Distances in the universe

But it is gravity pulling the rope down into a catenary shape.

It is the combination of the tension in the rope (here a fine strand) and gravity acting on the mass of the rope making the catenary shape.


You have provided NO explanation at all.

HOW does gravity pull the rope to bend into a catenary shape?

That rope will become a bending line.

How a does a bending rope transmit the forces applied on it?

Your bibliographical references DO NOT explain HOW gravity provides the magical force to bend a straight rope into a bending rope (catenary).

How about you stop trying to change the topic and deal with your massive failure?

What is the force X is applying to the rope (i.e. pulling on the rope with)? Is it A? -A? A-B? -A-B? Something else?

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.

Newton's third law says 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.


Then, YOU FAILED TO CORRECTLY APPLY THIS LAW TO THE TWO BOATS ON A LAKE EXAMPLE.


Then the net force on the rope from boat Y will be B.
Thus the total net force on the rope will be -A+B. There is no other force available to do anything to the rope.

Boat X is pulling with a force of -A (remember, it was your friends here that demanded the direction of the force be properly accounted for, so I fulfilled their wish accordingly).

Reaction force from the rope: A

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

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


Very simple.


SEE HOW YOU FAILED TO APPLY THE 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.


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.


Newton's third law dictates A=-B.
You are the one that is claiming Newton's third law is wrong.

Newton's third law is fine.

It is you who has provided a piece of crap analysis.


The RE requested that I introduce the proper labeling of forces (their direction) in the analysis.

Let us see now what happens when we apply this basic requirement to jack's 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.


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.


This is priceless.

Amazing.


jack's horrible piece of crap, with the forces correctly labeled, has reached the amazing conclusion that A=B.


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|.

The RE analysis leads directly to the ONLY case which can never be experienced in reality.


Newton's third law dictates A=-B.

It does no such thing.

Newton's third law simply states 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.


Your piece of crap analysis reached this conclusion:

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.

Another contradiction.

You said this: Newton's third law dictates A=-B.

But now, by your own priceless analysis, A=B!


Newton's third law has to be properly applied.

My equations work out perfectly, by contrast.

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


No contradictions whatsoever.

There's a massive contradiction. You say first that the force applied to the rope by the person on boat X is -A, and then you say it's -A-B. Can't have both.

One clarification: in your example, can the person move independently of the boat, or are his feet glued onto it?

There's a massive contradiction. You say first that the force applied to the rope by the person on boat X is -A, and then you say it's -A-B. Can't have both.

There is no contradiction because I said no such thing.


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 RE requested that I introduce the proper labeling of forces (their direction) in the analysis.

Let us see now what happens when we apply this basic requirement to jack's 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.


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.


This is priceless.

Amazing.


jack's horrible piece of crap, with the forces correctly labeled, has reached the amazing conclusion that A=B.


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.



Let us see now what happens further with jack's piece of thrash analysis.


The net force on boat x is -A.
The net force on boat y is A.
The net force on the string is A-A=0.

Let us apply the correct labeling of forces again.


The net force on boat x is A.
The net force on boat y is A.
The net force on the string is (-A - A)=0, so -2A=0.

What a piece of crap.

Not only is jack's analysis contradicting his basic requirement that A=-B, but also his forces now do not cancel each other to add up to zero; instead they add up to -2A!


Let us remember what jack said:

Newton's third law dictates A=-B.

But now, having properly labeled the forces his piece of crap analysis leads to this:

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.

It doesn't get any worse than this for the RE: a total and most direct contradiction.

...

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.
...

you still have to shown a prove for that.

ignoring fact does not change facts.

that you are ignoring to show us prove of your claim shows that you know that you are wrong.

Any two human beings, when forces will be analyzed at the most infinitesimal level (we are talking here about the 100,000,000,000,000,000,....th fraction of a Newton, will be pulling with DIFFERENT FORCES.

The basic RE requirement is now this (see my previous message):

A = B

Henry VIII and Queen Elizabeth I will apply different forces.

Donald Trump and Hillary Clinton will apply different forces.

So will you and any of your relatives, neighbors, countrymen.


The burden of proof IS ON YOU, NOT ME.

You have to prove that in each and every situation, the two forces will be TOTALLY AND EXACTLY THE SAME.


You can't even have this:

Force A = 100.000,000,001

and

Force B = 100.000,000,000


It won't satisfy the RE requirement as listed above.


