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Physics Question
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Chrono
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RE: Physics Question
quote:
Originally posted by CookieRevised
It's like saying x/y wouldn't be the same as x+10 / y+10.
1/2 != 11/12 :P

(A)

This post was edited on 02-27-2010 at 01:29 PM by Chrono.
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02-27-2010 01:29 PM
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RE: Physics Question
quote:
Originally posted by CookieRevised

quote:
Originally posted by vaccination
Lol cookie, perpetual motion is possible in a frictionless environment,
Perpetual motion machines aren't possible, not even in theory, not even in a frictionless environment since there should always be an equilibrium of energy. And a perpetual motion machine is exactly based on the nonequilibrium of energy (being it motion energy, heat energy or whatever energy): it produces more energy than you put into it, that's the whole definition of such machines.

The fact that it is said that a perpetual motion machine is not possible in practice is not _just_ because of friction, but mostly because you can not get more energy out of something than you put into it (while preserving its mass). friction is only a tiny part of the problem of perpetual motion machines. Cancelling that out and you would still have the same problems as before: you can not get more energy out of it than you have put into it.

So, forgetting the fact that it isn't possible to completely rule out friction (even with supercooled, superconductors or whatever), thus in a theoretical friction free world, the moment you want to extract energy out of such a perpetual motion machine means that you will eventually loose the stored energy in that machine. And therefore it isn't a real perpetual motion machine to begin with as it will eventually run out.


Why ignore half my post? -.-

quote:
Originally posted by CookieRevised

There is no more (or less) force pulling on the chain on the left side than there is on the right side, ergo, there is nothing to start movement, ergo there is no movement, even in a theoretical frictionless system, no matter what angle or length the two slopes are, as long as both ends are on the same horizontal plane. The system is always in equilibrium.

I'm (almost) sure that if you take those equations from above and work them further out you would see they are equal (you can in theory because there is always a direct relation between the various elements of the two sides (mass, length, angle, etc), by using ratios instead of real numbers)). The more I think about it the more I am convinced of that.

Uhm, yes there is. F = mgh. The force due to gravity is greater on the left side as there is more mass.

quote:
Originally posted by foaly
quote:
Originally posted by CookieRevised
You simply hang a second chain of the same type as the one you had on both the ends. This does _not_ change anything in the system since the chain would pull equally hard on both sides; aka the two opposite forces (one on the left end, the other on the right end) cancel eachother out. But, it is now a lot easier to see in pic4 why there wouldn't be a movement, not even in a frictionless environment; it's just a closed chain hanging on a irregular object; it is never going to move out of itself.

It would actually, you move the centerpoint of the mass of the chain...

I was just about to point this out, connecting the ends of the chain like you have in the image COMPLETELY changes to whole system and cannot be considered.

This post was edited on 02-27-2010 at 02:29 PM by vaccination.
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02-27-2010 02:20 PM
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RE: Physics Question
quote:
Originally posted by CookieRevised
I'm (almost) sure that if you take those equations from above and work them further out you would see they are equal (you can in theory because there is always a direct relation between the various elements of the two sides (mass, length, angle, etc), by using ratios instead of real numbers)). The more I think about it the more I am convinced of that.

I already showed this in both my posts. Maybe it's hard for you to understand that without diagrams. The equations do equate.

I'm still not entirely sure how the "green" chain helps explain how it would slip in neither direction besides just stating that it's obvious. It seems to me that the chain underneath does nothing as long as you assume that the top part of the chain doesn't move to start off with. You can't just claim that.
02-27-2010 06:52 PM
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CookieRevised
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RE: RE: Physics Question
quote:
Originally posted by foaly
quote:
Originally posted by CookieRevised
You simply hang a second chain of the same type as the one you had on both the ends. This does _not_ change anything in the system since the chain would pull equally hard on both sides; aka the two opposite forces (one on the left end, the other on the right end) cancel eachother out. But, it is now a lot easier to see in pic4 why there wouldn't be a movement, not even in a frictionless environment; it's just a closed chain hanging on a irregular object; it is never going to move out of itself.

It would actually, you move the centerpoint of the mass of the chain...

You do not add any more mass to one end than on the other end, therefore eventhough the centerpoint might be vertically lower, it stays in the same horizontal position and thus hasn't any influence on where the chain would move.

What you say is that if you add 10Kg on both ends the chain would suddenly behave differently than before. This is not the case. The chain will not move any differently than before.

Or to put it in another way, what you say is that a lighter chain is going to behave differently than a heavier chain, eventhough the lay on the exact same triangle in the exact same position. yes, the complete system (chain + block) is going to be heavier, but that does not influence the movement of the chain at all.

