
#37
Aug3110, 07:37 AM

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Your point is fallacious. Specifically, it is your qualification "ex" (meaning external) on the force is that is fallacious.
So, without fallacies: There is some larger, closed system of constant mass particles in which our somewhat arbitrary open system lives. Call the particles in our open system set A and the other particles set B. Whether a given particle is in set A or in set B varies with time. At some time t denote A(t) as the set of particles that are in set A at time t, B(t) as the set of particles in set B at time t. I'll split set B into two disjoint subsets, sets B_{}(t) and ΔB(t) such that B(t) = B_{}(t) ∪ ΔB(t). The reason for this partition of B(t) is that the subset ΔB(t) is about to join A(t): A(t+Δt) = A(t) ∪ ΔB(t) and B(t+Δt) = B_{}(t). The total momenta of the particles in sets A(t) and ΔB(t) are [tex]\begin{aligned} P_{A(t)}(t) &= \sum_{i\in A(t)} p_i(t) \\ P_{\Delta B(t)}(t) &= \sum_{j\in\Delta B(t)} p_j(t) \end{aligned}[/tex] Assuming our Δt is small, the momentum of particle i at times t and t+Δt are approximately related by p_{i}(t+Δt) = p_{i}(t) + F_{i}(t)Δt, where F_{i}(t) is the net force acting on particle i at time t. Expanding that net force into the contributions from each particle, [tex]\begin{aligned} P_{A(t)}(t+\Delta t) &\approx \sum_{i\in A(t)} \left( p_i(t) + \sum_{j\in A(t), j\ne i} F_ij(t) \Delta t + \sum_{j\in \Delta B(t)} F_ij(t) \Delta t + \sum_{j\in B(t)} F_ij \Delta t\right) \\ P_{\Delta B(t)}(t+\Delta t) &\approx \sum_{j\in \Delta B(t)} \left( p_j(t) + \sum_{i\in A(t)} F_ji(t) \Delta t + \sum_{i\in \Delta B(t), i\ne j} F_ji(t) \Delta t + \sum_{i\in B(t)} F_ji(t) \Delta t\right) \\ \end{aligned}[/tex] By Newton's third law, each of the following sums will vanish: [tex]\begin{aligned} &\sum_{i\in A(t)}\sum_{j\in A(t), j\ne i} F_ij(t) \\ &\sum_{j\in \Delta B(t)}\sum_{i\in \Delta B(t), i\ne j} F_ji(t) \\ &\sum_{i\in A(t)}\sum_{j\in \Delta B(t)} F_ij(t) + \sum_{j\in \Delta B(t)}\sum_{i\in A(t)} F_ji(t) \end{aligned}[/tex] Using this, and summing to form the total momentum of particles A at time t+Δt yields [tex]\begin{aligned} P_{A(t+\delta t)}(t+\Delta t) &= P_{A(t)}(t+\Delta t) + P_{\Delta B(t)}(t+\Delta t) \\ &\approx \sum_{i\in A(t)} \left( p_i(t) + \sum_{j\in B(t)} F_ij(t) \Delta t\right) + \sum_{j\in \Delta B(t)} \left( p_j(t) + \sum_{i\in B(t)} F_ji(t) \Delta t\right) \end{aligned}[/tex] The final term, [itex]\sum_{j\in \Delta B(t)}\sum_{i\in B(t)} F_ji(t) \Delta t[/itex], will be second order assuming that the set [itex]\Delta B(t)\to\Phi\,\text{as}\,\Delta t\to 0[/itex]. Define [tex]\begin{aligned} F_{\text{ext}}(t) &\equiv \sum_{i\in A(t)} \sum_{j\in B(t)} F_ij(t) \\ \Delta m_{\Delta B(t)}(t) &\equiv \sum_{j\in \Delta B(t)} m_j \\ v_e(t) &\equiv \frac 1 {m(t)} \sum_{j\in \Delta B(t)} p_j(t) \end{aligned}[/tex] Note that by conservation of mass, [tex]\Delta m_A(t) = \Delta m_{\Delta B(t)}(t)[/tex] With this the total momentum of particles A at time t+Δt becomes [tex] P_{A(t+\delta t)}(t+\Delta t) \approx \sum_{i\in A(t)} p_i(t) + F_{\text{ext}}(t) \Delta t + \Delta m_{A(t)}(t) v_e(t) [/tex] Subtracting the momentum of particles A at time t, dividing by Δt and taking the limit Δt→0 yields [tex] \frac{d}{dt}P_{A(t)}(t) = F_{\text{ext}}(t) + \frac{dm_A(t)}{dt}v_e(t) [/tex] 



