is converted into kinetic energy.
The front wheel will stik (at a different force than it needed to start slipping thus the hysteresis) and the process will repeat multiple times per second (smaller the energy storage the faster the cycle of charge/discharge). Again prove is in the video. You can build your own mechanism and you will observe the exact same thing.  It should not be needed to build the experiment if you understand physics including newton's 3'rd law as without energy storage an object will not move in the opposite direction of applied force. 

[IanB]

In the ideal world of diagram (a) there are no forces, therefore no analysis based on force balances or force pairs can exist.

To put this plainly, F1, F2, F3 and F4 are always, unavoidably, and exactly zero. It is not possible to make them anything other than zero.

Force balance analysis can only be applied to static systems (the branch of applied mathematics called "statics"). In picture (a) the wheeled assembly is moving freely from left to right and there is no resistance to motion for forces to exist.*

The physical simile is a free body moving in space at constant velocity, being pushed by a force labeled "F". It is clear that F must be zero, since a body moving at constant velocity in free space experiences no forces.

* Unless we introduce factors like friction in bearings.

[cbutlera]

But F1 and F2 are not a Newton’s third law force pair.  Carefully read again the paragraph from page 5 of the paper regarding the N2-NF misconception.  It addresses precisely this issue.  Have you now decided that you do not agree with the arguments presented in this paper?

What will you say that F1 = F2 are a consequence of if not Netwonds 3'rd law ?
Or are you saying F1 is not equal with F2 in my examples ?

If so please explain how F1 can be different from F2. What is the mechanism to allow that ?

Yes, F1 is not equal and opposite to F2 in your examples, except in the case where both are zero.

The vehicle in example a) and the cylinder in example b) both have an internal mechanism that guarantees that F1 and F2 will have different magnitudes (except in the case where both are zero).  This should be most obvious in example b).  Both pistons are exposed to the same fluid pressure, but they have different working surface areas, so as a consequence, F1 and F2 will have different magnitudes.

Sorry, it’s getting very late.  I need to sign off now.

[electrodacus]

In the ideal world of diagram (a) there are no forces, therefore no analysis based on force balances or force pairs can exist.

To put this plainly, F1, F2, F3 and F4 are always, unavoidably, and exactly zero. It is not possible to make them anything other than zero.

Force balance analysis can only be applied to static systems (the branch of applied mathematics called "statics"). In picture (a) the wheeled assembly is moving freely from left to right and there is no resistance to motion for forces to exist.*

The physical simile is a free body moving in space at constant velocity, being pushed by a force labeled "F". It is clear that F must be zero, since a body moving at constant velocity in free space experiences no forces.

* Unless we introduce factors like friction in bearings.

Why do you say in diagram a there can be no forces in ideal case ?
So you saying that appling F1 the assembly will just move freely in the opposite direction of applied force ? How will that ever make sense ?

What about case b) as that is simpler to visualize ?  Will there be possible to apply a force F1 ? and will F2 not be equal and opposite to F1?

[electrodacus]

Yes, F1 is not equal and opposite to F2 in your examples, except in the case where both are zero.

The vehicle in example a) and the cylinder in example b) both have an internal mechanism that guarantees that F1 and F2 will have different magnitudes (except in the case where both are zero).  This should be most obvious in example b).  Both pistons are exposed to the same fluid pressure, but they have different working surface areas, so as a consequence, F1 and F2 will have different magnitudes.

Sorry, it’s getting very late.  I need to sign off now.

As I suspected you have the same misunderstanding as the other about the mechanisms in the 3 examples.
I already shown in video that F1 = F2 and also significantly large (can be calculated by the amount the belt was stretched).

The internal mechanism in a) and b) can not do force amplification / multiplication in the way they are set up.
For that to be true the body of the vehicle in a) (blue part) will need to be connected rigidly to ground and same in case b) the cylinder will need to be connected to ground.
Yes fluid pressure is the same also F2 equal and opposite to F1.  Please do a google search for a torque multiplier and see if that works with just two points of contact as in my examples a) and b).   You will find out that all torque multipliers requires 3 points of contact in order to work. 

