# The Flat Earth Society

## Flat Earth Discussion Boards => Flat Earth Debate => Topic started by: MouseWalker on January 07, 2020, 06:54:48 PM

Title: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: MouseWalker on January 07, 2020, 06:54:48 PM
Quote
DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II

From a classic text on mechanics:

When science teachers are asked how does gravity work, they answer in this manner:

Gravity is a force.

Gravity is directed towards the center of the orbit i.e. the sun.

That makes gravity the centripetal force.
no the out word circular motion is the centripetal force, gravity is the string holding the ball in.
Quote
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.

not quite, is the results of inertia of the earth and gravity

Quote
Now, ball = earth

you = sun

tension in the string = gravity

Gravity is the reason one object orbits another. An analogy is swinging a ball on a string over your head. The string is like gravity, and it keeps the ball in orbit. If you let go of the string, the ball flies away from you. (Dr. Eric Christian, April 2011)
yes
Quote

http://scienceline.ucsb.edu/getkey.php?key=4569 (UCSB Science Line)

Centrifugal force acts on a rotating object in a direction opposite the axis of rotation. Imagine that you have a tennis ball tied to a string. If you swing the tennis ball on the string around in a circle, you would feel the ball tugging on the string. That is the centrifugal force on the ball. It is counteracted by tension in the string that you are holding. In this example, the tension force in the string is like the gravitational force between the earth and the sun. The ball doesn't get closer or farther from your hand. If you suddenly cut the string, the ball would go flying away, but that wont happen to the earth because of the sun's gravity.
yes
Quote
http://scienceline.ucsb.edu/getkey.php?key=4583

Forces can make something move or stop something from moving. For a planet in orbit around the sun, the string is invisible. That invisible string is the gravitational force between the Earth and the sun.

yes
Quote
Then, the Mass Attraction and General Relativity Attraction concepts are not viable models for the cause of gravity and inertia.
why?
Quote
Applying any "attractive" force model to the Earth Moon dynamic forces, we obtain this system:

The Earth’s attractive gravitation balances the orbital centrifugal force of the Moon.
The Moon’s attractive gravitation balances the orbital centrifugal force of the Earth.

the same force not two, from a from different point of view.
Quote

At first this may seem like an orderly and balanced attractive force system; however,... the following paradox exists. If the seat, source and cause of the "apparent" attraction forces are "internal" to each of the bodies...the attraction concept produces twice the force that is necessary to balance the centrifugal orbital forces of a planet moon system. The concept of "attraction" between bodies requires that the force “from” each separate body acts on the remote body,-- and equally on the originating body. Another example of a balanced system is a rope under tension; each end has an equal amount of opposing force. As noted by Newton's third law of motion, " To every action there is always an opposed  equal reaction".

This double force paradox is directly applicable to the "mass attraction",... the General Relativity “attraction” and all other attraction type concepts of gravity.

This example may help visualize the double force issue.

Let there be two rafts ( x and y )  freely floating on a clear calm lake with a rope between them.
Both rafts are still and are a rope length apart.
The man on (raft x) pulls on the rope which is attached to raft y.
Raft x will move toward raft y,… and raft y will move toward raft x.
Both rafts will receive equal and opposite force and motion.
It is not possible for (raft x) to remain still and be the source of the force.

I see no doubling divorces of forces
Quote
The Mass Attraction Models of Gravitation

The attraction concepts accept Newton's inverse square equation of gravity's force between two bodies as:
F = G x (M1 x M2) / r squared .
The surface gravity (g) for each of the bodies can be derived from the gravitational constant (G) and the mass and radius of the bodies. Using Newton's equation the g forces, allegedly "seated" in each of the "two" bodies acting on the other at a distance, can be calculated.

