Is the tension on an "ideal rope" (massless) the same at each end?

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rabinoz

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Agreed the thread needed splitting up and it is more a Technology, Science & Alt Science issue that a Flat Earth vs Globe issue, so here it is:

This thread needs to be split up.

Any comments?

A physics illiterate has the audacity to challenge me on the double forces of gravitation paradox?
No, I made no mention of any "double forces of gravitation paradox", but I do have "the audacity to challenge" you to the properties of an ideal rope.

Quote from: sandokhan
Somebody tell me this is a joke.
There is no such thing as an ideal rope: THE IDEALIZATION OF THE ROPE is AN as yet unaccounted for problem in modern physics.
The physicists who wrote up that paragraph need to wake up and do some experiments in the real world.
Did you do any experiments in the "real world"? I did!
I did not have a spare ship handy, but I had a 10 kg concrete block and a spring balance (a force measuring device).
I found the sliding friction of that block on a smooth concrete surface is 5 kg (so that's the one weak man in some of your analogies) and
I found that I can pull somewhere over 20 kg with one hand, let's call it 25 kg (I'm not strong, let's call that the strength of the strong man).

Now when I pull that 10 kg concrete block along as smoothly as I can with the spring balance at the block end of the rope it reads about 5 kg (a bit hard to keep it steady) and
when I pull that 10 kg concrete block along with the spring balance at my end of the very light rope it reads the same 5 kg.

I wonder why it doesn't read 25 kg on my end, after all, I can pull 25 kg with one hand!

Quote from: sandokhan
From one of the best texts ever on mechanics:


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

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

You could, however, got a text version of that book, "A System of Natural Philosophy" by John Lee Comstock and actually quoted your reference.
And you might in fairness quote 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 or A System of Natural Philosophy by John Lee Comstock pages 31,32

Para 112. Seems to sort out the equal forces.

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

Quote from: sandokhan
THE ROPE WILL TRANSMIT TWO DIFFERENT FORCES.
No, the rope will not transmit two different forces. Let's leave it here till you digest the material in your reference, "one of the best texts ever on mechanics".

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

So, any comments. Do you still want to carry on with "Andre the Giant and Shirley Temple", etc.

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Ichimaru Gin :]

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #1 on: May 13, 2018, 08:57:15 PM »
What is the Andre the Giant and Shirley Temple scenario?
I saw a slight haze in the hotel bathroom this morning after I took a shower, have I discovered a new planet?

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Bullwinkle

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #2 on: May 13, 2018, 09:16:09 PM »
What is the Andre the Giant and Shirley Temple scenario?

A low budget '70s porno.

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sandokhan

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #3 on: May 13, 2018, 10:04:21 PM »
rabinoz is a physics illiterate.

He is papering this forum with messages which would not be accepted anywhere else.

No one in his/her right mind would say that two different people who have different weights/strengths would pull with the very same force on each end of a rope.

They would get banned instantly if they tried to state something like this.

Yet here, rabinoz and jackblack are allowed to roam freely, LOWERING THE QUALITY OF THE CONTENT OF THIS FORUM.

Perhaps this was their goal from the very start, and certainly they have benefited from the lack of moderation and concern coming from the admin.


Let me describe the situation first, since rabinoz does not understand the experiment (no friction involved).

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. 

Now, let us place a person in each boat, they will pull on each end of the rope.

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: for example, in boat x we have Andre the Giant or Henry VIII, while in boat y we have Shirley Temple or Queen Elizabeth I.


Here is the RE analysis:

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

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


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.

The rope will transmit two different forces, A and B. The analysis is perfect and yields the correct result.

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.


rabinoz and jackblack must have their posting privileges removed at once from the upper forums: here we have two basket cases, two physics illiterates who are using anywhich miserable subterfuge to get their away around this forum and in every debate, only an insane person would believe that two persons of different weights/strengths would pull with the very same force on each end of a rope (the same meaning the very same all the way to the last decimal).

rabinoz is wasting everyone's time here with this crap, and no one is getting involved to get rid of his presence: THAT IS WHY WE NEED A TECHNICAL MODERATOR TO TAKE CARE OF THINGS LIKE THIS.

« Last Edit: May 13, 2018, 10:29:19 PM by sandokhan »

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sandokhan

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #4 on: May 13, 2018, 10:33:22 PM »
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.

Let's put your word to the test (you'll find out why I DID NOT include this quote in my reference to the text on mechanics):

The universal "law" of gravitational attraction tells us that the force exerted for each body, the Earth - Moon system, must be exactly the same.

Moon attractive force for Earth:

Fmn = 2.1096 x 1019 kgf

Fmn = [(G x Mmn)/d2] x Me

Earth attractive force for Moon:

Fe = 2.1096 x 1019 kgf

Fe = [(G x Me)/d2] x Mmn


Mmn = 7.349 x 1025 gm

Me = 5.9736 x 1027 gm

d = 3.7633 x 1010 cm


If a single counterexample could be found which defies the equation F=GMm/r 2, then that would mean that the forces exerted by the Earth and the Moon on each other, in the Newtonian mechanics context, must be different from one another.


DEPALMA SPINNING BALL EXPERIMENT:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg753387#msg753387

The law of universal gravitation totally defied: FOR THE SAME MASS OF THE STEEL BALLS, AND THE SAME SUPPOSED LAW OF ATTRACTIVE GRAVITY, THE ROTATING BALL WEIGHED LESS AND TRAVELED HIGHER THAN THE NON-ROTATING BALL.


