Distances in the universe

jack, you are trying to fool your readers.

A = -B is not the same as A = B.

If we label the force exerted by boat X, directed from the left, as -A, THEN THE FORCE EXERTED BY BOAT Y IS LABELED THE SAME, AS B (directed from the right).


Then, your own equations amount to a piece of crap.


So now, X, pulling with a force of -A, is equal and opposite to Y pulling with a force of B.

That is NOT what your equations say at all.


This was your original analysis:

The net force on boat x is -A.
The net force on boat y is -B.
The net force on the string is A+B.
As the string isn't moving, the net force on the string is 0, so A+B=0 so B=-A.

You labeled force A with the wrong direction. ONLY FORCE A.

Here is the correction.

Boat X is pulling with force -A (directed to the left).

Then the net force on boat X will be, according to jack, A.

Boat Y is pulling with force B (directed to the right).

The net force on boat Y will be -B.

The net force on the string will be: -A + B

As the string isn't moving, the net force on the string is 0, then -A + B = 0, so -A = -B, so A=B.


A AND B ARE NO LONGER OPPOSING REACTION FORCES.


It is no different to claiming that if X is pulling on the rope with a force of -A, the rope will pull back with a force of B, which is different to force A.

See how you are desperately trying to surreptitiously dodge the fact that your analysis is a piece of crap?


It makes all the difference: your conclusion now reads as A = B, and not, A = -B.

A AND B ARE NO LONGER OPPOSING REACTION FORCES.


Your analysis is a piece of crap jack.


That creates a quite explicit relationship between them of A=-B. There is no other option.

Not anymore.

Your analysis, properly labeled, reads now as follows:

The net force on the string will be: -A + B

As the string isn't moving, the net force on the string is 0, then -A + B = 0, so -A = -B, so A=B.


So if you take note that your "A" is really -A, you end up with this:
(-A)=B.
i.e. A=-B, just like before and just like Newton's law dictates.

The other way is to teak Newton's law slightly, for example, having the action force be -A, and the reaction force be B.
Then -A=-B or A=B.

YOU CAN'T HAVE IT BOTH WAYS.

With force A improperly labeled, we get A = -B.

With force A properly labeled, we get A = B.

Now, you no longer have A and B as OPPOSING REACTION FORCES.


Your analysis is a piece of crap.


If you instead have X applying a force of -A and Y applying a force of B, the requirement changes such as -A=-B, or A=B.

NOW, THEY ARE NO LONGER OPPOSING FORCES.


REMEMBER WHAT YOU WROTE EARLIER? HOW CAN YOU BE SO FORGETFUL?

Now, the claims that jack has made here regarding A = -B are really going to come back and destroy even further his "analysis".

In order for these to be valid action-reaction pairs, this requires A=-B

So the total list of forces I have:
X pulls on rope with force A=-B.
Rope pulls on X with force -A=B.
Rope pulls on Y with force A=-B.
Y pulls on rope with force -A=B.

Reality dictates that for the ideal case, A=-B.
If it doesn't then Newton's third law is violated.

As such, you MUST have A=-B. Anything else is NOT describing this situation.

But for the ideal case of a massless rope, reality dictates that |A|=|B|. This is summarised in the third law of motion:
For every action (A) there is an equal but opposite reaction (B). i.e. |A|=|B| (or more specifically, A=-B).
That is what reality demands.
So it isn't surprising that the RE analysis ends with reality.

Let me remind you of Newton's third law expressed in a simple way:
A=-B.
A is the action, B is the equal but opposite reaction.
The conclusion you are saying is a contradiction IS newton's third law.
So if my analysis is wrong, then Newton's third law is wrong.

It only so happens that the correct labeling of forces leads directly to the conclusion that A = B. A most troubling contradiction in view of the quotes referenced above.


You specifically stated:

A is the action, B is the equal but opposite reaction.

BUT NOW, A = B.

NO LONGER OPPOSING REACTION FORCES.


Your analysis is a piece of crap jack.


When I did label correctly force A, my analysis improved, a sure sign of a correct formula. No contradictions there.


It all comes down to the equations jack.

Yours lead to a magnificient contradiction.

You said that A = -B, nothing else would be correct.

BUT NOW A = B.

My equations work just fine.

Here is how the balance of forces are to be properly applied.

Two boats pulled toward each other on a lake.

Man from boat X is pulling with force A, directed to the left.

Man from boat Y is pulling with force B, directed to the right.

Forces A and B are, of course, of different magnitude.


What are the forces acting on boat X?

To the left we will have a negative direction.

Boat X will be acted upon by TWO FORCES: A (the reaction force on the action force -A) and B.


What are the forces acting on the left end side of the rope?

-A and -B.


What are the forces acting boat Y?

To the right we will have the positive direction.

Boat Y will be acted upon by two forces: -B (the reaction force on the action force B) and
-A.


What are the forces acting on the right end side of the rope?

A and B.


Net force on boat X: A + B

Net force on boat Y: -A - B


Net force on the string: [-A - B] + [A + B]


The string/rope will not move: [-A - B] + [A + B] = 0


All forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.



When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.


Boat Y is pulling on the rope and thus boat X with force B.

Reaction force from boat X on the rope: -B.


This is what you are missing.



You and your relatives, neighbours, countrymen, will ALWAYS pull WITH DIFFERENT FORCES.

ALWAYS.

No two persons can pull with the exactly same force, as your analysis requires: A = B.

My analysis takes this wonderful fact into account, forces A and B are never the same.

F=GMm/r^2.
This will be the same for each body, they must be EXACTLY the same.

I am sorry jack, the above law is another piece of crap.

I can bring here any number of experiments, Lamoreaux, DePalma, Biefeld-Brown, Allais WHICH DEFY THIS "LAW".

Newton dismissed this law as pure insanity.


No. It leads directly to the requirement for the ONLY cases which can be experienced in reality, A=-B.

Not anymore. YOUR ANALYSIS, PROPERLY LABELED, READS NOW: A = B.

NO LONGER OPPOSING REACTION FORCES.

And you still cannot have A = B exactly.

It won't happen in the real world.


You have X pulling on the rope with a force of A, but applying a force of A-B to the rope.
Does A=A-B? Only when B is 0.

You are acting delusional jack.

I never said any such thing.

What are the forces acting on boat X?

To the left we will have a negative direction.

Boat X will be acted upon by TWO FORCES: A (the reaction force on the action force -A) and B.


What are the forces acting on the left end side of the rope?

-A and -B.


What are the forces acting boat Y?

To the right we will have the positive direction.

Boat Y will be acted upon by two forces: -B (the reaction force on the action force B) and
-A.


What are the forces acting on the right end side of the rope?

A and B.


Net force on boat X: A + B

Net force on boat Y: -A - B


Net force on the string: [-A - B] + [A + B]


The string/rope will not move: [-A - B] + [A + B] = 0


All forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.



When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.


Boat Y is pulling on the rope and thus boat X with force B.

Reaction force from boat X on the rope: -B.


This is what you are missing.

rabinoz, you are dreaming.

We are talking here about AN INVERTED CATENARY.

One which, as if by pure magic, stays suspended in midair, on a spherical earth.


Let us reduce the distance from six miles, to 500 meters.

The curvature will then measure a few centimeters.


Let us now place a mold (a hollow plastic shell, 500 meters in length, in the shape of an inverted catenary, which matches the curvature of the round earth exactly) between the two boats.

The rope then will be placed on this plastic shell, and both sides will pull as usual.


Let us now REMOVE THAT MOLD/PLASTIC SHELL from underneath the rope.

On a flat earth, the rope will straighten right away.


HOW DO YOU EXPLAIN THE HYPOTHESIS THAT THE ROPE WILL NOW HANG IN MIDAIR IN THE SHAPE OF A BENDING LINE?

If we remove that plastic mold/shell, it is only by PURE MAGIC that the rope will assume a bending line shape, no other explanantion is possible.


Explain how the Earth attracts that rope to shape it into an inverted catenary.

Woah Jack, I can see how you earned that badge. I read the whole thing, and I agree. Sandy is essentially claiming that Newtons third law is false, as most of his claim does not abide by it.

I am not claiming that Newton's third law is false.

This alone shows your superficial analysis of this thread.


That badge worn by jack is pretty silly given his piece of crap analysis.


Please read carefully.

Here is his original analysis:

The net force on boat x is -A.
The net force on boat y is -B.
The net force on the string is A+B.
As the string isn't moving, the net force on the string is 0, so A+B=0 so B=-A.

He labeled force A with the wrong direction. ONLY FORCE A.

Here is the correction.

Boat X is pulling with force -A (directed to the left).

Then the net force on boat X will be, according to jack, A.

Boat Y is pulling with force B (directed to the right).

The net force on boat Y will be -B.

