jack, wake up!
Your equations AMOUNT TO NOTHING.
No. They show the reality of the situation.
IN THE RE ANALYSIS THE TWO BOATS HAVE TO MEET IN THE MIDDLE.
Yes, 2 boats, of the same mass (ignoring other force) will always meet in the middle, just like in reality (although that gets more complicated with different forces like resistance of motion through the water).
So no failure at all.
But don't worry, your "analysis" reaches the same conclusion, that the 2 boats meet in the middle.
So you are just objecting to your own analysis and claims.
There is no working around that.
And there is no need to.
The forces applied are the SAME in your analysis.
Yes, because I care about matching reality, which means the 2 forces will be the same.
That is the ONLY option you got: by hypothesis, the two boats have equal mass.
BUT DIFFERENT FORCES ARE APPLIED AT EACH END OF THE ROPE.
Again, this is your baseless claim which makes the situation impossible.
In reality, the 2 forces will be the same.
Your text book even agrees with me. 2 identical boats meet in the middle. It is only when you make one boat large and the other small that they don't.
So your book is in perfect agreement with me and is contradicting you.
If you think they can meet anywhere other than the middle, why don't you try justifying it?
Especially as the net force on the 2 boats are equal and opposite.
Your other message is useless.
Are you just saying that because it completely destroyed any chance of you refuting me?
Here is what you wrote:
Like I said, if you have an objection state which point it is, why you object, and what you think the correct answer is and why.
By the very hypothesis, A DOES NOT EQUAL B.
A cannot equal B.
It doesn't equal B. A=-B.
Of course, you will object to that as well. But as you are only objecting to point 16, I take that to mean you fully accept all prior points.
The issue then is that was point 15.
It shows quite clearly that A=-B, and that your baseless claim about the situation is completely wrong.
In reality, the forces are equal but opposite.
Stop just appealing to your bullshit baseless claim. It is wrong. I know it is wrong. This shows that it is wrong.
This is basically your argument:
"YOU SAID THIS.
I SAY THAT IS WRONG.
SO YOU ARE WRONG!!!."
You are just baselessly asserting that I am wrong without justifying it at all.
On the other hand, I have shown quite explicitly why your claim is wrong.
Go through the questions I asked, and tell me what the first one you disagree with and why.
The forces applied by you and disputeone are DIFFERENT to start with.
No. They aren't. Get that through your thick skull.
Continuing to repeat the same refuted bullshit doesn't make your case any better. It just makes you look more and more pathetic.
Do you accept that all force acting on boat X comes from the rope? (for this part I don't care if it is because X is pulling on the rope or Y is, the only important part is that the entirety of the force acting on boat x is being applied to boat X by the rope)
Your tag team partner, sigma, is trying to use same argument, as follows:
And you still refuse to answer.
Why is that? Is it because you know that doing so will show you to be full of shit?
But a rope can transmit more than one force.
No. It can't.
By the very nature of how it works, a rope anchored/pulled at 2 points (1 point at each end) in opposite directions will result in the rope being under tension and applying a force equal in magnitude to the tension in the rope.
That is the sole force the rope transmits.
It is very easy to prove this.
IF THE ROPE WAS TRANSMITTING ONLY ONE FORCE, AS JACK AND SIGMA (NOT TO MENTION RABINOZ) WOULD HAVE US BELIEVE, THEN THE TWO BOATS WOULD MEET IN THE MIDDLE. ALWAYS.
No. It is impossible to prove your baseless bullshit.
They only meet in the middle if they are identical. If they are different (e.g. different mass), they don't meet in the middle.
But that is also the result of your analysis.
Here is the proof.
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 RE's own analysis, assumed to be the very best that they can come up with, reaches the following conclusion:
As the string isn't moving, the net force on the string is 0, so A+B=0 so B=-A.
That is: |A|=|B|
If A = B, then the two boats will meet in the middle.
And here is what the proof is based upon your analysis (with some irrelevant crap removed and slightly corrected):
The net force on boat x is -A+B.
The net force on boat y is -B+A.
