Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II

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Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #30 on: January 18, 2020, 04:46:52 AM »
The equations are very clear and direct: you need to deal with real life situations where two different persons will apply two different forces on each end of the rope.
Your mean your fictional fantasy which in no way describes reality?
Your very argument shows that that fantasy of yours is impossible.
The only way for 2 people to apply a different force to the rope is for there to be a net force on the rope which then accelerates the rope.

What are the forces acting on the left end side of the rope?
-A and -B.
What are the forces acting on the right end side of the rope?
A and B.
DOUBLE THE FORCES needed in Newtonian mechanics!
No, exactly the forces needed, but said in a ridiculous way.
Notice a key part? It directly contradicts your claim, that the forces on each end of the rope are different.
This means that your claim, the very basis for your "paradox" is pure fiction.

What provides the force on the rope?
Is it the person pulling it (at a given end)?
That is the only thing pulling on the rope so it must be.
That means Person A, at the left side of the rope, pulls with a force of -A-B = -C.
That means Person B, at the right side of the rope, pulls with a force of A+B = C.
They both pull with a force equal in magnitude and opposite in direction.
Directly defying your entirely baseless claim.

If you wish to claim such garbage then prove that 2 people can pull on a rope with different forces, with no net force on the rope.
Go ahead and try, prove the impossible.


And again, THIS HAS NOTHING TO DO WITH GRAVITY!

Now, the Earth and the Moon should also start to travel towards each other: the very same mechanism applies, the supposed attractive force of gravity.
You mean they will orbit around their common barycenter, as they do.

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sandokhan

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Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #31 on: January 18, 2020, 05:45:18 AM »
You are trolling this forum.

Your mean your fictional fantasy which in no way describes reality?

Let us take a look at what you wrote.

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 hare-brained analysis leads to the ONLY SITUATION WHICH CANNOT EXIST IN REALITY: two persons which will apply exactly the same force.

Can't be.

No two persons can apply the very same force.

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 RE analysis leads to a total disaster, where the basic requirement is this |A|=|B|.

Which can NEVER be the case.

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.


Your analysis turns out to be PURE FANTASY.

But not mine.

I include from the very start the fact that force A does not equal force B.

The only way for 2 people to apply a different force to the rope is for there to be a net force on the rope which then accelerates the rope.

You are trolling this thread.

The net force on the rope is zero, yet there are two different forces being applied on each end.

The rope does not move.

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.

It directly contradicts your claim, that the forces on each end of the rope are different.

Another big fail from you.

THE FORCES ON EACH END ARE DIFFERENT, A AND B.

They have to be different since they are being applied by two different people.

However, the NET FORCE ON THE ROPE IS ZERO.

What provides the force on the rope?
Is it the person pulling it (at a given end)?
That is the only thing pulling on the rope so it must be.


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

-A and -B.


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

A and B.

That means Person A, at the left side of the rope, pulls with a force of -A-B = -C.
That means Person B, at the right side of the rope, pulls with a force of A+B = C.


Person A pulls with force -A, not -A + -B; person B pulls with force B, not A + B.

You totally screwed up.

As usual.

« Last Edit: January 18, 2020, 06:13:30 AM by sandokhan »

Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #32 on: January 18, 2020, 07:45:15 AM »
You say the forces can’t be balanced then try to balance the forces.

In what universe does that make any sense?

Are the forces balanced or not?

Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #33 on: January 18, 2020, 11:57:50 AM »
You are trolling this forum.
Your mean your fictional fantasy which in no way describes reality?
No, that would be you.
You keep appealing to your fictional fantasy where different people magically apply a different force to the rope.
Yet what do you conclude?

That the force on the rope from Person A pulling is -A-B, and the force on the rope from Person B pulling is equal in magnitude but opposite in direction, i.e. A+B.

i.e. your very hypothesis is wrong.

Yet rather than accept that your hypothesis is wrong (even though you have disproved it yourself), you just repeatedly reassert it.

Prove that your hypothesis can happen. Until you do, all you are showing is that your fantasy is wrong.

THE FORCES ON EACH END ARE DIFFERENT, A AND B.
What are the forces acting on the left end side of the rope?
-A and -B.
What are the forces acting on the right end side of the rope?
A and B.
Notice how you are directly contradicting yourself here?

You claim that the forces on each end of the rope are different, yet then show they are equal and opposite.

What is the force acting on the left side of the rope?
Is it -A, or is it -A-B?

Person A pulls with force -A, not -A + -B;
You totally screwed up.
As usual.
No, it is you screwing up, as usual.
What provides the force of -A-B to the rope?
The only thing attached to the rope at the left end is Person A.
Either they are pulling on the rope with a force of -A-B, or that is not the force on the rope.

That is why I asked you what is providing the force on the rope.

Again, you have 2 options:
Either the forces on each end of the rope and equal and opposite, or there is a net force on the rope.

There is no alternative.
Your hypothetical situation, where the force at each end is not equal and opposite, demands that there is a force on the rope.
Yet to "prove" a paradox you ignore that and pretend there is no force, which demands that the forces are equal and opposite.