You are out of luck on this one.

Any two human beings, when forces will be analyzed at the most infinitesimal level (we are talking here about the 100,000,000,000,000,000,....th fraction of a Newton, will be pulling with DIFFERENT FORCES.

The basic RE requirement is now this (see my previous message):

A = B

Henry VIII and Queen Elizabeth I will apply different forces.

Donald Trump and Hillary Clinton will apply different forces.

So will you and any of your relatives, neighbors, countrymen.


The burden of proof IS ON YOU, NOT ME.

You have to prove that in each and every situation, the two forces will be TOTALLY AND EXACTLY THE SAME.


You can't even have this:

Force A = 100.000,000,001

and

Force B = 100.000,000,000


It won't satisfy the RE requirement as listed above.


You are out of luck on this one.

that is the max force that the person can pull

at the end at the rope only the smaller force is pulling.

as i said show us a rope where there is two different forces at each end pulling by showing us as a real experiment.

the reality counts, not what you make up in your head.

Somebody earlier mentioned that if force A is applied on only one boat, the other boat will not move.

Uh, yes it will. In an environment with no friction, a force of -A will be applied on the other side.

So the equation A+-A would equal 0, making the rope not move whatsoever, which means both boats would move at the same rate towards each other.

Now, say force B was applied on one boat and force A was applied on the other. Then, the following equation would be treu where X is the distance the boat moved.

(A-B)-(B-A)=X

A-B would be the force applied on boat 1, and B-A would be the force applied on boat 2. Therefore, X would equal zero, and the rope would not move.

The RE requested that I introduce the proper labeling of forces (their direction) in the analysis.

Let us see now what happens when we apply this basic requirement to jack's 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.


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.


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.

There is no contradiction because I said no such thing.

You did. That's what you're saying. It's just that you're confused, and don't really understand what you're saying.

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

You can't have that, unless someone's losing grip, but I'll let it be for now, because I want to show you the contradiction in your method.

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

Which combine of course into a single force. Let's call it C. Let's also call the force on boat Y D. D and C are of equal magnitude. C is exerted by the feet of the person on the boat. Therefore there's a reaction to the force C, which is -C, that is applied on the person, to the direction that the person is pulling. THIS force is the force the person exerts to the rope. THIS force is the force with which the person pulls the rope. And it's equal in magnitude to the force with which the person on boat Y is pulling against the rope. So here you have your contradiction.

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

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

Exactly. Boat X is pulling on the rope with a force of magnitude B, and boat Y is pulling on the rope with a force of magnitude B. Equal. You can't have them be different, otherwise the rope would move, someone would lose grip, and that sort of stuff.

Any two human beings, when forces will be analyzed at the most infinitesimal level (we are talking here about the 100,000,000,000,000,000,....th fraction of a Newton, will be pulling with DIFFERENT FORCES.

You are wondering how it's possible for two humans to be pulling with the exact same force. So I'm asking you, how is it possible that you're pushing against the floor with your legs with exactly the same force as the floor is pushing you upwards? An even better example:


How is the acrobat on the top able to push downwards with her hands with exactly the same force as the one below her pushes her upwards with her feet?

BTW these are some pretty sweet acrobatics if you ask me.

Better yet: let's say you're in a skate park and you're pushing against Intikam's hands, and Intikam is pushing back against you. Do you realize why the force you are exerting on Intikam is exactly the same he is exerting on you? Do you understand why the force with which you're pushing him is exactly the same as the force with which he is pushing you?

There's a massive contradiction. You say first that the force applied to the rope by the person on boat X is -A, and then you say it's -A-B. Can't have both.

I never said any such thing.

Let me remind you.

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.


Which combine of course into a single force. Let's call it C. Let's also call the force on boat Y D. D and C are of equal magnitude.

When you use a single force you will reach a direct contradiction.

Remember this?

Boat X pulls the rope with force F
The rope pulls boat X with force -F.
The rope pulls boat Y with force F.
Boat Y pulls the rope with force -F.

The net force on boat X is -F.
The net force on boat Y is F.
The net force on the rope is F-F=0.


By the very hypothesis, A DOES NOT EQUAL B.

A cannot equal B.



Exactly. Boat X is pulling on the rope with a force of magnitude B, and boat Y is pulling on the rope with a force of magnitude B.

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

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


The situations that you have described next are DIFFERENT from the two boats on a lake scenario.