The flatards' own analysis, assumed to be the very best pathetic bullshit that they can come up with to try and deny reality, reaches the following conclusion:
|-A+B|=|-B+A|
If |-A+B|=|-B+A|, then the two boats will meet in the middle.
So congratulations, your own analysis shows they meet in the middle.
Now, jack, let me remind you of what your friend sigma said earlier where HE AGREES WITH ME.
Just because he agrees with a single point doesn't mean he agrees entirely.
Remember this?
The two boats will meet at some point between them.
Agreed for the boat scenario.
Yes. Do you?
He was stating that if you have 2 boats of different masses they will meet at some point between them.
He is not saying that if you have 2 identical boats that they can meet anywhere other than in the middle.
Stop lying about what people are saying.
You do agree that if the two identical boats pull with different forces, then THEY WILL MEET AT SOME POINT BETWEEN THEM, never exactly in the middle.
No. They won't.
They pretty much wont meet at all.
The rope will fly off one of the boats as it was not applying enough force to hold it there.
It is only if the 2 boats apply equal and opposite forces that the string remains.
If you go to reality instead, it then gets more complicated due to other forces acting.
Since by the very hypothesis forces A and B are of a different magnitude, the boats will never meet exactly in the middle.
Again, that is just your baseless claim, and your analysis shows it to be wrong.
You end up with the 2 boats pulling with the same magnitude force (which you don't explicitly state and refuse to answer questions on), and the same magnitude net force on the boats, resulting in them meeting in the middle.
But they do not.
Prove it.
You are yet to do this, and your analysis shows that they will.
Since jackblack needs to vent his fury at his miserable analysis having been thrashed thoroughly in this thread he will pull extremely hard: 200N (location: boat X).
You mean since jackblack is determined to keep beating this dead horse until it flees like pathetic coward or finally gets some life and starts mounting a rational defence, he will pull with 200 N.
disputeone is trolling along, he will pull with a force of 140N (location: boat Y).
And thus gets the rope pulled straight out of his hand as not enough force was applied to hold it there (in fact, he is just helping it along).
Of course they will meet not in the middle, but somewhere along the line of the rope.
Of course they will not meet in the middle, as the rope goes flying and they are no longer connected so they don't meet at all.
We quickly notice the stupendous contradiction which we reach in an instant of a second.
No, we quickly the notice the stupendous error you made by asserting that A and B are different in magnitude when they have to be the same.
There is no way that 200N + 140N = 0.
That's right, which means your situation is pure bullshit. It is physically impossible.
Let us remember that the foregoing RE analysis cannot be applied even in a hypothetical case where:
Force A = 100.000,000,000 N
Force B = 100.000,000,001 N
Even in this case B will not be equal to -A.
That's right, but again, that isn't the situation the RE analysis applies to.
In this case, both people are pulling on the rope in the same direction.
What this means is that there is no tension at all unless the rope goes past and attaches to something applying a force of -200 N.
Do you not understand that we are dealing with non-scalar quantities? Where direction (and thus sign) matters?
That if 2 people are pulling in opposite directions, then one force will be >0 and the other will be <0.
EQUATIONS DO NOT LIE.
But you do, and you use the equations you are lying about to help.
There is no way that 200N + 140N = 0.
And as equations don't lie, this shows that this situation is impossible.
A possible situation would be when A=200 N and B = -200 N.
All forces balance out perfectly.
No they don't.
You have an action reaction pair between the rope and X.
X is applying a force of A to the rope.
The rope is applying a force of A-B to boat X.
For them to balance you need these to be equal but opposite, i.e.:
-A=-A+B.
Or putting in your numbers:
-200 N = -200 N + 140 N.
They will never be equal.
140 N will never equal 0.
Thus you have unbalanced forces and a completely impossible situation.
So no, plenty of contradictions.
The forces applied at each end of the rope will ALWAYS be different.
Certainly you'd be pulling with a different force than, say, jack. Right?
No. You are completely wrong.
The forces applied at each end of a mass-less rope under tension will always be equal.
Certainly he would be pulling with a force equal but opposite me.
No.