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.

there is no doubling forces, only the point of reference is changing
Quote

Using: 1 ) Newton’s equation as given above, 2 ) basic arithmetic, 3 ) common logic and 4 ) the mechanics of force, it is shown that the assumed Earth and Moon seated forces are equal; and as a result;…"all attraction models" produce twice the force that is required to balance the centrifugal forces of orbit!
why? : no Dublin of forces occur
Quote

The General Relativity Model of Gravitation

The exact same paradox arises with the General Relativity (GR) concept of gravity. It postulates that Mass warps a hypothetical "fabric of spacetime" and the warped fabric of spacetime causes “attraction” of other masses. Since in the GR theory the seat of the attractive force is anchored within the center of the planet’s and moon’s positions, we would again have twice the force required to balance the orbital forces of the Earth Moon system.
why? :  you are seeing twice, where there is no twice;  no Dublin of forces occur
It is the same force, just viewed from a different pointe of view.
Quote

Stanley V. Byers
« Last Edit: May 12, 2018, 06:51:23 AM by sandokhan »

It's hard, you almost have it, but then you jump to the wrong conclusion,
And won't let go.

Title: Re: Not DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: MouseWalker on January 07, 2020, 09:50:52 PM
Quote
DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II

From a classic text on mechanics:

When science teachers are asked how does gravity work, they answer in this manner:

Gravity is a force.

Gravity is directed towards the center of the orbit i.e. the sun.

That makes gravity the centripetal force.
no the out word circular motion is the centripetal force, gravity is the string holding the ball in.
Quote
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.

not quite, is the results of inertia of the earth and gravity

Quote
Now, ball = earth

you = sun

tension in the string = gravity

Gravity is the reason one object orbits another. An analogy is swinging a ball on a string over your head. The string is like gravity, and it keeps the ball in orbit. If you let go of the string, the ball flies away from you. (Dr. Eric Christian, April 2011)
yes
Quote

http://scienceline.ucsb.edu/getkey.php?key=4569 (UCSB Science Line)

Centrifugal force acts on a rotating object in a direction opposite the axis of rotation. Imagine that you have a tennis ball tied to a string. If you swing the tennis ball on the string around in a circle, you would feel the ball tugging on the string. That is the centrifugal force on the ball. It is counteracted by tension in the string that you are holding. In this example, the tension force in the string is like the gravitational force between the earth and the sun. The ball doesn't get closer or farther from your hand. If you suddenly cut the string, the ball would go flying away, but that wont happen to the earth because of the sun's gravity.
yes
Quote
http://scienceline.ucsb.edu/getkey.php?key=4583

Forces can make something move or stop something from moving. For a planet in orbit around the sun, the string is invisible. That invisible string is the gravitational force between the Earth and the sun.

yes
Quote
Then, the Mass Attraction and General Relativity Attraction concepts are not viable models for the cause of gravity and inertia.
why?
Quote
Applying any "attractive" force model to the Earth Moon dynamic forces, we obtain this system:

The Earth’s attractive gravitation balances the orbital centrifugal force of the Moon.
The Moon’s attractive gravitation balances the orbital centrifugal force of the Earth.

the same force not two, from a from different point of view.
Quote

At first this may seem like an orderly and balanced attractive force system; however,... the following paradox exists. If the seat, source and cause of the "apparent" attraction forces are "internal" to each of the bodies...the attraction concept produces twice the force that is necessary to balance the centrifugal orbital forces of a planet moon system. The concept of "attraction" between bodies requires that the force “from” each separate body acts on the remote body,-- and equally on the originating body. Another example of a balanced system is a rope under tension; each end has an equal amount of opposing force. As noted by Newton's third law of motion, " To every action there is always an opposed  equal reaction".

This double force paradox is directly applicable to the "mass attraction",... the General Relativity “attraction” and all other attraction type concepts of gravity.

This example may help visualize the double force issue.

Let there be two rafts ( x and y )  freely floating on a clear calm lake with a rope between them.
Both rafts are still and are a rope length apart.
The man on (raft x) pulls on the rope which is attached to raft y.
Raft x will move toward raft y,… and raft y will move toward raft x.
Both rafts will receive equal and opposite force and motion.
It is not possible for (raft x) to remain still and be the source of the force.

I see no doubling divorces of forces
Quote
The Mass Attraction Models of Gravitation

The attraction concepts accept Newton's inverse square equation of gravity's force between two bodies as:
F = G x (M1 x M2) / r squared .
The surface gravity (g) for each of the bodies can be derived from the gravitational constant (G) and the mass and radius of the bodies. Using Newton's equation the g forces, allegedly "seated" in each of the "two" bodies acting on the other at a distance, can be calculated.