[IanB]

Why do you say in diagram a there can be no forces in ideal case ?

It's a diagram, so there are no frictional forces and no inertial forces. There are also no rigid bodies. Therefore no forces.

So you saying that appling F1 the assembly will just move freely in the opposite direction of applied force ? How will that ever make sense ?

You cannot apply F1 because there are no forces. If the belt is turning as in the picture then the wheels will be turning clockwise and the assembly will be moving to the right. No forces are required for this to be happening.

What about case b) as that is simpler to visualize ?  Will there be possible to apply a force F1 ? and will F2 not be equal and opposite to F1?

There is no need to discuss case (b) as that is different and therefore not relevant.

[electrodacus]

You cannot apply F1 because there are no forces. If the belt is turning as in the picture then the wheels will be turning clockwise and the assembly will be moving to the right. No forces are required for this to be happening.

There is no need to discuss case (b) as that is different and therefore not relevant.

I'm glad you say that case (b) is different since while it is not true it means you likely understand case (b).
Belt will not need to move in order to apply a force F1 to the mechanism in case (a).

All 3 case represent the same thing floating body/cylinder/GND and thus they can not do force or voltage amplification.
In case (c) output current can not be different from input current but you can still have a current different from zero same as you can have a force different from zero in both case (a) and case (b)

Maybe you can explain case (b).  Can you apply a force F1 in that case and can F2 be anything other than equal an opposite to F1 ?
And if so what is the problem there ? Why can you not have F2 say 2 * F1 ?

[IanB]

Belt will not need to move in order to apply a force F1 to the mechanism in case (a).

You cannot apply a force at F1 in case (a), because the wheel will turn, and by yielding reduce the force to zero.

All 3 case represent the same thing floating body/cylinder/GND and thus they can not do force or voltage amplification.

The first case is not floating as it has a wheel resting on the solid red block, and the other wheel is resting on the belt, therefore not floating either.

[electrodacus]

You cannot apply a force at F1 in case (a), because the wheel will turn, and by yielding reduce the force to zero.

The wheel will not turn unless the belt is elastic and the stretching of the belt allows the wheel to rotate while force will increase.
You are imagining that the axle friction is so high and that is the reason why the belt stretched so much in my experiment but that is not the reason (not even close).
The F1 wants to push the vehicle to the left while the left wheel wants to push the vehicle to the right with the force F2 equal and opposite to F1.
It is the same as case (b)
I think you are imagining that F1 only wants to turn the wheel and not push the vehicle in that same direction and that will be true if the belt was not connected and the wheel axle had no friction.
But due to the way the belt is connected it is basically a locked gearbox no different from having the wheels welded to the body.

The first case is not floating as it has a wheel resting on the solid red block, and the other wheel is resting on the belt, therefore not floating either.

The wheels can rotate independently of the body but not independent of each other due to belt.
The body drawn in blue is not connected to anything.  If I were to connect the body to ground right in between the red box and the treadmill then F2 can be 2xF1 (assuming 2:1 gear ratio) and so that back wheel can push the red box with a force twice as large relative to grout as the input F1 also relative to ground.
Same thing for case (b) where the cylinder body will need to be connected to ground.

You can not do force multiplication with just two points of contact and you need a minimum of 3 points in order to do that no matter the device.
Just have a search on google images for "torque multiplier wrench" and you will see that all of them require 3 points of contact in order to work.

[IanB]

You cannot apply a force at F1 in case (a), because the wheel will turn, and by yielding reduce the force to zero.

The wheel will not turn unless the belt is elastic and the stretching of the belt allows the wheel to rotate while force will increase.

