Newton's laws in variable mass systems

[cjl]

Thread split from [thread=424724]this thread[/thread].[/color]

No. The spray can is generating power at a constant rate. There is no extra work.

Not true.

The spray can is generating a constant force, not a constant power. As a result, the work done by the spray can depends on the distance it travels, and the power generated depends on the rate at which it is traveling. The farther the spray can travels, the more work is done.

 

[D H]

The spray can is generating a constant force, not a constant power. As a result, the work done by the spray can depends on the distance it travels, and the power generated depends on the rate at which it is traveling. The farther the spray can travels, the more work is done.

Careful there, cjl! You are mixing reference frames here. The spray can is generating a constant force in the inertial frame instantaneously co-moving with the person holding the spray can. It is not generating a constant force in all inertial frames.

 

[cjl]

Careful there, cjl! You are mixing reference frames here. The spray can is generating a constant force in the inertial frame instantaneously co-moving with the person holding the spray can. It is not generating a constant force in all inertial frames.

Yes it is. In all frames, the massflow from the spray can is the same, and the velocity difference between the spray can and the spray is the same. Therefore, in all frames, the force is the same.

 

[D H]

No, it's not. The force is d/dt(mv), and this a frame-dependent quantity when the mass is not constant (which it most certainly is not in the case of a rocket, or a spray can).

 

[Dickfore]

In non-relativistic mechanics it is not.

 

[D H]

Sure it is. The definition of force is F=dp/dt, not F=ma. In the case of a variable mass object the force acting on the object is a frame-dependent quantity.

 

[cjl]

Sure it is. The definition of force is F=dp/dt, not F=ma. In the case of a variable mass object the force acting on the object is a frame-dependent quantity.

Nope.

If the massflow from the can is constant, and the can's exhaust velocity is constant, then the force is constant. The acceleration will be variable if the can's mass is changing, but the force will not be variable.

(Specifically, using your definition, dp/dt = dm/dt*v. If dm/dt is constant, and the exhaust velocity is constant, then dp/dt is constant. So, if dp/dt is constant, then by your own definition, F is constant).

 

[D H]

(Specifically, using your definition, dp/dt = dm/dt*v. If dm/dt is constant, and the exhaust velocity is constant, then dp/dt is constant. So, if dp/dt is constant, then by your own definition, F is constant).

Wrong again.

Now you are not only mixing frames but you are mixing quantities! Your v here is the velocity of the exhaust relative to the vehicle (or in this case, the velocity of the spray relative to DaveC).

Do again without mixing things up. To avoid mixing things up, use

• m is the (time-varying) mass of the vehicle, including the yet-unburnt fuel.
• v is the velocity of the vehicle from the perspective of some inertial frame.
• p is the momentum of the vehicle from this frame, p=mv.
• u is the velocity of the exhaust relative to the vehicle.
• \dot m is the mass flow rate.

F=dp/dt is the definition of force, and it is very much a frame dependent quantity here.

 

[Dickfore]

Please state the transformation law for the force components from one Galilean frame of reference to another.

 

[cjl]

Wrong again.

Now you are not only mixing frames but you are mixing quantities! Your v here is the velocity of the exhaust relative to the vehicle (or in this case, the velocity of the spray relative to DaveC).

Do again without mixing things up. To avoid mixing things up, use

• m is the (time-varying) mass of the vehicle, including the yet-unburnt fuel.
• v is the velocity of the vehicle from the perspective of some inertial frame.
• p is the momentum of the vehicle from this frame, p=mv.
• u is the velocity of the exhaust relative to the vehicle.
• \dot m is the mass flow rate.

F=dp/dt is the definition of force, and it is very much a frame dependent quantity here.

Oh, I'm not the one mixing things up here.

Vehicle mass is irrelevant, so m vanishes.
Vehicle velocity is irrelevant, so v vanishes
Vehicle momentum is irrelevant, since both terms it is dependent on are irrelevant.
Exhaust velocity matters, so u is the first term out of your list that actually shows up.
Exhaust flow rate matters, so \dot m is the only other item on your list that actually shows up.

