Distances in the universe

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sandokhan

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Re: Distances in the universe
« Reply #270 on: May 02, 2017, 01:51:25 PM »
i say the same force.
now it up to you to proof me wrong with an experiment.
if you are really believe in your claim than do the test.


It can't be the same force.

You are forgetting that your friend here jack has done the unthinkable: to actually screw things up for the RE permanently.

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

It doesn't work.


TWO QUESTIONS FOR ALL OF THE RE AND FE:

What is the net force on boat X?

What is the net force on boat Y?


Here let me help you.

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


What then is the net force on boat Y?

What is the net force on the rope?

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sandokhan

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Re: Distances in the universe
« Reply #271 on: May 02, 2017, 02:02:29 PM »
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]


Since the string is not moving, the net force on the string is 0:

[A - B] + [B - A] = 0


All forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.
« Last Edit: May 02, 2017, 02:05:50 PM by sandokhan »

Re: Distances in the universe
« Reply #272 on: May 02, 2017, 02:04:17 PM »
i say the same force.
now it up to you to proof me wrong with an experiment.
if you are really believe in your claim than do the test.


It can't be the same force.

You are forgetting that your friend here jack has done the unthinkable: to actually screw things up for the RE permanently.

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

It doesn't work.


TWO QUESTIONS FOR ALL OF THE RE AND FE:

What is the net force on boat X?

What is the net force on boat Y?


Here let me help you.

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


What then is the net force on boat Y?

What is the net force on the rope?

you are still only claiming stuff.

proof it.

as i said, on that you write it doesn't make it a prove.


Re: Distances in the universe
« Reply #273 on: May 02, 2017, 02:28:15 PM »
It is not just boat Y that is pulling on the rope, and implicitly on boat X, BUT ALSO the man in boat X that is doing the pulling.
Yes, man in boat x is also pulling. That is how action/reaction pairs work.

Action-reaction pairs
Boat X is pulling with force A the rope/boat Y.
This is not part of a pair.
That is trying to apply a force to 2 different things.
Decide what X is pulling on, is it the rope or boat Y. You can't have both.

The rope will react with force A.
So is boat X pulling on the rope instead of boat Y?
As if it was boat Y, it should be boat Y pulling back.

The rest of your "pairs" are the same, not pairs at all.

These are the action-reaction pairs, clearly described, on the two boats/rope.
No they are not.
You are listing 2 receiving entities in the action part. That isn't how forces work.

If you want to have the rope transmit the force, that is fine, but then don't have boat x acting directly on boat y, instead have it act on the rope and in turn have the rope act on y.

Very simple.
So why do you still fail to provide the pairs?

The rope/string will transmit two simultaneous forces: for boat X, as an example, while the man is pulling with force A (and thus the rope will react with force A as well), boat Y will also apply A SEPARATE FORCE, force B.

Two different forces.

They cannot be mixed with one another.
No. They have to be the same, as that is how tension works.
A string under tension will pull on both ends with that tension, and whatever is attached to those ends pulls back with that tension.

Then, we have a huge problem regarding the Earth-Moon system.
Again, wait until you understand this problem before moving on to anything more complex, and there will be a step between these boats and the Earth-Moon system.

Alphabet soups, endless permutations/combinations of letters will not change one basic fact: there are two forces acting on boat X (and respectively, on boat Y).
I know it wont change it, because that isn't the fact.
The fact is there is a single force acting on boat X, that of the rope pulling on it.
In a real life situation boat X WILL NOT MOVE forward JUST because boat Y is pulling (force B); not at all.

Boat X will also thrust forward based on a second force, the man doing the pulling on the rope with force A.
[/quote]
As an action/reaction pair.
The man pulling in boat X is the reaction to the man pulling in boat Y.
The man pulling in boat X doesn't move boat X as that isn't applying a force to boat X.

Forces A and B, are DIFFERENT, of different magnitude, they cannot be mixed with one another as the RE are obviously trying to do in order to escape the dramatic consequences: a simple two boat, one rope example will create double the forces required by Newtonian mechanics.
No they aren't.
They are equal but opposite, just like what is required for an action/reaction pair.

You are trying to invent a problem which simply doesn't exist.

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sandokhan

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Re: Distances in the universe
« Reply #274 on: May 02, 2017, 02:34:58 PM »
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]


Since the string is not moving, the net force on the string is 0:

[A - B] + [B - A] = 0


All forces balance out perfectly.

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


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

Here is the way with more words, formatted like you have (but with the missing details. I will even highlight them in red for you):
Boat x is pulling on the string rope with force A. Reaction: the rope is pulling back on boat x with force -A.

AT THE SAME TIME, BOAT Y IS PULLING on the rope WITH FORCE B. Reaction: the rope is pulling boat y with force -B.

These are the 2 action/reaction pairs involved.
Boat x is pulling on the string, and so is boat y.
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.

Thus they can be rewritten as:
Boat x is pulling on the string rope with force A. Reaction: the rope is pulling back on boat x with force -A.

AT THE SAME TIME, BOAT Y IS PULLING on the rope WITH FORCE -AB. Reaction: the rope is pulling boat y with force AB.

And there we have it, 4 forces, 2 action/reaction pairs, and all perfectly balanced, all without any BS doubling of forces.

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.

Can you point out anything wrong with that?

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

They are of different magnitude.

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.


The correct FE equations are:

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]


Since the string is not moving, the net force on the string is 0:

[A - B] + [B - A] = 0


All forces balance out perfectly.