BIEFELD-BROWN EFFECT:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1852363#msg1852363


NIPHER EXPERIMENT:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1852840#msg1852840


LAMOREAUX EXPERIMENT:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1616174#msg1616174


ALLAIS EFFECT:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg760382#msg760382


TOTAL DEMOLITION OF STR/GTR:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg769750#msg769750


Hence, we have a most direct proof of the double forces of attractive gravitation paradox.

In the two boats on a lake example, connected by a rope, the forces exerted on each of end of the rope, forces A and B, will ALWAYS be different, which means that the balance of forces equation will include TWICE  the forces needed in the Newtonian system.

Net force on boat X: A + B

Net force on boat Y: -A - B


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



By contrast, the analysis based only on a single force being transmitted through the rope, leads to a most direct contradiction: |A|=|B|. But forces A and B can never be exactly the same.


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

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

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

« Last Edit: May 13, 2018, 10:35:07 PM by sandokhan »

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rabinoz

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #5 on: May 13, 2018, 10:38:46 PM »
rabinoz is a physics illiterate.

He is papering this forum with messages which would not be accepted anywhere else.

No one in his/her right mind would say that two different people who have different weights/strengths would pull with the very same force on each end of a rope.

I humbly suggest that you read you own reference more thoroughly before going off the deep end again.
Download from: A System of Natural Philosophy by John Lee Comstock pages 31,32 or A System of Natural Philosophy by John Lee Comstock pages 31,32

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.

So, any comments. Do you still want to carry on with "Andre the Giant and Shirley Temple", etc.

Just remember that you said  that this is  "one of the best texts ever on mechanics"!

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sandokhan

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #6 on: May 13, 2018, 10:48:17 PM »
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.

Let's put your word to the test (you'll find out why I DID NOT include this quote in my reference to the text on mechanics):

ALLAIS EFFECT

REFERENCE #1

CONFIRMATION OF THE ALLAIS EFFECT DURING THE 2003 SOLAR ECLIPSE:

http://www.acad.ro/sectii2002/proceedings/doc3_2004/03_Mihaila.pdf

(it also shows that the effect was confirmed during the August 1999 solar eclipse)


The title of the paper is as follows:

A NEW CONFIRMATION OF THE ALLAIS EFFECT
DURING THE SOLAR ECLIPSE OF 31 MAY 2003

"During the total solar eclipse of 11 August 1999, the existence of the Allais effect was confirmed."

The authors indicate that more measurements/experiments have to be undertaken during future solar eclipses.


REFERENCE #2

CONFIRMATION OF THE ALLAIS EFFECT DURING THE SEPT. 2006 SOLAR ECLIPSE:

http://www.hessdalen.org/sse/program/Articol.pdf

The title of the article is as follows:

A confirmation of the Allais and Jeverdan-Rusu-Antonescu effects
during the solar eclipse from 22 September 2006 , and the quantization
of behaviour of pendulum


"The experiments made with a paraconical pendulum during annular solar eclipse from 22 September 2006 confirm once again the existence of the Allais effect."


REFERENCE #3

CONFIRMATION OF THE ALLAIS EFFECT DURING THE 2008 SOLAR ECLIPSE:

http://ivanik3.narod.ru/Astrophiz/AnomSunEclip/pugarticleGoodey.pdf

Published in the Journal of Advanced Research in Physics


Given the above, the authors consider that it is an inescapable conclusion from our experiments that after the end of the visible eclipse, as the Moon departed the angular vicinity of the Sun, some influence exerted itself upon the Eastern European region containing our three sets of equipment, extending over a field at least hundreds of kilometers in width.

The nature of this common influence is unknown, but plainly it cannot be considered as gravitational in the usually accepted sense of Newtonian or Einsteinian gravitation.


We therefore are compelled to the opinion that some currently unknown physical influence was at work.


REFERENCE #4

The Allais pendulum effect confirmed in an experiment performed in 1961:

http://www.science-frontiers.com/sf074/sf074a05.htm


REFERENCE #5

Observations of Correlated Behavior of Two Light Torsion Balances and a Paraconical Pendulum in Separate Locations during the Solar Eclipse of January 26th, 2009:


http://www.researchgate.net/publication/235701910_Observations_of_Correlated_Behavior_of_Two_Light_TorsionBalances_and_a_Paraconical_Pendulum_in_Separate_Locationsduring_the_Solar_Eclipse_of_January_26th_2009

http://www.hindawi.com/journals/aa/2012/263818/

Published in the Advances in Astronomy Journal

Another independent confirmation has been obtained of the previously established fact that at the time of solar eclipses, a specific reaction of the torsion balance can be observed. During a solar eclipse, the readings of two neighboring TBs seem to be correlated. This fact demonstrates the nonaleatory character of the reactions of TBs. Consequently, the reaction of these devices is deterministic, not random. A solar eclipse is such a determinant, since upon termination of a solar eclipse, the correlation becomes insignificant. This conclusion is supported by the PP observations. The PP graph and the TB graphs showed obvious similarity, with the coefficient of correlation of these two independent curves being close to 1.