This is easily demonstrable by the fact that from any frame, the vehicle's exhaust will have a constant speed relative to the vehicle (at the instant that it is emitted). Since the fuel was originally traveling with the vehicle, this relative velocity u is also equal to the \DeltaV of the fuel. The fuel mass flow rate \dot m is then used to figure out the fuel's \dot p. This is fairly easily done through the equation \dot p = \dot m*\DeltaV. Since \DeltaV is constant (as we defined exhaust velocity to be constant), and \dot m is constant (as we defined massflow to be constant), the exhaust's \dot p will always be the same, as viewed from any frame. Since conservation of momentum holds, the vehicle's \dot p must be equal to -\dot p for the fuel. Since the fuel's \dot p is constant, the vehicle's \dot p is also constant, and therefore F is constant.


(Sorry for some of the strange formula appearances - I'm not that familiar with Latex)

 

[pallidin]

"Vehicle mass is irrelevant"?
I know of no such provision in any reputable contexts of physical laws.

 

[pallidin]

"Vehicle momentum is irrelevant"
Oh, please...

 

[pallidin]

That inertia or momentum is irrelevant is preposterous.
Those aspects exist in all mass-systems. Never been proven otherwise.

 

[D H]

Please state the transformation law for the force components from one Galilean frame of reference to another.

Short answer: Suppose inertial frame B is moving at velocity V wrt inertial frame A. The force (using F=dp/dt) on some object with non-constant mass m in frame B is

F_B = F_A + \dot mV

In short, force is not a galilean invariant.

Long answer: Dealing with variable mass systems are a bit tricky. The choices are

• Use F=dp/dt. This has the obvious advantage of being connected with the conservation laws. It has also the obvious disadvantage of not being a galilean invariant.
• Use F=ma. This is a galilean invariant, but it is no longer connected with the conservation laws. Using F=ma will lead to all kinds of problems. It is not valid for computing work, for example.
• Rewrite Newton's second law as F_{\text{ext}} = dp/dt - u\dot m. For example, see Halliday and Resnick. (Did you sleep through freshman physics?)
  
  There is a slight problem here with work and momentum. This is essentially F=ma again.
• Throw up your hands in disgust and claim that Newton's laws only apply to systems with constant mass. For example, see Plastino & Muzzio, Celestial Mechanics and Dynamical Astronomy, 53:3 (1992) http://articles.adsabs.harvard.edu//full/1992CeMDA..53..227P/0000227.000.html.

Oh, I'm not the one mixing things up here.

Oh yes you are.

Vehicle mass is irrelevant, so m vanishes.
Vehicle velocity is irrelevant, so v vanishes

'Nuff said. Well, almost 'nuff said.

This is easily demonstrable by the fact that from any frame, the vehicle's exhaust will have a constant speed relative to the vehicle (at the instant that it is emitted). Since the fuel was originally traveling with the vehicle, this relative velocity u is also equal to the \DeltaV of the fuel. The fuel mass flow rate \dot m is then used to figure out the fuel's \dot p. This is fairly easily done through the equation \dot p = \dot m*\DeltaV. Since \DeltaV is constant (as we defined exhaust velocity to be constant)

You are mixing frames. The exhaust velocity is only constant with respect to the vehicle, and the vehicle is accelerating. Arguing whether Newton's laws do or do not apply to objects with non-constant mass is, IMHO, being a bit too pedantic. Arguing whether Newton's laws do or do not apply in accelerating frames is a lot less pedantic. They don't -- at least not without the addition of fictitious forces.