No two persons can pull with exactly the same force as the RE analysis requires.

Let me remind you.

I don't want you to remind me anything. You've said the same things a million times already. Please make smaller posts and omit unnecessary stuff that you've already said.

When you use a single force you will reach a direct contradiction.

You do know that any two forces can be combined into single force, right? PLEASE don't tell me you want that explained as well.

By the very hypothesis, A DOES NOT EQUAL B.

I told you what happens with this hypothesis. The force on the rope isn't balanced. If two different forces are acting on two ends of the rope, there will be a non zero net force acting on the rope, and the rope will move. It's that simple. 

The situations that you have described next are DIFFERENT from the two boats on a lake scenario.

Of course they're different. I thought that maybe you'd understand them if they're different. But the principle is the same.

No two persons can pull with exactly the same force as the RE analysis requires.

Do you realize that "pulling" could also be considered providing resistance by gripping to the rope, right? You know how static friction works, right? Now think about the examples I gave you with the acrobats.

You say that A cannot equal B, when you specifically stated that A=B earlier. It cannot. A-B=B-A, representing the total forces acting on each boat separately. They truly are equal, showing that the forces acting on the string are equal, and the string will not move.

You say that A cannot equal B, when you specifically stated that A=B earlier. It cannot. A-B=B-A, representing the total forces acting on each boat separately. They truly are equal, showing that the forces acting on the string are equal, and the string will not move.

i think Sandy knows that the forces on both side are equal, he ignores to make a experiment himself that would proof it.
but than he would have to admit that his view of a flat earth without gravity is wrong.

I thought we were talking about a hypothetical setting with no friction, just to discuss the forces involved.

I thought we were talking about a hypothetical setting with no friction, just to discuss the forces involved.

it does not matter, in both cases the assumption from Sandy is wrong.

Quote

By the very hypothesis, A DOES NOT EQUAL B.

I told you what happens with this hypothesis. The force on the rope isn't balanced. If two different forces are acting on two ends of the rope, there will be a non zero net force acting on the rope, and the rope will move. It's that simple.

Um... no, it's not. The force of the rope is actually balanced, according to one of Newton's laws of motion:

For every action, there is an equal and opposite reaction.

So, the forces acting on boat X are A-B, with A being the force acted on boat X and -B being the equal, opposite force from the force on boat Y. The same goes for boat Y, which ends with the following equation:

A-B=B-A

So therefore, the rope WILL NOT move

Flat Earth is not Real.
The Earth is Spherical, not 100% sphere, just like the moon. If you find a telescope and view the moon, you can see the edge turning aside. The same as Earth.

It just seemed flat because you only see the surroundings you see but if you were on the tallest Building beside a sea, you could see ships far away coming out like it came from under the sea but really it is because the farther it is, the roundness shows up, just like how the Moon or Sun Sets/Rise.

Flat Earth is not Real.
The Earth is Spherical, not 100% sphere, just like the moon. If you find a telescope and view the moon, you can see the edge turning aside. The same as Earth.

It just seemed flat because you only see the surroundings you see but if you were on the tallest Building beside a sea, you could see ships far away coming out like it came from under the sea but really it is because the farther it is, the roundness shows up, just like how the Moon or Sun Sets/Rise.

Please find a different forum, we're arguing about something else right now.

There's a massive contradiction. You say first that the force applied to the rope by the person on boat X is -A, and then you say it's -A-B. Can't have both.

I never said any such thing.
Let me remind you.
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.

No two persons can pull with exactly the same force as the RE analysis requires.

Firstly, it is not a "RE analysis", it is a "Physics analysis"using Newton's Laws and some elementary mechanics.

But, this 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 even a helicopter,
  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.

An almost trivial example of this is a force limited fishing reel, such as
Now, can everyone point out if they agree with these basic points? If we can't even do that this can get nowhere.

I agree with everything you just said, but using our hypothetical scenario, our rope is massless and tensionless. We are simply trying to debate the forces acting on either boat, when two people pulling at different forces begin to pull on the rope, which means the rope will not move.

Somebody earlier mentioned that if force A is applied on only one boat, the other boat will not move.

That is because it wont.

Uh, yes it will. In an environment with no friction, a force of -A will be applied on the other side.

