*Through one rope.*

Why then do not both the Earth and Moon start moving toward each other, just like in the two boats connected by a rope example?

Because they are moving sideways.

The force doesn't cause them to move towards one another, it causes them to accelerate towards one another, which results in a curved path through space, i.e. an orbit.

But like I said, deal with the simple situation until you can understand it. Then move on to more complex ones.

The analogy between the two boats on a lake pulled by a rope and the Earth-Moon system is perfect.

...

Imagine a ball attached to a string and you are holding the other end of the string and moving your hand in such a way that the ball is in circular motion. Then tension in the string is centripetal force.

Notice the difference?

If not, when you finally understand the boat problem, we can move on to this and see if you can understand the difference then.

*If boat A pulls on the rope with force Fa, then the tension in the rope all along is Fa and boat B must also be pulling on the rope with Fa. It cannot be anything else.*

Brilliant.

It is only that boat Y is pulling with force B: a force of different magnitude than force A.

No. That is a physically impossible situation.

Boat Y needs to be pulling on the rope with a force equal and opposite that of boat X.

That is how tension in ropes work.

If the rope is pulling on boat Y with a force of A, then there needs to be a reaction force to complete the action/reaction pair of boat Y pulling back on the rope with force -A.

But boat Y is pulling on the rope with a force of B.

So boat Y is pulling on the rope with a force of B, and boat Y is pulling on the rope with a force of -A. But this is the same force.

Boat Y is pulling on the rope with a force of -A=B.

How will boat X thrust forward? Based on which forces?

What are the forces acting on boat X through the rope itself?

That would be the tension in the rope, which I have described as F or -F, and you have described as A and B.

Boat X is pulling on the rope with force F=A=-B.

The rope is pulling back with a force of -F=-A=B.

That is the only force pulling boat X.

If any other force in the string pulled boat X then boat X would need to pull back on the string with more than just that force.

Does boat X move forward as if nobody is pulling there at point X with force A? Just because boat Y is pulling with force B?

Do you mean as in if boat x was just tied to the rope so the boat pulls back with its reaction force or do you mean if no force was applied to the rope by boat x or anything on boat x at all?

If the latter, then no, as there is then no force acting on boat x.

Remember, for every action there is an equal but opposite reaction.

In order for anything to pull on X, X needs to pull back.

And if not, it is obvious that there will be TWO FORCES ACTING ON BOAT X.

No. It is obvious that there is an action/reaction pair for boat X.

The man doing the pulling on the rope with force A, and at the same time boat X will be pulled by boat Y with force B.

And the man in boat X doing the pulling isn't pulling on boat X thus that force is not acting on boat X and thus that force will not move boat X.

Does force B exerted by boat Y equal force A exerted by boat X to start with? Are the two men doing the pulling applying the SAME FORCE, or not? And if not, do you agree that force A does not equal force B?

Potentially not at the start, but once the situation is set up with tension in the rope, then yes, they are applying the same force.

If they weren't, the rope would fly out of one of their hands.

The two forces A and B ARE NOT EQUAL TO START WITH.

They are of DIFFERENT MAGNITUDE.

Yet you have reached the conclusion that they are equal, which they cannot be.

Proof by contradiction that your analysis is catastrophically wrong.

No. You have baselessly assumed that they cannot be equal (but opposite), when they must be.

This is not a proof by contradiction that we are wrong. It is a proof by contradiction that you are wrong.

They are of identical magnitude and opposite in sign, as backed up by Newton's third law, for ever action there is an equal but opposite reaction.

If they were not equal but opposite, then the rope would fly out of one of their hands and move entirely towards the other, severing the connection.

Or better said, your friend jack has, in a most unfortunate way for the RE.

You mean where I clearly demonstrated that they were the same force? That A=-B?

But the very hypothesis of the example/situation states that force A (exerted by boat X) and force B (applied by boat Y) ARE NOT EQUAL.

No. It isn't.

This example is merely 2 boats pulling themselves together with a force.

It is only in your delusional fantasy land (which has been disproven numerous times), that they are difference forces.

We have shown that they MUST be equal.

A does not equal B.

Yet, the "analysis" shown above signed jack (yes, you can blame him for this), says that A = B.

You have just been shown a proof by contradiction.

Not quite. You claimed |A|!=|B|.

I then analysed the situation, showing clearly that A=-B, and thus |A|=|B|.