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.

there is no doubling forces, only the point of reference is changing
Quote

Using: 1 ) Newton’s equation as given above, 2 ) basic arithmetic, 3 ) common logic and 4 ) the mechanics of force, it is shown that the assumed Earth and Moon seated forces are equal; and as a result;…"all attraction models" produce twice the force that is required to balance the centrifugal forces of orbit!
why? : no Dublin of forces occur
Quote

The General Relativity Model of Gravitation

The exact same paradox arises with the General Relativity (GR) concept of gravity. It postulates that Mass warps a hypothetical "fabric of spacetime" and the warped fabric of spacetime causes “attraction” of other masses. Since in the GR theory the seat of the attractive force is anchored within the center of the planet’s and moon’s positions, we would again have twice the force required to balance the orbital forces of the Earth Moon system.
why? :  you are seeing twice, where there is no twice;  no Dublin of forces occur
It is the same force, just viewed from a different pointe of view.
Quote

Stanley V. Byers
« Last Edit: May 12, 2018, 06:51:23 AM by sandokhan »

It's hard, you almost have it, but then you jump to the wrong conclusion,
And won't let go.
There was no angry in my intent.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Unconvinced on January 09, 2020, 05:50:52 AM
Centripetal means acting towards the centre.  Centripetal forces can be due to gravity or a physical structure.

Centrifugal forces act outwards, although they are really just momentum trying to go straight, so often aren’t considered real forces.

I’m not aware of any “double force paradox”.  It all adds up.

Some people here would do well to learn how to draw free body diagrams.  They make things like this so much easier to understand.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan on January 09, 2020, 06:32:02 AM
The doubles forces of attractive gravitation is one of the most devastating arguments against RE.

Complete demonstration, using FE and RE equations:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1905467#msg1905467
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Unconvinced on January 09, 2020, 07:16:19 AM
The doubles forces of attractive gravitation is one of the most devastating arguments against RE.

Complete demonstration, using FE and RE equations:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1905467#msg1905467

Your example is just plain wrong.

The men pulling on the rope must be pulling the same amount if nothing is moving. A=B

Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan on January 09, 2020, 07:40:58 AM
My example is perfect.

You haven't studied the situation being described at all.

RE equation leads to this: A = -B.

You wrote A=B.

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.

Here is the correct FE analysis.

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.

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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Yes on January 09, 2020, 07:54:18 AM
You appear to be mixing up your conventions.  Does a positive force represent a force to the left and a negative force represent a force to the right, or does a positive force represent attraction and a negative force represent repulsion?  It seems to change depending on if you're in space or on the ocean.

But they include TWICE THE FORCES NEEDED in the Newtonian system.
I've seen you make this claim before.  You have this bizarre idea that the equation of Newtonian gravity is wrong because you have to apply it twice which gets you double the force.  Don't do that.  Don't apply it twice.  Calculate f = (G m1 m2) / r2 once.  Also keep in mind that that equation is a description of non-relativistic gravitational mechanics, not an explanation of it.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan on January 09, 2020, 08:05:49 AM
My conventions are very clear.

You better not bring repulsive gravity into our discussion.

Because then you are going to have to explain this:

Two bodies are pulled to each other by an external pressure.

Let's see how Newton describes this force in the Principia:

“In attractions, I briefly demonstrate the thing after this manner. Suppose an obstacle is interposed to hinder the meeting of any two bodies A, B, attracting one the other: then if either body, as A, is more attracted towards the other body B, than that other body B is towards the first body A, the obstacle will be more strongly urged by the pressure of the body A than by the pressure of the body B, and therefore will not remain in equilibrium: but the stronger pressure will prevail, and will make the system of the two bodies, together with the obstacle, to move directly towards the parts on which B lies; and in free spaces, to go forwards in infinitum with a motion continually accelerated; which is absurd and contrary to the first law.”

the obstacle will be more strongly urged by the pressure of the body A

PRESSURE = PUSHING FORCE

ATTRACTION = PULLING FORCE

Newton's clear description again:

the obstacle will be more strongly urged by the pressure of the body A than by the pressure of the body B, and therefore will not remain in equilibrium: but the stronger pressure will prevail

Here is a chance to redeem yourself.