Your hypothesis directly contradicts reality.

You may as well start with a hypothesis that 1!=3/3, and then complain about REers showing it is wrong, and acting like we made some mistake when proving that 1=3/3, all because it contradicts your FALSE hypothesis.

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sandokhan

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Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #34 on: January 18, 2020, 12:10:13 PM »
That the force on the rope from Person A pulling is -A-B, and the force on the rope from Person B pulling is equal in magnitude but opposite in direction, i.e. A+B.

The force applied by person A, left side, is -A. Period.

The net force on the left side of the rope is -A + -B.

Your word tricks don't work with me.

You claim that the forces on each end of the rope are different, yet then show they are equal and opposite.

What is the force acting on the left side of the rope?
Is it -A, or is it -A-B?


Force A and force B will ALWAYS be different. They have to be, since they are being applied by two different persons.

The force acting on the left side of the rope is, as always, -A + -B (-A - B).

What provides the force of -A-B to the rope?
The only thing attached to the rope at the left end is Person A.


You mean you don't know?

A is pulling with -A (directed to the left) + -B (reaction force from force B).

That is why the two boats will start moving towards each other.

Again, you have 2 options:
Either the forces on each end of the rope and equal and opposite, or there is a net force on the rope.


The forces at each end, those applied by the persons, will always be different.

But the net force on the rope will be zero.

Sheer beauty.

A catastrophe for the RE.


Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #35 on: January 18, 2020, 12:42:19 PM »

The forces at each end, those applied by the persons, will always be different.

But the net force on the rope will be zero.

Sheer beauty.

A catastrophe for the RE.

Two different forces in opposite directions add up to a net force of zero?


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rabinoz

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Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #36 on: January 18, 2020, 12:51:29 PM »
That the force on the rope from Person A pulling is -A-B, and the force on the rope from Person B pulling is equal in magnitude but opposite in direction, i.e. A+B.

The force applied by person A, left side, is -A. Period.

The net force on the left side of the rope is -A + -B.
Stop your spamming with this silly non-existent "paradox". Repeating the same error does not magically make it correct.

An ideal massless rope is simply a connector so the forces on each end of the rope simply have to be equal in magnitude and in opposite directions.

And the tension force anywhere along the length will be the same as the magnitude of those forces.
If you would like the treatment of a rope with a finite mass and different forces at each end I'm sure we can oblige.

If there is nothing else connected to the rope that is all that is possible. If you must have such a trivial thing spelt out, look at this:

Mechanics - 1.2.4.2 - Ideal Rope by Bob Trenwith


Or another attempt:

Applied maths: The 'massless' rope by Adam Beatty




Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #38 on: January 18, 2020, 01:52:32 PM »
That the force on the rope from Person A pulling is -A-B, and the force on the rope from Person B pulling is equal in magnitude but opposite in direction, i.e. A+B.

The force applied by person A, left side, is -A. Period.

The net force on the left side of the rope is -A + -B.

Your word tricks don't work with me.
Then stop trying to use them.

Again, what is providing the force on the left side of the rope?
It is only being pulled by Person A.
That means the force on the left side of the rope comes entirely from person A pulling it.
That means it MUST be equal to the force applied by Person A.
There is no other option.

If you wish to disagree, tell me what entity is providing the force of -B on the left side of the rope.

Force A and force B will ALWAYS be different. They have to be, since they are being applied by two different persons.
Your own analysis shows that is not the case; that even when you have different people, they will apply equal and opposite forces to the rope, or there will be a net force on the rope.

What provides the force of -A-B to the rope?
The only thing attached to the rope at the left end is Person A.

A is pulling with -A (directed to the left) + -B (reaction force from force B).
And here you go admitting that person A is pulling on the rope with a force of -A-B, directly contradicting your hypothesis that they are only pulling with a force of -A.

Again, you have refuted yourself.

The forces at each end, those applied by the persons, will always be different.
But the net force on the rope will be zero.
A catastrophe for the RE.
No, a catastrophe for you, as you have clearly shown that cannot be the case.

Again, if you wish to assert such contradictory nonsense, you will need to prove it.

Linking to your prior threads where you have been refuted, will not help you.


Again, what force is person A applying to the rope?
Is it -A, or -A-B?

If the former, what is applying the force of -B to the rope, noting it cannot be person A in any way.
If the latter, then you have equal and opposite forces on the rope.

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Yes

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Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #39 on: January 18, 2020, 03:47:18 PM »
Sandokhan: please demonstrate your paradox with a free body diagram.
It doesn't have to look good, it can be done with MS Paint and a mouse.  That's fine.  I just want to see the forces on each important component: the two boats and the rope.

I suspect if you create these diagrams, you will find a mistake in your argument.  And then you will be all the wiser for it.
Signatures are displayed at the bottom of each post or personal message. BBCode and smileys may be used in your signature.