In particular, we wonder how any physical momentum can be transferred to our instrument during a solar eclipse. Gravity can hardly suffice as an explanation even for understanding the results of the PP measurements. The gravitational potential grows slowly and smoothly over a number of days before eclipse and then declines smoothly afterwards without any sudden variations, but we see relatively short-term events. Moreover, gravity is certainly not applicable to the explanation of the results of the TB observations, since the TB is not sensitive to changes in gravitational potential.

The cause of the time lag between the response of the device in Suceava and the reactions of the devices in Kiev also remains unknown. What can be this force which acts so selectively in space and time?

The anomalies found, that defy understanding in terms of modern physics, are in line with other anomalies, described in a recently published compendium “Should the Laws of Gravitation be reconsidered?” [14].


REFERENCE #6

Precise Underground Observations of the Partial Solar Eclipse of 1 June 2011 Using a Foucault Pendulum and a Very Light Torsion Balance

Published in the International Journal of Astronomy and Astrophysics Journal


http://www.researchgate.net/publication/235701885_Precise_Underground_Observations_of_the_Partial_Solar_Eclipse_of_1_June_2011_Using_a_Foucault_Pendulum_and_a_Very_Light_Torsion_Balance

http://file.scirp.org/Html/3-4500094_26045.htm

http://www.scirp.org/journal/PaperInformation.aspx?PaperID=26045


Simultaneous observations of the solar eclipse on 06/01/2011 were carried out using a Foucault pendulum and a torsion balance. The instruments were installed in a salt mine, where the interference was minimal. Both instruments clearly reacted to the eclipse. We conclude that these reactions should not be considered as being gravitational effects.

REFERENCE #7

Dr. Erwin Saxl experiment (1970)

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1629054#msg1629054

Published in the Physical Review Journal

Saxl and Allen went on to note that to explain these remarkable eclipse observations, according to "conventional Newtonian/Einsteinian gravitational theory," an increase in the weight of the pendumum bob itself on the order of ~5% would be required ... amounting to (for the ~51.5-lb pendulum bob in the experiment) an increase of ~2.64 lbs!

This would be on the order of one hundred thousand (100,000) times greater than any possible "gravitational tidal effects" Saxl and Allen calculated (using Newtonian Gravitational Theory/ Relativity Theory).



A TOTAL DEFIANCE OF NEWTONIAN MECHANICS.

For the same masses/corresponding distances of the Earth, Sun and the Moon, during the Allais experiment, the pendulum's direction of rotation changed from clockwise to counterclockwise, at the end of the eclipse it resumed its normal direction of rotation.

In order to arrive at an explanation, M. Allais considered a wide range
of known periodic phenomena, including the terrestrial tides, variations in
the intensity of gravity, thermal or barometric effects, magnetic variations,
microseismic effects, cosmic rays, and the periodic character of human
activity. Yet, on close examination, the very peculiar nature of the
periodicity shown by the change in azimuth of the pendulum forced the
elimination of all of these as cause.


Dr. Maurice Allais:

In both cases, with the experiments with the anisotropic
support and with those with the isotropic support, it is found
that the amplitudes of the periodic effects are considerably
greater than those calculated according to the law of gravitation,
whether or not completed by the theory of relativity.
In the case of the anisotropic support, the amplitude of
the luni-solar component of 24h 50m is about twenty million
times greater than the amplitude calculated by the theory of
universal gravitation.

In the case of the paraconical pendulum with isotropic
support, this relation is about a hundred million.


In other words, the pendulum motions Allais observed during his two eclipses – 1954 and 1959 -- were physically IMPOSSIBLE … according to all known “textbook physics!”


"Allais used the phrase “a brutal displacement” … to describe the “sudden, extraordinary backwards movement” of the pendulum his laboratory chief had seen (and carefully recorded!), even while not knowing its “mysterious” cause ... until later that same afternoon.

Here (below) is what those “anomalous eclipse motions” in Allias’ pendulum looked like; this graphic, adapted from Scientific American, depicts the mechanical arrangement of Allais’ unique paraconical pendulum (below – left).

The three vertical panels to its right illustrate the pendulum’s “highly anomalous motions” -- recorded during two partial solar eclipses to cross Allais’ Paris laboratory in the 1950’s (the first in 1954, the second in 1959); the phase of each eclipse that corresponded with these “anomalous motions,” is depicted in the last three vertical strips (far right)."




"This normal, downward-sloping trend is abruptly REVERSED!

From there, things rapidly got even more bizarre--

As the pendulum’s azimuth motion continues in an accelerating, COUNTER-clockwise direction … for the next 45 minutes; then, after peaking, the pendulum motion REVERSES direction (moving clockwise again …), only to reverse BACK again (counterclockwise!) … briefly [as the Moon reaches “mid-eclipse” (the central green line)] -- before abruptly reversing once more, accelerating again in a CLOCKWISE direction … before eventually “bottoming out” … parallel to the ORIGINAL “Foucault/Earth rotation” downward-sloping trend line!"

HERE ARE THE PRECISE CALCULATIONS INVOLVING THE ALLAIS EFFECT:




Dr. Maurice Allais:

With regard to the validity of my experiments, it seems
best to reproduce here the testimony of General Paul Bergeron,
ex-president of the Committee for Scientific Activities for
National Defense, in his letter of May 1959 to Werner von
Braun:

"Before writing to you, I considered it necessary to
visit the two laboratories of Professor Allais (one 60
meters underground), in the company of eminent
specialists – including two professors at the Ecole
Polytechnique. During several hours of discussion, we
could find no source of significant error, nor did any
attempt at explanation survive analysis.