The net force on the string will be: -A + B

As the string isn't moving, the net force on the string is 0, then -A + B = 0, so -A = -B, so A=B.


A AND B ARE NO LONGER OPPOSING REACTION FORCES.



How then can you write something like: I can see how you earned that badge.


jack specifically stated:

A is the action, B is the equal but opposite reaction.

BUT NOW, A = B.

NO LONGER OPPOSING REACTION FORCES.


jack's analysis is a piece of crap.



Perhaps you like this sort of crap.


Here is the correct analysis.

My equations work just fine.

Here is how the balance of forces are to be properly applied.

Two boats pulled toward each other on a lake.

Man from boat X is pulling with force A, directed to the left.

Man from boat Y is pulling with force B, directed to the right.

Forces A and B are, of course, of different magnitude.


What are the forces acting on boat X?

To the left we will have a negative direction.

Boat X will be acted upon by TWO FORCES: A (the reaction force on the action force -A) and B.


What are the forces acting on the left end side of the rope?

-A and -B.


What are the forces acting boat Y?

To the right we will have the positive direction.

Boat Y will be acted upon by two forces: -B (the reaction force on the action force B) and
-A.


What are the forces acting on the right end side of the rope?

A and B.


Net force on boat X: A + B

Net force on boat Y: -A - B


Net force on the string: [-A - B] + [A + B]


The string/rope will not move: [-A - B] + [A + B] = 0


All forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.



When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.


Boat Y is pulling on the rope and thus boat X with force B.

Reaction force from boat X on the rope: -B.


This is what almost everyone here is missing.



My equations work.

jack's equations reach a contradiction immediately which means they are nothing but a piece of crap.

So who defeated who?

You might want to change your previous statement...

Quote from: Babushka on May 10, 2017, 05:40:50 PM

Woah Jack, I can see how you earned that badge. I read the whole thing, and I agree. Sandy is essentially claiming that Newtons third law is false, as most of his claim does not abide by it.

I am not claiming that Newton's third law is false.

You clearly are.

Newton's third law:

When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.


Let us now apply this law to the two boats on a lake situation.


Here is how the balance of forces are to be properly applied.

Two boats pulled toward each other on a lake.

Man from boat X is pulling with force A, directed to the left.

Man from boat Y is pulling with force B, directed to the right.

Forces A and B are, of course, of different magnitude.


What are the forces acting on boat X?

To the left we will have a negative direction.

Boat X will be acted upon by TWO FORCES: A (the reaction force on the action force -A) and B.


What are the forces acting on the left end side of the rope?

-A and -B.


What are the forces acting boat Y?

To the right we will have the positive direction.

Boat Y will be acted upon by two forces: -B (the reaction force on the action force B) and
-A.


What are the forces acting on the right end side of the rope?

A and B.


Net force on boat X: A + B

Net force on boat Y: -A - B


Net force on the string: [-A - B] + [A + B]


The string/rope will not move: [-A - B] + [A + B] = 0


All forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.




By contrast, the RE analysis turns out to be a piece of crap.


Please read carefully.

Here is jack's original analysis:

The net force on boat x is -A.
The net force on boat y is -B.
The net force on the string is A+B.
As the string isn't moving, the net force on the string is 0, so A+B=0 so B=-A.

He labeled force A with the wrong direction. ONLY FORCE A.

Here is the correction.

Boat X is pulling with force -A (directed to the left).

Then the net force on boat X will be, according to jack, A.

Boat Y is pulling with force B (directed to the right).

The net force on boat Y will be -B.

The net force on the string will be: -A + B

As the string isn't moving, the net force on the string is 0, then -A + B = 0, so -A = -B, so A=B.


A AND B ARE NO LONGER OPPOSING REACTION FORCES.


By his very own analysis we get: A = B.


But he requires that A = -B.


It doesn't get any worse than this for your side, disputeone...

sandokhan, you are a broken record. Give it a rest. No one believes or cares for your BS theories. Can we please get back to the original topic. There was actually some interesting information and discussions before your stupid boats and ropes nonsense

I agree. I think everyone else is pretty much in agreement  here. Sandman will never change so we might as well move on.

jack, you are trying to fool your readers.

You go completely astray when you claim that the tension on a massless rope can be different at each end.
A rope can only connect a force from one end to the other. Even if the rope has (a reasonable) mass there is little difference if the two ends of the rope are at the same level.

This means that the two people cannot pull with different forces. Just because the stronger man has the strength to pull 350 N does not mean that he is able to pull 350 N in every circumstance. If the weaker man can only apply 200 N to the rope then that is the tension in the rope - there's nothing that the strong man can do, except to drag the rope through the hands of the weak man.

An example of applied force being limited is a drag force limited game fishing reel, such as Find a ".pdf" of the manual in Shimano DENDOUMARU 9000 Beast Master Instruction Manual.

That reel can limit the drag to a maximum of 25 kg, or 245N, so it can be set to limit at your weak man's strength.
A line commonly used with such a rod is rated "PE8" or 80lb (356 N) breaking strain. So setting the reel to limit the drag to 200 N is quite reasonable.

Now, that marlin (our "strong man") can pull with a much larger force, but let's limit him to 350 N to fit.
That marlin would like nothing better that to break the line and escape, but the game fisherman (out weak man) sets the drag setting of his rod to 200 N.
This means that even though the marlin is strong enough exert a force of 350 N on the line (our "rope") the tension on the line is limited by the drag setting to 200 N, the fishing line is safe and the marlin finally tires out.

That's about as simple an analogy as I can work out - and it involves boats, sort of!

But, you clearly did not understand the following, so read it again:

---

This is never going to be resolved until we sort out some basic points. Newton's Laws are probably OK so let's look at men and ropes.

Primer for men and ropes

• This is a "ROPE", "R" is for "ROPE":
  An ideal rope is massless, of infinite strength and zero elasticity.
  An ideal rope can transmit a force and change its direction, for example using an ideal pulley.
  But
  Quote

  . . . . . the tension is the same throughout its length, so that the rope won't have infinite acceleration. This means that if I draw a free body diagram of a infinitely small piece of the rope, I will have to show two tension forces of same magnitude and in opposite directions, cancelling each other.
  Physics Stack Exchange, Interaction between an ideal pulley and an ideal rope

  Hence if
  the force applied to the left end is F_W, the tension in the left end is T_L = F_W,
  the tension in the right end is T_R = T_L and the force on the right end is F_R = T_R.
  
  Hence F_R = R_L.
• This is a "WEAK MAN", "W" is for "WEAK MAN":
  The "weak man" actually pulls on the rope with a force F_W, though may may be able to pull with up to 200 N.
  
  It makes no difference whether the WEAK MAN is on hard ground, ice, in a boat or even a helicopter,
  if the WEAK MAN pulls with a force of F_W he pulls with a force of F_W.
  I would have thought that axiomatic!
  
• This is a "STRONG MAN", "S" is for "STRONG MAN":
  The "strong man" actually pulls on the rope with a force F_S, though may may be able to pull with up to 350 N.
  
  It makes no difference whether the STRONG MAN is on hard ground, ice, in a boat or is even a fish,
  if the STRONG MAN pulls with a force of F_S he pulls with a force of F_S.
  I would have thought that axiomatic!

The crux of this whole "discussion" is "the tension on an ideal rope is the same throughout its length".
Though another issue is that even though a person is able to pull with a certain force, they do not necessarily pull with that force. In many situations, it may not be possible, because of other constraints to do that.
This might arise when lifting a mass at a constant speed or pulling against a winch limited to a certain force.

Now, do you agree with those poimts. If not, where do we differ?

Quote from: sandokhan on May 10, 2017, 10:51:30 PM

Quote from: Babushka on May 10, 2017, 05:40:50 PM

Woah Jack, I can see how you earned that badge. I read the whole thing, and I agree. Sandy is essentially claiming that Newtons third law is false, as most of his claim does not abide by it.

I am not claiming that Newton's third law is false.

You clearly are.

Newton's third law:

When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.


Let us now apply this law to the two boats on a lake situation.


Here is how the balance of forces are to be properly applied.

Two boats pulled toward each other on a lake.

Man from boat X is pulling with force A, directed to the left.

Man from boat Y is pulling with force B, directed to the right.

Forces A and B are, of course, of different magnitude.

(Emphasis mine.)

Do you honestly not understand that your statement directly contradicts Newtons third law?

Why does this keep happening to me?

Edit. Quote.

disputeone, you are scientifically literate, are you not?

How then can you write something like this:

Newton's third law:

When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

Forces A and B are, of course, of different magnitude.


Newton's third law works fine applied AT EACH END OF THE ROPE.

However, the NET FORCE ON THE STRING/ROPE WILL BE EQUAL TO ZERO.