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rabinoz

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Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #40 on: January 18, 2020, 04:13:01 PM »
Cut the crap.

https://www.theflatearthsociety.org/forum/index.php?topic=75798.0

https://www.theflatearthsociety.org/forum/index.php?topic=71192.msg1931730#msg1931730
No you cut the crap. There's nothing there to help you in the slightest!

For example, you might read this again:
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Quote from: sandokhan
From one of the best texts ever on mechanics:


That quote you give does not give any experiments, it's only a "thought experiment", though I have no disagreement with it.

It's a bit old, but so is Newton's work.
I won't argue and in fact, though a bit dated, that book seems to agree with what I've been saying all along.

You could, however, got a text version of that book, "A System of Natural Philosophy" by John Lee Comstock.
And you might in fairness quoted a bit more of you reference:

Quote from: John Lee Comstock
    109. It is easy to conceive, that if a man in one boat pulls at a rope attached to another boat, the two boats, if of the same size, will move towards each other at the same rate;
but if the one be large and the other small, the rapidity with which each moves will be in proportion to its size, the large one moving with as much less velocity as its size is greater.
    110. A man in a boat pulling a rope attached to a ship, seems only to move the boat, but that he really moves the ship is certain, when it is considered, that a thousand boats pulling in the same manner would make the ship meet them half way.
    111. It appears, therefore, that an equal force acting on bodies containing different quantities of matter, move them with different velocities, and that these velocities are in an inverse proportion to their quantities of matter.
    112. In respect to equal forces, it is obvious that in the case of the ship and single boat, they were moved towards each other by the same force, that is, the force of a man pulling by a rope. The same principle holds in respect to attraction, for all bodies attract each other equally, according to the quantities of matter they contain, and since all attraction is mutual, no body attracts another with a greater force than that by which it is attracted.
    113. Suppose a body to be placed at a distance from the earth, weighing two hundred pounds; the earth would then attract the body with a force equal to two hundred pounds, and the body would attract the earth with an equal force, otherwise their attraction would not be equal and mutual.
Another body weighing ten pounds, would be attracted with a force equal to ten pounds, and so of all bodies according to the quantity of matter they contain; each body being attracted by the earth with a force equal to its own weight, and attracting the earth with an equal force.


From: A System of Natural Philosophy by John Lee Comstock pages 31,32

Para 112. Seems to sort out the equal forces.

But I realise now why you selected that little bit from John Lee Comstock's book - the rest of it totally ruins all your arguments against Newtonian Gravitation, etc, etc.
And remember that this is "From one of the best texts ever on mechanics"!
So please read the chapter on GRAVITY from p. 24, the very chapter you quote from!


Quote from: sandokhan
THE ROPE WILL TRANSMIT TWO DIFFERENT FORCES.

No, the rope will not transmit two different forces. Let's leave it here till you digest the material in your reference, "one of the best texts ever on mechanics".

May I use other parts of that splendid reference in future debates?

Please explain why you cherry-picked, as you so often do, that little but from "From one of the best texts ever on mechanics" and ignored the but that I added!

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sandokhan

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Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #41 on: January 18, 2020, 10:34:45 PM »
The rope will transmit both forces, A and B.

It has to, since both boats will start to move towards each other.

Person A is pulling with force -A (directed to the left). Person B is pulling with force B (directed to the right).

"To every action there is always an opposed  equal reaction"

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.

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.


A total dismissal of the attractive gravitation concept.

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rabinoz

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Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #42 on: January 18, 2020, 10:55:26 PM »
The rope will transmit both forces, A and B.
A rope cannot "transmit both forces, A and B". Please get even remotely connected to reality!
Go and devise a real-life experiment and check it yourself.

Why haven't you even responded to my simple "thought experiment"?

My 4WD, in low range, has a draw-bar capability of about 2000 kg but a small kid on his trolley can only pull about 10 kg.

What strength rope do I need to connect the boy, pulling as hard as he possibly can, to the 4WD?
Is it:
  • 10 kg + a bit for safety,
  • 2000 kg + a bit for safety or
  • 2010 kg + a bit for safety?

Stop trolling the site with impossible pseudo-scientific claptrap.

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sandokhan

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Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #43 on: January 18, 2020, 11:00:48 PM »
You are trolling the upper forums, since you have nothing to add here.

Both boats will start to move towards each other: this means the rope is transmitting BOTH FORCES, A and B.

The equations work out perfectly, a clear sign of their correctness.

The rope will transmit both forces, A and B.

It has to, since both boats will start to move towards each other.

Person A is pulling with force -A (directed to the left). Person B is pulling with force B (directed to the right).

"To every action there is always an opposed  equal reaction"

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.

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.


A total dismissal of the attractive gravitation concept.

Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #44 on: January 19, 2020, 01:57:24 AM »
The rope will transmit both forces, A and B.
It has to, since both boats will start to move towards each other.
Yes, just like if you want to do it the simple way and have one force from person A and one from person B, force A and B need to be equal and opposite to get a net force of 0 in the rope.

"To every action there is always an opposed  equal reaction"
Forces A and B are, of course, of different magnitude.
No, they are, of course, of EQUAL MAGNITUDE!

Your analysis even showed that.