 

[Dickfore]

Short answer: Suppose inertial frame B is moving at velocity V wrt inertial frame A. The force (using F=dp/dt) on some object with non-constant mass m in frame B is

F_B = F_A + \dot mV

This is wrong. Your invalid conclusion is a consequence of trying to use the definition F = dp/dt. This equation has F_net on the left side and that it strictly holds for point particles. Mechanics teaches us that 2nd Newton's Law is not a definition of force. Forces are defined independently of it and depend on the nature of the bodies that interact and their relative position and velocity. For example, Newton's Law of universal Gravitation:

<br /> \mathbf{F} = -G \, \frac{m_{1} \, m_{2}}{r^{2}_{1 2}} \, \hat{\mathbf{r}}_{1 2}<br />

 

[D H]

Mechanics teaches us that 2nd Newton's Law is not a definition of force.

Good one! Tell me another. I have deadlines galore and humorous posts such as that do help.

 

[Dickfore]

If you think this is humorous, then your deadlines are probably related to some accounting project.

 

[cjl]

"Vehicle mass is irrelevant"?
I know of no such provision in any reputable contexts of physical laws.

Rocket motors do not change in thrust just because you attach them to a different vehicle. The force they generate is purely a function of their massflow and their exhaust velocity. Hence, vehicle mass and velocity are irrelevant.

(Specifically, F = \dot mV_e for any rocket motor, in which \dot m is the fuel mass flow rate, and V_e is the exhaust velocity)

 

[cjl]

You are mixing frames. The exhaust velocity is only constant with respect to the vehicle, and the vehicle is accelerating. Arguing whether Newton's laws do or do not apply to objects with non-constant mass is, IMHO, being a bit too pedantic. Arguing whether Newton's laws do or do not apply in accelerating frames is a lot less pedantic. They don't -- at least not without the addition of fictitious forces.

Of course if you're in the vehicle frame, you have to add fictitious (inertial) forces, but that's somewhat irrelevant to my point here. The rocket motor (since that's effectively what we're talking about here) is a constant thrust device in any frame. If you ignore drag, and operate in an inertial frame, this means that the net force applied to an object by a rocket motor is constant (assuming the mass of the rocket motor is negligible). In a non-inertial frame, there will be other forces, yes, but the rocket motor is still generating the same force as it always was, even in a non-inertial frame.

 

[D H]

If you think this is humorous, then your deadlines are probably related to some accounting project.

Actually, my deadlines are very closely related to what has become the topic of this thread. This stuff is my job. In other words, "Why, yes, I am a rocket scientist."

The reason your post was humorous is because it is exactly contrary to the modern interpretation of Newton's second law. This modern interpretation is that Newton's second law is not truly a scientific law. Instead, Newton's second law is definitional. In particular, it defines the concepts of momentum and force.

 

[D H]

Specifically, F = \dot mV_e for any rocket motor, in which \dot m is the fuel mass flow rate, and V_e is the exhaust velocity

That is correct only if you toss the definition of force as F=dp/dt and use m·dv/dt=F_ext+u·dm/dt in its place, where u is relative velocity of expelled material. Defining F_reaction≡u·dm/dt let's one simply use F=ma, even for a system with non-constant mass. This form is admittedly very useful as the basis for the equations of motion of a rocket. It is however absolutely useless for computing things like work precisely because it throws out the connection with the conservation laws.

 

[Cleonis]

the modern interpretation of Newton's second law. This modern interpretation is that Newton's second law is not truly a scientific law. Instead, Newton's second law is definitional. In particular, it defines the concepts of momentum and force.

I think the interpretation you mention here may have been modern at some point in time, but it's been overtaken.
(This is far away from the original topic of this thread. DH, if you prefer that this discussion is restarted in a thread of its own, please say so. )


I will start with a general, sweeping statement. Then I will discuss a particular example, claiming that the lesson from that example is universally valid.

All laws of physics have a dual character, in that they define the concepts they use, and make statements about these concepts.
The upshot is that it's never a case of either/or. Newton's laws are definitions and true physical laws.


Now discussion of a historical example:
The definition of the concept of electric resistance is Ohm's law: R = V/I . There is no such thing as first defining the physics of electric resistance, and then proceed to discover Ohm's law R = V/I.