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

Re: Distances in the universe
« Reply #275 on: May 02, 2017, 02:36:59 PM »
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]


Since the string is not moving, the net force on the string is 0:

[A - B] + [B - A] = 0


All forces balance out perfectly.

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

You claim that the force in the rope is 0?
That would mean that the rope can be very thin even not existing.

Wow you are way of from the reality.
Keep on living in you imaginary world, but stop claiming this kind of Bullshit as true, somebody even could believe you.

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sandokhan

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Re: Distances in the universe
« Reply #276 on: May 02, 2017, 02:41:13 PM »
You seem to be having a problem with prepositions.

You claim that the force in the rope is 0?

I used the preposition "on", not "in".

Re: Distances in the universe
« Reply #277 on: May 02, 2017, 03:28:36 PM »
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.

Re: Distances in the universe
« Reply #278 on: May 02, 2017, 03:29:15 PM »
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]


Since the string is not moving, the net force on the string is 0:

[A - B] + [B - A] = 0


All forces balance out perfectly.

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

You claim that the force in the rope is 0?
That would mean that the rope can be very thin even not existing.

Wow you are way of from the reality.
Keep on living in you imaginary world, but stop claiming this kind of Bullshit as true, somebody even could believe you.
The net force on the rope, not the tension in the rope.
This means the rope isn't moving towards either boat.

Re: Distances in the universe
« Reply #279 on: May 02, 2017, 03:34:39 PM »
And Sandy, just for you, a shorter version, just going based on the key points:

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]
Simplifying these by using F, where F=A-B:
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]

Thus, in this situation, X is being pulled by the string with a force of -F, thus to complete the action/reaction pair, X must be pulling on the string with a force of F.
Similarly, Y is being pulled by the string with a force of F, and thus must be pulling on the string with a force of -F.
Thus the net force on the string would be F-F, which matches the above.

Notice how X is pulling with a force of F, which is equal but opposite the force from Y (-F), just like I claimed before, and going directly against what you repeatedly asserted?

Also notice how X is pulling with a force of F=A-B, not a force of A like you repeatedly asserted?
Notice how you have contradicted yourself twice?
In what the force X is pulling on the string with, and with the forces from X and Y being different or equal in magnitude.

In both cases your analysis shows your initial assumption/claim was wrong, and in both cases it fully supports me.

Good job refuting yourself.

Re: Distances in the universe
« Reply #280 on: May 02, 2017, 04:19:58 PM »
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]


Since the string is not moving, the net force on the string is 0:

[A - B] + [B - A] = 0


All forces balance out perfectly.

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

Hey I can do that to.  Let's add force c,d,e, and f which have different magnitude than a.

The net force on boat X will be: -a+b-c+d-e+f
The net force on boat Y will be: a-b+c-d+e-f
The net force on the rope will be: (-a+b-c+d-e+f)+(a-b+c-d+e-f) = 0

All the forces balance out perfectly
But they include 6 TIMES "THE FORCES NEEDED in the Newtonian system", whatever that means, lol.

Since there is no physical significance of breaking what is a single tension (pulling) force acting on the rope into a,b,c,d,e, and f:

let a-b+c-d+e-f = A

The net force on Boat X will be : -A
The net force on Boat Y will be : A
The net force on the rope will be: A-A=0

All forces balance out perfectly
But they include only what is needed and only what is physically significant

Now what is this nonsense about "TWICE THE FORCES NEEDED"?


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sandokhan

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Re: Distances in the universe
« Reply #281 on: May 02, 2017, 08:59:58 PM »
jack, you have lost big in this thread.

You are not even capable of thinking clearly.

You really need some help.


Here is how jack's twisted logic works:

You have baselessly assumed that they cannot be equal (but opposite), when they must be.

jack, the truth is starring you in the face: forces A and B are of different magnitude. Not the same.

This thread has turned into a complete defeat for you.


Have you lost it totally?

FORCES A AND B ARE OF DIFFERENT MAGNITUDE TO START WITH.

Different men, different strengths, different force of pulling.


How in the world can you write something like this?


They are not equal, get it straight through your head.

Otherwise, people here are going to believe that you are not all there.


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

Using your twisted logic, you reached the worst kind of contradiction: where your conclusion drastically violates the hypothesis of the problem.

It doesn't get any worse than this.


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.


jack, you have lost touch with reality since you are unable to accept defeat.

Do you realize how wrong you are?


How can A = B when the two forces are of different magnitude to start with?

A does not equal B.

But to you it is no problem at all.

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.


I told you that you most definitely have lost touch with the real world.

The hypothesis is most directly true: FORCES A AND B ARE NOT EQUAL.

You used the wrong net force on boat X to start with.

In your twisted little logic, then you proceeded to substitute at will, until you reached the conclusion that A = B.


You couldn't be more wrong!


Your analysis shows that the RE way of doing things is totally wrong.


Here are the correct equations, where we take into account the TWO FORCES ACTING ON BOAT X RESPECTIVELY ON BOAT Y:

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]



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?


You have totally lost touch with reality, jack.

It is that simple.


I always claimed that there are TWO FORCES acting on boat X: -A AND B.

ALWAYS.


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.


They cannot be equal since by the very hypothesis, A does not equal B.

Since you have reached the wrong conclusion, something is very wrong with your analysis: you did not count the net force on boat X right.

But I did.

I do not have to substitute anything to get the final result.

Everything works out fine.