Again, quit with the BS, quit with the distractions.

Tell us what is applying the force of -B to the left side of the rope.
Is it the person on the left? Or is it pure magic?

The only thing which can possibly transfer the force is Person A.

But that means person A is pulling on the rope with a force of -A-B.
Likewise that means person B is pulling on the rope with a force of A+B.
Each person pulls the rope with a force which is equal in magnitude, directly contradicting your false hypothesis.

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rabinoz

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Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #45 on: January 19, 2020, 02:06:56 AM »
You are trolling the upper forums, since you have nothing to add here.
I guess you can't answer my question, got that!

Quote from: sandokhan
Both boats will start to move towards each other:
Yes, they will.

Quote from: sandokhan
this means the rope is transmitting BOTH FORCES, A and B.
No, the tension in the rope must be the same along its whole length and it constrains the applied forces to be equal in magnitude.

Quote from: sandokhan
The equations work out perfectly, a clear sign of their correctness.
No it is not "a clear sign of their correctness" if those equations omit vital conditions and yours do.
They ignore constraints imposed by the rope.

Quote from: sandokhan
The rope will transmit both forces, A and B.

It has to, since both boats will start to move towards each other.
Sort of but those forces must be exactly equal in magnitude

Now, would you please explain why you chose to ignore this from you own reference, the one you called "From one of the best texts ever on mechanics:"?
Quote from: John Lee Comstock
    112. In respect to equal forces, it is obvious that in the case of the ship and single boat, they were moved towards each other by the same force, that is, the force of a man pulling by a rope.
From: A System of Natural Philosophy by John Lee Comstock pages 31,32
That, from you own reference, says exactly what I've been saying all along.

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sandokhan

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Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #46 on: January 19, 2020, 02:35:10 AM »
I guess you can't answer my question, got that!

You do not even have a question.

You are trying to bamboozle your readers, as usual.

You presented a situation where there is friction present (4WD + trolley), tryint to trick your readers to go along.

There is no friction present between the Earth and the Moon.

That is why the example with the two boats/rafts on a lake, where there is very little friction, is the analogy.

No, the tension in the rope must be the same along its whole length and it constrains the applied forces to be equal in magnitude.

Sure, but the applied forces cannot be equal.

You have a pro wrestler on one side of the rope and yourself on another. Are the forces being applied equal?

In respect to equal forces...

Force A does not equal force B.

No it is not "a clear sign of their correctness" if those equations omit vital conditions and yours do.
They ignore constraints imposed by the rope.


You still don't get it. The equations either work or they do not. Mine work perfectly.


 force A and B need to be equal and opposite to get a net force of 0 in the rope.

The net forces on the rope are equal and opposite (A + B) and (-A -B).

Forces A and B are very different. They have to be.

Your analysis even showed that.


See my previous answer.

Tell us what is applying the force of -B to the left side of the rope.
Is it the person on the left? Or is it pure magic?


Don't you understand that everyone here is now laughing at you?

"To every action there is always an opposed  equal reaction"

The only thing which can possibly transfer the force is Person A.

No.

Both persons are pulling.

But that means person A is pulling on the rope with a force of -A-B.
Likewise that means person B is pulling on the rope with a force of A+B.
Each person pulls the rope with a force which is equal in magnitude


Completely wrong!

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

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

Net force on the left side of the rope: -A + (-B) (force applied by person A + reaction force on force B)

Net force on the right side of the rope: A + B (reaction force on force A + force applied by person B)

Net force on the rope: 0

Perfect analysis.

Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
« Reply #47 on: January 19, 2020, 02:58:49 AM »
I guess you can't answer my question, got that!
You do not even have a question.
Then what is this?
What strength rope do I need to connect the boy, pulling as hard as he possibly can, to the 4WD? Is it:
    That sure seems like he has a question.
    But that isn't the only question you can't answer.
    You also completley ignored mine.

    Again, what applies the force of -B to the rope on the left side?

    It seems the only thing available is person A, but you want to claim that person A is pulling with a force of just -A, not -A-B, all so you can pretend that the people on each end can magically pull with a different force.

    Sure, but the applied forces cannot be equal.
    Again, your own analysis shows that it must be.
    You have a force of -A-B applied to one end of the rope and a force of A+B applied to the other, equal but opposite.

    Your own argument shows you are wrong.

    You have a pro wrestler on one side of the rope and yourself on another. Are the forces being applied equal?
    Is there a net force on the rope?
    If not, then yes, the forces are equal.
    If so, then the forces are not equal.

    See, typically what would happen when there is a massive difference in strength is the rope is pulled out of one of your hands, because you cannot apply enough force to keep the net force on the rope as 0.

    You still don't get it. The equations either work or they do not. Mine work perfectly.
    Only when you have an equal and opposite force on each end of the rope.
    Try it without that.

    Actually present the hypothetical situation you claim:
    Person A pulls with a force of -A.
    There is nothing else to pull on the rope from the left side, so the total force on the rope from the left side is -A.
    Yes, there is a reactionary force, that is the rope pulling on person A with a force of A.