Yes, if one boat is pulling with a force of A, the other boat will be applying a force of -A to the rope, if it is attached to the rope.
Thus you won't have a situation where just force A is applied to one boat.
Instead you have the situation where one boat is pulling with A while the other is pulling with -A.

If only one side is applying a force to the rope, then the other side CANNOT be applying any force to the rope, and thus by Newton's third law, the rope cannot be applying any force to the boat and thus the rope will merely move off the boat without taking the boat with it.

So the equation A+-A would equal 0, making the rope not move whatsoever, which means both boats would move at the same rate towards each other.

Again, only if one boat is applying a force of A, and the other is applying a force of -A.
If it is just one side applying a force of A, then the rope has a net force of A and it will move.

Now, say force B was applied on one boat and force A was applied on the other. Then, the following equation would be treu where X is the distance the boat moved.
(A-B)-(B-A)=X

Again, it wouldn't be.
If force A is applied by one boat and B was applied by the other, then the net force on the rope would be A+B, and thus the rope would move.

You are describing a situation where one boat applies a force of A-B and the other applies a force of B-A.

A-B would be the force applied on boat 1, and B-A would be the force applied on boat 2. Therefore, X would equal zero, and the rope would not move.

And A-B would be the force applied by boat 2 and B-A would be the force applied by boat 1. Thus you don't have one boat pulling with force A and the other pulling with force B.

One boat would only be putting a force of A on the rope, this is true, but also a force of -B would be applied to the same side from the equal opposite force from the other side.

And as far as just force A on one side and no force on the other, we're assuming the rope is tied to the other boat.

Newton's third law says 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.

Yes, I know Newton's third law.
2 simultaneous forces such that A=-B.

Then, YOU FAILED TO CORRECTLY APPLY THIS LAW TO THE TWO BOATS ON A LAKE EXAMPLE.

No. I correctly applied it, ending up with the conclusion that A=-B, just as the law dictates.

On the other hand, you completely violated it, saying that X is pulling on the rope with a force of A, but the rope is pulling back with a force of -A+B (or with your new ones, X is pulling with a force of -A and the rope is pulling back with a force of A+B.

Are A and -A+B (or -A and A+B) equal and opposite? No. As such, you are violating Newton's third law.
The only way out of it is if you admit that X is actually pulling on the rope with a force of A-B (or -A-B in your new labels).
So do you admit that X is actually pulling on the rope with a force of A-B thus violating the description of the scenario, or do you admit that you are violating Newton's third law as you have unbalanced action-reaction pairs?

remember, it was your friends here that demanded the direction of the force be properly accounted for, so I fulfilled their wish accordingly.

No. You didn't.
You didn't need to change A to -A.
A can be negative.
For example, Henry is pulling on the rope with a force of A, where A=-350 N.
There is no need to have A be negative.

However making A negative does show your dishonesty even more, as it shows your double counting more easily, as you continually have expressions of the form A+B, almost like A and B are the same thing and you are just counting them twice.

Boat X is pulling with a force of -A.

And if this is the case, then the only force X is capable of applying to the rope is -A. In order for X to apply any other force to the rope, it needs to be pulling with that force.

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

So do you admit X is pulling on the rope with a force of -B in addition to the force of -A?
If not, X CANNOT be exerting a force of -B in addition to the force of -A on the rope.

You must have either -A is that reaction force and thus -A=-B, or you have X pulling on the rope with a force of -A-B.
Anything else is a violation of Newton's laws of motions with forces appearing by magic.

SEE HOW YOU FAILED TO APPLY THE THIRD LAW?

No. I applied it correctly, considering 2 bodies and ensuring their force was equal and opposite.
X is pulling on the rope with force A, then the rope is pulling on X with force -A.
In order to satisfy Newton's third law, the rope CANNOT be pulling on X with a force of anything other than -A.
If the rope pulls on X with a force of -A+B, then X MUST pull on the rope with a force of A-B.

For any 2 entities, the force one applies to the other must be equal but opposite the force the other applies to it.
i.e. A=-B. There is no other way.

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's right, no magic invention of forces at all.
X is exerting a force on the rope, thus the rope is exerting an equal but opposite force on X. It doesn't magically double the forces.

This is what you are missing.

No. I am missing nothing.
What you are missing is that Newton's third law doesn't magically make new forces appear.
If X is pulling the rope with a force of A, that is all X is applying to the rope.
There is no distinction between A being a force which then has a reaction force, or A being a reaction force.
There is no distinction between an action force and a reaction force.
It is impossible to tell the 2 apart.