The belt is not elastic and there is no stretching. This is an ideal diagram. There are no physical effects such as stretching, friction or inertia. Everything is perfect and ideal.

You are imagining that the axle friction is so high and that is the reason why the belt stretched so much in my experiment but that is not the reason (not even close).

No friction. No stretching. Perfectly ideal.

The F1 wants to push the vehicle to the left while the left wheel wants to push the vehicle to the right with the force F2 equal and opposite to F1.
It is the same as case (b)

There is no F1 because there is no force. No force because no friction, no inertia, nothing to resist any attempted force applied. No resistance, no force.

I think you are imagining that F1 only wants to turn the wheel and not push the vehicle in that same direction and that will be true if the belt was not connected and the wheel axle had no friction.
But due to the way the belt is connected it is basically a locked gearbox no different from having the wheels welded to the body.

No locked gearbox as clearly seen from the diagram and mathematical analysis.
You can build it and see. If you want to avoid stick-slip energy storage, use cogs and toothed gears so that no slipping is possible. The assembly will simply move to the right with no force required if the belt moves as indicated.

The first case is not floating as it has a wheel resting on the solid red block, and the other wheel is resting on the belt, therefore not floating either.

The wheels can rotate independently of the body but not independent of each other due to belt.

Precisely.

The body drawn in blue is not connected to anything.  If I were to connect the body to ground right in between the red box and the treadmill then F2 can be 2xF1 (assuming 2:1 gear ratio) and so that back wheel can push the red box with a force twice as large relative to grout as the input F1 also relative to ground.
Same thing for case (b) where the cylinder body will need to be connected to ground.

There are no forces, therefore no force multiplication.

You can not do force multiplication with just two points of contact and you need a minimum of 3 points in order to do that no matter the device.
Just have a search on google images for "torque multiplier wrench" and you will see that all of them require 3 points of contact in order to work.

There are no forces, therefore no force multiplication.

[electrodacus]

No locked gearbox as clearly seen from the diagram and mathematical analysis.
You can build it and see. If you want to avoid stick-slip energy storage, use cogs and toothed gears so that no slipping is possible. The assembly will simply move to the right with no force required if the belt moves as indicated.

There is a difference between math and physics.

I build it and saw this gearbox is locked. You can see for yourself in this video  https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0
Force is applied by the paper to the right wheels (input wheels). As you can see there is no rotation and the entire vehicle is dragged in the direction that the force is applied.
The applied force is equal with the frictional force at the back wheels and it is significant (much larger than what it will be required if the gearbox was not locked).  This is exactly the same design as the on in diagram (a). 

[IanB]

I build it and saw this gearbox is locked. You can see for yourself in this video  https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0

Your video clearly shows the wheels turning and the vehicle moving in the manner predicted. The paper goes one way, the vehicle goes the other way.

[electrodacus]

I build it and saw this gearbox is locked. You can see for yourself in this video  https://odysee.com/@dacustemp:8/stick-slip-removed-from-front-wheels:0

Your video clearly shows the wheels turning and the vehicle moving in the manner predicted. The paper goes one way, the vehicle goes the other way.

Did you watched the video ?  I'm referring to first 15 seconds.
The last 9 seconds is a different setup where right wheels (input wheels) will slip and that allows the vehicle to move in the opposite direction of applied force by taking advantage of the energy storage and stick slip histeresis.

So what is the reason the wheels are locked while vehicle is being dragged in those first 15 seconds ?

[IanB]

So what is the reason the wheels are locked while vehicle is being dragged in those first 15 seconds ?

You built it wrong. If the wheels slip it is not working properly. Put some weight on it so the wheels can't slip.

[electrodacus]

You built it wrong. If the wheels slip it is not working properly. Put some weight on it so the wheels can't slip.

The gearbox is locked so if you want to make the vehicle move you have two choices. Either the left wheels slip and so you drag the vehicle in the direction of applied force (first 15 seconds) or you allow right wheels to slip and then you have the charging and discharging of the energy storage triggered by the stick slip hysteresis fast enough that you will not even notice without slow motion video.