"I should also tell you that during the last two years,
more than ten members of the Academy of Sciences and
more than thirty eminent personalities, specialists in
various aspects of gravitation, have visited both his
laboratory at Saint-Germain, and his underground
laboratory at Bougival.

"Deep discussions took place, not only on these
occasions, but many times in various scientific contexts,
notably at the Academy of Sciences and the National
Center for Scientific Research. None of these discussions
could evolve any explanation within the framework of
currently accepted theories."


This letter confirms clearly the fact that was finally
admitted at the time - the total impossibility of explaining the
perceived anomalies within the framework of currently
accepted theory.



IF YOU WANT ANYONE TO BELIEVE YOU THAT ALL BODIES ATTRACT EACH OTHER EQUALLY, YOU'LL HAVE TO EXPLAIN THE ALLAIS EFFECT.


Dr. Maurice Allais:

In both cases, with the experiments with the anisotropic
support and with those with the isotropic support, it is found
that the amplitudes of the periodic effects are considerably
greater than those calculated according to the law of gravitation,
whether or not completed by the theory of relativity.
In the case of the anisotropic support, the amplitude of
the luni-solar component of 24h 50m is about twenty million
times greater than the amplitude calculated by the theory of
universal gravitation.

In the case of the paraconical pendulum with isotropic
support, this relation is about a hundred million.



Given the above, the authors consider that it is an inescapable conclusion from our experiments that after the end of the visible eclipse, as the Moon departed the angular vicinity of the Sun, some influence exerted itself upon the Eastern European region containing our three sets of equipment, extending over a field at least hundreds of kilometers in width.

The nature of this common influence is unknown, but plainly it cannot be considered as gravitational in the usually accepted sense of Newtonian or Einsteinian gravitation.


We therefore are compelled to the opinion that some currently unknown physical influence was at work.


ANOTHER HUGE WIN FOR THE FE!

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rabinoz

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #7 on: May 14, 2018, 01:30:41 AM »
<< Irrelevant material deleted >>
The topic happens to be "Is the tension on an "ideal rope" (massless) the same at each end?"

If you want to debate your gravitation hypotheses then make your own thread.

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Shifter

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #8 on: May 14, 2018, 01:42:46 AM »
<< Irrelevant material deleted >>
The topic happens to be "Is the tension on an "ideal rope" (massless) the same at each end?"

If you want to debate your gravitation hypotheses then make your own thread.

Given the topic and who you are baiting with this thread that's a pretty obnoxious post rab. Why don't you just knock it off and stop the shit posting? You dont own the monopoly on this subject

Quoting someone and then 'deleting it' is just being a dick. If you don't want to debate, than why are you even here?

Let me get your logic straight anyway rab. So Arnold Schwarzenegger could pull from one end of the rope and and toddler pulling on the other. Same force you say?  ??? ??? ::) ::) ::)


 
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RIP rabinoz. Forum legend

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rabinoz

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #9 on: May 14, 2018, 02:06:13 AM »
What is the Andre the Giant and Shirley Temple scenario?

You don't want to know, but if you insist then this might fill you in on the whole soap opera:
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
If Andre the Giant and Shirley Temple each pull on the different ends of a rope, obviously the forces applied will be different.

However, now the FE have to deal with this:
Quote from: JackBlack on May 11, 2017, 04:25:46 AM
Quote from: sandokhan on May 11, 2017, 04:03:14 AM
If Henry VIII and Queen Elizabeth I would be pulling, each located at one end of that rope, would those forces be the same? Certainly not.
For a massless rope, it must be.
And here is all the detail on Henry VIII and Queen Elizabeth I: Re: Distances in the universe « Reply #569 on: May 11, 2017, 09:03:14 PM »
Refer to the original posts if you need more.

Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #10 on: May 14, 2018, 02:40:56 AM »
In my humble opinion (a physics illiterate’s opinion) you guys are discussing two different things.

1: the two forces of the actors pulling the rope.
2: the combined force of said actors transmitted by said rope.

Or am I, a physics Mr. Bean, missing something here?
Be gentle

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Master_Evar

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #11 on: May 14, 2018, 03:30:29 AM »
In my humble opinion (a physics illiterate’s opinion) you guys are discussing two different things.

1: the two forces of the actors pulling the rope.
2: the combined force of said actors transmitted by said rope.

Or am I, a physics Mr. Bean, missing something here?

Yes. The the force that each actor pulls on the rope with is equal, otherwise the rope would accelerate in some direction. If you have one strong person and one weak person pulling on a rope, and the rope is not accelerating, then both of them are pulling on the rope with the same force. This can mean one of three things:
1. The stronger person is not pulling with full strength.
2. The stronger person is not strong enough that he can overcome the grip and weight of the weaker person.
3. The stronger person is already dragging the weaker person along as quickly as possible.

The combined forces of the actors pulling on a rope that is not accelerating will always be 0 (given an ideal rope that doesn't weigh anything, a rope that weighs something will require both actors to lift it up, creating a net force straight up).

It's really as simple as f=ma. Is the rope accelerating? If not, then the force is 0 and both actors pull with equal and opposite forces.
Math is the language of the universe.

The inability to explain something is not proof of something else.

We don't speak for reality - we only observe it. An observation can have any cause, but it is still no more than just an observation.

