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## What exactly is the reactive centrifugal force (split)

 Quote by stevendaryl But in general, there are more than two objects. So suppose we have three objects, $O_1, O_2$ and $O_3$, with $O_2$ in the center. Object $O_1$ exerts a force only on $O_2$, and $O_2$ exerts forces on $O_1$ and $O_3$. Object $O_3$ exerts a force only on $O_2$ In this setup, it's not the case that the change in momentum of $O_1$ is equal and opposite to the change in momentum of $O_2$, because $O_1$ is not the only object acting on $O_2$. It's certainly true that the total change in momentum is zero, if you add up the changes for all three objects. But they don't come in action/reaction pairs. To me, what comes in pairs is "the force on $O_1$ due to $O_2$" and "the force on $O_2$ due to $O_1$"
If there are three bodies interacting, it is much more difficult to analyse forces because the duration of the interactions between bodies may differ. What one can easily determine, however, are the changes in motion that result from the interactions. For three interacting bodies, A, B and C, if body A experiences a change in motion (momentum), bodies B and C must experience a change in motion that, combined, is equal in magnitude to the change in A's motion but opposite in direction.

AM

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 Quote by DaleSpam Andrew Mason, please provide a mainstream scientific reference for Newton's 3rd law being as you describe, equal and opposite net forces or equal and opposite changes in momentum.
Do you consider Newton to be mainstream?

Newton's explanation of the third law uses both the concept of a force and of change in motion. He begins by talking of a body pressing on another, the latter pressing back on the first (a finger pressing a stone and the stone pressing back on the finger), and then says:
If a body impinges upon another, and by its force changes the motion of the other, that body also (because of the equality of the mutual pressure) will undergo an equal change, in its own motion, toward the contrary part. The changes made by these actions are equal, not in the velocities but in the motions of the bodies; that is to say, if the bodies are not hindered by any other impediments. For, as the motions are equally changed, the changes of the velocities made toward contrary parts are reciprocally proportional to the bodies.
It is interesting that Newton speaks of "bodies ... not hindered by any other impediments". One can argue whether he was equating action with "force on an otherwise unimpeded body", but that is not an unreasonable interpretation.

With respect to the astronaut space-station example, the third law says that a body undergoing centripetal change in motion due to contact with another unimpeded body causes that other body to have an equal change of motion in the opposite direction. So if we are talking about the astronaut undergoing a centripetal change in motion due to contact with the space station/other astronaut (taken as one body), the third law "reaction" is a change in motion of the space station/other astronaut in the opposite direction. That change in motion always points to the centre of rotation.

AM

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 Quote by stevendaryl Hmm. Let's see if that's true. Suppose you have three objects 1, 2, 3. There are correspondingly 6 forces:$\stackrel{\rightarrow}{F_{12}}$ = force of 1 on 2 $\stackrel{\rightarrow}{F_{21}}$ = force of 2 on 1 $\stackrel{\rightarrow}{F_{13}}$ = force of 1 on 3 $\stackrel{\rightarrow}{F_{31}}$ = force of 3 on 1 $\stackrel{\rightarrow}{F_{23}}$ = force of 2 on 3 $\stackrel{\rightarrow}{F_{32}}$ = force of 3 on 2 The fact that the situation is static means that the force on any object is zero. So we have:$\stackrel{\rightarrow}{F_{12}} + \stackrel{\rightarrow}{F_{32}} = 0$ $\stackrel{\rightarrow}{F_{13}} + \stackrel{\rightarrow}{F_{23}} = 0$ $\stackrel{\rightarrow}{F_{21}} + \stackrel{\rightarrow}{F_{31}} = 0$ That's three constraints. I don't see how you can derive that $\stackrel{\rightarrow}{F_{12}} + \stackrel{\rightarrow}{F_{21}} = 0$
If it is static, all the forces on each object must sum to 0. So:

$\vec{F}_{12} + \vec{F}_{32} + \vec{F}_{13} + \vec{F}_{23} + \vec{F}_{21} + \vec{F}_{31} = 0$

You get that from the first law.

If $\vec{F}_{12} + \vec{F}_{32} = 0$ and if 2 is not accelerating then either $\vec{F}_{12} = \vec{F}_{32} = 0$ or there are other forces acting on 2.

In that case you would have to write:

$\vec{F}_{12} + \vec{F}_{32} + \vec{F}_{13} + \vec{F}_{23} + \vec{F}_{21} + \vec{F}_{31} + \sum F_{i(ext)} = 0$, which is again just the first law.

Take a simpler case of a magnet (body 1) and an iron bar (body 2) stuck to it by the magnet's magnetic force. The total force of the magnet on the bar is 0 and the total force of the bar on the magnet is 0. If it were otherwise, there would be acceleration.