#38
Aug3110, 07:58 AM

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#39
Aug3110, 08:16 AM

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Please do stop with the fallacious arguments.




#40
Aug3110, 08:29 AM

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Why don't you derive your equation? State your assumptions and your definitions. Then use your equation to derive (a) the work performed on a variable mass system and (b) the time derivative (power) of the variable mass system's kinetic energy. 



#41
Aug3110, 08:41 AM

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#42
Aug3110, 08:47 AM

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I don't know. You tell me. I've done math while all you have done is to use fallacious arguments. The equation in post #36 is yours, not mine. Define your terms and derive that result. Then use that result to calculate (a) the work performed on the variable mass system and (b) the time derivative of the system's kinetic energy.




#43
Aug3110, 08:56 AM

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It seems we have misunderstood each other. The equation I posted in #36 was in repsonse to your post #35. The point I tried to make is that that equation is incorrect for open systems, which you yourself demonstrated with the derivation in step #37.
In conclusion, you were wrong in post #14 where you stated a force transformation law. 



#44
Aug3110, 09:18 AM

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There is no error in post #14. I derived it in post #50. I challenged you to find the flaw and all you could do was use fallacious reasoning. I haven't the foggiest idea what you are going on about now. Please elaborate.
And use some math instead of fallacious arguments this time. 



#45
Aug3110, 09:27 AM

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lol, what ever you say. I don't want to quarrel and get a ban. If someone reads this thread and knows Physics, they will draw a conclusion for themselves.




#46
Aug3110, 09:33 AM

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#47
Aug3110, 09:44 AM

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The distinction between a geometric system, and a material system is crucial to understand when dealing with Newton's second law, and what forces act upon.
A geometric system is simply a mathematically defined region within space that we designate as our system, and particles/mass may well flow into, and out of that region. Forces do NOT act upon a mere spatial region, they act upon material particles CONTAINED within that region. Thus, F=dp/dt is perfectly valid, as long as we are talking about a material system, i.e, which is defined by consisting of the same particles throughout time (and in the Newtonian world, thus have fixed mass). We may perfectly well formulate a Newton's 2.law for geometric systems, and it is highly useful, for example by calculating the reaction force on a tube section through which the fluid passes. The rate of change of momentum within a geometric system consists not only of the effects of forces acting upon (momentarily) contained particles, but also that less momentum may flow into the region than leaves it, or vice versa (momentum itself being carried by massed particles). The following thread goes into the details: http://www.physicsforums.com/showthread.php?t=72176 



#48
Aug3110, 09:46 AM

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There are three camps in classical physics regarding Newton's laws,
The last camp has withdrawn from the argument. The first two camps will arise at the same results if they are careful about their math. 



#49
Aug3110, 09:49 AM

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Question: What form of Newton's Second Law is used in Hydrodynamics?




#50
Aug3110, 09:54 AM

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Let us have an object O moving at constant velocity U; no part of it is subject to any force whatsoever. I can perfectly well construct a system containing solely of parts of this object, yet that system will experience a constant increase in momentum, with nothing resembling of a force is acting upon any part of object O. 



#51
Aug3110, 09:55 AM

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http://en.wikipedia.org/wiki/Cauchy_momentum_equation 



#52
Aug3110, 10:00 AM

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Sure the chain rule work, as long as you use it properly. Leibniz' rule of differentiation of an integral with moving boundaries is a fancy version of the chain rule, and that is the one to use here. 



#53
Aug3110, 10:23 AM

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#54
Aug3110, 10:24 AM

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