    Then at the other end, you have person B pulling on the rope with a force of B. Again, that is the only force on the rope from that side.
    The reactionary force is the rope pulling on person B with a force of -B.

    The net force on the rope is B-A.

    That means one of 2 things:
    A and B are equal, or there is a net force on the rope.

    There is no alternative.

    If you wish to disagree, you need to tell us what is applying the force of -B to the rope from the left side.
    It can't be person A, as they are just pulling with a force of -A.
    It can't by anything else as nothing else is pulling on the rope.

    Again, your hypothesis is wrong.
    You have refuted it yourself.
    Repeatedly asserting the same false hypothesis will not help your case.

    Tell us what is applying the force of -B to the left side of the rope.
    Is it the person on the left? Or is it pure magic?

    Don't you understand that everyone here is now laughing at you?
    And there you go with more pathetic avoidance.

    Stop with the distractions and insults and just answer the question.
    What is applying a force of -B to the left side of the rope?
    Until you answer that, all you have is a failed hypothesis.

    Both persons are pulling.
    Person B is on the right side of the rope, not the left.
    They cannot pull the left side of the rope to the left.
    As I said above, the reactionary force for that is the rope pulling person B.
    It is not the rope being pulling in the other direction.

    Do you actually understand action-reaction pairs?

    Person A is pulling with force -A (directed to the left).
    Again, your analyis requires them to be pulling with a force of -A-B.

    Net force on the left side of the rope: -A + (-B) (force applied by person A + reaction force on force B)
    No, the reaction force from person B pulling on the rope is a force on person B.

    Try again.

    Tell us what on the left side of the rope is pulling on the rope with a force of -B.

    Anything else is just a pathetic distraction from your complete failure.[/list]

    *

    sandokhan

    • Flat Earth Sultan
    • Flat Earth Scientist
    • 4904
    Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
    « Reply #48 on: January 19, 2020, 03:32:26 AM »
    You still don't seem to understand what is going on.

    You were given the chance to present your side of the story.

    Here is what you wrote:

    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.


    Do you understand the conclusion of your hare-brained analysis?

    If not, here it is:

    The RE analysis leads to a total disaster, where the basic requirement is this |A|=|B|.

    Which can NEVER be the case.

    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.


    By hypothesis, force A can never equal exactly force B.

    Which means your conclusion is wrong.


    Let us suppose now, that only one of the rafts/boats does the pulling (that is, side A will have the rope attached to it, no person A pulling).

    Person B pulls on the rope from the right.

    What are the forces on the left side of the rope (in boat A)?

    -B (reaction force on force B).


    Now let us bring person A back.

    Both persons are pulling now, force A does not equal force B.

    What are the forces on the left side of the rope now?

    Yes, person A is pulling with force -A (to the left) BUT ALSO person B is pulling.

    Reaction force is the SAME as in the previous situation: -B.

    Then, the net force on the left side of the rope is now: -A + -B, or -A -B.

    Very simple to understand.


    *

    rabinoz

    • 24886
    • Real Earth Believer
    Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
    « Reply #49 on: January 19, 2020, 05:03:59 AM »
    I guess you can't answer my question, got that!

    You do not even have a question.
    Yes I did!

    My 4WD, in low range, has a draw-bar capability of about 2000 kg but a small kid on his trolley can only pull about 10 kg.

    What strength rope do I need to connect the boy, pulling as hard as he possibly can, to the 4WD?
    Is it:
    • 10 kg + a bit for safety,
    • 2000 kg + a bit for safety or
    • 2010 kg + a bit for safety?
    Quote from: sandokhan
    You are trying to bamboozle your readers, as usual.

    You presented a situation where there is friction present (4WD + trolley), trying to trick your readers to go along.
    Friction has nothing to do with the problem! So answer that question.

    Quote from: sandokhan
    There is no friction present between the Earth and the Moon.
    That is why the example with the two boats/rafts on a lake, where there is very little friction, is the analogy.
    No, friction has nothing to do with the case!
    The whole problem is that YOU will not face the plain simple fact that the forces on the ends of an ideal rope must be exactly the same magnitude!

    Quote from: sandokhan
    No, the tension in the rope must be the same along its whole length and it constrains the applied forces to be equal in magnitude.
    Sure, but the applied forces cannot be equal.
    Sorry but even your own "one of the best texts ever on mechanics" disagrees with YOU! Look:
    Now, would you please explain why you chose to ignore this from you own reference, the one you called "From one of the best texts ever on mechanics:"?
    Quote from: John Lee Comstock
        112. In respect to equal forces, it is obvious that in the case of the ship and single boat, they were moved towards each other by the same force, that is, the force of a man pulling by a rope.
    From: A System of Natural Philosophy by John Lee Comstock pages 31,32
    That, from you own reference, says exactly what I've been saying all along.
    Quote from: sandokhan
    You have a pro wrestler on one side of the rope and yourself on another. Are the forces being applied equal?
    Yes, the forces MUST be equal! If the weak man can only pull 20 kg that is the tension on the rope.
    The pro-wrestler might be capable of pulling 200 kg but if the tension in the rope is 20 kg[1] then that is all the force he can apply.