There is no such thing as measuring electric resistance directly; the observables are current strength, I, and electromotive force, V. Also, in the early years of investigation of electrics there were a number of different ways in usage of how to gage electromotive force, and none of them was the same as the modern one. The modern way of gaging electromotive force is designed to be as linear as possible, but in the early days there was no way of knowing which of the methods was linear and which wasn't. However, it was noticed that with some definitions of electromotive force Ohm's law obtained, and with other definitions it didn't (or with much more deviation from the law). This had an influence on how the concept of electromotive force was defined: the scientists came to favor definitions of electromotive force for which Ohm's law obtained. In other words, the law was used to define the concept of electromotive force.

So, is Ohm's law just circular reasoning? No, it isn't. Over time it became increasingly clear that a material's electric resistance can be predicted on the basis of its structural properties alone. The metal contains a large population of electrons that are so free that they can flow through the metal like a fluid or a gas.

Historically, Ohm's law was intuited on the basis of very little evidence, and subsequently Ohm's law influenced the way that the concept of electromotive force was defined. Over time Ohm's law grew from strength to strength, gaining support in ways that were entirely independent from its first conception.

Back to sweeping statements:
This dual character applies for all laws of physics. Each law of physics serves both as law of physics and as operational definition of how the data must be organized so that the law obtains.

No circular reasoning
This does not mean the laws of physics are circular reasoning. The laws of physics are not circular reasoning: that should be obvious to anyone.

 

[Dickfore]

Actually, my deadlines are very closely related to what has become the topic of this thread. This stuff is my job. In other words, "Why, yes, I am a rocket scientist."

The reason your post was humorous is because it is exactly contrary to the modern interpretation of Newton's second law. This modern interpretation is that Newton's second law is not truly a scientific law. Instead, Newton's second law is definitional. In particular, it defines the concepts of momentum and force.

Actually, I don't care about the credentials of someone who posts over the Internet. Please give reference where this "modern definition of Newton's second law" is given.

 

[Cleonis]

Please give reference where this "modern definition of Newton's second law" is given.

I cannot speak for DH, but for later ideas I can give a source that dates to the 1970's. (This material has influenced me enormously.)

In the book 'Gravitation' by Misner, Thorne and Wheeler.
(paragraph 12.3):

Point of principle: how can one write down the laws of gravity and properties of spacetime first (paragraph 12.1) and only afterward (here) come to grip with the nature of the coordinate system and its nonuniqueness? Answer: (a quotation from paragraph 3.2, slightly modified) "Here and elsewhere in science, as emphasized not least by Henri Poincaré, that view is out of date which used to say: 'define your terms before you proceed'. All the laws and theories of physics, including Newton's laws of gravity, have this deep and subtle character that they both define the concepts they use (here Galilean coordinates) and make statements about these concepts"

Contrariwise, the absence of some body of theory, law and principle deprives one of the means properly to define or even use concepts. Any forward step in human knowledge is truly creative in this sense: that theory concept, law, and measurement —forever inseparable—are born into the world in union.

So ideas like this were in circulation in the 70's

 

[Dickfore]

That's all nice, but it is irrelevant to the discussion at hand.

 

[D H]

Please give reference where this "modern definition of Newton's second law" is given.

From Thornton & Marion, Classical Dynamics of Particles and Systems, Fifth Edition, 2004 (uses of emphasis are theirs, not mine):

Therefore, Newton's Secont Law can be expressed as

\mathbf F = \frac{d\mathbf p}{dt} = \frac{d}{dt}(m\mathbf v)

The definition of force becomes complete and precise only when "mass" is defined. Thus the First and Second Laws are not really "laws" in the usual sense; rather they may be considered definitions. Because length, time, and mass are concepts normally already understood, we use Newton's First and Second Laws as the operational definition of force. Newton's Third Law, however, is a law. It is a statement concerning the real physical world and contains all of the physics in Newton's laws of motion.​

2004 modern enough?