Newton's third law is fine.

Not with your analysis.
With your analysis you have X pulling on the rope with a force of -A and the rope pulling back with a force of A+B.

The only way for Newton's third law to be fine is if the rope is pulling back with a force of A, or if X is pulling with a force of -A-B, which would then violate the situation.

The RE requested that I introduce the proper labeling of forces (their direction) in the analysis.
Let us see now what happens when we apply this basic requirement to jack's piece of crap analysis.

Yes, when you were using numbers, claiming the force one is applying is 350 N and the force the other is applying is 125 N.
In this case you have both pushing the rope in the same direction.

My perfectly sound analysis is not crap in any way and easily meets these requirements.

Notice a key thing I said (which is just Newton's third law?
A=-B. i.e. one person is pulling with a force of say 350 N, while the other is pulling with a force of -350 N. One is positive, one is negative. Thus they can both be pulling on the rope. No problem at all.

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

Yes, by changing A to -A, you just get that.
So now, X, pulling with a force of -A, is equal and opposite to Y pulling with a force of B.
Still holding to Newton's third law of motion, still having no net force on the string.

reached the amazing conclusion that A=B.

Yes, has reached the completely unsurprising conclusion that is Newton's third law of motion.

It takes a single counterexample to show this.

And so far all you have done is baselessly asserted that they exist.

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.
It is just baselessly asserting pure bullshit.

The single counterexample needs to be physically possible.

By the very hypothesis, forces A and B are not equal.
They are of different magnitude.

Not according to Newton's third law of motion.
That creates a quite explicit relationship between them of A=-B. There is no other option.
Last time I checked, 2 forces that are equal in magnitude but opposite in direction are equal in magnitude, not different in magnitude.

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

Not when pulling on the same rope in opposite directions with no one else there while holding the rope under tension.

Forces A and B WILL ALWAYS BE DIFFERENT.

Yes, one will be positive and one will be negative, such that A=-B.
That is the requirement to hold a rope in tension. The tension in the rope will be the magnitude of the forces. T=|A|=|B|.
If the forces are different you no longer have a rope under tension (not unless you move away from the ideal case and consider the mass of the string, but you aren't doing that in your analysis).

Force A can never equal force B.

So you are claiming that for an action there cannot be an equal but opposite reaction?

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|.

And it wouldn't be the ideal case of the mass-less string. Instead, you would have to consider the mass of the string and realise that X isn't just pulling boat Y, it is also pulling the string.
In this case, as B is greater Y will be pulling the string towards them.
The net force on X will be A. The net force on Y will be -B. The net force on the string will be B-A. This will result in a slight acceleration of the string.

Note the situation you are trying to model, the gravitational attraction between the 2 objects.
This force is given by the equation:
F=GMm/r^2.
This will be the same for each body, they must be EXACTLY the same.

So the RE analysis works fine for a massless string, equivalent to the situation you are trying to model. On the other hand, your analysis fails miserably and has you double counting forces.

The RE analysis leads directly to the ONLY case which can never be experienced in reality.

No. It leads directly to the requirement for the ONLY cases which can be experienced in reality, A=-B.
Your analysis leads directly to a conclusion which will only hold when A=B=0.
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.
So for any non-zero value of A or B, you have a contradiction.

For example, in the case above, Henry is pulling on the rope with a force of -350 N, which contradicts your claim that the net force on the rope from X is -475 N.
Last time I checked, -350 does not equal -475.

Do you think they are equal?

If not, then how can you claim there is no contradiction?
If X is pulling with a force of A, then the only force applied to the rope will be -A.

You said this: Newton's third law dictates A=-B.
But now, by your own priceless analysis, A=B!

No, by your manipulation of my analysis.
A=-B, when X is pulling with a force of A and Y is pulling with a force of B.
You are changing it so X is instead pulling with a force of -A.
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.

Newton's third law has to be properly applied.

As I did.
You have X pulling the rope with force A, and the rope pulling X with a force B.
As per Newton's third law, A=-B. That is the only option (which is equivalent to changing A to -A)

Instead you have X pulling the rope with force -A, and the rope pulling with force A+B.
So by Newton's third law, A=A+B. Does that make sense to you?

The other option is an even more direct contradiction, where you have forces appear by magic, where X is pulling the rope with a force of -A, but the net force on the rope from X is -A-B.
Does -A=-A-B?
Only when B is 0.

