Can Newton's Laws of Motion be simplified?


by Lamarr
Tags: laws, motion, newton, simplified
aaaa202
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Nov20-12, 05:33 PM
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Andrew Mason: I don't like your derivation of the third law from Galilean relativity. Suppose that we have a system of three particles A,B,C. Then the condition that the momentum is conserved in the system is as you say:
(Fab+Fac+Fba+Fbc+Fca+Fcb)*dt=0
But I don't see how setting Fxy = -Fyx is the only solution to that equation. How about for instance: Fac = -Fab, Fba=-Fbc,Fca=-Fcb? why couldn't I assume that. It would be a wicked world yes, but I don't see a problem with it.
Andrew Mason
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Nov20-12, 10:22 PM
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Quote Quote by aaaa202 View Post
Andrew Mason: I don't like your derivation of the third law from Galilean relativity. Suppose that we have a system of three particles A,B,C. Then the condition that the momentum is conserved in the system is as you say:
(Fab+Fac+Fba+Fbc+Fca+Fcb)*dt=0
But I don't see how setting Fxy = -Fyx is the only solution to that equation. How about for instance: Fac = -Fab, Fba=-Fbc,Fca=-Fcb? why couldn't I assume that. It would be a wicked world yes, but I don't see a problem with it.
The interaction times are not the same for all the interactions.

If, during an interval Δt, a, b and c interact with each other, the times of interactions between a and b, between a and c and between b and c will almost always differ. You have to analyse each interaction separately. When you do that, the following will always hold true in Galilean Relativity:

(Fab + Fba)Δtba + (Fac + Fca)Δtac + (Fbc + Fcb)Δtbc = 0

In order for this relationship to apply at ALL times during the interaction, the force pairs must be equal in magnitude and opposite in direction (ie. each force pair sums to 0).

AM


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