When in doubt; sources!

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sandokhan

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #12 on: May 14, 2018, 04:04:46 AM »
The the force that each actor pulls on the rope with is equal, otherwise the rope would accelerate in some direction. If you have one strong person and one weak person pulling on a rope, and the rope is not accelerating, then both of them are pulling on the rope with the same force.

Completely wrong.

The rope is not going anywhere, and YET both persons in the boats would pull on each end of the rope with different forces.

The forces will ALWAYS be different, they have to be since we are dealing with different individuals.

Here is the mathematical demonstration.

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. 

Now, let us place a person in each boat, they will pull on each end of the rope.

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: for example, in boat x we have Andre the Giant or Henry VIII, while in boat y we have Shirley Temple or Queen Elizabeth I.

Here is the correct 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.

The rope will transmit two different forces, A and B. The analysis is perfect and yields the correct result.

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.


Why are the RE so afraid to admit that the two persons in the boat will pull WITH DIFFERENT FORCES?

Because we can immediately make the analogy to the Earth - Moon system.

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

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

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



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rabinoz

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #13 on: May 14, 2018, 05:19:21 AM »
<< Irrelevant material deleted >>
The topic happens to be "Is the tension on an "ideal rope" (massless) the same at each end?"

If you want to debate your gravitation hypotheses then make your own thread.
Given the topic and who you are baiting with this thread that's a pretty obnoxious post rab. Why don't you just knock it off and stop the shit posting? You dont own the monopoly on this subject
I am not "baiting". Sandokhan suggested the thread "Dyson–Harrop satellites" needed to be split, so I split it.
No, I "dont own the monopoly on this subject" and the subject is "Is the tension on an 'ideal rope' (massless) the same at each end?" not "gravitation".

Quote from: Shifter
Quoting someone and then 'deleting it' is just being a dick. If you don't want to debate, than why are you even here?
His whole massive post was about "gravitation" and not the topic, "Is the tension on an 'ideal rope' (massless) the same at each end?". So I fail to see why I should debate "gravitation" here.

Quote from: Shifter
Let me get your logic straight anyway rab. So Arnold Schwarzenegger could pull from one end of the rope and and toddler pulling on the other. Same force you say?  ??? ??? ::) ::) ::)
Yes, I claim that the forces must be the same.
While Arnold Schwarzenegger could pull with a much greater force than the toddler, the rope constrains them to be the same.
Arnold Schwarzenegger is able to pull with a much greater force than the toddler, here he is constrained to the maximum force the toddler can apply.

If they were pulling on each of the same rope, then the force must be the same. That is the basic property of an "ideal rope".
A rope simply links the two together.

Or consider that my car has a draw-bar pull in 1st gear at 20 kph of about 1000 kg force but a 200 kg trailer only needs about 20 kg force to drag it on a smooth surface.
Imagine the car being attached to the trailer by a rope (I know that soft tows are illegal but . . . ) so:
  • What force is the car applying to the rope?
  • What force is the trailer applying to the rope?
I would say that while that car is capable of applying a 1000 kg force if needed, here only 20 kg force is needed, so that is the tension everywhere akong the rope.

As I quoted in an earlier post the forces on each end of an ideal massless rope are must be the same in magnitude and opposite in direction.

MIT puts it fairly well:
Quote
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.
From: MIT Scripts, Tension interaction

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sandokhan

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #14 on: May 14, 2018, 05:36:18 AM »
I claim that the forces must be the same.

Your claim is worthless.

It is only that you can prove that counts, and what you can prove at the present time is nothing at all.

Your dodging of the problem at hand is duly noted: your examples involve FRICTION.

As I quoted in an earlier post the forces on each end of an ideal massless rope are must be the same in magnitude and opposite in direction.

And so they are:

Net force on boat X: A + B

Net force on boat Y: -A - B


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



Here is the mathematical demonstration that your claim IS WORTHLESS

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. 

Now, let us place a person in each boat, they will pull on each end of the rope.

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: for example, in boat x we have Andre the Giant or Henry VIII, while in boat y we have Shirley Temple or Queen Elizabeth I.

Here is the correct 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 on 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.

The rope will transmit two different forces, A and B. The analysis is perfect and yields the correct result.

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.


Your worthless claim has been refuted.

« Last Edit: May 14, 2018, 05:39:30 AM by sandokhan »

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rabinoz

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #15 on: May 14, 2018, 05:51:45 AM »
The the force that each actor pulls on the rope with is equal, otherwise the rope would accelerate in some direction. If you have one strong person and one weak person pulling on a rope, and the rope is not accelerating, then both of them are pulling on the rope with the same force.

Completely wrong.

The rope is not going anywhere, and YET both persons in the boats would pull on each end of the rope with different forces.

The forces will ALWAYS be different, they have to be since we are dealing with different individuals.
If the rope is "the rope is not going anywhere" it is not accelerating.
So if the rope is massless, it follows that the net force on the rope must be zero.
The net force on the rope is the difference in the forces. Hence the forces must be equal, so the strong man is not able to exert his full strength without the rope accelerating.
You might read Is the tension on an "ideal rope" (massless) the same at each end? « Reply #11 on: Today at 08:30:29 PM »

Look, sandokhan, it is not the slightest bit of use repeating the same old illogical argument.
Quote from: sandokhan
Here is the mathematical demonstration.

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. 