Your boat would simply thrust forward, as would his boat.
Nope. That requires him to apply the force to the boat, he would need to apply all 200 N.
If he can't, then the rope flies out of his hand.
Remember, you are on water, not on land.
So?
Inertia alone is enough to make it impossible. You don't need the added friction of the land.
A simple example, get a rope and tape something heavy too it. Now spin it around going faster and faster. Eventually you spin it too fast and the tape is no longer able to apply the force needed to accelerate the object and the rope comes out of the tape.
I never said anything about the masses of the two men in those boats. I said the boats were identical, that's all.
And in my analysis the mass of the men was included in the boats. mX is not just the mass of the boat itself, it is the mass of the boat and everything in it, including the person or people pulling the rope.
So if they boats are identical but the people aren't, then mX!=mY and thus they do not meet in the middle.
So do you agree that 2 identical boats with identical people being pulled together by a rope will always meet in the middle?
All of you here are missing the most important point.
Forces A and B can never be the same.
No. We understand fulling that that "point" is pure bullshit.
A and B must always be equal and opposite. Otherwise there is a net force on the rope and it goes flying out of someone's hand.
Never.
ALWAYS
FORCES A AND B ARE OF DIFFERENT MAGNITUDE BY HYPOTHESIS.
No, by baseless claim.
If you wish to make that your hypothesis, then fine, your hypothesis is bullshit and doesn't match reality at all. It is a failed hypothesis that any sane person would have discarded long ago.
If you are pulling with a force of 200N, and disputeone is pulling with a force of 140N, then A does not equal B.
And the situation doesn't exist in reality.
Go ahead and test with a measuring device your strength down to the 100,000,000th fraction of Newton against all the people in the world.
You won't find an exact match. Absolutely guaranteed.
The limit of your strength, or just when you are trying to apply it to the same rope?
If the latter, then yes, you will, every time you ever measure it it will be an exact match.
The simplest way to measure it is with a spring scale.
This works quite simply, you hook it onto something and pull. It doesn't even matter which way you put it. It tells you how much force you are applying (typically in kgf or lbf). And guess what? Because of how it works, it means you don't even need to bother with the rope. You can just each grab an end and pull.
As it only has one scale it needs to be the same. There isn't a separate measurement for how much each side is pulling, because they need to be equal.
Go try it.
See if you can find any situation where you don't end up with a perfect match.
This is what we are talking about here, what the RE require.
Yes, which is also what reality requires.
This is the absurd requirement of a failed theory.
No. As it matches reality, it is a good theory, which works.
My analysis suffers from no such requirements. It can even handle the theoretical case where the two forces are perfectly matched down to the 100,000,000th fraction of a Newton.
No. It fails, spectacularly.
With you ignoring direction you get a result of if the 2 are pulling with an equal force, the boats don't move at all.
But it remains unbalanced and full of contradiction.
By contrast, the RE's own analysis CANNOT EVEN HANDLE a basic infinitesimal difference at the level of a 100,000,000th fraction of a Newton. A totally failed theory.
Yes, because that doesn't happen in reality, showing how the theory matches reality, so a perfectly fine theory.
jack is pulling with 200 N force (force A).
disputeone is pulling with a force of 140 N (force B).
Does force A equal force B?
Certainly not.
That is why the RE analysis amounts to pure thrash.
No. That is why your completely delusional scenario amounts to pure trash, because it isn't realistic at all.
In order for it to be real, the 2 forces need to be the same.
So? That means the reaction force from me pulling will be in the same direction as the force disputeone is applying.
That means they would be in the same direction.
But, at the same time, disputeone IS ALSO PULLING ON JACK'S BOAT.
That was the "force disputeone is applying"
So no, not "at the same time, blah"
I already covered that.
Regardless, that was being applied to your delusional, completely unrealistic scenario.
A second force applied.
Totally unaccounted for by the bumbling RE analysis.
Because it doesn't exist in reality.
Yet the RE analysis requires that |A|=|B| which is a complete nonsense.
No. Your delusional scenario has |A|!=|B| which makes it completely nonsense.