From the pages of the same Principia:

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

Please describe the forces acting on the rope.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Yes on January 09, 2020, 09:14:15 AM
Please describe the forces acting on the rope.
Lucky for me, you asked for the easy part.  Let's make it a little easier still and assume the rope has no plastic or elastic characteristics.  The exact scenario isn't clear, so I'll describe two.  First, let's say the stone is fixed.  Here, the rope is under tension and not moving, so the forces acting on the rope net to zero.  Relevant components include a pull from the horse, a pull from the stone, and a pull from gravity.  All of these must necessarily be balanced, since the rope is not moving.  Second scenario, the horse is dragging the stone forward.  This case is similar to the first, except that there is a net force in the direction of the horse.  However, I only took a statics course in college, not a dynamics course, so I'm not certain of the terminology to describe that the force isn't constant and does not result in a constant acceleration. I'll let you fill me in on that.

Meanwhile, here's the fun part: this has nothing to do with gravity.  There is no gravitational tension between bodies.  Our sun isn't swinging planets around on a tether like an out-of-control yo-yo trick.  Read what Newton has to say about that on that page you cited.  Consider his analogy with a magnet.  "So the gravitation between the earth and its parts is mutual."

Also, in case you're hung up on Netwon's language of a pressure, I'd recommend updating your understanding of gravity to something in line with the last hundred years.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan on January 09, 2020, 09:20:54 AM
You did not describe the forces acting on the rope.

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

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

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

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

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

Double the forces needed in the Newtonian description of mechanics.

Meanwhile, here's the fun part: this has nothing to do with gravity.  There is no gravitational tension between bodies.  Our sun isn't swinging planets around on a tether like an out-of-control yo-yo trick.

When science teachers are asked how does gravity work, they answer in this manner:

Gravity is a force.

Gravity is directed towards the center of the orbit i.e. the sun.

That makes gravity the centripetal force.

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.

Now, ball = earth

you = sun

tension in the string = gravity

Gravity is the reason one object orbits another. An analogy is swinging a ball on a string over your head. The string is like gravity, and it keeps the ball in orbit. If you let go of the string, the ball flies away from you. (Dr. Eric Christian, April 2011)

http://scienceline.ucsb.edu/getkey.php?key=4569 (UCSB Science Line)

Centrifugal force acts on a rotating object in a direction opposite the axis of rotation. Imagine that you have a tennis ball tied to a string. If you swing the tennis ball on the string around in a circle, you would feel the ball tugging on the string. That is the centrifugal force on the ball. It is counteracted by tension in the string that you are holding. In this example, the tension force in the string is like the gravitational force between the earth and the sun. The ball doesn't get closer or farther from your hand. If you suddenly cut the string, the ball would go flying away, but that wont happen to the earth because of the sun's gravity.

http://scienceline.ucsb.edu/getkey.php?key=4583

Forces can make something move or stop something from moving. For a planet in orbit around the sun, the string is invisible. That invisible string is the gravitational force between the Earth and the sun.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Yes on January 09, 2020, 10:00:39 AM
You did not describe the forces acting on the rope.

Double the forces needed in the Newtonian description of mechanics.
Don't you find it a little suspicious that you're the only person to claim this in hundreds of years?  Wouldn't it be far more likely that you are misunderstanding something?  Just because you're a smart guy doesn't mean you don't make mistakes.  Humility is the first step to learning.

tension in the string = gravity
Tension in a rope that is fixed on one end applies a force of the swinging body in the direction of the center.  In this picture, the free body diagram of a swinging body is indeed analogous to the free body diagram of an orbiting body.  But don't take this analogy too far.  For example, if the orbit of a satellite changes to increase its distance from earth, the force of gravity does not increase correspondingly.  There is no string attaching the satellite to the earth, after all.  There are equal forces acting on the string, but there are not forces acting on the gravity that holds a body in orbit.  So no force doubling is required.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan on January 09, 2020, 10:06:04 AM
There are equal forces acting on the string, but there are not forces acting on the gravity that holds a body in orbit.  So no force doubling is required.

But there is:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1909690#msg1909690

We have been left certain clues along the way, from the group of people who wrote the works attributed to Newton and Huygens, which point out the sheer fallacy of the law of universal gravitational attraction:

“That gravity should be innate, inherent, and essential to matter, so that one body can act upon another at a distance through a vacuum without the mediation of anything else, by and through which their action and force may be conveyed from one to another, is to me so great an absurdity that I believe no man, who has in philosophical matters a competent faculty of thinking, can ever fall into it.”