You have already seen what jack's analysis, which you do accept as being valid, has led to: a total mess, a most direct contradiction.

That analysis is based only on a single force.

It does not work.

Forces A and b will always be different.


Please read carefully.

We are applying now Newton's third law to each end of the rope.


Here is how the balance of forces are to be properly applied.

Two boats pulled toward each other on a lake.

Man from boat X is pulling with force A, directed to the left.

Man from boat Y is pulling with force B, directed to the right.

Forces A and B are, of course, of different magnitude.


What are the forces acting on boat X?

To the left we will have a negative direction.

Boat X will be acted upon by TWO FORCES: A (the reaction force on the action force -A) and B.


What are the forces acting on the left end side of the rope?

-A and -B.


What are the forces acting boat Y?

To the right we will have the positive direction.

Boat Y will be acted upon by two forces: -B (the reaction force on the action force B) and
-A.


What are the forces acting on the right end side of the rope?

A and B.


Net force on boat X: A + B

Net force on boat Y: -A - B


Net force on the string: [-A - B] + [A + B]


The string/rope will not move: [-A - B] + [A + B] = 0


All forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.



Newton's third law was applied CORRECTLY at each end of the rope.

The final equations confirm this view/analysis.

The net force on the string/rope will be zero.

disputeone, you are scientifically literate, are you not?

How then can you write something like this:

Quote from: sandokhan on May 10, 2017, 11:26:49 PM

Newton's third law:

When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

Forces A and B are, of course, of different magnitude.

You know that was a quote from you right?

Why does this keep happening to me?

Edit. Added quote tag.

This means that the two people cannot pull with different forces.

On a lake, two men situated in two boats, SURE THEY CAN.


Primer for men and ropes

You are on land now. Totally different situation, where now we have to deal with friction.


A rope can only connect a force from one end to the other.

Let us see what happens when you have it your way.

Here is jack's original analysis:

The net force on boat x is -A.
The net force on boat y is -B.
The net force on the string is A+B.
As the string isn't moving, the net force on the string is 0, so A+B=0 so B=-A.

He labeled force A with the wrong direction. ONLY FORCE A.

Here is the correction.

Boat X is pulling with force -A (directed to the left).

Then the net force on boat X will be, according to jack, A.

Boat Y is pulling with force B (directed to the right).

The net force on boat Y will be -B.

The net force on the string will be: -A + B

As the string isn't moving, the net force on the string is 0, then -A + B = 0, so -A = -B, so A=B.


A AND B ARE NO LONGER OPPOSING REACTION FORCES.


jack specifically stated:

A is the action, B is the equal but opposite reaction.

BUT NOW, A = B.

NO LONGER OPPOSING REACTION FORCES.


jack's analysis is a piece of crap.



marlin and the reel example

In your example, the marlin cannot break free.

Thus we have a situation entirely similar to this: a large boat is pulling on a smaller boat, the rope being tied to the second, smaller boat.

Boat X (the large boat) is pulling only with 200 N (certainly it could pull with a much larger force).

Boat Y (the smaller boat) is exerting a force of 350 N on the rope.


Let us now analyze your example using both sets of equations.



Boat X is pulling with force -200 N (directed to the left).

Then the net force on boat X will be, according to jack, 200 N.

Boat Y is pulling with force 350 N (directed to the right).

The net force on boat Y will be -350 N.

The net force on the string will be: -200 + 350

As the string isn't moving, the net force on the string is 0, then -200 + 350 = 0, so -200 = -350, so 200=350.


rabinoz, does 200 = 350?

Certainly not.


Let us now apply the CORRECT set of equations.

We are applying now Newton's third law to each end of the rope.


Here is how the balance of forces are to be properly applied.

Boat X is pulling with force A, directed to the left.

Boat Y is pulling with force B, directed to the right.

Forces A and B are, of course, of different magnitude.


What are the forces acting on boat X?

To the left we will have a negative direction.

Boat X will be acted upon by TWO FORCES: 200 N (the reaction force on the action force -200 N) and B = 350 N


What are the forces acting on the left end side of the rope?

-200 and -350.


What are the forces acting on boat Y?

To the right we will have the positive direction.

Boat Y will be acted upon by two forces: -350 (the reaction force on the action force B) and -200.


What are the forces acting on the right end side of the rope?

A and B: A = 200, B = 350


Net force on boat X: A + B = 200 + 350

Net force on boat Y: -A - B = -200 - 350


Net force on the string: [-200 - 350] + [200 + 350]


The string/rope will not move: [-200 - 350] + [200 + 350] = 0


All forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.

disputeone, you are scientifically literate, are you not?

Sandokhan, you are scientifically literate, are you not?

How then can you write something like this?
"Man from boat X is pulling with force A, directed to the left.
Man from boat Y is pulling with force B, directed to the right.
Forces A and B are, of course, of different magnitude."

An ideal rope can transmit a force and change its direction, for example using an ideal pulley.
But

. . . . . the tension is the same throughout its length, so that the rope won't have infinite acceleration. This means that if I draw a free body diagram of a infinitely small piece of the rope, I will have to show two tension forces of same magnitude and in opposite directions, cancelling each other.
Physics Stack Exchange, Interaction between an ideal pulley and an ideal rope

Hence if
the force applied to the left end is F_W, the tension in the left end is T_L = F_W,
the tension in the right end is T_R = T_L and the force on the right end is F_R = T_R.

Your previous paragraphs apply to a rope being pulled on land.

There we have to deal with friction.

As simple as this.

Now, read my previous message where I analyze your marlin/reel example using both sets of equations.

Two men can push or pull with different forces obviously.

If there were two ropes, and the men pulled on one rope each, of course the forces would be different.

The problem is that the men are both pulling on the same rope, therefore the rope has one foce (tension) on it. There's no getting past this.

When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

I have neither the desire nor the patience to argue this further with you. Also learn to use quotes man it's really easy and makes your posts legible.

www.bbcode.org

The problem is that the men are both pulling on the same rope, therefore the rope has one foce (tension) on it. There's no getting past this.

Let's see what happens when you have it your way.


Here is jack's original analysis:

The net force on boat x is -A.
The net force on boat y is -B.
The net force on the string is A+B.
As the string isn't moving, the net force on the string is 0, so A+B=0 so B=-A.

He labeled force A with the wrong direction. ONLY FORCE A.

Here is the correction.

Boat X is pulling with force -A (directed to the left).

Then the net force on boat X will be, according to jack, A.

Boat Y is pulling with force B (directed to the right).

The net force on boat Y will be -B.

The net force on the string will be: -A + B

As the string isn't moving, the net force on the string is 0, then -A + B = 0, so -A = -B, so A=B.


A AND B ARE NO LONGER OPPOSING REACTION FORCES.


jack specifically stated:

A is the action, B is the equal but opposite reaction.

BUT NOW, A = B.

NO LONGER OPPOSING REACTION FORCES.


jack's analysis is a piece of crap.


Using ONLY ONE FORCE, we reach a direct contradiction.

Therefore this kind of analysis is a piece of thrash.


You need to include BOTH FORCES acting on each end of the rope to obtain a correct result.

This means that the two people cannot pull with different forces.

On a lake, two men situated in two boats, SURE THEY CAN.

Rubbish!

Primer for men and ropes
You are on land now. Totally different situation, where now we have to deal with friction.

Ground, water or air, who cares! Friction has no effect of how much force a man can exert.
You said that the weak man could pull only 200 N and that's is the only force that can be applied to his boat.
I could redraw the diagram with the men in boats and it would make no difference at all to the argument.

If the weak man can pull only 200 N on his end of the rope, that is the tension in the rope and the force on his boat.
If the strong man tries to put more tension on the rope it must slip through the weak man's figures.

See my "marlin fishing" analogy.

Let us remember what I wrote earlier.


But the forces applied on each end will always be different.

It takes a single counterexample to show this.

By the very hypothesis, forces A and B are not equal.

They are of different magnitude.


Donald Trump will pull with a different force than Hillary Clinton.

Henry VIII will pull with a different force than Queen Elizabeth I.

You will pull with a different force than any of your relatives, neighbors, countrymen.


Forces A and B WILL ALWAYS BE DIFFERENT.

Force A can never equal force B.

Even if we had, as an example, force A = 100.000,000,000,021 N and force B = 100.000,000,000,034 N, it would still NOT satisfy the RE requirement which is this: A=B.

You can't even have this:

Force A = 100.000,000,001

and

Force B = 100.000,000,000


So jack wrote this:

The single counterexample needs to be physically possible.

Your friend rabinoz, HAS JUST PROVIDED THE PERFECT PHYSICAL SETTING WHERE NOW WE HAVE TWO DIFFERENT FORCES ACTING ON EACH END OF THE ROPE.