So yes, this is a proof by contradiction.

It shows that your claim is wrong. It shows that |A|=|B|, that the 2 forces are equal (in magnitude, but opposite in sign), and that your baseless assumption that they are different is completely wrong.

So no, it doesn't disprove mechanics.

It shows that in this situation force A and force B have to be equal but opposite.

Therefore, something is missing.

What is missing is the fact that there ARE TWO FORCES acting on boat X.

No, what is missing is the fact that you are completely wrong and refuse to admit it and have no way to rationally defend your BS to try to show you aren't wrong.

Remember, the analysis done above is the standard/official line. Yet, in the case of two boats pulling on a rope, it fails in a most catastrophic way.

No. It works perfectly, showing that you are completely wrong.

It doesn't get any better than this. He has just equated A with B, when by hypothesis A does not equal B.

Yes, I am showing your hypothesis is wrong.

Is this too hard for you to understand?

Do you not know how a proof by contradiction works?

Here is a brief summary:

Assume statement !S is true.

Do logical/mathematical analysis on !S.

Reach a contradiction.

Thus !S is false.

Thus S is true.

(You can also make the assumption at a later point).

I did the same (well technically, I didn't make the assumption but it can be made as there are 2 variables).

Assume the 2 forces are different in magnitude (i.e. not equal).

Do logical/mathematical analysis.

Arrive at conclusion that 2 forces are equal in magnitude.

Thus have reached a contradiction.

Thus the assumption (that the 2 forces are different in magnitude) is false.

Thus the negation of that assumption is true, i.e. the 2 forces are equal in magnitude.

What don't you understand about that?

Your first question makes no sense at all.

No two boats can meet exactly in the middle since the force applied by the two men will necessarily be different in magnitude (just like in the Earth - Moon system).

Yet the claim clearly states that 2 boats of the same size meet in the middle.

And your analysis clearly shows the forces acting in the case of the Earth moon system are equal.

Your friend Jack has ALREADY done such an "examination", remember?

It leads to the most disastrous contradiction of them all.

It is only disastrous for you as it shows you to be wrong.

The application of the third law is what comes into question.

There are two forces acting on boat X.

What is the net force on boat X?

What is the net force on boat Y?

No. There is a single force acting on each.

For boat X the net force is -A=B.

For boat Y the net force is A=-B.

They are equal but opposite. Just like you would expect for the third law.

Yet you are completely incapable of doing such an analysis, as you are incapable of showing the action/reaction pairs.

*If the men are of different strength then the maximum amount of pulling force on the rope will be limited by the weaker man.*

Beyond this limit the weaker man will no longer be able to hold onto the rope (assuming he is holding it only by his hands).

On the lake no such thing would happen.

Only on land.

No, it happens anywhere there is resistance to his motion.

On the lake, he needs to transmit the force to the boat.

If he isn't strong enough, he will either be pulled off the boat, or the rope will be pulled out of his hands.

Even if he was just sitting there in a friction-less vacuum, he still has his own inertia resisting the motion.

If he is only capable of applying a force of 1000 N to the rope, and the other person pulls with a force of 2000 N, then the rope will fly out of his hands because of his own inertia, because he is not capable of using his hands to transmit that 2000 N of force to his body.

By hypothesis, the two forces are DIFFERENT, boat X will pull with force A which is different in magnitude than force B (applied by boat Y).

And of course you have your friend jack to thank him for his brilliant analysis where he equated A and B reaching a most stupendous contradiction.

No, not a most stupendous contradiction.

A rational, analytical, mathematical contradiction of your hypothesis to show your hypothesis is wrong, and show that the 2 forces are equal in magnitude.

Your friend jack, using his analysis, has managed to demonstrate that the Earth and the Moon will meet, since his equations lead to the most blatant contradiction.

No. My analysis only shows a contradiction of your claim. It doesn't contradict a correct analysis of the situation, something you are yet to do.

Regardless, that doesn't address the key issue raised against your claim that they should move towards each other.

By the way... here are the numbers on the Earth - Moon system.

https://www.theflatearthsociety.org/forum/index.php?topic=70349.msg1905037#msg1905037

I may not be a genius, but those last 2 numbers, (the force on the moon, 2.1096E+19 kgf, and the force on Earth, 2.1096E+19 kgf) look very similar. Some would even say they are THE SAME.

