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What exactly is the reactive centrifugal force (split) 
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#37
Jan3113, 06:16 PM

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[itex]\vec{F}_{12} + \vec{F}_{32} + \vec{F}_{13} + \vec{F}_{23} + \vec{F}_{21} + \vec{F}_{31} = 0[/itex] You get that from the first law. If [itex]\vec{F}_{12} + \vec{F}_{32} = 0[/itex] and if 2 is not accelerating then either [itex]\vec{F}_{12} = \vec{F}_{32} = 0[/itex] or there are other forces acting on 2. In that case you would have to write: [itex]\vec{F}_{12} + \vec{F}_{32} + \vec{F}_{13} + \vec{F}_{23} + \vec{F}_{21} + \vec{F}_{31} + \sum F_{i(ext)} = 0[/itex], 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. AM 


#38
Jan3113, 06:33 PM

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AM 


#39
Jan3113, 06:38 PM

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#40
Jan3113, 06:45 PM

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AM 


#41
Jan3113, 09:14 PM

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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 [itex]F_{12}=F_{21}[/itex] but your version would be something like [itex]dp_1/dt =  dp_2/dt[/itex]. 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...sThirdLaw%2F 


#42
Jan3113, 09:20 PM

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#43
Feb113, 01:18 AM

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


#44
Feb113, 01:33 AM

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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. 


#45
Feb113, 03:18 AM

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#46
Feb113, 06:14 AM

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"With respect to the astronaut spacestation 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? AM 


#47
Feb113, 06:40 AM

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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 


#48
Feb113, 06:40 AM

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


#49
Feb113, 06:42 AM

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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. 


#50
Feb113, 07:00 AM

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As I said, in a threebody problem, there are 6 twobody interaction forces: [itex]\vec{F_{12}}, \vec{F_{21}}, \vec{F_{13}}, \vec{F_{31}}, \vec{F_{23}}, \vec{F_{32}}[/itex] where [itex]\vec{F_{IJ}}[/itex] means the force of object [itex]I[/itex] on object [itex]J[/itex]. If it's a static situation, the sum of the forces on each object is zero. So we conclude: [itex]\vec{F_{12}} = \vec{F_{32}}[/itex] [itex]\vec{F_{13}} = \vec{F_{23}}[/itex] [itex]\vec{F_{21}} = \vec{F_{31}}[/itex] Then Newton's third law (interpreted in terms of equal and opposite forces) implies [itex]\vec{F_{12}} = \vec{F_{21}}[/itex] [itex]\vec{F_{13}} = \vec{F_{31}}[/itex] [itex]\vec{F_{23}} = \vec{F_{32}}[/itex] These 6 equations have the solution [itex]\vec{F_{12}} = \vec{F_{21}} = \vec{F_{13}} = \vec{F_{31}}= \vec{F_{23}} = \vec{F_{32}} = 0[/itex] But I don't see how you can get that conclusion without invoking the force version of the 3rd law. 


#51
Feb113, 07:03 AM

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Every atom in that space station/2 astronaut example is accelerating. There are a gazillion tension forces. The total force on each atom is the atom's mass multiplied by its centripetal acceleration. AM 


#52
Feb113, 07:10 AM

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Please find a clear reference. Now the mathematical expression you are looking for is [itex]dp_i/dt=\Sigma dp_{j \ne i}/dt[/itex]. You need to find somewhere that states not just that this expression is true, but that this is Newton's 3rd law. 


#53
Feb113, 07:29 AM

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#54
Feb113, 09:04 AM

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AM 


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