So if you think I build it wrong you do it and make sure the input wheels do not slip then let me know how it works.

[IanB]

So if you think I build it wrong you do it and make sure the input wheels do not slip then let me know how it works.

I know it works and I know how it works. I have shown you it working with an animation. Animations are ideal, they cannot have any sticking or slipping or any other non-ideal behavior. Animations are perfect, and they work perfectly.

Please stop creating these pointless threads.

[electrodacus]

I know it works and I know how it works. I have shown you it working with an animation. Animations are ideal, they cannot have any sticking or slipping or any other non-ideal behavior. Animations are perfect, and they work perfectly.

Please stop creating these pointless threads.

Anything you want can work in an animation without a physics model. Same is true with math.
You will be surprised when you will try the real model. You will observe the same thing I showed in my video.
You had no explanation for why the gearbox is locked in those 15 seconds while the vehicle is dragged in the direction of applied force. That is because you do not understand the physics involved and specifically Newton's 3'rd law.

[cbutlera]

...
The internal mechanism in a) and b) can not do force amplification / multiplication in the way they are set up.
For that to be true the body of the vehicle in a) (blue part) will need to be connected rigidly to ground and same in case b) the cylinder will need to be connected to ground.
Yes fluid pressure is the same also F2 equal and opposite to F1.
...

OK, let’s examine your example b) more closely.

Assume that your diagram b) represents the state of the system at time t = 0, and at that time, all of the parts of the system are at rest.

Also assume that a force is applied to the lever in the direction toward the cylinder starting at time t=0 such that:

F1 = 0 for t < 0
F1 = X for t >= 0, where X is a constant.

What will happen to F2?  How do you think that the system will evolve?

[electrodacus]

OK, let’s examine your example b) more closely.

Assume that your diagram b) represents the state of the system at time t = 0, and at that time, all of the parts of the system are at rest.

Also assume that a force is applied to the lever in the direction toward the cylinder starting at time t=0 such that:

F1 = 0 for t < 0
F1 = X for t >= 0, where X is a constant.

What will happen to F2?  How do you think that the system will evolve?

If X is less than needed to move the red box then F2 will have the same value and opposite direction to F1

If the body of the cylinder was to be connected to ground then F1 will be applied between the small piston and the cylinder body that is rigidly connected to ground and then F2 can be say 2 * F1 as it will be pushing relative to the cylinder body that is now referenced to ground and not relative to F1

As long as cylinder body is floating as in diagram (b) F2 can not be anything other than equal and opposite of F1.

[cbutlera]

OK, let’s examine your example b) more closely.

Assume that your diagram b) represents the state of the system at time t = 0, and at that time, all of the parts of the system are at rest.

Also assume that a force is applied to the lever in the direction toward the cylinder starting at time t=0 such that:

F1 = 0 for t < 0
F1 = X for t >= 0, where X is a constant.

What will happen to F2?  How do you think that the system will evolve?

If X is less than needed to move the red box then F2 will have the same value and opposite direction to F1

If the body of the cylinder was to be connected to ground then F1 will be applied between the small piston and the cylinder body that is rigidly connected to ground and then F2 can be say 2 * F1 as it will be pushing relative to the cylinder body that is now referenced to ground and not relative to F1

As long as cylinder body is floating as in diagram (b) F2 can not be anything other than equal and opposite of F1.

Yes, lets assume that the red box is fixed to the ground and unable to move.

So are you saying that the system will reach a static equilibrium with F1 = F2 = X and with the cylinder block remaining stationary?

[electrodacus]

Yes, lets assume that the red box is fixed to the ground and unable to move.

So are you saying that the system will reach a static equilibrium with F1 = F2 = X and with the cylinder block remaining stationary?

Yes that is exactly what I'm saying.