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]


I'm sorry, you had it your way, with the F force: you reached a most direct contradiction of the hypothesis itself.


Here is how your analysis went.

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.


Using your way, YOU REACHED A CONTRADICTION which violates the hypothesis, that forces A and B are not equal.

FORCES A AND B ARE OF DIFFERENT MAGNITUDE TO START WITH.


Only my analysis takes this crucial fact into account.

I do not have to make substitutions to reach a direct conclusion.


TWO FORCES ACT ON BOAT X: -A AND B.

TWO FORCES ACT ON BOAT Y: -B AND A.


The net force on the rope will be:

[A - B] + [B - A]

All of the forces balance out perfectly.

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



The net force on boat X will be: -a+b-c+d-e+f

It does not work out like that.

There are only two forces acting on boat X: -A AND B.

No other forces. Period.
« Last Edit: May 02, 2017, 10:48:29 PM by sandokhan »

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rabinoz

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Re: Distances in the universe
« Reply #282 on: May 02, 2017, 10:19:11 PM »

FORCES A AND B ARE OF DIFFERENT MAGNITUDE TO START WITH.
It is not possible for this to be true. If they are pulling on the same rope they are each pulling with the tension in the rope.
If the rope is assumed massless, then the tension in the rope is exactly the same at each point in the rope, including the ends.

Quote from: sandokhan
Different men, different strengths, different force of pulling.
Yes, "Different men, different strengths", different capsbilities but not necessarily "different forces of pulling".
Just because one man is capable of pulling 700 N and the other capable of pulling only 500 N, does not that they pull those forces  all times.
Consider the top diagram with man A, the "weaker one", pulling the rope attached to a wall. Obviously he can apply his full strength of 500 N.
The tension in the rope is 500 N as is the force on the wall.
Now, in the lower diagram, try as he might, man A cannot pull more than 500 N, as that is his limit.
As before, the tension in the rope is 500 N as is the force applied to man B, the strong man.
The equal and opposite reaction force to the tension in the rope at the right hand end is still 500 N.
Even though man B is capable of pulling 700 N, he cannot do that here without over-exerting man A!

The tension in the rope is limited to the lesser of the capabilities of the two men.

But when you get to the earth-moon system, it's not even a matter of two forces of different strengths.
The moon-earth gravitational force is simply Fm-E = (G x mm x ME)/d2
There are no two forces of different strengths.

PS I certainly hope that you never try to design a suspending bridge.

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sandokhan

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Re: Distances in the universe
« Reply #283 on: May 02, 2017, 10:44:07 PM »
rabinoz, you are not helping the RE faction.

Not at all.


Not even if those two men in boats X and Y were twins, their applied forces could not be the same.


Force A is different, totally different than force B.

They cannot be the same.

Not at all. Not ever.


Different men, different forces applied to the rope.

It is that simple.


Then you have a huge problem, since the twisted RE analysis leads to a most direct contradiction.

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.


Using only a single force on boat X, we reach the worst kind of contradiction.

A does NOT equal B, by hypothesis. Those two forces are not of the same magnitude.

Yet, by using the RE's own analysis we reach a direct contradiction.


As I have said from the very start, boat X and respectively boat Y, will be acted upon by TWO FORCES:

TWO FORCES ACT ON BOAT X: -A AND B.

TWO FORCES ACT ON BOAT Y: -B AND A.


The net force on the rope will be:

[A - B] + [B - A] = 0

All of the forces balance out perfectly.

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


No wild substitutions, no contradictions to be reached.



But when you get to the earth-moon system, it's not even a matter of two forces of different strengths.

But it is a matter of having the double number of forces.

Remember this?

https://www.theflatearthsociety.org/forum/index.php?topic=70349.msg1905030#msg1905030


Boat X has two forces acting upon it: -A and B.

So will the Earth.

So will the Moon.

Within the "attraction" concepts:

From Earth, the concept requires that Earth's gravity is attracting the Moon; and an equal Earth anchored “attraction” force is pulling the Earth toward the Moon.

From the Moon, the Moon's gravity is attracting the Earth; and this Moon seated force is equally pulling the Moon toward the Earth.

Earth attracts the Moon, BUT ALSO an equal Earth anchored “attraction” force is pulling the Earth toward the Moon.

The Moon attract the Earth, BUT ALSO this Moon seated force is equally pulling the Moon toward the Earth.
 
There are FOUR FORCES INVOLVED HERE.

"All attraction models" produce twice the force that is required to balance the centrifugal forces of orbit!

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disputeone

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Re: Distances in the universe
« Reply #284 on: May 02, 2017, 11:32:46 PM »
Leave my buddy Newton alone.

For all the law is fulfilled in one word, even in this.

The reason I am consistently personally attacked here.
https://www.theflatearthsociety.org/forum/index.php?topic=69306.msg1960160#msg1960160

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Shifter

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Re: Distances in the universe
« Reply #285 on: May 02, 2017, 11:42:41 PM »
rabinoz, you are not helping the RE faction.

Not at all.


Not even if those two men in boats X and Y were twins, their applied forces could not be the same.


Force A is different, totally different than force B.

They cannot be the same.

Not at all. Not ever.


Different men, different forces applied to the rope.

It is that simple.


Then you have a huge problem, since the twisted RE analysis leads to a most direct contradiction.

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.


Using only a single force on boat X, we reach the worst kind of contradiction.

A does NOT equal B, by hypothesis. Those two forces are not of the same magnitude.

Yet, by using the RE's own analysis we reach a direct contradiction.