If you add 10Kg on both the sides of a scale in balance, that scale isn't going to move, it would stay where it is. If you add 10Kg on both sides in an unbalanced system, that unbalanced system isn't going to flip to one side any faster than it was before or all of a sudden flip to the other side. Aka: adding an equal force to both sides of a balance system (which might be in balance or not, doesn't matter) is not going to have ANY influence at all on the balance of that system.

quote:
Originally posted by Chrono
quote:
Originally posted by CookieRevised
It's like saying x/y wouldn't be the same as x+10 / y+10.
1/2 != 11/12 :P

(A)
sorry, I didn't meant "/" as in an divided by symbol. "<=>" or whatever might have been a better symbol; x and y are those formulas provided by Chancer and co above (which I explained in that same post). You can also think of x and y as both ends of a scale and the "/" as the tipping point.

quote:
Originally posted by vaccination
Why ignore half my post? -.-
I didn't ignore anything. In the reply I gave I replied to every bit of your post. See second paragraph (talking about "not possible in real life") and third paragraph (talking about superconductors etc).

quote:
Originally posted by vaccination
quote:
Originally posted by CookieRevised

There is no more (or less) force pulling on the chain on the left side than there is on the right side, ergo, there is nothing to start movement, ergo there is no movement, even in a theoretical frictionless system, no matter what angle or length the two slopes are, as long as both ends are on the same horizontal plane. The system is always in equilibrium.

I'm (almost) sure that if you take those equations from above and work them further out you would see they are equal (you can in theory because there is always a direct relation between the various elements of the two sides (mass, length, angle, etc), by using ratios instead of real numbers)). The more I think about it the more I am convinced of that.

Uhm, yes there is. F = mgh. The force due to gravity is greater on the left side as there is more mass.
No there isn't. You have exactly the same additional force pulling on the right side as you have the same additional mass on the right side! Hence the system does NOT change by adding the bottom chain!

quote:
Originally posted by gif83
I already showed this in both my posts. Maybe it's hard for you to understand that without diagrams. The equations do equate.
Why using 'difficult' equations and 'hard' to understand diagrams when you might solve it with simple logic.

quote:
Originally posted by gif83
I'm still not entirely sure how the "green" chain helps explain how it would slip in neither direction besides just stating that it's obvious.
dunno... because it is obvious* (without the use of equations and too much thinking)?
*A closed chain isn't going to start moving out of itself all of a sudden. reason: see previous posts.

quote:
Originally posted by gif83
It seems to me that the chain underneath does nothing as long as you assume that the top part of the chain doesn't move to start off with. You can't just claim that.
Yes I can claim that. Because if the bottom part isn't going to do anything, as you say, than why should the top part move at all as they are connected... Implying that the top part moves implies automatically that the bottom part would also move, which it does not. And that is the whole essence.

This post was edited on 02-27-2010 at 08:19 PM by CookieRevised.
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02-27-2010 07:53 PM
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RE: Physics Question
quote:
Originally posted by CookieRevised
Yes I can claim that. Because if the bottom part isn't going to do anything, as you say, than why should the top part move at all as they are connected... Implying that the top part moves implies automatically that the bottom part would also move, which it does not. And that is the whole essence.

What I meant was the bottom chain doesn't move ONLY IF the top doesn't move. you are trying to work out whether or not the top part of the chain moves. You can't say the top does't move because the bottom does as the bottom's movement is dependent on what's going on on top.

I guess I found equations to be far less brain hurt :p
02-27-2010 08:14 PM
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RE: RE: Physics Question
quote:
Originally posted by gif83
What I meant was the bottom chain doesn't move ONLY IF the top doesn't move. you are trying to work out whether or not the top part of the chain moves.

You can't say the top does't move because the bottom does as the bottom's movement is dependent on what's going on on top.
Since the chains ARE connected you can turn it the other way around too, that is pure logic and exactly the whole essence (and beauty?) of this method.

Neither the top or the bottom can move as long as the bottom or the top will not move. No need for equations, diagrams and maths (well except for: If A is B, then B is A).

quote:
Originally posted by gif83
I guess I found equations to be far less brain hurt :p
bleh, as long as we agree the top chain wouldn't move :p

This post was edited on 02-27-2010 at 08:27 PM by CookieRevised.
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02-27-2010 08:25 PM
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RE: Physics Question
quote:
Originally posted by CookieRevised
Neither the top or the bottom can move as long as the bottom or the top will not move. No need for equations, diagrams and maths (well except for: If A is B, then B is A)

Agreed. However, as of such there is nothing so far I have seen that says either side will not move. If you say the top side will not move then the bottom chain is pointless. if you say the bottom chain won't move, how can you say that without knowing about the top?

I'm probably just being really dense here.
02-27-2010 08:36 PM
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RE: Physics Question
quote:
Originally posted by CookieRevised
quote:
Originally posted by foaly
quote:
Originally posted by CookieRevised
You simply hang a second chain of the same type as the one you had on both the ends. This does _not_ change anything in the system since the chain would pull equally hard on both sides; aka the two opposite forces (one on the left end, the other on the right end) cancel eachother out. But, it is now a lot easier to see in pic4 why there wouldn't be a movement, not even in a frictionless environment; it's just a closed chain hanging on a irregular object; it is never going to move out of itself.