    Now stop spamming the upper fora with your calculations.
    They might be correct but they are based on an incorrect and impossible premise and so are useless!

    Look, I am capable of lifting about 50 kg but when I lift a 10 kg mass on a rope I am only applying a force 10 kg to that rope.
    Go and measure it yourself, there's no friction involved!

    Now please believe what all the references, including your "one of the best texts ever on mechanics", say! YOU are wrong - get used to it!

    [1] I made a bad typo and wrote 200 kg where I intended to write 20 kg - sorry about that.
    « Last Edit: January 19, 2020, 12:55:45 PM by rabinoz »

    *

    sandokhan

    • Flat Earth Sultan
    • Flat Earth Scientist
    • 4904
    Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
    « Reply #50 on: January 19, 2020, 05:29:10 AM »
    You are trolling this thread.

    I thought that your servant already addressed my answers to your questions.

    You are confusing the forces applied on each end with the NET FORCES ON EACH END OF THE ROPE.

    Two very different concepts/things.

    Force A cannot equal force B.

    Yes, the forces MUST be equal! If the weak man can only pull 20 kg that is the tension on the rope.
    The pro-wrestler might be capable of pulling 200 kg but if the tension in the rope is 200 kg then that is all the force he can apply.


    You are dodging the issue.

    How can 20 kg be equal to 200 kg (pulling forces)? They cannot.

    You must take into account this basic fact: force A cannot and will not equal force B.

    See how you are trying to cheat yourself and your readers?

    The textbook on mechanics refers TO TOTAL FORCES, NET FORCES.

    NOT THE APPLIED FORCES.

    Do you understand the difference? It seems not, or if you do, you are trolling this thread.


    Let us suppose now, that only one of the rafts/boats does the pulling (that is, side A will have the rope attached to it, no person A pulling).

    Person B pulls on the rope from the right.

    What are the forces on the left side of the rope (in boat A)?

    -B (reaction force on force B).


    Now let us bring person A back.

    Both persons are pulling now, force A does not equal force B.

    What are the forces on the left side of the rope now?

    Yes, person A is pulling with force -A (to the left) BUT ALSO person B is pulling.

    Reaction force is the SAME as in the previous situation: -B.

    Then, the net force on the left side of the rope is now: -A + -B, or -A -B.

    Very simple to understand.


    Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
    « Reply #52 on: January 19, 2020, 12:21:19 PM »
    You still don't seem to understand what is going on.
    No, I understand quite well.
    You are setting up a scenario which is literally impossible, and then proceeding to prove it is impossible.

    I understand that my analysis shows your scenario to be impossible.
    But that is simply because it is.

    Remember, the only one who is saying the force applied by person A and person B is different, is you.
    You are yet to establish that is the case at all.

    Instead, you directly contradict by showing that the force acting on the left side of the rope, which only person A can apply, is (-A-B), while the force on the right side of the rope, which only person B can apply, is (A+B), i.e. the 2 people pull with a force equal in magnitude but opposite in direction.

    This is not difficult to grasp at all.

    Again, either they both pull with the same force, or there is a net force on the rope which accelerates the rope.

    If you wish to disagree, you need to tell us what magical entity provides the force of -B on the left side of the rope and A on the right side of the rope. You are yet to even begin doing this.

    And no, it isn't a reaction force from person B pulling on the rope. The reaction force for that is the rope pulling on person B.

    The action-reaction pair is not person B pulling on the right side of the rope so a force magically appears on the left side.
    If that was the case, it would be impossible to move any object.

    Here is the proper way to analyse it.
    Instead of removing the person and have the raft apply the force instead, remove the person, and break the link.
    Now there is nothing on the left side of the rope.
    Person B pulls on the rope with a force of B, the rope pulls on person B with a force of -B. That is the action-reaction pair.
    Then there are no forces acting on the left side of the rope.
    This means there is a net force of B on the rope, and the rope accelerates.

    Yet according to your insanity, there is magically a force of -B acting on the left side of the rope even though there is literally nothing to put it there, and thus the net force on the rope is magically 0 and it magically can't move.

    That is the nonsense your analysis leads to.

    Now again, what applies the force of -B to the left side of the rope.
    You claim it isn't person A (as that would destroy your claim).
    It is not any reactionary force from person B.
    There is nothing else to apply the force.

    So what applies this force?

    *

    rabinoz

    • 24886
    • Real Earth Believer
    Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
    « Reply #53 on: January 19, 2020, 01:14:21 PM »
    I thought that your servant already addressed my answers to your questions.
    It you are addressing ME, then, I'm sorry I have no servants.

    Now answer my very practical and real-life question!

    My 4WD, in low range, has a draw-bar capability of about 2000 kg but a small kid on his trolley can only pull about 10 kg.