 

[Dickfore]

From Thornton & Marion, Classical Dynamics of Particles and Systems, Fifth Edition, 2004 (uses of emphasis are theirs, not mine)

https://www.amazon.com/Classical-Dy..._top_cm_cr_acr_txt?ie=UTF8&showViewpoints=1

Nice reviews. Just goes to show that latest date != modern. After the death of the author Marion, Thornton took over the editing process, and, apparently, had done a terrible job.

In any case, this transformation formula:

F_B = F_A + \dot mV

is incorrect.

 

[D H]

After the death of the author Marion, Thornton took over the editing process, and, apparently, had done a terrible job.

Wrong again! You asked for something recent, I gave you something recent. The very same text, with the exact same emphases, appears in my first edition (1965; yep, it dates me).

 

[Dickfore]

I asked for a reference for the "modern interpretation of Newton's Laws", not recent.

Also, please derive Newton's law of universal gravitation from Newton's second law only.

 

[D H]

In any case, this transformation formula:

F_B = F_A + \dot mV

is incorrect.

O RLY?

Assume frame B is moving at a constant velocity V wrt frame A. Using F=dp/dt=d(mv)/dt as the definition of force (which is the context for which you asked me to provide the transformation), the force in frames A and B on object with time-varying mass is

\begin{aligned}<br /> \mathbf F_A &amp;= \frac{d}{dt}(m\,\mathbf v_A) \\<br /> \mathbf F_B &amp;= \frac{d}{dt}(m\,\mathbf v_B)<br /> \end{aligned}

where v_A and v_B are the velocities of the object as expressed in /observed in frames A and B. Velocity transforms additively:

\mathbf v_B = \mathbf v_A + \mathbf V

With this,

\begin{aligned}<br /> \mathbf F_B &amp;= \frac{d}{dt}(m,\mathbf v_B) = \frac{d}{dt}(m(\mathbf v_A+\mathbf V)) \\<br /> &amp;= \frac{d}{dt}(m\mathbf v_A) + \frac{d}{dt}(m\mathbf V) \\<br /> &amp;= \mathbf F_A + \dot m \mathbf V<br /> \end{aligned}<br />

 

[Dickfore]

O RLY?

Assume frame B is moving at a constant velocity V wrt frame A. Using F=dp/dt=d(mv)/dt as the definition of force (which is the context for which you asked me to provide the transformation), the force in frames A and B on object with time-varying mass is

\begin{aligned}<br /> \mathbf F_A &amp;= \frac{d}{dt}(m\,\mathbf v_A) \\<br /> \mathbf F_B &amp;= \frac{d}{dt}(m\,\mathbf v_B)<br /> \end{aligned}

where v_A and v_B are the velocities of the object as expressed in /observed in frames A and B. Velocity transforms additively:

\mathbf v_B = \mathbf v_A + \mathbf V

With this,

\begin{aligned}<br /> \mathbf F_B &amp;= \frac{d}{dt}(m,\mathbf v_B) = \frac{d}{dt}(m(\mathbf v_A+\mathbf V) \\<br /> &amp;= \frac{d}{dt}(m\mathbf v_A) + \frac{d}{dt}(m\mathbf V) \\<br /> &amp;= \mathbf F_A + \dot m \mathbf V<br /> \end{aligned}<br />

YARLY! Imagine a collection of non-interacting balls traveling all uniformly relative to an inertial reference frame with a velocity V. If you mentally isolate a smaller and smaller subset of them, according to your formula, it seems there is a force acting on this subset. But, by definition, this force can not come from the neighboring points. Where does this force come from then?

 

[D H]

Instead of using logical fallacies, Dick (post #49, red herring; post #51, appeal to ridicule), why don't you try to find the math error in post #50?

 

[Dickfore]

Find the math error in post #50?