As I have said from the very start, boat X and respectively boat Y, will be acted upon by TWO FORCES:

TWO FORCES ACT ON BOAT X: -A AND B.

TWO FORCES ACT ON BOAT Y: -B AND A.


The net force on the rope will be:

[A - B] + [B - A] = 0

All of the forces balance out perfectly.

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


No wild substitutions, no contradictions to be reached.



But when you get to the earth-moon system, it's not even a matter of two forces of different strengths.

But it is a matter of having the double number of forces.

Remember this?

https://www.theflatearthsociety.org/forum/index.php?topic=70349.msg1905030#msg1905030


Boat X has two forces acting upon it: -A and B.

So will the Earth.

So will the Moon.

Within the "attraction" concepts:

From Earth, the concept requires that Earth's gravity is attracting the Moon; and an equal Earth anchored “attraction” force is pulling the Earth toward the Moon.

From the Moon, the Moon's gravity is attracting the Earth; and this Moon seated force is equally pulling the Moon toward the Earth.

Earth attracts the Moon, BUT ALSO an equal Earth anchored “attraction” force is pulling the Earth toward the Moon.

The Moon attract the Earth, BUT ALSO this Moon seated force is equally pulling the Moon toward the Earth.
 
There are FOUR FORCES INVOLVED HERE.

"All attraction models" produce twice the force that is required to balance the centrifugal forces of orbit!

I tried reading your post. I really did, but it was just full of nonsensical rubbish and assumptions it became too hard! Can you dumb it down in a succinct formula? Much like Newton's Laws, lets call them Sandokhan's Laws. What are the basic principles of whatever it was you were just talking about?

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rabinoz

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Re: Distances in the universe
« Reply #286 on: May 03, 2017, 12:07:43 AM »
rabinoz, you are not helping the RE faction.
So says the smartest man in the world who can't understand a simple physics problem!
Quote from: sandokhan
Not even if those two men in boats X and Y were twins, their applied forces could not be the same.
Force A is different, totally different than force B.
They cannot be the same.
Not at all. Not ever.
Different men, different forces applied to the rope.
No, never!

If the rope is assumed massless, then the tension in it is the same all along its length.
Any nett force would accelerate the rope, but since it is supposedly massless, that acceleration would be infinite.
And Einstein wrote of the infinite:
....
“Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.”
― Albert Einstein

So, please explain in less than 10 pages how it can be any different.

Or read Physics Stack Exchange, Nature of tension in a massless rope.
And, no I did not bother looking that up before earlier posts - all I looked for was suitable diagrams.

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sandokhan

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Re: Distances in the universe
« Reply #287 on: May 03, 2017, 01:12:09 AM »
Newton's third law is fine.

No problem.

It has to be applied correctly to the two boats pulled by a rope situation.


The RE had the chance to make their case.

It turned out to be a total disaster.


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.

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.


A total catastrophe.


Forces A and B are of a different magnitude, in fact they could never be equal at all.

By the very hypothesis, A DOES NOT EQUAL B.

A cannot equal B.

Yet, by using the twisted RE logic, using only a single force acting on boat X (respectively on boat Y), the analysis reaches a point where the absolute value of A equals the absolute value of B. A most direct contradiction of the hypothesis.


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]


Since the string is not moving, the net force on the string is 0:

[A - B] + [B - A] = 0


In the correct FE analysis, the TWO FORCES acting on boat X are included perfectly and correctly.

The man in boat X is pulling on the rope, while at the same time boat Y is pulling on that same rope with force B. The correct analysis must take these facts into account.

A perfect demonstration that there are indeed two forces acting on boat X, respectively on boat Y: the equations work out in total balance, no wild substitutions are to be made, no contradiction is to be reached at all.

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rabinoz

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Re: Distances in the universe
« Reply #288 on: May 03, 2017, 01:53:17 AM »
Forces A and B are of a different magnitude, in fact they could never be equal at all.
By the very hypothesis, A DOES NOT EQUAL B.
A cannot equal B.
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]
Since the string is not moving, the net force on the string is 0:
[A - B] + [B - A] = 0
Please show on a diagram what you are talking about and what the tension in the rope is.

The way you describe it makes no sense at all.

Re: Distances in the universe
« Reply #289 on: May 03, 2017, 01:57:32 AM »
jack, you have lost big in this thread.

You are not even capable of thinking clearly.

You really need some help.
I have continually refuted you. Why don't you cut out the pathetic insults, it clearly isn't working.



Here is how jack's twisted logic works:

You have baselessly assumed that they cannot be equal (but opposite), when they must be.
No, that is how my completely sane logic works, as well as the logic of pretty much everyone.
You are claiming/assuming they are of different magnitude. You have no basis for this claim.
As such, it is a baseless assumption.

What is twisted there?

jack, the truth is starring you in the face: forces A and B are of different magnitude. Not the same.
Repeating the same baseless assumption won't magically make it true.

The truth is right there staring you in the face, a truth you have somewhat acknowledged multiple times by quoting my proof that A=-B.
The forces are equal but opposite, just like you would expect for an action/reaction pair.

FORCES A AND B ARE OF DIFFERENT MAGNITUDE TO START WITH.
PROVE IT!!!

Different men, different strengths, different force of pulling.
Sure, they can have different strengths, that doesn't mean the force will necessarily be different.

A strong person can lift up a 1L bottle of water without applying a giganewton of force.