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 Quote by A.T. It is the interpretation of Newton's third that you will find in every modern physics book. It doesn't matter how you call this Law ("3rd law" or "generalized 3rd law"). All your hand waving cannot change the fact that: - It is a law of physics. - It states two equal but opposite forces. - It applies to static cases too. - It means that my astronauts exert a centrifugal reaction force on the station
We agree that there is a force exerted on the space station by the astronaut that is a reaction to the force of the space station causing the astronaut to experience centripetal acceleration. All I am saying is that that reaction force causes the space station centre of mass to undergo centripetal acceleration toward the centre of rotation. So the reaction to the centripetal change of motion of the astronaut is a centripetal change of motion of the thing that is acting on the astronaut. And that is what the third law says must occur.

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 Quote by stevendaryl I agree with you that thinking of equal and opposite forces works better than thinking of equal and opposite changes of motion, but in the case you are talking about, the change in momentum of one of the masses is equal and opposite to the vectorial sum of the changes of momentum of the other two masses. That might be what AM meant.
Exactly. See my post #135 above.

AM

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 Quote by DaleSpam Yes, this is correct. Note that the gravitational force (mg) is not equal to the change in momentum (ma=0). So your correct statement here contradicts your own incorrect statement elsewhere.
The reason there is no change in momentum is because the body and the earth are each exerting a total force of 0 on each other. If it were otherwise, there would be an acceleration.

 Consider three identical masses, each connected by identical springs of neglible mass to both of the others and starting at rest from a position where the springs are all equally stretched. Each object is experiencing a change in motion and there is no other body experiencing an equal and opposite change in motion.
True, but there is a change in motion of the other two that combines to be equal and opposite to the change in motion of the first. The third law says that you cannot have just one body changing its motion. See my post #135 above.

AM

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 Quote by Andrew Mason It is interesting that Newton speaks of "bodies ... not hindered by any other impediments". One can argue whether he was equating action with "force on an otherwise unimpeded body", but that is not an unreasonable interpretation.
That phrase seems to contradict your position and support the mainstream interpretation. I do not think that this reference supports your interpretation.

Also, this may be a translation error. Other translations are pretty clear on the standard interpretation. "Whatever draws or presses another is as much drawn or pressed by that other. If you press a stone with your finger, the finger is also pressed by the stone. "
http://www.marxists.org/reference/su.../en/newton.htm

Do you have any mainstream scientific reference that clearly supports your unorthodox stance? To ensure clarity, it would be best if it expressed the law mathematically. I.e. the usual version is something like $F_{12}=-F_{21}$ but your version would be something like $dp_1/dt = - dp_2/dt$. Can you find a reference that actually expresses the third law that way?

Here are some clear references for the usual way:
http://en.wikipedia.org/wiki/Newton%....27s_third_law
http://www.phy6.org/stargaze/Snewton3.htm
http://www.ck12.org/concept/Newtons-...s-Third-Law%2F

Mentor
 Quote by Andrew Mason The reason there is no change in momentum is because the body and the earth are each exerting a total force of 0 on each other. If it were otherwise, there would be an acceleration.
Agreed. Since the total force is zero then, according to your interpretation, there is no third law pair for either the normal force or the gravitational force.

 Quote by Andrew Mason True, but there is a change in motion of the other two that combines to be equal and opposite to the change in motion of the first.
But that isn't what the third law says, even in your interpretation as expressed above. The Newton quote certainly doesn't lend support to this weird three-body version of the third law. You are just making things up as you go along, and contradicting yourself and your sources.

 Quote by Andrew Mason All I am saying is that that reaction force causes the space station centre of mass to undergo centripetal acceleration toward the centre of rotation.
And you are wrong as usual. The space station centre of mass has no acceleration toward the centre of rotation. The space station centre of mass is static at the centre of rotation.

Needless to say that even if you were right, it would be still irrelevant. It would not change the fact that the contact force Frcf acts centrifugally on the wall.

 Quote by Andrew Mason Do you consider Newton to be mainstream?
Your interpretation of Newton is not mainstream.

But even if you were right about what he initially meant centuries ago in his Latin prose, it would still be irrelevant. Laws of physics are often generalized over the course of time. Today we have a 3rd law that applies to forces regardless of accelerations.

 Quote by DaleSpam your version would be something like $dp_1/dt = - dp_2/dt$.
Expressed that way, it would be a rather useless law, that applies only in a special case where no other forces are acting (or cancel exactly). It would not be a law, but rather a case specific rule. A law in physics is usually a generally applicable rule.
 Quote by DaleSpam Can you find a reference that actually expresses the third law that way?
I'm sure many school books use the simplest possible example (two bodies, one interaction) to introduce the 3rd Law. But if Andrew goes a few pages further, he should find examples where there are more forces acting.