An example of applied force being limited is a drag force limited game fishing reel, such as

SHIMANO FORCEMASTER 9000 ELECTRIC REEL
If you pull too hard, you either break the line or rip the hook out and bingo, one big marlin joins the "ones that got away
and are this big.

Find a ".pdf" of the manual in Shimano DENDOUMARU 9000 Beast Master Instruction Manual.

That reel can limit the drag to a maximum of 25 kg, or 245N, so it can be set to limit at your weak man's strength.
A line commonly used with such a rod is rated "PE8" or 80lb (356 N) breaking strain. So setting the reel to limit the drag to 200 N is quite reasonable.

Now, that marlin (our "strong man") can pull with a much larger force, but let's limit him to 350 N to fit.

Vanuatu Fishing

In the example provided by rabinoz, the marlin cannot break free.

Thus we have a situation entirely similar to this: a large boat is pulling on a smaller boat, the rope being tied to the second, smaller boat.

Boat X (the large boat) is pulling only with 200 N (certainly it could pull with a much larger force).

Boat Y (the smaller boat) is exerting a force of 350 N on the rope.


Let us now analyze this example using both sets of equations.



Boat X is pulling with force -200 N (directed to the left).

Then the net force on boat X will be, according to jack, 200 N.

Boat Y is pulling with force 350 N (directed to the right).

The net force on boat Y will be -350 N.

The net force on the string will be: -200 + 350

As the string isn't moving, the net force on the string is 0, then -200 + 350 = 0, so -200 = -350, so 200=350.


Does 200 = 350?

Certainly not.


Let us now apply the CORRECT set of equations.

We are applying now Newton's third law to each end of the rope.


Here is how the balance of forces are to be properly applied.

Boat X is pulling with force A, directed to the left.

Boat Y is pulling with force B, directed to the right.

Forces A and B are, of course, of different magnitude.


What are the forces acting on boat X?

To the left we will have a negative direction.

Boat X will be acted upon by TWO FORCES: 200 N (the reaction force on the action force -200 N) and B = 350 N


What are the forces acting on the left end side of the rope?

-200 and -350.


What are the forces acting on boat Y?

To the right we will have the positive direction.

Boat Y will be acted upon by two forces: -350 (the reaction force on the action force B) and -200.


What are the forces acting on the right end side of the rope?

A and B: A = 200, B = 350


Net force on boat X: A + B = 200 + 350

Net force on boat Y: -A - B = -200 - 350


Net force on the string: [-200 - 350] + [200 + 350]


The string/rope will not move: [-200 - 350] + [200 + 350] = 0


All forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.

I'm going to try to stop repeating myself by refuting your repititions.

jack, you are trying to fool your readers.

A = -B is not the same as A = B.

Thanks for showing your dishonesty once again.
I never said it was.

You have 2 different descriptions of the one problem. To make it even clearly, unlike you who is intentionally trying to confuse people, I will use different letters.
My first analysis simplified a bit:
X is pulling with a force of A, Y is pulling with a force of B.
Thus the net force ont he rope is A+B.
Thus A=-B.

The second option, which is what you are trying to change it in to:
X is pulling with a force of -E, Y is pulling with a force of B.
Thus the net force ont he rope is -E+B.
Thus E=B.

In order to compare the 2 equations, you need to note what each variable is.
You can't just say "This has A, so does this, so these "A"s must be the same.
But E is not a direct replacement of A. Instead, you go from the force X applies being A, to the force X applies being -E.
This means that E=-A.
So subbing that in gives us -A=B, which is the same as saying A=-B.

No contradiction, no problem.
In both cases, you have the force applied by boat X being equal and opposite the force applied by boat Y.

In order for your claims to be true, you need to claim that A=-A.
That is you need to claim that X pulling with a force of A is the same as X pulling with a force of -A.
Are you going to claim that?

You labeled force A with the wrong direction. ONLY FORCE A.

No. I didn't.
A is a placeholder.
It can be positive, it can be negative, it can be 0.
With it just represented as A, it has no sign and thus has no direction.

In the case of your scenario, where say Henry, on boat X is pulling with a force of -350 N, A=-350 N.

A does not need to be a positive number.

A AND B ARE NO LONGER OPPOSING REACTION FORCES.

That's right.
But now X isn't applying a force of A, it is applying a force of -E (which you wish to call -A, but I will not allow anymore due to the confusion you are trying to make).
-E and B are opposing reaction forces as required by Newton's third law.

See how you are desperately trying to surreptitiously dodge the fact that your analysis is a piece of crap?

Again, you are the one dodging here.
I have repeatedly defended my argument and refuted yours.
So far all you have done is baselessly asserted that impossible situations are possible, outright lied about things (or made an extremely dishonest representation), and have just repeated the same refuted crap.
You are unable to answer simple yes or no questions as they would expose the contradictions in your analysis.

YOU CAN'T HAVE IT BOTH WAYS.

Unlike you, I can, just not at the same time/with the same analysis.
With one way, X is pulling with a force of A.
With the other, X is pulling with a force of -E (which you wish to call -A).
In the first case, A=-B, as the opposing reaction forces are A and B.
In the second case, -E=-B, as the opposing reaction forces are -E and B.

Stop trying to treat the 2 different "A"s as if they are the same.
To further this point, stop calling the second one A, instead, call it E.
So with this situation, X is not pulling with -A, X is pulling with a force of -E.
Do you think you can do that, or will that make it to hard for you to come up with lies to spout to try and defend your claims or refute mine?

Yours lead to a magnificient contradiction.

No. Mine does not lead to a contradiction, not in the slightest. The only thing it contradicts is your blatant lies, such as your lies claiming that 2 entities can pull on the rope with forces that are not equal and opposite, and your blatant lies about the analysis where you pretend A is the same -A (or with the new labelling, where you pretend A=E).

BUT NOW A = B.

No. Now E=B, where A=-E.
The A you have in that equation is not the same as the one in A=-B.
Again, to avoid further confusion/lies, call it E from now on. (and after this post I will make no more clarifications regarding that substitution to you and may start editing the posts to show it correctly, especially when you are quoting me).

My equations work just fine.

No. They don't.
You have X applying a force of -E (to avoid any further confusion or your pathetic lies about the analysis, I am no longer using A when the force X applies is deemed to be a negative pro-numeral such as -E) to the rope, and the rope magically recieving a force of -E-B.
How can the rope receive a force of -E-B from X is X is only applying a force of -E?
It requires -E=-E-B.
You also have a complete violation of Newton's third law in a similar manner.
You have the X applying a force of -E to the rope, but the rope applies a force of E+B.
By Newton's third law, these forces must be equal and opposite.
Thus -(-E)=E+B or E=E+B or B=0.

The only way to avoid a contradiction with your analysis is if both E and B are 0.

The other option which will solve these problems is to instead have X applying a force of -E-B to the rope, but the situation demands it applies a force of -E.

So no matter what, as long as A or B is non-zero, you have a contradiction, something you are yet to address.

Boat X will be acted upon by TWO FORCES: A (the reaction force on the action force -A) and B.

But it can't.
X can only be acted upon by a single force, E.
Newton's third law dictates this.
You must have equal and opposite forces existing between any 2 entities.
So if X applies a force of -E to the rope, the rope MUST apply a force of EXACTLY E. Anything else is a direct violation of Newton's third law.

What are the forces acting on the left end side of the rope?
-A and -B.

Again, this is impossible.
X is the only thing acting on the rope, and by the very description of the situation (which you like to call a hypothesis for some reason), X is applying a force of -E to the rope, not -E-B.
As such, the forces acting on the left side of the rope will be -E, not -E-B.
The only way for it to be -E-B is if X is pulling with a force of -E-B, directly violating the description of the situation.

All forces balance out perfectly.

No. The don't.
The only way for them to balance out perfectly is if they are 0, or if X is actually pulling with a force of -E-B, directly contradicting the situation.

But they include TWICE THE FORCES NEEDED in the Newtonian system.

Yes, because you are counting them twice.
Just realise that E=B (By Newton's third law, for this situation which has a massless rope which thus can't have a net force applied to it)
That means you have X pulling with a force of -B, yet magically the rope is receiving a force from X of -B-B.
Similarly, the rope is magically exerting a force of B+B on X.
You are literally counting the forces twice.

That is why your analysis includes twice the forces.

Alternatively, you can consider it as only counting half the force X is applying to the rope in the original description of the situation.

When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

And it doesn't matter which is which.
There is no distinction between an action and a reaction force.
You pulling on the rope because you are pulling on it is no different to you pulling on the rope because someone else is pulling on it.

Boat Y is pulling on the rope and thus boat X with force B.
Reaction force from boat X on the rope: -B.
This is what you are missing.

No, it isn't.
I fully accept that.
The reaction force from boat X on the rope is -B. This is the force B is pulling on the rope with. This is the force -E.