Similarly, someone with greater strength can apply less force to pull the rope.
The weaker person will be the limiting factor, where if they can't provide/channel enough force, the rope will fly out of their hand or they will fly off the boat.

How in the world can you write something like this?
BECAUSE IT IS TRUE!!!

They are not equal, get it straight through your head.
THEN PROVE THEY AREN'T.
So far all you have done is assert they aren't.

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

Using your twisted logic, you reached the worst kind of contradiction: where your conclusion drastically violates the hypothesis of the problem.
No. I reached one of the best kinds of conclusions, a conclusion which completely solves the problem, with no paradox or contradiction.

The only thing I contradicted was your baseless assumption.
All that means is your assumption was wrong.

The actual "hypothesis" of the situation was that there were 2 people in 2 boats.
The one in boat X is pulling on a rope with force A.
The one in boat Y is pulling on  the same rope with force B.
There are no other entities around to influence the situation.

Based upon this, A=-B, as I showed repeatedly.

It only contradicts your baseless claim, thus it is only catastrophic for you.

It doesn't get any worse than this.
Somewhat correct. It can't get much worse for you. Especially after you basically refuted yourself.

jack, you have lost touch with reality since you are unable to accept defeat.
I'll make you a deal, you actually defeat me, and I will accept defeat.
So far you have just repeatedly gotten your ass handed to you. I'm not the one being defeated here.

Do you realize how wrong you are?
For this specific situation?
Somewhere between 0 and 1%.
Do you realise how wrong you are?
Around 90-100%

How can A = B when the two forces are of different magnitude to start with?
That is the whole point, they aren't different in magnitude. How about you stop baselessly claiming that.
That is where a key part of your BS analysis of this problem comes from, your baseless assumption.

A does not equal B.
That is right. A=-B. Notice the MINUS!!!
They are equal but opposite.

But to you it is no problem at all.
Yes, because I discarded your bullshit, baseless claim.

Why should your baseless claim be a problem for me?

The hypothesis is most directly true: FORCES A AND B ARE NOT EQUAL.
No. It is most definitely false, as shown by my proof by contradiction.
There is no escaping that.
You also effectively disproved it with your analysis, showing that X doesn't pull with A, but with A-B, which is also equal but opposite to what Y pulls with, so the real force is equal but opposite, completely contradict your baseless claim.

So no, your baseless claim is not true, in any way.

You used the wrong net force on boat X to start with.
No. I used the correct one.
You used the wrong, claiming that it is magically experiencing a force of -A+B, while it is only providing a force of A, a complete defiance of Newton's laws of motion and objective reality.

In your twisted little logic, then you proceeded to substitute at will, until you reached the conclusion that A = B.
Yes, I substitute to have a single force of X pulling on the rope rather than 2, to accurately show how much force X is pulling on the rope with, and then I get A=-B, exactly as you would expect from an honest, rational analysis.

You couldn't be more wrong!
Stop just asserting bullshit.
If you think I am wrong point out exactly what I did wrong, quoting what I said, correcting it explicitly (by effectively copying and pasting what I said but with correction) and then explain why.

Your analysis shows that the RE way of doing things is totally wrong.
No. My analysis shows that your method is completely wrong, because it will reach a contradiction (or several), and a correct method does not produce double the force.

Here are the correct equations, where we take into account the TWO FORCES ACTING ON BOAT X RESPECTIVELY ON BOAT Y:
Then it is incorrect, as there is only a single force acting on boat X, the force of the rope pulling boat X in reaction to boat X pulling the rope. There is no other force. There cannot be any other force.
The only entity capable of acting on X is the rope.
Any force the rope applies to X must be matched in an action/reaction pair by X applying a force to the rope.
As such, if boat X applies a force of A to to the rope, then the rope applies a force of -A to boat X. There is no possibility for any other force.


The net force on boat X will be: -A + B
And all that force must come from the rope pulling on boat X.
This means there is an action-reaction pair between boat X and the rope such that:
boat X applies a force of A-B to the rope, and
the rope applies a force of -A+B to boat X.

This means boat X is pulling on the rope with a force of A-B, not A like you claim.

The net force on boat Y will be: -B + A
Again, provided by the rope.
Just like the above, this means boat Y will be applying a force to the rope of B-A, not B like you claim.

Also note that this is -(A-B). This means the force boat Y is applying to the rope is equal but opposite the force boat X is applying to the rope, i.e. boat X and boat Y are pulling the rope with forces that are equal in magnitude, not different in magnitude like you claim.

Thus, not only are you contradicting yourself (your baseless claim that the forces are different), you are also contradicting the very basis of the hypothesis/situation you are trying to model.

You have totally lost touch with reality, jack.

It is that simple.
Yes, my analysis is that simple, yet you still either don't understand it or are unwilling to admit you were wrong.

I always claimed that there are TWO FORCES acting on boat X: -A AND B.
Yes, you repeatedly baselessly asserted that, without any rational backing.
I repeatedly pointed out that you were wrong, and that in reality A=-B, thus it is the same force.

Since you have reached the wrong conclusion, something is very wrong with your analysis: you did not count the net force on boat X right.
No. I reached the correct conclusion, based upon an accurately, rational, honest analysis of the situation, free from baseless claims about it (like |A|!=|B|).
My conclusion and reasoning was also entirely consistent, both internally (i.e. with itself) and with the situation, unlike yours which contradicts both.

But I did.
No. You didn't.
You counted the same force twice, resulting in you falsely concluding that the force is double what you expect.