    What strength rope do I need to connect the boy, pulling as hard as he possibly can, to the 4WD?
    Is it:
    • 10 kg + a bit for safety,
    • 2000 kg + a bit for safety or
    • 2010 kg + a bit for safety?
    Now, either answer it or admit you are quite out of touch with reality.

    Quote from: sandokhan
    You are confusing the forces applied on each end with the NET FORCES ON EACH END OF THE ROPE.
    Not in the slightest!
    But you have been trying to do that for years with claims like: "The Deluge occurred some 310 years ago."

    Quote from: sandokhan
    Two very different concepts/things.

    Force A cannot equal force B.
    Of course, they can and in this situation the fact that they are connected by a rope constraints them to be equal!
    Don't you have any understanding of what a constraint means?

    Quote from: sandokhan
    Yes, the forces MUST be equal! If the weak man can only pull 20 kg that is the tension on the rope.
    The pro-wrestler might be capable of pulling 200 kg but if the tension in the rope is 20 kg << I mistyped thas as 200 kg >> then that is all the force he can apply.

    You are dodging the issue.
    How can 20 kg be equal to 200 kg (pulling forces)? They cannot.
    I never ever claimed that "20 kg can be equal to 200 kg"! Please read what I said and what all the references say.

    The whole point is that simply because your "pro-wrestler might be capable of pulling 200 kg does not mean that he pulls 200 kg in every situation.

    Quote from: sandokhan
    You must take into account this basic fact: force A cannot and will not equal force B.

    See how you are trying to cheat yourself and your readers?
    I'm cheating nobody! Of course, force A can and will be equal force B if there is a constraint forcing that and here there is, THE ROPE!.

    Quote from: sandokhan
    The textbook on mechanics refers TO TOTAL FORCES, NET FORCES.

    NOT THE APPLIED FORCES.
    Read you own reference again!
    Quote from: John Lee Comstock
        112. In respect to equal forces, it is obvious that in the case of the ship and single boat, they were moved towards each other by the same force, that is, the force of a man pulling by a rope.
    From: A System of Natural Philosophy by John Lee Comstock pages 31,32
    Read the exact words written and not those you wish were there "they were moved towards each other by the same force, that is, the force of a man pulling by a rope".

    If you carry on this ridiculous path I'll be force to do a dynamic analysis of the whole situation and show how pathetic you claims are.

    In closing please explain exactly what is wrong with MIT's explanation of "Tension interaction".
    Quote from: MIT
    Tension interaction
    Ropes, wires, strings, etc. are commonly used to provide force in everyday situations. A force provided by a rope or string is generally called a tension force.
    Tension as a Force

    Properties of Ideal Ropes
    When discussing ropes, strings, etc. in this course, it will generally be that they have zero mass. In this case, their behavior is fairly simple. The important aspects can be summarized with two simple rules:

        A segment of a massless rope can only exert a tension force if it is stretched between two points of contact with other objects.
        If a massless rope is stretched between two points of contact with other objects, the tension force exerted by a given segment of the rope on the objects on either side will be equal in size and will point directly along the rope segment.

    Tension as Constraint
    Tension does not have an associated force law like gravitational or elastic restoring forces do. Instead, tension acts as a constraint. It will take on whatever value is necessary to keep the objects joined by the rope at the same separation.
    « Last Edit: January 19, 2020, 02:45:14 PM by rabinoz »

    Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
    « Reply #54 on: January 19, 2020, 01:25:37 PM »
    My bad, missed this part before.

    You are trolling this thread.
    No, that is still you, still ignoring simple questions and your own conclusions which show your scenario is impossible.

    You are confusing the forces applied on each end with the NET FORCES ON EACH END OF THE ROPE.
    There is no net force on each end of the rope. There is the net force on the rope, and the force applied at each end.

    You are wanting to pretend that you can apply a force to a rope, to have this force exist on the rope, without actually applying a force.

    Perhaps we should also be asking you to draw a simple diagram showing where these forces are.

    How can 20 kg be equal to 200 kg (pulling forces)? They cannot.
    That is right, they can't.
    But according to you, it doesn't matter how strong person B is or how weak person A is, person A will easily be able to hold the rope against the strength of person B.

    You are suggesting that pro-wrestler, or even a truck won't be able to pull a rope from a baby's hand.

    Here is a nice simple scenario for you:
    You have a baby, held in place by something, like a chair where they are strapped in, so they can't escape.
    You have a truck, with no trailers at all so its full force can go into pulling the rope.
    Now you have a rope between them.

    To put some numbers onto it, lets say the baby pulls with a force of 1 N, and the truck pulls with a force of 1 MN.

    What do you think will happen?
    Option 1 - The forces will be magically doubled, with this baby being able to hold back the truck even though it can only apply a force of 1 N.
    Option 2 - The baby, unable to counter the force of the truck, has the rope slide through and out of its hands, with the truck and rope then accelerating away?

    I know what any sane person would pick.
    Yet your analysis requires that this baby can hold back the truck. Do you not realise just how insane that is?

    Perhaps to better demonstrate it is instead of the baby, you have a piece of string, with a breaking strength of 1 N.
    So the maximum force this can apply is 1 N.
    Again, what happens?
    Does the string break and the truck and rope accelerate, or can this piece of string hold back a truck?