There is no math error. There is a physical error:

\begin{aligned}<br /> \mathbf F_A &amp;= \frac{d}{dt}(m\,\mathbf v_A) \\<br /> \mathbf F_B &amp;= \frac{d}{dt}(m\,\mathbf v_B)<br /> \end{aligned}

together with

\begin{aligned}<br /> &amp;= \dot m <br /> \end{aligned}<br />

 

[Cleonis]

From [...] Marion, Classical Dynamics of Particles and Systems, [...] :

[...]
The definition of force becomes complete and precise only when "mass" is defined. Thus the First and Second Laws are not really "laws" in the usual sense; rather they may be considered definitions. Because length, time, and mass are concepts normally already understood, we use Newton's First and Second Laws as the operational definition of force. Newton's Third Law, however, is a law. It is a statement concerning the real physical world and contains all of the physics in Newton's laws of motion.​

I question Jerry B. Marion's philosophy here.

An example: it is often argued - and validly, in my opinion - that the third Law implies that centrifugal force isn't a force; centrifugal force is not part of a Newtonian force-pair, as defined by the third Law.

The third Law, like all other physical laws, serves as operational definition of the concepts it makes statements about.

It appears that Marion's underlying premise is one of dichotomy: that physics statements are either a definition, or a physical law. That whole dichotomy-attitude needs to be dropped. Physics laws are both: operational definitions and laws.

 

[D H]

There is no math error. There is a physical error:

Another good one! You're just chock full of humor today, aren't you?

You can find tons of physics texts that define F as F=dp/dt, starting with Newton. That said, you can find others that define F as F=ma. You can also find others that claim that Newton's laws, strictly speaking, apply only to particles, so it doesn't matter which definition you use.

The advantage of using F=dp/dt lies in the connection to the conservation laws. The disadvantage is the force acting on a mass-varying object is frame-dependent with this definition. The advantage of using F=ma is that whether mass is constant or varying, this definition makes force frame independent. The disadvantage is the lost connection with the conservation laws. Using F=ma in conjunction with work is a bad idea for variable mass systems. It will lead to incorrect results.

 

[Dickfore]

You can also find others that claim that Newton's laws, strictly speaking, apply only to particles, so it doesn't matter which definition you use.

This is the point. Please derive

<br /> \Sigma\mathbf{F}_{ex} = \frac{d \mathbf{P}}{d t}<br />

for an (open) system with variable mass.

 

[D H]

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

\begin{aligned}<br /> P_{A(t)}(t) &amp;= \sum_{i\in A(t)} p_i(t) \\<br /> P_{\Delta B(t)}(t) &amp;= \sum_{j\in\Delta B(t)} p_j(t)<br /> \end{aligned}

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,

\begin{aligned}<br /> P_{A(t)}(t+\Delta t) &amp;\approx<br /> \sum_{i\in A(t)} \left(<br /> p_i(t) +<br /> \sum_{j\in A(t), j\ne i} F_ij(t) \Delta t +<br /> \sum_{j\in \Delta B(t)} F_ij(t) \Delta t +<br /> \sum_{j\in B(t)} F_ij \Delta t\right) \\<br /> P_{\Delta B(t)}(t+\Delta t) &amp;\approx<br /> \sum_{j\in \Delta B(t)} \left(<br /> p_j(t) +<br /> \sum_{i\in A(t)} F_ji(t) \Delta t +<br /> \sum_{i\in \Delta B(t), i\ne j} F_ji(t) \Delta t +<br /> \sum_{i\in B(t)} F_ji(t) \Delta t\right) \\<br /> \end{aligned}

By Newton's third law, each of the following sums will vanish:

\begin{aligned}<br /> &amp;\sum_{i\in A(t)}\sum_{j\in A(t), j\ne i} F_ij(t) \\<br /> &amp;\sum_{j\in \Delta B(t)}\sum_{i\in \Delta B(t), i\ne j} F_ji(t) \\<br /> &amp;\sum_{i\in A(t)}\sum_{j\in \Delta B(t)} F_ij(t) +<br /> \sum_{j\in \Delta B(t)}\sum_{i\in A(t)} F_ji(t)<br /> \end{aligned}