Using your way, YOU REACHED A CONTRADICTION which violates the hypothesis, that forces A and B are not equal.
Again, that isn't the hypothesis, that is your baseless claim.
The hypothesis is that boat X pulls with a force of A, and boat Y pulls with a force of B. There is no requirement for them to be different. An honest, rational analysis concludes that A=-B.
Your analysis requires A=B=0, which still contradicts your claim that they are different.

Only my analysis takes this crucial fact into account.
It isn't a fact. It is pure bullshit.
It contradicts reality.
Even your analysis contradicted it.

TWO FORCES ACT ON BOAT X: -A AND B.
No. A single force does. -A=B.
They are the same force.
If you like, you can represent this force as -F, and then break it apart into -A+B, but then X still applies a force of F to the rope, not A.

All of the forces balance out perfectly.
No. They don't. Not with your dishonest bullshit.
You claim X pulls on the rope with a force of A, yet if the rope is pulling on boat X with a force of -A+B, boat X MUST pull on the rope with a force of A-B. There is no way around that. So no, it doesn't balance.
You are missing a term of -B in the force X applies to the rope (or you have an extra term of +B in the force the rope applies).
Similarly you are missing a term of -A in the force Y applies to the rope.
So no, it doesn't balance.

In order for it to balance you need -A=-A+B, and you need -B=-B+A.
Does that look balanced to you?
Just in case you can't tell, you can simplify it by adding A or B to both sides.
This gives you:
0=B and 0=A
The only way to balance that is if A=B=0 (which again goes against your claim), but that would mean there is no force being applied.

So no, it isn't balanced.

But they include TWICE THE FORCES NEEDED in the Newtonian system.
Yes, because you are counting them twice/only counting half.


It does not work out like that.
That's right. It doesn't. So why do you think your way does?

There are only two forces acting on boat X: -A AND B.
No. There is only a single force acting on boat X. -A=B.

No other forces. Period.
That's right. No other forces, just -A=B.

Re: Distances in the universe
« Reply #290 on: May 03, 2017, 01:59:58 AM »
Sandy, do us a favour and fill in the blanks;

Code: [Select]
The net force on boat X (which is applied by the rope) is ___
Thus, boat X is pulling the rope with a force of ___

The net force on boat Y (which is applied by the rope) is ___
Thus, boat Y is pulling the rope with a force of ___

Thus the net force on the rope is ___+___=___

Can you do that?

Don't try doing any bullshit or rewriting it yourself.
Copy that list exactly as it is, and replace the ___ with values.
You can use your A and B and + and - and brackets and so on.

Re: Distances in the universe
« Reply #291 on: May 03, 2017, 02:01:24 AM »
I tried reading your post. I really did, but it was just full of nonsensical rubbish and assumptions it became too hard! Can you dumb it down in a succinct formula? Much like Newton's Laws, lets call them Sandokhan's Laws. What are the basic principles of whatever it was you were just talking about?

Basically it goes something like this:
For every action (at least in the directions Sandy decides) there is a double but opposite reaction.

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sandokhan

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Re: Distances in the universe
« Reply #292 on: May 03, 2017, 02:18:11 AM »
jack, you need medical help.

Urgently.

You are unable to digest your total loss.


PROVE IT!!!

The proof is in the pulling.

The man in boat X will pull on the rope with force A.

There is no way that a second person, in boat Y, will be able to pull WITH THE EXACT SAME FORCE.

It won't happen, it cannot happen.

The man in boat Y will pull on the rope with force B.


Two different forces, of a different magnitude.


Having no clue as to how to correctly apply Newton's third law to this problem, you proceeded as follows:

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.

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.



Your final result is this: the absolute value of A equals the absolute value of B.

Your final result could not be more wrong.

More disastrous.

But to you it is not a problem.

Which means you urgently need some help.


Using your own analysis, you reached the conclusion that A and B are the same forces.

By the very hypothesis, these two forces are of a different magnitude.


They are different forces, not to be confused with one another.


It is all over for you jack.


Your analysis led to a catastrophic contradiction for the RE, thanks to you.


My analysis is very simple, and for the first time, it takes into account the TWO FORCES ACTING ON EACH BOAT.


TWO FORCES ACT ON BOAT X: -A AND B.

TWO FORCES ACT ON BOAT Y: -B AND A.


The net force on the rope will be:

[A - B] + [B - A] = 0

All of the forces balance out perfectly.

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


No wild substitutions, no contradictions to be reached.


This means boat X is pulling on the rope with a force of A-B, not A like you claim.

Are you well jack?

I never claimed it was just A.

I always claimed that there two forces acting on boat X.


You cannot have A=-B, since this conclusion contradicts the hypothesis: we have two different forces to start with.

FORCES A AND B ARE OF A DIFFERENT MAGNITUDE: the two men doing the pulling with apply a different force each, not the same.

Your bumbling analysis led to a humongous contradiction.

That means your analysis based on a single force acting on boat X is wrong.


My analysis on the other hand is perfect.

TWO FORCES ACT ON BOAT X: -A AND B.

TWO FORCES ACT ON BOAT Y: -B AND A.


The net force on the rope will be:

[A - B] + [B - A] = 0

All of the forces balance out perfectly.

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


No wild substitutions, no contradictions to be reached.

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disputeone

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Re: Distances in the universe
« Reply #293 on: May 03, 2017, 02:27:42 AM »
FORCES A AND B ARE OF DIFFERENT MAGNITUDE TO START WITH.

sandokhan, I usually enjoy your posts, but you are making an ass of yourself.