    You must take into account this basic fact: force A cannot and will not equal force B.
    Again, THAT IS BASIC FICTION!
    If you want to claim it as a fact, you need to prove it.
    So far every analysis has shown that is fiction, not fact.
    The only way for A and B to be different in magnitude is if there is a net force on the rope (or if you go crazy and try to separate them into 2 separate forces and then have those separate forces be different).

    See how you are trying to cheat yourself and your readers?
    You are setting up a hypothetical scenario which is literally impossible. All so you can pretend there are magically doubled forces.

    Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
    « Reply #55 on: January 20, 2020, 11:13:11 AM »
    You are trolling this thread.

    I thought that your servant already addressed my answers to your questions.

    You are confusing the forces applied on each end with the NET FORCES ON EACH END OF THE ROPE.

    Two very different concepts/things.

    Force A cannot equal force B.

    Yes, the forces MUST be equal! If the weak man can only pull 20 kg that is the tension on the rope.
    The pro-wrestler might be capable of pulling 200 kg but if the tension in the rope is 200 kg then that is all the force he can apply.


    You are dodging the issue.

    How can 20 kg be equal to 200 kg (pulling forces)? They cannot.

    You must take into account this basic fact: force A cannot and will not equal force B.

    See how you are trying to cheat yourself and your readers?

    The textbook on mechanics refers TO TOTAL FORCES, NET FORCES.

    NOT THE APPLIED FORCES.

    Do you understand the difference? It seems not, or if you do, you are trolling this thread.


    Let us suppose now, that only one of the rafts/boats does the pulling (that is, side A will have the rope attached to it, no person A pulling).

    Person B pulls on the rope from the right.

    What are the forces on the left side of the rope (in boat A)?

    -B (reaction force on force B).


    Now let us bring person A back.

    Both persons are pulling now, force A does not equal force B.

    What are the forces on the left side of the rope now?

    Yes, person A is pulling with force -A (to the left) BUT ALSO person B is pulling.

    Reaction force is the SAME as in the previous situation: -B.

    Then, the net force on the left side of the rope is now: -A + -B, or -A -B.

    Very simple to understand.
    I will change the vessels at each end
    They will be saucers, the children use to slide down a Snow Hill, we'll be on a hockey rink, essentially friction less. The rope will be an ideal rope, having no weight of its own.
    There is a marker at the center of the rope.
    Child A on the left.
    Child B on the right.
    Both children have the same weight.
    Start with both children in the opposite cage.

    Child B pulls in 10 feet of the road.
    Question what happens?
    1. Child A moves 10 feet to the right.
    2. Child B moves 10 feet to the left.
    3 Child A moves 5 feet to the right and
       Child B moves 5 feet to the left.
    Has the center point changed from mid point of the rink?

    step 2: Child a pulls in 20 feet
    1. Child A moves 20 feet to the right.
    2. Child B moves 20 feet to the left.
    3 Child A moves 10 feet to the right and
       Child B moves 10 feet to the left.
    Has the center point changed from mid point of the rink?

    step 3: by adding weights to child B we have doubled his weight,  Essentially doubling his mass.
    if Child A pulls in 20 feet what happens?
    This is where you get to answer the question.

    if Child B pulls in 20 feet what happens?
    This is where you get to answer the question.
    Does the center point changed from mid point of the rink?

    If both children pull in 20 feet?
    What then?


    The point that I am trying make is: it does not matter which child is doing the polling, the motion is the same.

    The the universe has no obligation to makes sense to you.
    The earth is a globe.

    *

    rabinoz

    • 24886
    • Real Earth Believer
    Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
    « Reply #56 on: January 20, 2020, 12:47:53 PM »
    You are trolling this thread.
    Sandokhan's favourite response when he has nothing rational to say so just spams the thread with the same old . . . . and then accuses us of "You are trolling this thread".

    Quote from: MouseWalker
    The point that I am trying make is: it does not matter which child is doing the polling, the motion is the same.
    If you look at Sandokhan's current work in:
    DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX VII at Advanced Flat Earth Theory « Reply #667 » you'll find he's preparing a whole new assault against such a simple thing as the "properties of a rope".

    *

    sokarul

    • 16775
    • Discount Chemist
    Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
    « Reply #57 on: January 20, 2020, 06:06:11 PM »
    You are trolling this thread.
    Sandokhan's favourite response when he has nothing rational to say so just spams the thread with the same old . . . . and then accuses us of "You are trolling this thread".

    Quote from: MouseWalker
    The point that I am trying make is: it does not matter which child is doing the polling, the motion is the same.
    If you look at Sandokhan's current work in:
    DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX VII at Advanced Flat Earth Theory « Reply #667 » you'll find he's preparing a whole new assault against such a simple thing as the "properties of a rope".

    Teh jet engines work on rotating mass!!!111!!111111!!!
    Sokarul

    ANNIHILATOR OF  SHIFTER

    Run Sandokhan run