Using this, and summing to form the total momentum of particles A at time t+Δt yields

\begin{aligned}<br /> P_{A(t+\delta t)}(t+\Delta t) &amp;= P_{A(t)}(t+\Delta t) + P_{\Delta B(t)}(t+\Delta t) \\<br /> &amp;\approx<br /> \sum_{i\in A(t)} \left(<br /> p_i(t) +<br /> \sum_{j\in B(t)} F_ij(t) \Delta t\right) +<br /> \sum_{j\in \Delta B(t)} \left(<br /> p_j(t) +<br /> \sum_{i\in B(t)} F_ji(t) \Delta t\right)<br /> \end{aligned}

The final term, \sum_{j\in \Delta B(t)}\sum_{i\in B(t)} F_ji(t) \Delta t, will be second order assuming that the set \Delta B(t)\to\Phi\,\text{as}\,\Delta t\to 0.

Define

\begin{aligned}<br /> F_{\text{ext}}(t) &amp;\equiv \sum_{i\in A(t)} \sum_{j\in B(t)} F_ij(t) \\<br /> \Delta m_{\Delta B(t)}(t) &amp;\equiv \sum_{j\in \Delta B(t)} m_j \\<br /> v_e(t) &amp;\equiv \frac 1 {m(t)} \sum_{j\in \Delta B(t)} p_j(t)<br /> \end{aligned}

Note that by conservation of mass,

\Delta m_A(t) = \Delta m_{\Delta B(t)}(t)

With this the total momentum of particles A at time t+Δt becomes

<br /> P_{A(t+\delta t)}(t+\Delta t) \approx<br /> \sum_{i\in A(t)} p_i(t) + F_{\text{ext}}(t) \Delta t + \Delta m_{A(t)}(t) v_e(t)<br />

Subtracting the momentum of particles A at time t, dividing by Δt and taking the limit Δt→0 yields

<br /> \frac{d}{dt}P_{A(t)}(t) = F_{\text{ext}}(t) + \frac{dm_A(t)}{dt}v_e(t)<br />

 

[Dickfore]

<br /> \frac{d}{dt}P_{A(t)}(t) = F_{\text{ext}}(t) + \frac{dm_A(t)}{dt}v_e(t)<br />

So, you couldn't derive it. Nice.

 

[D H]

Please do stop with the fallacious arguments.

 

[D H]

So, you couldn't derive it. Nice.

Of course not. Your equation is wrong in the context of F=dp/dt.

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.

 

[Dickfore]

Of course not. Your equation is wrong in the context of F=dp/dt.

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.

How can I derive a wrong equation?

 

[D H]

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.

 

[Dickfore]

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.

 

[D H]

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.

 

[Dickfore]

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.

 

[afallingbomb]

This may be of use here:

http://articles.adsabs.harvard.edu//full/1992CeMDA..53..227P/0000227.000.html

 

[arildno]

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:
https://www.physicsforums.com/showthread.php?t=72176

 

[D H]

This may be of use here:

http://articles.adsabs.harvard.edu//full/1992CeMDA..53..227P/0000227.000.html

I already posted a link to the same article in [post= 2858679]post #14[/post].

There are three camps in classical physics regarding Newton's laws,

1. F=dp/dt=d(mv)/dt. With this definition of force, that F=m·dv/dt+v·dm/dt is a straightforward application of the chain rule. Dick: Does the chain rule not apply in physics?
2. F=ma, even for variable mass systems. The changes in state that result from mass variations are just another external force. With this definition the equations of motion fall straight out of Newton's 2nd law. An unfortunate side effect of this definition is the loss of connection to the conservation laws. Using this definition of force to compute work is invalid.
3. F=dp/dt, F=ma: Potato/patato. Newton's laws only apply to point masses with constant mass (e.g., Plastino & Muzzio). To this camp, there is no such thing as a variable mass system in the context of 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.

 

[Dickfore]

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

 

[arildno]

F=ma, even for variable mass systems. The changes in state that result from mass variations are just another external force. With this definition the equations of motion fall straight out of Newton's 2nd law. An unfortunate side effect of this definition is the loss of connection to the conservation laws. Using this definition of force to compute work is invalid.

This is a meaningless position.

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.