Every action has an equal and opposite reaction.

You have to debunk Newtons third law before continuing your argument.

Edit.

jack, you need medical help.

Urgently.

You are unable to digest your total loss.

What? Because he is for all intents an purposes beating you with a big stick in the corner, you call him mentally ill?

Why do people do this?
« Last Edit: May 03, 2017, 02:30:27 AM by disputeone »
For all the law is fulfilled in one word, even in this.

The reason I am consistently personally attacked here.
https://www.theflatearthsociety.org/forum/index.php?topic=69306.msg1960160#msg1960160

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sandokhan

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Re: Distances in the universe
« Reply #294 on: May 03, 2017, 02:56:28 AM »
d1, please follow the technical details very closely.

I am not disputing Newton's third law, as you wrongly accused me of doing.


Every action has an equal and opposite reaction.

Exactly.

Precisely.


There are two forces acting on boat X: the man doing the pulling himself, and the pulling from boat Y: they occur at the same time.

That is why they have to be taken into account.


Here is the analysis provided by jack.

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.

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.


Based on his analysis, he reached the conclusion that A = -B.


The man in boat X will pull on the rope with force A.

There is no way that a second person, in boat Y, will be able to pull WITH THE EXACT SAME FORCE.

It won't happen, it cannot happen.

The man in boat Y will pull on the rope with force B.


Two different forces, of a different magnitude.


He has defeated himself, using his own bumbling analysis.

It seems you are incapable, d1, of following this discussion, for you have reached the wrong conclusion also.


Using his own analysis, your RE friend reached the conclusion that A and B are the same forces.

By the very hypothesis, these two forces are of a different magnitude.


They are different forces, not to be confused with one another.



Can you understand this much?


Now, here is my correct analysis, taking into consideration BOTH FORCES ACTING ON BOAT X, AND THE TWO FORCES ACTING ON BOAT Y:

TWO FORCES ACT ON BOAT X: -A AND B.

TWO FORCES ACT ON BOAT Y: -B AND A.


The net force on the rope will be:

[A - B] + [B - A] = 0

All of the forces balance out perfectly.

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


No wild substitutions, no contradictions to be reached.



jackblack is a classic case of cognitive dissonance: he cannot digest his own loss.


FORCES A AND BY ARE OF A TOTALLY DIFFERENT MAGNITUDE TO START WITH: OUR OWN HYPOTHESIS.

HIS BUMBLING ANALYSIS LED TO HIS CONCLUSION THAT A = -B.

A TOTAL CONTRADICTION.

What is going on with your own mind d1, to actually make a post where claim that someone else is winning?


If A DOES NOT EQUAL B to start with, and he reached the conclusion that A = - B, and to him this is not a problem, does he need medical help or not?


Re: Distances in the universe
« Reply #295 on: May 03, 2017, 02:58:47 AM »
You are unable to digest your total loss.
It is pretty hard to digest something which doesn't exist.
I haven't received a loss at all here.

PROVE IT!!!

The proof is in the pulling.

The man in boat X will pull on the rope with force A.

There is no way that a second person, in boat Y, will be able to pull WITH THE EXACT SAME FORCE.

It won't happen, it cannot happen.
So rather than prove it, you just baselessly assert it. Good job.

Now care to try for some actual proof?

Newtons third law indicates that if you have 2 people pulling a rope creating tension in the rope, they must be pulling on the rope with equal but opposite forces. Otherwise the rope will fly out of one of their hands.

So reality indicates that not only can they, they MUST pull with equal but opposite force.

Two different forces, of a different magnitude.
Yes, that is the baseless claim you are yet to prove.

Having no clue as to how to correctly apply Newton's third law to this problem, you proceeded as follows:
No. By correctly applying Newton's third law, something you seem incapable of doing.

Your final result is this: the absolute value of A equals the absolute value of B.

Your final result could not be more wrong.
No. My final result couldn't be more correct, as that is the correct answer to this problem.

But to you it is not a problem.
Again, that is because I didn't baselessly assume pure bullshit like you did.
I am fine with the 2 forces being equal, because there is NOTHING preventing them from being equal and that is the correct answer.

Which means you urgently need some help.
No, which means I can honestly and accurately analyse a situation rather than discarded the correct answer because of preconceived bullshit.

By the very hypothesis, these two forces are of a different magnitude.
No. By your baseless claim, which you are yet to back up in any way.

The hypothesis was merely that boat X was pulling with force A, and boat Y was pulling with force B.
My analysis is entirely consistent with that.

Your analysis isn't, at least not when using the 3rd law.

Your analysis led to a catastrophic contradiction for the RE, thanks to you.
Again, it only contradicts your baseless claim.
On the other hand, your analysis completely contradicts the hypothesis and your claim.

My analysis is very simple, and for the first time, it takes into account the TWO FORCES ACTING ON EACH BOAT.
You mean it counts the same force twice.

All of the forces balance out perfectly.
No. They don't.
I never claimed it was just A.
[/quote]
So with what force is X pulling on the rope?

That means your analysis based on a single force acting on boat X is wrong.
No. It means it doesn't go based upon your baseless bullshit.

My analysis on the other hand is perfect.
Except that it counts the forces twice.