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 Quote by A.T. And you are wrong as usual. The space station centre of mass has no acceleration toward the centre of rotation. The space station centre of mass is static at the centre of rotation.
Are you including both astronauts in the space station? The space station I refer to includes only the other astronaut. Read my #136 post:
"With respect to the astronaut space-station example, the third law says that a body undergoing centripetal change in motion due to contact with another unimpeded body causes that other body to have an equal change of motion in the opposite direction. So if we are talking about the astronaut undergoing a centripetal change in motion due to contact with the space station/other astronaut (taken as one body), the third law "reaction" is a change in motion of the space station/other astronaut in the opposite direction. That change in motion always points to the centre of rotation."
Here is a simple question: If the reaction force to the centripetal acceleration of astronaut 1 is a centrifugal static force (ie it does not result in any contrary change in motion of the thing that is accelerating astronaut 1) what is the force that causes the centripetal acceleration of the centre of mass of the space station/other astronaut?

 Needless to say that even if you were right, it would be still irrelevant. It would not change the fact that the contact force Frcf acts centrifugally on the wall.
We agree on the force and we agree on the direction. And all I am saying is that since this force causes the rest of the mass to which the astronaut is connected to undergo acceleration toward the centre of rotation, it is a true centripetal force. Calling it centrifugal is a misnomer.

AM

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 Quote by DaleSpam Agreed. Since the total force is zero then, according to your interpretation, there is no third law pair for either the normal force or the gravitational force. But that isn't what the third law says, even in your interpretation as expressed above. The Newton quote certainly doesn't lend support to this weird three-body version of the third law. You are just making things up as you go along, and contradicting yourself and your sources.
It is not a weird three-body version of the third law. Body A exerts forces on both Body B and Body C. Body C and Body B exert equal and opposite forces on Body A. But in order to analyse the interaction using forces you need to know the instantaneous direction and duration of these forces and whether the bodies are rigid or deform during the interaction etc. It is not easy to analyse using forces.

The third law says that the action (which we can take to be A's action on B and C) is accompanied by an equal and opposite reaction (the action of B and C on A). So we can say that the change in motion of A is accompanied by equal and opposite changes in motion of B and C. This is true at all times before during and after the interaction.

AM

 Quote by Andrew Mason You get that from the first law. If $\vec{F}_{12} + \vec{F}_{32} = 0$ and if 2 is not accelerating then either $\vec{F}_{12} = \vec{F}_{32} = 0$ or there are other forces acting on 2.
I don't get that, at all. From $\vec{F}_{12} + \vec{F}_{32} = 0$, you get $\vec{F}_{12} = - \vec{F}_{32}$, not that
$\vec{F}_{12} = \vec{F}_{32} = 0$

The fact that 2 is not accelerating is what tells us that $\vec{F}_{12} + \vec{F}_{32} = 0$.

 Quote by Andrew Mason The space station I refer to includes only the other astronaut.
Nice try. Why not include both astronauts in the space station, so we have just one object, no interactions and finally no reactive centrifugal force.

Sorry, you don't get to decide how people divide the system into parts. In my example there are 3 objects: station & 2 astronauts. This is a perfectly valid way to analyze the scenario. So deal with it, and don't try to change it.
 Quote by Andrew Mason ...acceleration...
Irrelevant. The force is there, regardless of acceleration. In the rotating frame there is no acceleration at all, but there still is a force on the wall pointing outwards: centrifugal.

 Quote by Andrew Mason It is not a weird three-body version of the third law. Body A exerts forces on both Body B and Body C. Body C and Body B exert equal and opposite forces on Body A. But in order to analyse the interaction using forces you need to know the instantaneous direction and duration of these forces and whether the bodies are rigid or deform during the interaction etc. It is not easy to analyse using forces.
But with the "force" interpretation of the third law, it is easy to analyze the interaction, at least in the special case (which is most relevant in most applications of Newton's laws) in which all forces are either (1) instantaneous two-body interactions, or (2) contact forces.

As I said, in a three-body problem, there are 6 two-body interaction forces:

$\vec{F_{12}}, \vec{F_{21}}, \vec{F_{13}}, \vec{F_{31}}, \vec{F_{23}}, \vec{F_{32}}$

where $\vec{F_{IJ}}$ means the force of object $I$ on object $J$. If it's a static situation, the sum of the forces on each object is zero. So we conclude:

$\vec{F_{12}} = -\vec{F_{32}}$
$\vec{F_{13}} = -\vec{F_{23}}$
$\vec{F_{21}} = -\vec{F_{31}}$

Then Newton's third law (interpreted in terms of equal and opposite forces) implies
$\vec{F_{12}} = -\vec{F_{21}}$
$\vec{F_{13}} = -\vec{F_{31}}$
$\vec{F_{23}} = -\vec{F_{32}}$

These 6 equations have the solution
$\vec{F_{12}} = \vec{F_{21}} = \vec{F_{13}} = \vec{F_{31}}= \vec{F_{23}} = \vec{F_{32}} = 0$

But I don't see how you can get that conclusion without invoking the force version of the 3rd law.

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