Huygens dismissed the attraction concept:

”Concerning the cause of the flux given by M. Newton, I am by no means satisfied [by it], nor by all the other theories that he builds upon his principle of attraction, which to me seems absurd, as I have already mentioned in the addition to the Discourse on Gravity. And I have often wondered how he could have given himself all the trouble of making such a number of investigations and difficult calculations that have no other foundation than this very principle."
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Unconvinced on January 09, 2020, 11:09:05 AM
My example is perfect.

You haven't studied the situation being described at all.

RE equation leads to this: A = -B.

You wrote A=B.

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.

Here is the correct FE analysis.

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.

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.

What you’ve done here is introduce a spurious additional term.

Lets go back a step.  Two men are holding each end of a rope.  First man pulls with force A.  To stay put, the second man has to resist this force, which you are calling a reaction force.

But what does resisting this force feel like?  It feels like pulling, doesn’t it?  So each man is pulling with force A.

Maybe the second man then gives it some extra welly, with an additional force B so both men are now pulling with A+B.  That just means both men are now pulling harder.

Whatever.  Both men pull the same amount.

Or, let’s cut the rope in half and stick a force gauge in the middle.  The gauge reads 50N.  That’s the force between the two men, that’s what they are both pulling.

This is analogous to the force of gravity.  The force between the two objects.  No doubling required.

In my first post I mentioned free body diagrams.  Honestly, they help.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Yes on January 09, 2020, 11:17:02 AM
Huygens dismissed the attraction concept
Huygens lived in the 1600s, bro.

In my first post I mentioned free body diagrams.  Honestly, they help.
I second the motion.

Sandokhan, if you're unfamiliar with free body diagrams, or if you're not confident on how to create or use them, there are many good and reputable sites online and videos on YT that will help you.  I have no doubt you'll pick up the technique easily.  I highly encourage you to take advantage of these tools.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan on January 09, 2020, 11:17:38 AM
So each man is pulling with force A.

No.

On one end of the rope we place Queen Elizabeth I, on the other end we have Henry VIII.

Not the same force.

It can NEVER be the same. No two persons can pull with the same force.

Both men pull the same amount.

No.

We have two forces: A and B.

Always different.

Or, let’s cut the rope in half and stick a force gauge in the middle.  The gauge reads 50N.

Doesn't matter.

You still have TWO FORCES acting on each end of the rope.

DOUBLE the forces required by Newtonian mechanics.

Both boats will start to move toward each other.

Then, both the Moon and the Earth should also start to move toward each other.

That is why the double forces of attractive gravitation paradox is so devastating.

Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Yes on January 09, 2020, 12:09:21 PM

Check it out, it's good stuff.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Unconvinced on January 09, 2020, 12:39:07 PM
So each man is pulling with force A.

No.

On one end of the rope we place Queen Elizabeth I, on the other end we have Henry VIII.

Not the same force.

It can NEVER be the same. No two persons can pull with the same force.

Both men pull the same amount.

No.

We have two forces: A and B.

Always different.

Or, let’s cut the rope in half and stick a force gauge in the middle.  The gauge reads 50N.

Doesn't matter.

You still have TWO FORCES acting on each end of the rope.

DOUBLE the forces required by Newtonian mechanics.

Both boats will start to move toward each other.

Then, both the Moon and the Earth should also start to move toward each other.

That is why the double forces of attractive gravitation paradox is so devastating.

If Henry VIII and Elizabeth I are having a tug of war and neither is moving, they are in equilibrium and are pulling the same amount by definition.

If Liz is too strong for Henry, she pulls him over onto his fat arse.

Simple.

PS. I like the way you are not only trying to debunk gravity, but probably the most fundamental principle in engineering.  Interesting that all the world’s engineers have had their calculations wrong all this time, but things still work and not one of them has noticed such a basic problem.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: JackBlack on January 09, 2020, 01:25:50 PM
The doubles forces of attractive gravitation is one of the most devastating arguments against RE.
Except it is all in your head, and has been repeatedly refuted.
You weren't even able to address basics of how forces on a rope act, so how do you hope to deal with gravity?