That is what you are missing. This is the same force. It is not 2 separate forces acting independently.

You and your relatives, neighbours, countrymen, will ALWAYS pull WITH DIFFERENT FORCES.

No. Not when pulling a massless string. In that case we will always pull with the exact same magnitude force.
The easiest way to show this is to remove the string entirely and just have them pull on each other.
In this case they are pulling each other with an equal and opposite force as required by Newton's third law.

It is only when you go to real situations, where the rope has mass, which can allow it to have a net force on it and allow the tension to vary along it that you can start to pull with different forces.

I gave you the analysis for that as well (or something similar to it), and you just ignored it. Why was that? Did you not like how it showed you to be completely wrong?

Here it is again (again, with new letters to avoid any possible confusion):
2 boats, X and Y, masses mX and mY respectively, with X on the left (negative position) and Y on the right (positive position).
String of mass mS, with all the mass in the centre of the string for ease of analysis.
All on top of a frictionless lake (so superfluid helium-3? I recommend bring a coat It's a bit chilly, around 2 degrees if I recall correctly.)
X is pulling the rope with a force of -G (see, I'm even being nice and trying to help you understand by having this one be negative).
Y is pulling the rope with a force of H.
The rope is pulling back on X with a force of G, Thus X will be accelerated at a rate of G/mX.
The rope is pulling back on Y with a force of -H. Thus Y will be accelerated at a rate of -H/mY.
The net force on the string is H-G. Thus the string will be accelerated at a rate of (H-G)/mS.

No problem at all.
Notice how now we don't reach the conclusion that G=H?
That is because now the string can be accelerated.
We can also note the tension in the string.
On the left side, it will be |G|. On the right side, it will be |H|.
In reality, the tension in the string will vary along it's length, starting out at |G| on the left and finishing as |H| on the right.

My analysis takes this wonderful fact into account, forces A and B are never the same.

Except for the exact situation you are trying to use this analogy to solve, where the force from gravity will be exactly the same and the string will have no mass at all.

F=GMm/r^2.
This will be the same for each body, they must be EXACTLY the same.
I am sorry jack, the above law is another piece of crap.

No, it isn't. That is just another of your baseless claims.

I can bring here any number of experiments, Lamoreaux, DePalma, Biefeld-Brown, Allais WHICH DEFY THIS "LAW".

Except none actually defy this law.
At best, they show there is some other law at work as well.
For example, the Casmir Effect (by Lamoreaux) has nothing at all to do with gravity.
The Casmir effect is a result of virtual particles due to quantum fluctuations, where 2 plates will restrict the various particles which can exist between the plates, resulting in a force pushing the plates together.
(that's right, something can come from nothing, but if you put something there among the nothing, you reduce what can come from nothing).

It is based upon AREA, not mass like gravity is.
If Gravity was based upon area, then if you took 2 almost identical objects, where one was made of aluminium and the other made of steel, the aluminium one would fall much faster than steel one. It would accelerate at roughly 3 times the rate, as they have the same force acting on them (due to the same area), but a different mass.

But lets not get into this now. I don't want to change the topic. Perhaps once we have dealt with your blatant lies about the laws of motion we can move on to that.

Newton dismissed this law as pure insanity.

No, he dismissed action at a distance with no medium or intermediary as insanity.
We have space itself taking up that role.

And you still cannot have A = B exactly.
It won't happen in the real world.

Like I told you before, go get a spring scale (or 2), get a helper (or 2) put the spring scale (or both in series) between you and one of your helpers, and have you both pull, until you are at a steady state.
See what the force on each scale reads.
Now switch the scales (or people) and try again.
Also note the error of the scales.
It will be the same within error.
In all cases it will be.

A single scale reflects this best.

Or eve better, just go take a basic mechanics class.

You have X pulling on the rope with a force of A, but applying a force of A-B to the rope.
Does A=A-B? Only when B is 0.
You are acting delusional jack.
I never said any such thing.

No, you did. You said X is applying a force of A, which now is -E.
But you said the force on the left side of the rope (which is the force coming from X) is A-B which is now -E-B.
So you have X pulling the rope with a force of A, but magically applying a force of A-B.

What are the forces acting on the left end side of the rope?
-A and -B.

And look, you said it again.
The only place these can come from is boat X.
This means X must be applying a force of -E-B to the rope.
But X is just pulling with a force of -E.

rabinoz, you are dreaming.
We are talking here about AN INVERTED CATENARY.

No. WE are not.
YOU are the only one making such insane claims.

But like I said, deal with your ignorance on the force between the ropes before changing topic.

Newton's third law:
When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

That is a somewhat poor understanding of it as shown by your blatant misuse.
This reaction force is not necessarily a separate force.
For example, if you have X pulling the rope, there can't be another force from X on the rope. The force X is pulling on the rope with is the X side of the action/reaction pair.

Here is an actual quote from Newton:

Law III: To every action there is always opposed an equal reaction: or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.

I think the latter part says it best:
The mutation actions of 2 bodies upon each other are always equal but in opposite direction.

That means if X is applying a force of -E to the rope, and the rope is applying a force of B to X, then these forces must be equal but in opposite directions, i.e. -E=-B.
It is not saying you magically add additional forces.

Perhaps this is an even better quote based upon what we are discussing now:

If a horse draws a stone tied to a rope, the horse (if I may so say) will be equally drawn back towards the stone: for the distended rope, by the same endeavour to relax or unbend itself, will draw the horse as much towards the stone, as it does the stone towards the horse, and will obstruct the progress of the one as much as it advances that of the other.

i.e if you get a rope attached to another object (which can be another person) that object will pull you back equal to what you are pulling it.
i.e. A=-B=-E.

So yes, what you are saying goes directly against Newton's third law.

The single counterexample needs to be physically possible.
Your friend rabinoz, HAS JUST PROVIDED THE PERFECT PHYSICAL SETTING WHERE NOW WE HAVE TWO DIFFERENT FORCES ACTING ON EACH END OF THE ROPE.

No. He didn't.
He provided a case where you or the fish can apply such a great force that you snap the rope.
It is made such that if you or the marlin tries to pull with a force greater than that specified by the setting of the reel, the reel will unwind itself, relaxing that force, in the ideal case, instantly, such that you can never apply a force greater than that.

So no, he has provided a case where the 2 forces are equal, and that they are limited.

Does 200 = 350?
Certainly not.

That's right. Thus your situation is physically impossible as it violate's Newtons' third law of motion.

The problem is that the men are both pulling on the same rope, therefore the rope has one foce (tension) on it. There's no getting past this.
Let's see what happens when you have it your way.

You do that, but make sure it is my way!

Here is jack's original analysis:
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
As the string isn't moving, the net force on the string is 0, then -A + B = 0, so -A = -B, so A=B.


A AND B ARE NO LONGER OPPOSING REACTION FORCES.

If there was no rope and the two were simply pulling on each other A and B would be the reaction forces.
If a man pulls on a wall with force A, the reaction force applied by the wall  is equal in magnitude to A, but opposite in direction.

Putting a rope in between makes no difference. A rope simply transmits force.

jack specifically stated:

A is the action, B is the equal but opposite reaction.

BUT NOW, A = B.

NO LONGER OPPOSING REACTION FORCES.

jack's analysis is a piece of crap.

Using ONLY ONE FORCE, we reach a direct contradiction.

No contradiction. The equal tensions at each end of the rope are the reaction forces.

Therefore this kind of analysis is a piece of thrash.

You need to include BOTH FORCES acting on each end of the rope to obtain a correct result.

No! If you have both forces at each end, you have both pulling with the sum of the forces.
This is contrary to the original hypothesis which stated that the "weak man" can only pull 200 N an the "strong man" only pull 350 N.

You have them each pulling 550 N.

So, try again.

jack, you are really desperate.


Do you really think that your alphabet soup kind of analysis holds with your readers?


The second option, which is what you are trying to change it in to:

Not me. Your own RE friends requested that all of the forces be properly labeled.


X is pulling with a force of -E, Y is pulling with a force of B.
Thus the net force ont he rope is -E+B.
Thus E=B.

By your own hypothesis, E must be equal to -B.

By your own analysis, we get E = B.

A total contradiction.


But E is not a direct replacement of A. Instead, you go from the force X applies being A, to the force X applies being -E.
This means that E=-A.
So subbing that in gives us -A=B, which is the same as saying A=-B.

See how you are trying to manipulate the intelligence of your readers?

IF A = -B, THEN BY YOUR OWN WORDS, SINCE E = -A, WE WILL GET: -E = -B, SO THAT E NOW EQUALS B, CONTRADICTING YOUR OWN HYPOTHESIS, WHERE YOU REQUIRE THAT -E = B.