So no, your equations fail miserably.

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

No, it isn't.
As X is pulling with force -A, and Y is pulling with force B, the net force on the string will be -A+B.
In order for the net force to be as you claim, you need X pulling with a force of -A-B, and Y pulling with a force of A+B, directly violating the description of the situation.

There's a massive contradiction. You say first that the force applied to the rope by the person on boat X is -A, and then you say it's -A-B. Can't have both.
There is no contradiction because I said no such thing.

Yes, you have repeatedly refused to admit that.
But the situation clearly described X as pulling the rope with a force of -A (at least now that you have changed the sign of the force).
But your analysis dictates that X is applying a force of -A-B to the rope.

But the only way for X to be applying a force to the rope is by pulling on it. That means your analysis dictates that X is pulling the rope with a force of -A-B.

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

This is claiming that X is pulling on the rope with a force of -A-B. If X isn't pulling with this force, then how is X applying this force to the rope?

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

Yes, because you are counting the forces twice.

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, so if the net force on X from the rope is A+B, then X must be pulling on the rope with a force of -A-B, but that goes directly against the description of the situation where X is pulling with a force of -A.

Let us apply the correct labeling of forces again.

If you are going to replace A with -A, you need to do that in all cases.

This is the fixed version:
The net force on boat x is A.
The net force on boat y is -A.
The net force on the string is (-A +A)=0, so 0=0.

Not only is jack's analysis contradicting his basic requirement that A=-B

Not when you do it honestly, and realise that by substituting -A into A, you need to change the requirement to -A=-B.

but also his forces now do not cancel each other to add up to zero; instead they add up to -2A!

Only when you replace half the A's with -A.
When you do it properly then you have -A+A=0

Let us remember what jack said:

Quote from: JackBlack on May 10, 2017, 05:01:22 AM

Newton's third law dictates A=-B.

Yes, when you have X applying a force of A and Y applying a force of B.
That is because the force applied by one must be equal and opposite the force applied by the other.
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.

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.

No it doesn't. Not when you note that this is equivalent to my claim.
You have just changed A to -A.
If you take note of that and apply it to my conclusion:
A=-B (my claim)
-A=-B (subbed in your substitution)
A=B (simplified).

So no contradiction, no problem.
It just appears like one because of your dishonesty.

It would be akin to me saying that your analysis originally had the net force on X be -A+B, but now it is A+B. Is that a contradiction?
Not when you consider that the latter is a result of substituting -A into A for the former.

But I understand, you need to resort to this blatant dishonesty otherwise your analysis would be exposed as bullshit straight away.
You grasping at whatever straws you can to try and make a pathetic attempt (your best attempt) at justifying the mountains of bullshit you have spouted and "refuting" the facts we have provided.

Any two human beings, when forces will be analyzed at the most infinitesimal level (we are talking here about the 100,000,000,000,000,000,....th fraction of a Newton, will be pulling with DIFFERENT FORCES.

Thus exerting a net force on the rope.
Remember, you are trying to model gravity.
With gravity, F=GMm/r^2 directed towards the other object. That makes them equal by definition.
So they will be pulling with IDENTICAL forces.

In a real situation, you have slight variations and you have a string/rope with mass which can thus be accelerated.
Either the forces average out over time, or you have one pull in more rope and them not meet in the centre.

The burden of proof IS ON YOU, NOT ME.

No. You are claiming there is a contradiction and that |A|!=|B|. As such, the burden is on you to show that is possible for the ideal case of a massless rope.

But as you seem to want to go to a real situation, here it is, still slightly unreal, the 2 boats have masses MX and MY, the rope has a mass of m, and it is located right in the centre of the rope, with the rest of the rope being massless:

X will apply a force of A to the rope (taking note that A can be negative, you don't need to make the force -A, you can leave it as A).
This means the rope will pull back with a force of -A.
Y will apply a force of B to the rope.
Thus the rope will pull back with a force of -B.

Thus the net force on X will be -A, and it will thus accelerate at a rate of -A/MX.
The net force on Y will be -B and it will thus accelerate at a rate of -B/MX.
The net force on the rope will be A+B, and it will thus accelerate at a rate of (A+B)/m.

The tension in the rope will vary. In reality it will vary across the entire rope, but in our example to keep it simple, the tension on the X side of the rope will be |A|. The tension on the Y side will be |B|.