And of course, rather than link to the debate, where you were repeatedly refuted before you fled, you link to your compilation of dishonesty.

Complete demonstration, using FE and RE equations:
https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1905467#msg1905467
Yes, a complete demonstration that your hypothesis is wrong. YOU CANNOT HAVE DIFFERENT FORCES ON THE ROPE!

Person at one end pulls on the rope with a force of A, person a the other end pulls with B.
Thus A=-B.
This is the ONLY option.

You even reach effectively the same result in your analysis, you just label the forces differently.
Instead of A, you have A+B.
This is the force that the the person holding the rope at boat x is applying to the rope.
Instead of B, you have -A-B.

It isn't including twice the forces, it is just relabelling the forces.

It is the exact same setup, just with different labels, which again shows that the 2 forces are equal and opposite, and that you cannot have a situation like this where one person pulls on the rope with some force, and the person at the other end pulls with a different force (i.e. different in magnitude).

What you are doing is equivalent to starting with the hypothesis that 1=2 (or that 3/3 != 1), and then complaining that people show that that hypothesis is wrong.

Your paradox is based upon a blatant rejection of reality, where you complain that people show your rejection of reality is wrong.

Your false hypothesis can never happen if the net force on the rope is 0.
The only way to have it happen is if you go to a rope with mass which is accelerated. Then there is a net force on the rope and the rope moves.
Otherwise, there cannot be any net force on the rope and thus you must have A+B=0. There is no alternative.

If you wish to claim there is tell us how this extra force is magically applied to the rope?

Is it the person pulling on the rope applying a force of A+B?
Or do they only pull with a force of A, with the rope just magically getting the force of B from nothing at all?
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: rabinoz on January 09, 2020, 02:53:25 PM
So each man is pulling with force A.
No.
On one end of the rope we place Queen Elizabeth I, on the other end we have Henry VIII.
Not the same force.
If they are pulling on each end of an ideal massless "rope" the forces cannot be different - end of story!

Quote from: sandokhan
It can NEVER be the same. No two persons can pull with the same force.
But if Henry VIII is stronger than Queen Elizabeth I she simply gets pulled along. What's so hard about that?

Throw the rope away and have Henry VIII grab Queen Elizabeth I by the hands. What happens then? She gets pulled over.

Have you ever been involved in a tug-of-war? If one team is stronger than the other they simply pull the weaker team along.

But in all these cases the force on the rope is equal to the lesser of the capabilities of those doing the pulling.

Quote from: MIT
Tension interaction (https://scripts.mit.edu/~srayyan/PERwiki/index.php?title=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.

Tension and Energy
Tension is a non-conservative force, and therefore has no associated potential energy. When tension is internal, however, it is a non-dissipative force, performing zero net work on the chosen system. The reason is that an ideal rope cannot stretch, which guarantees that the two interacting objects will undergo the same displacement. Thus, the work done on the two objects will cancel by Newton's Third Law.

Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: MaNaeSWolf on January 17, 2020, 02:36:41 AM
discussions with sandokhan is like arguing with someone who traveled 1000 years to the future, and still refuses to believe that scientific progress and understanding have improved. He just ignores everything from the last few hundred years.

I wonder how he manages to use the internet.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Platonius21 on January 17, 2020, 07:21:39 PM
I think when folks like Sandokhan and Wise were babies they were accidentally given an anti-science vaccine instead of a measles vaccine. You can't convince them of anything, they are immune.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: HattyFatner on January 17, 2020, 10:19:03 PM
There is no basis for discussion here.

Without an acceptance of a global earth and gravity, of course there will be a paradox with observable reality and the theoretical sciences.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: rabinoz on January 17, 2020, 11:34:28 PM
There is no basis for discussion here.

Without an acceptance of a global earth and gravity, of course there will be a paradox with observable reality and the theoretical sciences.
But gravitation, as mass appearing to attract mass, has been measured hundreds of times to far higher precision than any flat-Earther has ever managed.
Not only that but it has been qualitatively demonstrated hundreds of time.

But you cannot deny that "something" holds you down on the Earth with a force equal to your weight - what is that "something"?

But all that is irrelevant to the topic. Did you read this part of the OP?
Quote
This example may help visualize the double force issue.