With one way, X is pulling with a force of A.
With the other, X is pulling with a force of -E (which you wish to call -A).
In the first case, A=-B, as the opposing reaction forces are A and B.
In the second case, -E=-B, as the opposing reaction forces are -E and B.

Have you lost your mind jack?

Let us see what happens using your newly founded opposing forces -E and B.


Boat X is pulling with force -E (directed to the left).

Then the net force on boat X will be, according to jack, E.

Boat Y is pulling with force B (directed to the right).

The net force on boat Y will be -B.

The net force on the string will be: -E + B

As the string isn't moving, the net force on the string is 0, then -E + B = 0, so -E = -B, so E=B.


E AND B ARE NO LONGER OPPOSING REACTION FORCES.


By your own hypothesis, you started with this requirement: -E = B.

Your own analysis leads to this: E = B.

A total contradiction!

Your analysis is a piece of crap jack.

jack, do you understand the meaning of an inverted catenary?

You have a rope being pulled by two boats, distance = 500 meters.

On a flat earth, the rope will be straight.

On a spherical earth, the rope will become a bending line, an inverted catenary.

As simple as this.


How does that rope stay in the shape of a bending line, suspened in midair? By pure magic?



If a horse draws a stone tied to a rope, the horse (if I may so say) will be equally drawn back towards the stone: for the distended rope, by the same endeavour to relax or unbend itself, will draw the horse as much towards the stone, as it does the stone towards the horse, and will obstruct the progress of the one as much as it advances that of the other.

So yes, what you are saying goes directly against Newton's third law.

It does not.

Even here there will be two forces acting on each end of the rope.

You had your way, where you only used one force to analyze this type of situation.

It led to a direct contradiction.

By contrast, my analysis leads to no such contradictions.


X end of the rope: horse is pulling with force -A, force A reacting on the horse, the stone is exerting through the rope a force B on the horse.

Forces acting on the rope at the X end: -A and -B (reaction forces)

Y end of the rope: -B, while the horse is pulling with force -A

Forces acting on the rope at the Y end: A and B

As simple as this.


By the way, this last example constitutes ANOTHER situation where different forces will be applied at end of the rope and where your piece of crap analysis leads to a direct contradiction, while my analysis works beautifully.

Do you really think that your alphabet soup kind of analysis holds with your readers?

I know that it may be confusing, but it is far less confusing than your dishonest ones where you are just using the same letters again and again.
If from the start you said the force X is applying needs to be negative so it is -E, instead of just pretending it is -A, you would then reach a conclusion of E=B, and thus not have the most recent mountain of crap you have been going on about.
It would be easily pointed out that -E=A, thus if E=B, -A=B.

Do you really think your dishonest tactics of pretending 2 different "A"s are the same will work?

The second option, which is what you are trying to change it in to:
Not me. Your own RE friends requested that all of the forces be properly labeled.

No. You. I am fine with it being A, as are all the other rational people here.
The issue was never it being A and B.
The issue was you having both A and B be positive.
If you have A be negative (e.g. -350 N) and B be positive (e.g. 350 N) there is no issue.

X is pulling with a force of -E, Y is pulling with a force of B.
Thus the net force ont he rope is -E+B.
Thus E=B.

By your own hypothesis, E must be equal to -B.

No. By Newton's third law and my analysis, A=-B, and E=B.
Never have I ever indicated that E=-B. It was A that equals -B.
This is because you have X pulling with a force of A, vs X pulling with a force of -E.
A and E are not the same. Instead A=-E.

See, this is why I am demanding you use -E instead of -A, so you can no longer pull this dishonest bullshit and pretend A=E, when it is actually A=-E.

By your own analysis, we get E = B.
A total contradiction.

No contradiction.
On one hand you have A=-B
On the other you have E=B, which is the same as -E=-B.
We note that -E=A and sub that in, then the second becomes A=-B.
No contradiction at all.

Just your pathetic lies.

But E is not a direct replacement of A. Instead, you go from the force X applies being A, to the force X applies being -E.
This means that E=-A.
So subbing that in gives us -A=B, which is the same as saying A=-B.
See how you are trying to manipulate the intelligence of your readers?

Nope. I'm not manipulating anything.
The original situation demanded X pulls with a force of A and Y pulls with a force of B.
In that case, A and B make an action-reaction pair and thus A=-B.

In your new situation you have X pulls with a force of -E and Y pulls with a force of B. (Note that A has been replaced by -E, i.e. A=-E).
In this case -E and B are an action-reaction pair and thus -E=-B.

You are the one trying to manipulate things.
You are changing the description of the situation and then pretending A=E, when it clearly doesn't.

IF A = -B, THEN BY YOUR OWN WORDS, SINCE E = -A, WE WILL GET: -E = -B, SO THAT E NOW EQUALS B, CONTRADICTING YOUR OWN HYPOTHESIS, WHERE YOU REQUIRE THAT -E = B.

Where have I ever required that -E=B?
It has always been -A=B, not -E.

Boat X is pulling with force -E (directed to the left).
Boat Y is pulling with force B (directed to the right).
E AND B ARE NO LONGER OPPOSING REACTION FORCES.

That's right. Do you know why?
Because X isn't pulling with a force of E, it is pulling with a force of -E.
Do you notice the difference?
What this means is that now -E and B are an action-reaction pair. i.e. -E=-B.

By your own hypothesis, you started with this requirement: -E = B.

No. I started with the requirement that A=-B.
Subbing in the fact that -E=A, this gives you -E=-B and thus E=B.
There is no contradiction at all.

jack, do you understand the meaning of an inverted catenary?

Yes. It has nothing to do with reality.
Like I said, deal with your ignorance about the forces between the boat before moving on.

If a horse draws a stone tied to a rope, the horse (if I may so say) will be equally drawn back towards the stone: for the distended rope, by the same endeavour to relax or unbend itself, will draw the horse as much towards the stone, as it does the stone towards the horse, and will obstruct the progress of the one as much as it advances that of the other.

So yes, what you are saying goes directly against Newton's third law.

It does not.

Even here there will be two forces acting on each end of the rope.

No, there isn't.
You have the horse pulling with some force, and the stone/rope pulling back with an equal and opposite force.

By the way, this last example constitutes ANOTHER situation where different forces will be applied at end of the rope and where your piece of crap analysis leads to a direct contradiction, while my analysis works beautifully.

No, it doesn't. As per Newton's third law, the horse and stone are pulling with equal but opposite forces. There is no magic doubling of forces.

jack, you are truly delusional.

Do you really think that your tricks substituting various letters in your piece of crap analysis will work with me or with your readers?

Think again.

Let us now go back to your original piece of crap analysis.

The net force on boat x is -A.
The net force on boat y is -B.
The net force on the string is A+B.
As the string isn't moving, the net force on the string is 0, so A+B=0 so B=-A.

You told your readers specifically:

A is the action, B is the equal but opposite reaction.

You reached the following conclusion: B=-A.


You also specified:

For every action (A) there is an equal but opposite reaction (B). i.e. |A|=|B| (or more specifically, A=-B).


But your force was mislabeled.


Boat X is pulling with force -A.

Here is the correction.

Boat X is pulling with force -A (directed to the left).

Then the net force on boat X will be, according to jack, A.

Boat Y is pulling with force B (directed to the right).

The net force on boat Y will be -B.

The net force on the string will be: -A + B

As the string isn't moving, the net force on the string is 0, then -A + B = 0, so -A = -B, so A=B.


A AND B ARE NO LONGER OPPOSING REACTION FORCES.


And still you have the same problem as before: your analysis applies only to the case where A = B.

That is, where the forces applied at each end of the rope must be exactly the same.


Which can never be the case.


You deem yourself to be expert on Newton's third law, yet you are unable to understand the most basic facts about this law.


Your analysis leads to the case where you claim both forces applied must be the same.


In reality both forces will always differ, even if by an infinitesimal amount.

That is why you need an analysis which takes this very fact into account.

Now a specific example.

A large boat is pulling on a smaller boat, the rope being tied to the second, smaller boat.

Boat X (the large boat) is pulling only with 200 N (certainly it could pull with a much larger force).

Boat Y (the smaller boat) is exerting a force of 350 N on the rope.


Let us now analyze this example using both sets of equations.



Boat X is pulling with force -200 N (directed to the left).

Then the net force on boat X will be, according to jack, 200 N.

Boat Y is pulling with force 350 N (directed to the right).

The net force on boat Y will be -350 N.

The net force on the string will be: -200 + 350

As the string isn't moving, the net force on the string is 0, then -200 + 350 = 0, so -200 = -350, so 200=350.


Does 200 = 350?

Certainly not.


Let us now apply the CORRECT set of equations.

We are applying now Newton's third law to each end of the rope.


Here is how the balance of forces are to be properly applied.

Boat X is pulling with force A, directed to the left.

Boat Y is pulling with force B, directed to the right.