Re: Distances in the universe
« Reply #296 on: May 03, 2017, 03:00:35 AM »
Like I asked before:

Fill in the blanks:
Code: [Select]
The net force on boat X (which is applied by the rope) is ___
Thus, boat X is pulling the rope with a force of ___

The net force on boat Y (which is applied by the rope) is ___
Thus, boat Y is pulling the rope with a force of ___

Thus the net force on the rope is ___+___=___

Think you can manage that?

*

sandokhan

  • Flat Earth Sultan
  • Flat Earth Scientist
  • 4904
Re: Distances in the universe
« Reply #297 on: May 03, 2017, 03:09:50 AM »
Newtons third law indicates that if you have 2 people pulling a rope creating tension in the rope, they must be pulling on the rope with equal but opposite forces. Otherwise the rope will fly out of one of their hands.

Yes.

Of course.


Now, here is my correct analysis, taking into consideration BOTH FORCES ACTING ON BOAT X, AND THE TWO FORCES ACTING ON BOAT Y:

TWO FORCES ACT ON BOAT X: -A AND B.

TWO FORCES ACT ON BOAT Y: -B AND A.


The net force on the rope will be:

[A - B] + [B - A] = 0


All of the forces balance out perfectly.

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


No wild substitutions, no contradictions to be reached.


Now, here is your own 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.

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.


Based on own analysis, you reached the conclusion that A = -B.


The man in boat X will pull on the rope with force A.

There is no way that a second person, in boat Y, will be able to pull WITH THE EXACT SAME FORCE.

It won't happen, it cannot happen.

The man in boat Y will pull on the rope with force B.


Two different forces, of a different magnitude.


Forces A and B are of a different magnitude, in fact they could never be equal at all.

By the very hypothesis, A DOES NOT EQUAL B.

A cannot equal B.

Yet, by using the twisted RE logic, using only a single force acting on boat X (respectively on boat Y), the analysis reaches a point where the absolute value of A equals the absolute value of B. A most direct contradiction of the hypothesis.



It is your very own analysis that has defeated you jack.

You are history here.


You have claimed that the absolute value of force A equals the absolute value of force B.

In total contradiction to the hypothesis, which states that forces A and B are not the same.


My analysis, which does take into account ALL OF THE FORCES, suffers from no such contradictions.

Here it is:


TWO FORCES ACT ON BOAT X: -A AND B.

TWO FORCES ACT ON BOAT Y: -B AND A.


The net force on the rope will be:

[A - B] + [B - A] = 0

All of the forces balance out perfectly.

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


No wild substitutions, no contradictions to be reached.



Re: Distances in the universe
« Reply #298 on: May 03, 2017, 03:36:11 AM »
Sandokhan I forgot to ask you when you replied to my threat a few months ago (it's a little too late to reply in that one I think) do you have any of your own written works on FE?
Flat vs. Round is a distraction aimed at covering the REAL truth.

The donut-shaped Earth is the only way, the light and salvation.

Then you will know the truth, and the truth will set you free. [John 8:32]

Re: Distances in the universe
« Reply #299 on: May 03, 2017, 04:12:35 AM »
Newtons third law indicates that if you have 2 people pulling a rope creating tension in the rope, they must be pulling on the rope with equal but opposite forces. Otherwise the rope will fly out of one of their hands.

Yes.

Of course.
Thus, whatever force X is pulling on the rope with, which you have described as A, is equal but opposite whatever Y is pulling on the rope with, which you have called B.
Thus A=-B.
QED.

Good job refuting yourself again.

Now, here is my correct analysis, taking into consideration BOTH FORCES ACTING ON BOAT X, AND THE TWO FORCES ACTING ON BOAT Y:
Enough of repeating the same refuted bullshit again and again.

Fill in the blanks:
Code: [Select]
The net force on boat X (which is applied by the rope) is ___
Thus, boat X is pulling the rope with a force of ___

The net force on boat Y (which is applied by the rope) is ___
Thus, boat Y is pulling the rope with a force of ___

Thus the net force on the rope is ___+___=___

Until you do that, your analysis is pure bullshit.

All of the forces balance out perfectly.

But they include TWICE THE FORCES NEEDED in the Newtonian system.
Sure, the ones you have shown balance, but you leave out other forces, such as the force X is pulling on the rope with, i.e. A.
In order for it to balance, that must be equal but opposite to the force the rope is pulling on the boat with, i.e. you need this equality to hold:
-A=-A+B.
Do you really think that is balanced?

That is also why you get twice the force. You double it/ignore half of it.

No wild substitutions, no contradictions to be reached.
Except when analysed according to Newton's third law, you do have a contradiction.
If the net force on X from the rope is -A+B, then X needs to apply a force of A-B to the rope, not the A you claim.
If the force X applies to the rope is A-B, and the force Y applies is B-A, then they are equal and opposite, contradicting your baseless assumption.
So yes, you do have contradictions.

Based on own analysis, you reached the conclusion that A = -B.
Yes. Based upon my analysis I reached a conclusion consistent with reality.

There is no way that a second person, in boat Y, will be able to pull WITH THE EXACT SAME FORCE.
Due to how forces work, there is. The person doesn't even need to be a person.
An inanimate object is capable of doing it.

Two different forces, of a different magnitude.
PROVE IT OR STOP REPEATING THE SAME REFUTED BULLSHIT!

By the very hypothesis, A DOES NOT EQUAL B.
No. Not the very hypothesis. Your baseless claim.

It is your very own analysis that has defeated you jack.
No. I didn't make the baseless claim that |A|!=|B|. That was you.
My analysis defeated you, not me.
But don't worry, I'm not alone. Your analysis defeated yourself as well.