Now, let us place a person in each boat, they will pull on each end of the rope.

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: for example, in boat x we have Andre the Giant or Henry VIII, while in boat y we have Shirley Temple or Queen Elizabeth I.

Here is the correct 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.
Sure, they balance out perfectly, but you now have Andre the Giant and Shirley Temple each pulling with a force of magnitude A + B.
And that is not what you postulated.

Quote from: sandokhan

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.

The rope will transmit two different forces, A and B. The analysis is perfect and yields the correct result.

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.

Why are the RE so afraid to admit that the two persons in the boat will pull WITH DIFFERENT FORCES?
We are not afraid to admit anything that is true, but your hypothesis is simply not true and it has nothing to do with gravitation.

Quote from: sandokhan

Because we can immediately make the analogy to the Earth - Moon system.

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

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

"All attraction models" produce twice the force that is required to balance the centrifugal forces of orbit!
No! the Earth attracting the Moon and the Moon attracting the Earth are the same force - action and reaction if you like.

And there is no way that I am going to budge from that position because it is correct.

You are not the only one that has this problem, read: Quora: Child and man pull on opposite ends of a rope .Is the force applied to rope by the child is smaller or same compare to the force applied by the man?

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rabinoz

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #16 on: May 14, 2018, 06:00:54 AM »
I claim that the forces must be the same.
Your claim is worthless.

It is only that you can prove that counts, and what you can prove at the present time is nothing at all.

Your dodging of the problem at hand is duly noted: your examples involve FRICTION.
You might have the courtesy of indicating who you are answering. How else can we check back on what you are claiming?
If it's this post, I fail to see any mention of FRICTION:
The force that each actor pulls on the rope with is equal, otherwise the rope would accelerate in some direction. If you have one strong person and one weak person pulling on a rope, and the rope is not accelerating, then both of them are pulling on the rope with the same force. This can mean one of three things:
1. The stronger person is not pulling with full strength.
2. The stronger person is not strong enough that he can overcome the grip and weight of the weaker person.
3. The stronger person is already dragging the weaker person along as quickly as possible.

The combined forces of the actors pulling on a rope that is not accelerating will always be 0 (given an ideal rope that doesn't weigh anything, a rope that weighs something will require both actors to lift it up, creating a net force straight up).

It's really as simple as f=ma. Is the rope accelerating? If not, then the force is 0 and both actors pull with equal and opposite forces.
And if it's not that post, what are you talking about?

What is the point of saying the same thing over and over?

You've said it, nobody here or anywhere else believes you - get used to it!

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sandokhan

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #17 on: May 14, 2018, 06:06:01 AM »
Your tricks do not work with me.

While Arnold Schwarzenegger could pull with a much greater force than the toddler, the rope constrains them to be the same.
Arnold Schwarzenegger is able to pull with a much greater force than the toddler, here he is constrained to the maximum force the toddler can apply.


But indeed Arnold DOES PULL WITH A MUCH GREATER FORCE. This force has to show up in the final result: for the RE analysis, |A|=|B|, which is impossible. The force exerted by Arnold is much greater than that exerted by a child.

Only the correct analysis includes BOTH FORCES, of different magnitude, exerted by Arnold and by the child.

So if the rope is massless, it follows that the net force on the rope must be zero.
The net force on the rope is the difference in the forces. Hence the forces must be equal, so the strong man is not able to exert his full strength without the rope accelerating.


THE TOTAL FORCES WILL BE EQUAL, NOT THE INDIVIDUAL FORCES EXERTED BY EACH PERSON.

Sure, they balance out perfectly, but you now have Andre the Giant and Shirley Temple each pulling with a force of magnitude A + B.

Completely wrong.

Please read:

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: for example, in boat x we have Andre the Giant or Henry VIII, while in boat y we have Shirley Temple or Queen Elizabeth I.


Forces A and B ARE OF DIFFERENT MAGNITUDE, AS THEY SHOULD BE: we have two different individuals pulling with different forces, this will always be the case.

WHILE ANDRE IS PULLING WITH FORCE A, AND SHIRLEY IS PULLING WITH FORCE B, the TOTAL FORCE on each end of the rope/string will be accounted as follows:

Here is the correct 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.


You sound very confused rabinoz.

You are unable, either intentionally or unintentionally, TO UNDERSTAND THE DIFFERENCE between the forces exerted by each person in the boat on each end of the rope, and THE TOTAL FORCES ON EACH END OF THE ROPE.


PULLING FORCES: A AND B

TOTAL FORCES ON EACH END OF THE STRING/ROPE:

-A and -B (FOR BOAT X) AND A AND B (FOR BOAT Y)


Quit quoting examples which involve FRICTION!!!

The Earth-Moon system and the two boats on a lake system are frictionless.


No! the Earth attracting the Moon and the Moon attracting the Earth are the same force - action and reaction if you like.

Please read the above analysis.

Substitute the Earth for boat X and the Moon for boat Y, and the force of gravity for the string.

EXACTLY THE SAME SITUATION, FOUR FORCES INVOLVED, A TOTAL DEFIANCE OF THE NEWTONIAN SYSTEM.

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Master_Evar

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #18 on: May 14, 2018, 06:23:57 AM »
What are the forces acting on boat X?
A. It's just A.

What are the forces acting boat Y?

-B. It's just -B.