It would actually, you move the centerpoint of the mass of the chain...

You do not add any more mass to one end than on the other end, therefore eventhough the centerpoint might be vertically lower, it stays in the same horizontal position and thus hasn't any influence on where the chain would move.

What you say is that if you add 10Kg on both ends the chain would suddenly behave differently than before. This is not the case.

That is not what I'm saying at all. Because you lower the balance of the chain way below the tumble point, you make it easier for the chain to balance...

What is probably even more important is that by adding the chain you make a circle, this makes the chain have to pull itself down when sliding one way or another...

I'll add a picture in a sec showing how ludicrous it is that you could just add a chain and expect it to be the same situation....
[Image: attachment.php?pid=989119]

Without the green chain the chain will obviously fall left, with it, it won't...
Although the blue and red chain aren't on the same horizontal height as the examples you gave... The theory behind it is still the same... You can't just connect the chain... it makes it a totally different object, that behaves completely differently

.png File Attachment: chains.png (2.21 KB)
This file has been downloaded 297 time(s).

This post was edited on 02-27-2010 at 08:54 PM by foaly.
02-27-2010 08:43 PM
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vaccination
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RE: Physics Question
Cookie, connecting the chains is NOT the same as adding a 10kg weight to either side. Lets say you connect a 20kg chain to the ends of the other chain (ie: the green chain in your diagrams has a mass of 20kg) this is not evenly distributed to 10kg at each end of the initial chain, like it would if you simply attached a 10kg mass to each end, instead the majority of the 20kg mass of the new chain falls downwards and to the middle, meaning that an extra 20kg of mass is now in effect on the original chain at the point where the two slopes meet.

Anyway, if you can't understand how f= mgh works, I would just stop posting in this thread, since you've mostly posted nonsense, I'm afraid. I mean, three of your four diagrams don't even match the criteria of the experiment(ignoring the absurdity of this random extra chain you've decided to magic up).

This post was edited on 02-27-2010 at 09:01 PM by vaccination.
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02-27-2010 09:00 PM
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RE: RE: Physics Question
quote:
Originally posted by foaly
That is not what I'm saying at all. Because you lower the balance of the chain way below the tumble point, you make it easier for the chain to balance...
Easier or harder is irrelevant. The thing would still be in the exact same state. The center of gravity has got nothing todo in all of this.

0.999999999999999999 is still lower than 1. It is not because it is 'closer' to one than 0.5 that it isn't lower anymore.

quote:
Originally posted by foaly
What is probably even more important is that by adding the chain you make a circle, this makes the chain have to pull itself down when sliding one way or another...
no. It pulls itself down _equally_ on both side, aka it cancels itself out! It does _not_ pull harder on one side than on the other side; it pulls equally hard on _both_ sides.


quote:
Originally posted by foaly
I'll add a picture in a sec showing how ludicrous it is that you could just add a chain and expect it to be the same situation....
[Image: attachment.php?pid=989119]
Your drawing is a complete different system than what SonicSam has drawn. My drawings in my previous post are of the exact same system as what SonicSam has drawn. Notice that in the system of SonisSam the end points lay on the same horizontal plane.

hence:
quote:
Originally posted by foaly
....The theory behind it is still the same...
But you are using a different system than SonicSam, not me.

quote:
Originally posted by vaccination
Cookie, connecting the chains is NOT the same as adding a 10kg weight to either side.
Sorry, but it certainly is. Look again at SonicSam's drawing...

quote:
Originally posted by vaccination
Lets say you connect a 20kg chain to the ends of the other chain (ie: the green chain in your diagrams has a mass of 20kg) this is not evenly distributed to 10kg at each end of the initial chain
Of course it is not 10Kg pulling on either side, but it IS evenly distributed. Hence there is no extra force pulling more on one side than it is on the other! That is basic (vector) math.

x     y
\   /
  \ /
   |
  #20Kg


quote:
Originally posted by vaccination
...like it would if you simply attached a 10kg mass to each end, instead the majority of the 20kg mass of the new chain falls downwards and to the middle, meaning that an extra 20kg of mass is now in effect on the original chain at the point where the two slopes meet.

Anyway, if you can't understand how f= mgh works, I would just stop posting in this thread, since you've mostly posted nonsense, I'm afraid.
I'm sorry, but what I posted is 100% correct and I even apply that science in every day work when I need to fly spot lights and scenery in the theater (hence the reason how I came up with that extra chain)

quote:
Originally posted by vaccination
I mean, three of your four diagrams don't even match the criteria of the experiment (ignoring the absurdity of this random extra chain you've decided to magic up).
They _all_ match exactly the criteria...
And again, the extra chain does not have _any_ influence on the balance of the system.

This post was edited on 02-27-2010 at 09:16 PM by CookieRevised.
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02-27-2010 09:06 PM
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