Happy, now we have a situation much closer to reality, where the sting, by virtue of having mass, can now have a net force acting upon it and be accelerated.

If we now let M1 and M2 be equal, and |B|>|A| we note that boat Y will move more and they will not meet in the middle. We also note that the middle of the string will end up on boat Y.

You have to prove that in each and every situation, the two forces will be TOTALLY AND EXACTLY THE SAME.

No. You have to prove that in the ideal case where the string is massless, the forces can be different.

You can't even have this:
Force A = 100.000,000,001
and
Force B = 100.000,000,000

That's right, because in the ideal case of the massless string, that means you have a net force on the string and thus it will be accelerated at an infinite rate.

It won't satisfy the RE requirement as listed above.

Because it is physically impossible.

I agree with everything you just said, but using our hypothetical scenario, our rope is massless and tensionless. We are simply trying to debate the forces acting on either boat, when two people pulling at different forces begin to pull on the rope, which means the rope will not move.

The problem is that that is a contradiction.
If they are pulling with different forces, then there will be a net force on the rope, which as it is massless will result in infinite acceleration.

One boat would only be putting a force of A on the rope, this is true, but also a force of -B would be applied to the same side from the equal opposite force from the other side.

No, it wouldn't.
If the boat is only putting a force of A on the rope, then the entire force on the rope from this side is A, not A-B. In order for it to be A-B, you need it to be putting a force of A-B onto the rope.

This force of A IS this force of -B.
They are the same force, you are counting them twice.

And as far as just force A on one side and no force on the other, we're assuming the rope is tied to the other boat.

It doesn't matter how it is.
In order for the rope to apply a force of A to that boat, that boat must be applying a force of -A to the rope.
You cannot have a force being applied just on one side and having both boats move.
Even if the rope is just tied to the boat, the boat can exert a force on the rope. You don't need a sentient individual there pulling on the rope.

But it is gravity pulling the rope down into a catenary shape.

It is the combination of the tension in the rope (here a fine strand) and gravity acting on the mass of the rope making the catenary shape.

You have provided NO explanation at all.
HOW does gravity pull the rope to bend into a catenary shape?

Is your internet broken?

That rope will become a bending line.
How a does a bending rope transmit the forces applied on it?
Your bibliographical references DO NOT explain HOW gravity provides the magical force to bend a straight rope into a bending rope (catenary).

Whether it be it gravitation, push aether something-or-other or witchcraft,
If  Robert Hooke understood it in the 1670s, and its equation was derived by Leibniz, Huygens and Johann Bernoulli in 1691,
You could even look it up in Wikipedia, see .

It is easier to see how loose chains hang first.
Wilipedia, Catenary.
Look up the bit

Analysis
Model of chains and arches
In the mathematical model the chain (or cord, cable, rope, string, etc.) is idealized by assuming that it is so thin that it can be regarded as a curve and that it is so flexible any force of tension exerted by the chain is parallel to the chain. The analysis of the curve for an optimal arch is similar except that the forces of tension become forces of compression and everything is inverted. An underlying principle is that the chain may be considered a rigid body once it has attained equilibrium. Equations which define the shape of the curve and the tension of the chain at each point may be derived by a careful inspection of the various forces acting on a segment using the fact that these forces must be in balance if the chain is in static equilibrium.
<< . . . . . . .  you need to see the rest  . . . . . . .  >>

There are plenty of other references.

The weight of a hanging rope or chain has to be supported by the vertical components tension at each end.
If we want to keep the sag small the angle the rope falls away at the end is quite small, and so the horizontal component will be much higher than the vertical component.


Every "tow-rope" forms a catenary. The very long ones towing large ships can have quite a lot of sag.
Also the design of long spans of HV power cables is always a compromise between the very high force need to keep the sag to an acceptable amount and the weight and strength of the cables needed.
This is why many long span HV power cable use a steel core for strength and aluminium (light weight and high conductivity) outside for the conductors.

Hope you can work out "bending ropes" now.

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.

However, just one thing. A force of A exerted by boat X will result in a force of -A being applied on the Boat Y end. This does not mean that boat Y is exerting a force of
-A on the rope.

UPDATE: Ok I think i understand now. The force on one end of the rope is not A-B, it is just A because A includes the -B force as well.

Plus we're getting a free lesson on action reaction force pairs, which is neato.