Let there be two rafts ( x and y )  freely floating on a clear calm lake with a rope between them.
Both rafts are still and are a rope length apart.
The man on (raft x) pulls on the rope which is attached to raft y.
Raft x will move toward raft y,… and raft y will move toward raft x.
Both rafts will receive equal and opposite force and motion.
It is not possible for (raft x) to remain still and be the source of the force.
More detail can be found in Technology, Science & Alt Science / Re: Is the tension on an "ideal rope" (massless) the same at each end? « Message by sandokhan on May 16, 2018, 07:04:50 AM » (https://www.theflatearthsociety.org/forum/index.php?topic=75798.msg2059786;topicseen#msg2059786)
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: JackBlack on January 18, 2020, 01:25:44 AM
There is no basis for discussion here.

Without an acceptance of a global earth and gravity, of course there will be a paradox with observable reality and the theoretical sciences.
Nope, his fake paradox has absolutely nothing to do with gravity and instead is just simple Newtonian physics.
Look at the examples he is using, with a rope, with nothing at all to do with gravity.

It is just his misunderstanding of physics or intentionally lying about it.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan on January 18, 2020, 01:56:13 AM
The double forces of gravitation paradox is very real.

It is devastating for the RET.

Very simple to understand.

Now, the complete demonstration that indeed there will be two forces acting on boat X, and two forces acting on boat Y.

Two boats on lake, boat X and boat Y, are being pulled toward each other using a single rope by the two men on each boat.

The force applied by the first man is force A.

The force applied by the second man is force B.

They are of different magnitude to start with, A does not equal B.

Here is how the RE analysis goes, reaching a most profound 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.

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.

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.

Here is the correct FE analysis.

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.

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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: JackBlack on January 18, 2020, 02:40:21 AM
The double forces of gravitation paradox is very real.

Again, what is the net force on the rope at Side A?
What is it at side B?
Notice how these are equal and opposite?
That means the very basis of your claim IS FANTASY!

There are no twice the forces involved, just you inventing a hypothetical situation which cannot describe reality at all and results in a direct contradiction of your hypothesis, directly refuting yourself.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Unconvinced on January 18, 2020, 03:00:53 AM
Wow.  A cut and paste from a few posts up.  Is that a new record?

What’s the point in saying the exact same thing again?  Several people have pointed out where you went wrong.

Can I make this any easier for you to get?  I’ll try.

You said at the beginning that the forces on either side can’t be equal, then set about balancing the forces so they are equal on both sides.

So which is it?  Do the forces need to be balanced or not?

Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: rabinoz on January 18, 2020, 03:01:44 AM
The double forces of gravitation paradox is very real.

It is devastating for the RET.

Very simple to understand.

Now, the complete demonstration that indeed there will be two forces acting on boat X, and two forces acting on boat Y.

Two boats on lake, boat X and boat Y, are being pulled toward each other using a single rope by the two men on each boat.

The force applied by the first man is force A.

The force applied by the second man is force B.

They are of different magnitude to start with, A does not equal B.
If the two men are connected by a massless rope the forces applied to each end must be equal in magnitude because a massless rope is simply a connector.

End of story!

Quote from: sandokhan
Here is how the RE analysis goes, reaching a most profound contradiction:
No it doesn't!

The twin men might be capable of applying very different forces but that is immaterial.

But in all these cases the force on the rope cannot be more than the lesser of the capabilities of those doing the pulling.

Quote from: MIT
Tension interaction (https://scripts.mit.edu/~srayyan/PERwiki/index.php?title=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.

Tension and Energy
Tension is a non-conservative force, and therefore has no associated potential energy. When tension is internal, however, it is a non-dissipative force, performing zero net work on the chosen system. The reason is that an ideal rope cannot stretch, which guarantees that the two interacting objects will undergo the same displacement. Thus, the work done on the two objects will cancel by Newton's Third Law.

MIT makes sense but your "DOUBLE FORCES . . . . . PARADOX" is quite illogical.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan on January 18, 2020, 03:08:49 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.

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!

This means that both boats will start to travel towards each other.

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.

What? Massless rope?

Cut the crap.