Forces A and B are, of course, of different magnitude.


What are the forces acting on boat X?

To the left we will have a negative direction.

Boat X will be acted upon by TWO FORCES: 200 N (the reaction force on the action force -200 N) and B = 350 N


What are the forces acting on the left end side of the rope?

-200 and -350.


What are the forces acting on boat Y?

To the right we will have the positive direction.

Boat Y will be acted upon by two forces: -350 (the reaction force on the action force B) and -200.


What are the forces acting on the right end side of the rope?

A and B: A = 200, B = 350


Net force on boat X: A + B = 200 + 350

Net force on boat Y: -A - B = -200 - 350


Net force on the string: [-200 - 350] + [200 + 350]


The string/rope will not move: [-200 - 350] + [200 + 350] = 0


All forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.

Here it is again (again, with new letters to avoid any possible confusion):
2 boats, X and Y, masses mX and mY respectively, with X on the left (negative position) and Y on the right (positive position).
String of mass mS, with all the mass in the centre of the string for ease of analysis.
All on top of a frictionless lake (so superfluid helium-3? I recommend bring a coat It's a bit chilly, around 2 degrees if I recall correctly.)
X is pulling the rope with a force of -G (see, I'm even being nice and trying to help you understand by having this one be negative).
Y is pulling the rope with a force of H.
The rope is pulling back on X with a force of G, Thus X will be accelerated at a rate of G/mX.
The rope is pulling back on Y with a force of -H. Thus Y will be accelerated at a rate of -H/mY.
The net force on the string is H-G. Thus the string will be accelerated at a rate of (H-G)/mS.

Wrong example.


My equations state quite clearly that the net force on the string will be zero.

Equations do not lie.

Here is how the balance of forces are to be properly applied.

Two boats pulled toward each other on a lake.

Man from boat X is pulling with force A, directed to the left.

Man from boat Y is pulling with force B, directed to the right.

Forces A and B are, of course, of different magnitude.


What are the forces acting on boat X?

To the left we will have a negative direction.

Boat X will be acted upon by TWO FORCES: A (the reaction force on the action force -A) and B.


What are the forces acting on the left end side of the rope?

-A and -B.


What are the forces acting boat Y?

To the right we will have the positive direction.

Boat Y will be acted upon by two forces: -B (the reaction force on the action force B) and
-A.


What are the forces acting on the right end side of the rope?

A and B.


Net force on boat X: A + B

Net force on boat Y: -A - B


Net force on the string: [-A - B] + [A + B]


The string/rope will not move: [-A - B] + [A + B] = 0


All forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.



When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.


Boat Y is pulling on the rope and thus boat X with force B.

Reaction force from boat X on the rope: -B.


This is what you are missing.



The net force on the string will be zero.

Moreover, we have been through this before: the "massless" rope approximation.

The tension in a rope HAS TO vary across its length in reality, but this is overlooked in modern physics.

So it is assumed the mass of a rope is negligible, and using the massless approximation won't cause too big of an error.

Do you understand these basic things?

The rope can be stretchy/springy.

This is as yet an unaccounted for problem in modern physics.


Certainly the rope HAS TO TRANSMIT BOTH FORCES APPLIED AT EACH END, SINCE THE NET FORCES, IN THE CORRRECT ANALYSIS WORK OUT FINE.



If Henry VIII and Queen Elizabeth I would be pulling, each located at one end of that rope, would those forces be the same? Certainly not.

If Donald Trump and Hillary Clinton would be pulling, each located at one end of that rope, would those forces be the same? Certainly not.

If Albert Einstein and Marie Curie would be pulling, each located at one end of that rope, would those forces be the same? Certainly not.


Forces A and B can never be the same.


This is what you put before your readers jack.

As the string isn't moving, the net force on the string is 0, so A+B=0 so B=-A.


That is: |A|=|B|


Do you fully understand the meaning of an equal sign?

|A|=|B|

No exceptions.

No other possibilities.

ONLY ONE CASE IS TO BE TAKEN INTO CONSIDERATION, UNDER JACK'S PIECE OF THRASH ANALYSIS: |A|=|B|.


But, in reality that could never be the case.


Again, do you fully understand the meaning of an equal sign?


Even if boat X was pulling with a force 100.000,000,000,000,021 N and boat Y was pulling with a force of 100.000,000,000,000,334 N, the requirements of jack's piece of garbage analysis COULD NEVER BE FULFILLED.


In each and every instant/situation where two people pull on a rope, two on a lake, the FORCES APPLIED WILL BE DIFFERENT.


This means the RE analysis, the piece of thrash posted by jack, IS INAPPLICABLE TO ANY REAL LIFE SITUATION.

Do you really think that your tricks substituting various letters in your piece of crap analysis will work with me or with your readers?

You, I highly doubt it. You seem to be too fixed on ignoring reality. But for other people, it does show your dishonesty far more clearly.
With making the force A and -E instead of A and -A, it is much clearer that the 2 are different, and means your analysis of my arguments doesn't work at all, so now you are reduced to blatantly lying about them.

You reached the following conclusion: B=-A.

Yes, I know that. So far you have been unable to refute it.

You also specified:
For every action (A) there is an equal but opposite reaction (B). i.e. |A|=|B| (or more specifically, A=-B).

Yes, I know that. That is Newton's third law.

But your force was mislabeled.

No it wasn't.
One is a positive force one is a negative force.

If you think they are mislabelled, feel free to explain why rather than continually assert that they are.
Remember, A and B can be positive or negative (but you must have one of each).
For example, you can have A being -350 N and B being 350 N. Or you can have it the other way around (now with boat X on the right for it to be pulling), and thus A is 350 N and B is -350 N.

So if A is -350 N and thus is a force directed to the left, how is it mislabelled?

Here is the correction.

No, here is the blatant lie about it, where you mix up boat X pulling with a force of A and -E.

Boat X is pulling with force -E.
Boat Y is pulling with force B.
As the string isn't moving, the net force on the string is 0, then -E + B = 0, so -E = -B, so E=B.
E AND B ARE NO LONGER OPPOSING REACTION FORCES.

Again, that is correct.
This isn't surprising at all.
As you have boat X pulling with a force of -E and boat Y pulling with a force of B, the action-reaction pair will be the forces -E and B. They will not be E and B.

What this means is that -E=-B, thus E=B.

And still you have the same problem as before: your analysis applies only to the case where E = B.

Yes, I have the same "problem" as before, where my analysis only works for the physically possible cases, where the forces obey Newton's third law. It wont work for the impossible cases where the massless rope gets a net force applied to it and accelerates to infinity and beyond.

In reality both forces will always differ, even if by an infinitesimal amount.

Not for the situation we are describing.
They will differ for the case of a rope with mass which has the tension vary across it.

This is not the situation we are describing. But I have provided that analysis.

Now a specific example.
A large boat is pulling on a smaller boat, the rope being tied to the second, smaller boat.
Boat X (the large boat) is pulling only with 200 N (certainly it could pull with a much larger force).
Boat Y (the smaller boat) is exerting a force of 350 N on the rope.

And yet again it is an impossible situation where the forces violate Newton's third law.

Does 200 = 350?
Certainly not.

That's right, it doesn't. This indicates the situation is physically impossible.

Let us now apply the CORRECT set of equations.

You mean your blatantly incorrect one?

We are applying now Newton's third law to each end of the rope.
Boat X will be acted upon by TWO FORCES: 200 N (the reaction force on the action force -200 N) and B = 350 N

And there you go throwing Netwon's third law out the window.
By Newton's third law, the force X is applying (-200 N) must be equal and opposite the force the rope is applying (350 N + 200 N).
i.e. 250=550.
Is that possible?
Certainly not.
As such, your analysis results in a direct violation of Newton's third law of motion.

-200 and -350.

But the only place this can come from is boat X.
Boat X is applying a force of -200 N to the rope.
Thus your analysis requires -200 N (the force X is applying to the rope)is equal to-200 N -350 N (the force the rope is receiving from X).
i.e. -200=-550.
Is that possible?
Certainly not.

As such, your situation is entirely impossible.

Equations don't lie, and they show your situation to be impossible.

Unlike my analysis which just requires the situation to be physically possible and have A=-B (or E=B), your analysis requires they are both 0.

Here it is again (again, with new letters to avoid any possible confusion):
2 boats, X and Y, masses mX and mY respectively, with X on the left (negative position) and Y on the right (positive position).
String of mass mS, with all the mass in the centre of the string for ease of analysis.
All on top of a frictionless lake (so superfluid helium-3? I recommend bring a coat It's a bit chilly, around 2 degrees if I recall correctly.)
X is pulling the rope with a force of -G (see, I'm even being nice and trying to help you understand by having this one be negative).
Y is pulling the rope with a force of H.
The rope is pulling back on X with a force of G, Thus X will be accelerated at a rate of G/mX.
The rope is pulling back on Y with a force of -H. Thus Y will be accelerated at a rate of -H/mY.
The net force on the string is H-G. Thus the string will be accelerated at a rate of (H-G)/mS.