The feet of the people pulling the rope is the only things exerting a force on the boats. The rope is not in direct contact with the boats, so the rope does not exert a force on the boats. Person A is not in direct contact with boat Y, so person A cannot directly exert a force on boat Y. Same with person B vs boat X.

So, the net force on boat X is A, so it accelerates to the right.
And the net force on boat Y is -B, so it accelerates to the left.

The net force on the rope is B - A, so given that A =/= B then the rope will be accelerating.
Math is the language of the universe.

The inability to explain something is not proof of something else.

We don't speak for reality - we only observe it. An observation can have any cause, but it is still no more than just an observation.

When in doubt; sources!

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sandokhan

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #19 on: May 14, 2018, 06:46:19 AM »
You should change your name to novice_evar.

The string/rope is connected to each person who is holding their respective end of the rope, and each person is standing on that boat on a lake.

Person X is directly in contact with the other person through the forces exerted on the rope, the other person is standing on boat Y.

Do not play tricks like these with me, you are wasting your time.

The RE analysis runs as follows:

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

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


Only a NOVICE would make such a foolish mistake.


By definition, force A does not equal force B, THEY ARE OF DIFFERENT MAGNITUDE.

The force exerted by Andre the Giant will be greater than that exerted by Shirley Temple.

In your example, FORCES A AND B HAVE TO BE EQUAL FROM THE START, which is never possible, you'll always have two different people with two different forces.


Here is the correct analysis.

Two boats pulled toward each other on a lake.

Man from boat X is pulling with force A, directed to the left: HE IS ACTING ON BOAT Y THROUGH THE OTHER PERSON WHO IS PULLING ON THE OTHER END OF THE ROPE.

Man from boat Y is pulling with force B, directed to the right: HE IS ACTING ON BOAT X THROUGH THE OTHER PERSON WHO IS PULLING ON THE OTHER END OF THE ROPE.

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.


REMEMBER that both boats will move in the water ACCORDING TO BOTH FORCES EXERTED THROUGH THE ROPE BY EACH PERSON.

Take a look at your faulty analysis:

So, the net force on boat X is A, so it accelerates to the right.

You are forgetting THAT BOAT Y IS PULLING WITH FORCE B ON THE ROPE THUS THE MOVEMENT OF BOAT X HAS TO TAKE INTO ACCOUNT THE INFLUENCE OF THIS FORCE.

And the net force on boat Y is -B, so it accelerates to the left.

BOAT X IS ACTING ON BOAT Y THROUGH THE ROPE, THROUGH THE FORCE A exerted by the person in boat X.


Yes, change your name to novice_evar.


The net force on the rope is B - A

Completely wrong.

Here is the correct analysis, since we HAVE TWO DIFFERENT FORCES TO DEAL WITH ON EACH SIDE 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.


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.

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Twerp

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #20 on: May 14, 2018, 06:52:44 AM »
Goodness Sandy, buy a couple of scales, attach them to either end of a rope, get a kid and a strong adult to pull on the scales, read both sets of scales. Either the scales read the same or they read differently. Problem solved. Let us know your results.
“Heaven is being governed by Devil nowadays..” - Wise

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sandokhan

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #21 on: May 14, 2018, 07:29:26 AM »
Let us now run a full analysis of the forces acting on the string/rope and on each person standing on his respective boat.


Two boats pulled toward each other on a lake, using a single rope.

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.


To the left we will have a negative direction.


What are the forces acting on person X?

Person 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 on person Y?

To the right we will have the positive direction.

Person 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 person X: A + B

Net force on person 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 with force A, while at the same time boat Y is pulling on that same rope with force B: forces A and B are of different magnitude. The correct analysis must take these facts into account.


The total forces on each end of the rope must be the same in magnitude and opposite in direction:

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


The string/rope will not move: [-A - B] + [A + B] = 0



However, THE INDIVIDUAL FORCES APPLIED ON EACH END BY EACH PERSON WILL BE DIFFERENT, AS THEY HAVE TO BE, SINCE THEY ARE EXERTED BY TWO DIFFERENT PEOPLE.

« Last Edit: May 14, 2018, 07:35:18 AM by sandokhan »

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Master_Evar

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #22 on: May 14, 2018, 07:45:47 AM »
Force does not add up like you describe sandokhan. Here's a demonstration of why your logic is illogical:

Imagine if I had a 1kg weight attached to another 1kg weight via a string, and lift up one of the weights.
My hand is lifting 2 kg spread evenly among the 2 weights, so each weight would have 1kg of force lifting them up from my hand alone. But the free-hanging weight is also subject to the reaction force of the weight in my hand, which is 1 kg. Therefore the free-hanging weight of 1kg is being held up by 2kg worth of force.

Math is the language of the universe.

The inability to explain something is not proof of something else.

We don't speak for reality - we only observe it. An observation can have any cause, but it is still no more than just an observation.

When in doubt; sources!

Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #23 on: May 14, 2018, 07:57:23 AM »
Goodness Sandy, buy a couple of scales, attach them to either end of a rope, get a kid and a strong adult to pull on the scales, read both sets of scales. Either the scales read the same or they read differently. Problem solved. Let us know your results.

Or even better, buy 1 scale and attach it to exact center point of the rope and let the kid and adult pull. Read the scale. Now move the scale away from the center point of the rope and let them pull again. read the scale. Now move the scale to the other side of the center point. Pull and read.