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

https://www.theflatearthsociety.org/forum/index.php?topic=71192.msg1931730#msg1931730
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: rabinoz on January 18, 2020, 04:38:52 AM
The equations are very clear and direct:
The questions might be but your logic is pure crap. Equations are useless if they do not represent a real system

Quote from: sandokhan
you need to deal with real life situations where two different persons will apply two different forces on each end of the rope.
No, it's you who refuse to "to deal with real life situations".

Two different persons simply cannot apply two different forces on each end of the rope - it is quite impossible!

This has been explained to you numerous time but you won't believe us and you won't believe MIT.

Look 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?

Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: JackBlack 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 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.

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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan on January 18, 2020, 05:45:18 AM
You are trolling this forum.

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.

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.

Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Unconvinced 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?
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: JackBlack on January 18, 2020, 11:57:50 AM
You are trolling this forum.
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.

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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan 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.

Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Unconvinced 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?

Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: rabinoz 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

Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan on January 18, 2020, 01:32:59 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
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: JackBlack 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.

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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Yes on January 18, 2020, 03:47:18 PM
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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: rabinoz 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:
(https://image.ibb.co/mv4ZMJ/double7.jpg)

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 (https://ia802606.us.archive.org/26/items/asystemnaturalp10comsgoog/asystemnaturalp10comsgoog.pdf)

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!
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan 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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: rabinoz 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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan 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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: JackBlack 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!

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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: rabinoz 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 (https://ia802606.us.archive.org/26/items/asystemnaturalp10comsgoog/asystemnaturalp10comsgoog.pdf)
That, from you own reference, says exactly what I've been saying all along.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan 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 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.

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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: JackBlack 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.

You have refuted it yourself.

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]
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan 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.

Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: rabinoz 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 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 (https://ia802606.us.archive.org/26/items/asystemnaturalp10comsgoog/asystemnaturalp10comsgoog.pdf)
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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sandokhan on January 19, 2020, 05:29:10 AM

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.

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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: Yes on January 19, 2020, 06:15:13 AM
Slide 22 from https://slideplayer.com/slide/9162748/ :
"The tension T⃗ is the force that the rope exerts on the box.  Thus T⃗ and F⃗box on rope are an action/reaction pair and have the same magnitude."
(https://slideplayer.com/slide/9162748/27/images/22/Ropes+The+box+is+pulled+by+the+rope%2C+so+the+box%E2%80%99s+free-body+diagram+shows+a+tension+force+..jpg) (https://slideplayer.com/slide/9162748/)

"A massless rope held at both ends will exert a force of the same size (size T in the FBD's drawn here) on the object at each end. The forces will point directly along the rope as shown in the free body diagrams of the two people pulling the rope in the example below:"
(https://scripts.mit.edu/~srayyan/PERwiki/images/e/ea/Unit1Mod6Rope2c.png) (https://scripts.mit.edu/~srayyan/PERwiki/index.php?title=Module_3_--_Tension_Force)

Slide 44 from https://slideplayer.com/slide/4722645/ :
(https://images.slideplayer.com/15/4722645/slides/slide_44.jpg) (https://slideplayer.com/slide/4722645/)

Similar:
(https://i.pinimg.com/originals/9f/7b/09/9f7b09b08768ae0ed4c7510627f47b61.png)

Sandokhan, please understand this is not controversial physics.  Have the humility it takes to admit your own misunderstanding and you will be able to learn.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: JackBlack 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.

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?
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: rabinoz on January 19, 2020, 01:14:21 PM
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.

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.
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 (https://ia802606.us.archive.org/26/items/asystemnaturalp10comsgoog/asystemnaturalp10comsgoog.pdf)
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 (https://scripts.mit.edu/~srayyan/PERwiki/index.php?title=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.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: JackBlack on January 19, 2020, 01:25:37 PM
My bad, missed this part before.

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).

You are setting up a hypothetical scenario which is literally impossible. All so you can pretend there are magically doubled forces.
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: MouseWalker on January 20, 2020, 11:13:11 AM

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.

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.

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.

Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: rabinoz on January 20, 2020, 12:47:53 PM
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 » (https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg2230897#msg2230897) you'll find he's preparing a whole new assault against such a simple thing as the "properties of a rope".
Title: Re: Debating: DOUBLE FORCES OF ATTRACTIVE GRAVITATION PARADOX II
Post by: sokarul on January 20, 2020, 06:06:11 PM