Wrong example.


My equations state quite clearly that the net force on the string will be zero.

Yes, your equations do state that, and guess what? They are wrong.
The only way for the net force on the string to be 0 is if the 2 boats are pulling it with equal but opposite force.
If they are not, then there will be a net force on the string and the string will be accelerated.

You can't have it both ways.
In a real situation the forces will differ at each end and thus there will be a net force on the rope.

Equations do not lie.

But you do, and you have done so repeatedly.

Here is how the balance of forces are to be properly applied.

You mean where you leave out the mass of the string, and just assert that there will always be a net force of 0?

What you are claiming is that if you get 1000 ships pulling a boat in one direction, and a single ship pulling it in the other, that boat wont move, regardless of what force is applied by the ships.
Do you really want to go down that path?

But they include TWICE THE FORCES NEEDED in the Newtonian system.

Again, that is because you are counting them twice.
For example, while X pulls on the rope with a force of A=-E=-B, you have the rope magically receive a force of 2A (or -2B or -2E) and boat X receive a force of -2A (or 2B or 2E) from the rope.
It isn't that surprising that if you count the forces twice, you end up with twice the force.

When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

Yes, and that isn't a magic extra force.

Moreover, we have been through this before: the "massless" rope approximation.

The tension in a rope HAS TO vary across its length in reality, but this is overlooked in modern physics.

That's right, and that is why I provided my analysis where the rope has a mass and thus can have a net force be applied to it, but you just rejected that.

So it is assumed the mass of a rope is negligible, and using the massless approximation won't cause too big of an error.

Yes, and thus it is assumed that the difference in magnitude of the forces is negligible and using the massless approximation where the forces are equal in magnitude wont cause too big of an error.

Certainly the rope HAS TO TRANSMIT BOTH FORCES APPLIED AT EACH END, SINCE THE NET FORCES, IN THE CORRRECT ANALYSIS WORK OUT FINE.

The correct analysis only works out fine when A=-B=-E. Anything else results in a physical impossibility for a massless rope.
The force applied at each end is the tension in the rope.
For a massless rope, it cannot change throughout the length of the rope.

For a real rope, with mass, in a real situation, the tension will vary, so if you had one end pulling with a force of -350 N and the other pulling with a force of 250 N, the rope will have a tension of 350 N at one end and 250 N at the other end. Only the lower force of 250 N will be transmitted through the rope. The rest, the remaining 100 N, will be a net force on the rope causing the rope to accelerate.

Do you fully understand the meaning of an equal sign?

Yes. Do you fully understand the meaning of the massless rope approximation and how it varies from reality?

ONLY ONE CASE IS TO BE TAKEN INTO CONSIDERATION, UNDER JACK'S PIECE OF THRASH ANALYSIS: |A|=|B|.

Yes, one case, which is physically possible, with a massless rope. Try getting that.

Instead you just repeatedly appeal to physically impossible situations that do nothing to disprove my analysis.

Even if boat X was pulling with a force 100.000,000,000,000,021 N and boat Y was pulling with a force of 100.000,000,000,000,334 N, the requirements of jack's piece of garbage analysis COULD NEVER BE FULFILLED.

And that couldn't happen in reality with a massless rope.
Only in reality, where the rope has a mass, the tension varies accross it and it is experiencing a net force of 0.000000000000313 causing it to be accelerated.
If the rope was massless, then it would need to accelerate at an infinite rate.

In each and every instant/situation where two people pull on a rope, two on a lake, the FORCES APPLIED WILL BE DIFFERENT.

And the rope wont be massless and thus a net force will apply to it, and thus the massless rope approximation wont work, and thus you wont end up with A=-B and no net force on the rope.
Instead you get a net force on the rope of A+B.

jack, slow down.

You are demanding of your readers to believe in the impossible.

Here is what you wrote.

That's right, it doesn't. This indicates the situation is physically impossible.

You have constructed your own little universe, your own reality, where two different persons, of different strengths, are to pull with the very same forces on both ends of the rope.

But those two persons will always pull with a different force, no matter what you say.


You will pull with a different force, than say, your relatives, your neighbours, or your own countrymen.


Your analysis requires this: that the absolute value of A equal the absolute value of B.


In my simple example we have boat X pulling with 200 N, while the second boat will exert a force of 350 N.


Using your analysis we reach the point where 200 = 350.

A contradiction.


My equations, by contrast, work just fine.

No contradiction.

Both forces are included.


And there you go throwing Netwon's third law out the window.
By Newton's third law, the force X is applying (-200 N) must be equal and opposite the force the rope is applying (350 N + 200 N).
i.e. 250=550.
Is that possible?
Certainly not.
As such, your analysis results in a direct violation of Newton's third law of motion.

Nothing was thrown out the window jack.

What are the forces acting on boat X?

To the left we will have a negative direction.

Boat X will be acted upon by TWO FORCES: 200 N (the reaction force on the action force -200 N) and B = 350 N


What are the forces acting on the left end side of the rope?

-200 and -350.


What are the forces acting on boat Y?

To the right we will have the positive direction.

Boat Y will be acted upon by two forces: -350 (the reaction force on the action force B) and -200.


What are the forces acting on the right end side of the rope?

A and B: A = 200, B = 350


Net force on boat X: A + B = 200 + 350

Net force on boat Y: -A - B = -200 - 350


Net force on the string: [-200 - 350] + [200 + 350]


The string/rope will not move: [-200 - 350] + [200 + 350] = 0


All forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.


These equations do not lie jack.

The net force on the string will be zero, as required.


But the only place this can come from is boat X.
Boat X is applying a force of -200 N to the rope.
Thus your analysis requires -200 N (the force X is applying to the rope)is equal to-200 N -350 N (the force the rope is receiving from X).
i.e. -200=-550.
Is that possible?
Certainly not.

As such, your situation is entirely impossible.

No such thing happened.

Please read carefully, if you can.


What are the forces acting on boat X?

To the left we will have a negative direction.

Boat X will be acted upon by TWO FORCES: 200 N (the reaction force on the action force -200 N) and B = 350 N


What are the forces acting on the left end side of the rope?

-200 and -350.


What are the forces acting on boat Y?

To the right we will have the positive direction.

Boat Y will be acted upon by two forces: -350 (the reaction force on the action force B) and -200.


What are the forces acting on the right end side of the rope?

A and B: A = 200, B = 350


Each and every force is being taken into account, using Newton's third law.

By hypothesis, the forces applied by the two boats will differ in magnitude, the very case your own analysis cannot handle.

The net force on the string will be zero, as required.

HAD THERE BEEN ANY CONTRADICTIONS, THE FINAL NET FORCE ON THE ROPE EQUATION WOULD HAVE REFLECTED ANY POSSIBLE INCONSISTENCIES.

If they are not, then there will be a net force on the string and the string will be accelerated.

Guess what?

My equations have different forces applied at each end, and the net force on the string will be zero.

Net force on boat X: A + B = 200 + 350

Net force on boat Y: -A - B = -200 - 350


Net force on the string: [-200 - 350] + [200 + 350]


The string/rope will not move: [-200 - 350] + [200 + 350] = 0


You have to apply Newton's third law AT BOTH ENDS OF THE ROPE.

This is what you are missing.


You tried the BS with the massless rope before it did not work out, it won't work now either.

The tension in a rope HAS TO vary across its length in reality, but this is overlooked in modern physics.

So it is assumed the mass of a rope is negligible, and using the massless approximation won't cause too big of an error.

Do you understand these basic things?

The rope can be stretchy/springy.

This is as yet an unaccounted for problem in modern physics.


Certainly the rope HAS TO TRANSMIT BOTH FORCES APPLIED AT EACH END, SINCE THE NET FORCES, IN THE CORRRECT ANALYSIS WORK OUT FINE.



If Henry VIII and Queen Elizabeth I would be pulling, each located at one end of that rope, would those forces be the same? Certainly not.

If Donald Trump and Hillary Clinton would be pulling, each located at one end of that rope, would those forces be the same? Certainly not.

If Albert Einstein and Marie Curie would be pulling, each located at one end of that rope, would those forces be the same? Certainly not.


Forces A and B can never be the same.


Here is how nicely my equations account for all of these forces:

Net force on boat X: A + B

Net force on boat Y: -A - B


Net force on the string: [-A - B] + [A + B]


The string/rope will not move: [-A - B] + [A + B] = 0


All forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.


By contrast, you are requiring that |A|=|B|.

In the real world, the two persons will apply different forces on each end of the rope.