Gather results, take into account the weight/mass of the rope and compare results. My prediction is that scale will not read a larger force when it is attached closer to the adult, nor will it show a smaller force when closer to the kid.

Be gentle

*

sandokhan

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #24 on: May 14, 2018, 08:11:53 AM »
evar and lamaface, cut the crap.

Here is the situation in question:

Two boats pulled toward each other on a lake, using a single rope.

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.


Replace now person X with the Earth, person Y with the Moon, and the rope with attractive gravitation.


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.


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

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

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


Two boats pulled toward each other on a lake, using a single rope.

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.


To the left we will have a negative direction.


What are the forces acting on person X?

Person 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 on person Y?

To the right we will have the positive direction.

Person 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 person X: A + B

Net force on person 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 with force A, while at the same time boat Y is pulling on that same rope with force B: forces A and B are of different magnitude. The correct analysis must take these facts into account.


The total forces on each end of the rope must be the same in magnitude and opposite in direction:

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


The string/rope will not move: [-A - B] + [A + B] = 0


However, THE INDIVIDUAL FORCES APPLIED ON EACH END BY EACH PERSON WILL BE DIFFERENT, AS THEY HAVE TO BE, SINCE THEY ARE EXERTED BY TWO DIFFERENT PEOPLE.


The analysis is perfect, everything works out fine.

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Master_Evar

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #25 on: May 14, 2018, 09:07:53 AM »
Sandokhan, can you tell me what went wrong in this case:

"Imagine if I had a 1kg weight attached to another 1kg weight via a string, and lift up one of the weights.
My hand is lifting 2 kg spread evenly among the 2 weights, so each weight would have 1kg of force lifting them up from my hand alone. But the free-hanging weight is also subject to the reaction force of the weight in my hand, which is 1 kg. Therefore the free-hanging weight of 1kg is being held up by 2kg worth of force."

I'm using the exact same logic as you are using in the boat example. In fact, I can perfectly translate one situation into the other. So, do you realise what is going wrong here?
Math is the language of the universe.

The inability to explain something is not proof of something else.

We don't speak for reality - we only observe it. An observation can have any cause, but it is still no more than just an observation.

When in doubt; sources!

*

sandokhan

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #26 on: May 14, 2018, 09:57:42 AM »
evar, you do not have the experience required to debate this topic. That much is obvious. Since you have seen that my calculation is perfect, and that the total forces on each end of the rope are the same in magnitude and opposite in direction while the individual forces applies on each end are different, you are trying now a different venue, hoping for the best that it will work. Your RE buddies tried pretty much the same thing, once they understood that my analysis is correct.

Please indicate to the readers any errors in the calculation I provided. If you cannot, any new example, which is of a different type than the two boats linked by a rope on a lake, will have to wait indefinitely.

Here is the calculation.

Two boats pulled toward each other on a lake, using a single rope.

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.


To the left we will have a negative direction.


What are the forces acting on person X?

Person 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 on person Y?

To the right we will have the positive direction.

Person 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 person X: A + B

Net force on person 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 with force A, while at the same time boat Y is pulling on that same rope with force B: forces A and B are of different magnitude. The correct analysis must take these facts into account.


The total forces on each end of the rope must be the same in magnitude and opposite in direction:

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


The string/rope will not move: [-A - B] + [A + B] = 0


However, THE INDIVIDUAL FORCES APPLIED ON EACH END BY EACH PERSON WILL BE DIFFERENT, AS THEY HAVE TO BE, SINCE THEY ARE EXERTED BY TWO DIFFERENT PEOPLE.


The analysis is perfect, everything works out fine.

« Last Edit: May 14, 2018, 09:59:45 AM by sandokhan »

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Master_Evar

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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #27 on: May 14, 2018, 10:05:03 AM »
Man you're bad at reading.

Can you figure out what is going wrong here:
"Imagine if I had a 1kg weight attached to another 1kg weight via a string, and lift up one of the weights.
My hand is lifting 2 kg spread evenly among the 2 weights, so each weight would have 1kg of force lifting them up from my hand alone. But the free-hanging weight is also subject to the reaction force of the weight in my hand, which is 1 kg. Therefore the free-hanging weight of 1kg is being held up by 2kg worth of force."

(Spoilers: The reason why above situation doesn't work is the exact same reason why are wrong with your "calculations")
Math is the language of the universe.

The inability to explain something is not proof of something else.

We don't speak for reality - we only observe it. An observation can have any cause, but it is still no more than just an observation.

When in doubt; sources!

*

sandokhan

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  • Flat Earth Scientist
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Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #28 on: May 14, 2018, 10:36:06 AM »
evar, so far you have not been able to show any errors in the simple example I provided.

You cannot draw attention to a new kind of example/situation, if you cannot find any errors in my analysis.

The trick you are employing is an old one, it is surprising that you'd try something like this with me.

Any new kinds of examples, involving force diagrams, will have to wait until you do one thing: show your readers any errors in my calculation.

The fact that you are drawing attention away from the discussion at hand means one thing only: you are unable to find any errors in my analysis. Until you do so, there's nothing else to discuss here, my calculation stands correct.


Re: Is the tension on an "ideal rope" (massless) the same at each end?
« Reply #29 on: May 14, 2018, 11:46:16 AM »
Aren’t you forgetting to take tension force into account? If you put that into your equation, it balances out in a Newtonian system. At least in my head when mentally drawing a force diagram.
Be gentle