That sure seems to go against your BS claim that they can't be.

So good job refuting yourself, yet again.

The RE have already done their calculations: they lead to a most obvious contradiction.

Again, they only contradict your claim, showing your claim is false. They do not contradict the reality of the situation.

He used THE SAME FORCES and he reached the worst contradiction of them all.

No. I showed they were equal but opposite.

Without the assumption that they are not equal but opposite, there is no contradiction.

If you think there is a contradiction in my analysis without the assumption that A and B are different in magnitude, please show it.

So far the only contradiction you have come up with is that I showed that A=-B, while you said |A|!=|B|. But that isn't me contradicting myself, that is me showing you are wrong.

It doesn't work.

Yes, it doesn't work with your baseless assumption.

The net force on boat X will be: **-A + B**

And where does this force come from? The rope? I can't see anywhere else it can come from.

So that means that the rope is pulling on X with a force of -A+B.

But as per Newton's third law, that means boat X is pulling on the rope with a force of A-B, not A like you claim.

So which is it?

Is X pulling on the rope (and thus boat Y) with a force of A, or a force of A-B?

See, this is why you get your "double forces" BS, because you are literally counting the forces twice.

This also counts as a proof by contradiction, but rather than contradicting our assumption and analysis, you have contradicted your own, showing that your assumption and/or analysis is wrong.

Your conclusion, that X pulls on the rope with a force of A-B, contradicts your assumption that A pulls on the rope with a force of A (and similar goes for boat Y), so it shows your analysis or assumption was wrong.

You can correct your analysis, like I did, but that would still show your assumption that |A|!=|B| is wrong.

Or you can correct your analysis and have X pull on the rope with a force of A-B, but then your assumption that the 2 pull with forces of different magnitudes is still wrong.

I have to do all of your homework.

The net force on boat X will be: **-A + B**

The net force on boat Y will be: **-B + A**

The net force on the string will be **[A - B] + [B - A]**

You mean the homework I already did for you and showed you were completely wrong, in several different ways?

Lets simplify things a bit, rather than using A and B, lets use another force, call it F, where F=A-B.

This means:

The net force on boat X will be:

**-F**The net force on boat Y will be:

**F**The net force on the string will be

**[F] - [F]**Notice how that is exactly what I said was happening?

Notice how you have just shown that I was completely right?

Just in case you don't, lets analyse it a bit more:

The net force on boat X will be:

**-F**This means the rope is pulling on boat X with a force of -F, and thus X is pulling on the rope with a force of F (not A like you claim).

The net force on boat Y will be:

**F**This means the rope is pulling on boat X with a force of F, and thus Y is pulling on the rope with a force of -F (not B like you claim).

The net force on the string will be

**[F] - [F]**This is composed of the already discussed forces, X pulling with a force of F, and Y pulling with a force of -F.

Notice how X and Y are pulling with forces F and -F respectively? Not A and B like you claim?

All forces balance out perfectly.

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

Yes, because you have counted them twice.

When you count every force twice, it will still balance.

However, now instead of X pulling on the rope with A like you claimed, it is pulling on the rope with A-B.

The other way of saying it is that you are pretending X is only pulling on the rope with half the force X is actually pulling on the rope.

By contrast/comparison, here is the catastrophic RE analysis:

You mean the accurate one, which shows the reality of the situation, where instead of X pulling on the rope with a force of A-B, it pulls on the rope with a force of A like you originally claimed, and that Y pulls on the rope with a force that is equal and opposite, which your conclusion still showed?

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

They are of different magnitude.

Yes, your hypothesis where that those 2 forces were not equal in magnitude.

I showed that that is wrong, and that A=-B.

What's your refutation of that conclusion? Basically this:

"NO!!! I SAID THEY WERE DIFFERENT!!! BELIEVE ME!!! YOU'RE WRONG!!!!"

The RE analysis has lead to the worst contradiction possible, where the errors come in full view: A has to be equated to B, when in reality A does not equal B.

No. The RE analysis only contradicts your assumption that the magnitude of A and B are different. Without that baseless assumption it works fine.

It shows that A=-B. You are yet to show that your baseless assumption is true.

The correct FE equations are:

The net force on boat X will be: **-A + B**

Which contradicts your claim that X is pulling on the rope with a force of A, instead you have it pulling on the rope with a force of A-B.