Meaning of the terms in the formula of the net external force

[brochesspro]

TL;DR

My question inquires about the meaning of each of the terms present in the formula for the net external force on a system and uses rocket propulsion as an example to further clarify the concept.

The mathematical representation of the net external force on a system(obtained from Newton's second law) is $\vec F_{net} = \frac {d\vec P}{dt}$, which is the rate of change of linear momentum of the system. If we substitute $\vec P = m\vec v$ into the formula for force and differentiate, we get:$$\vec F_{net} = m\frac{d\vec v}{dt} + \frac {dm}{dt}\vec v.$$
What do the terms $m$, $\frac{d\vec v}{dt}$, $\frac {dm}{dt}$, $\vec v$ mean exactly? Could you use rocket propulsion as an example to explain the meaning of these terms? I have an idea of what each of these could be in this case, please check whether they are right or wrong. Here, the system I am choosing is the total mass of the rocket and the ejected gas and the term "rocket" refers to the rocket and the gas present in the rocket.
$m$ is the total mass of the rocket at any given instant.
$\frac{d\vec v}{dt}$ is the rocket's acceleration at a given instant in time.
$\frac {dm}{dt}$ is the rate at which the gas is ejected out of the rocket to propel it.
$\vec v$ is the rocket's velocity at any given instant(I am especially not sure of this one, but the same goes for the others.).
Thank you for answering.

 

[Lnewqban]

Please, see:
https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

 

[jbriggs444]

The mathematical representation of the net external force on a system(obtained from Newton's second law) is $\vec F_{net} = \frac {d\vec P}{dt}$, which is the rate of change of linear momentum of the system. If we substitute $\vec P = m\vec v$ into the formula for force and differentiate, we get:$$\vec F_{net} = m\frac{d\vec v}{dt} + \frac {dm}{dt}\vec v.$$

Suppose that we have a railroad hopper car containing 50,000 kg of sand. It coasts down the tracks at 10 meters per second. As it moves, sand pours from the bottom at a rate of 100 kg per second.

Does the loss of sand from the system count as a "force"?

 

[brochesspro]

Suppose that we have a railroad hopper car containing 50,000 kg of sand. It coasts down the tracks at 10 meters per second. As it moves, sand pours from the bottom at a rate of 100 kg per second.

Does the loss of sand from the system count as a "force"?

It does not, as the sand will retain its momentum from when it was in the railroad hopper, and the speed of the hopper will remain constant. But what does that have to do with anything?
And also, I wanna change the question a bit, the system only consists of the rocket and not the ejected gas, thanks.

 

[jbriggs444]

It does not, as the sand will retain its momentum from when it was in the railroad hopper, and the speed of the hopper will remain constant. But what does that have to do with anything?

The momentum of the rairoad car plus contents is decreasing. Surely that means that it is subject to an external net force. Newton's second law which you have invoked says so. Is that not exactly the situation that you want to explore?

Let us not contemplate a high speed exhaust stream until you understand how to deal with a zero speed exhaust stream.

 

[brochesspro]

The momentum of the rairoad car plus contents is decreasing. Surely that means that it is subject to an external net force. Newton's second law which you have invoked says so. Is that not exactly the situation that you want to explore?

Let us not contemplate a high speed exhaust stream until you understand how to deal with a zero speed exhaust stream.

I think that I see what you mean. The ejecta must have some velocity relative to the hopper for there to be a net force on the hopper, am I right? Can we understand this using the second law of motion? The net external force is zero, and the mass of the railroad car at a time $t$ could be taken as $m = m_0 -rt$, where $r$ is the rate at which the sand falls in terms of mass per unit time(I take $r$ to be positive here.). $\frac {dm}{dt}$ is equal to $-r$, since there is a decrease in mass. Now, if the $v$ term were zero, everything would be fine, it would make sense if $v$ was the speed of the sand with respect to the hopper or the speed of the hopper with respect to the sand. Do you think this makes sense?

 

[A.T.]

It does not, as the sand will retain its momentum from when it was in the railroad hopper...

That's exactly why the loss of sand constitutes a transfer of momentum out of the system. And "force", in a more general sense, can be seen as the rate of momentum transfer.

 

[brochesspro]

That's exactly why the loss of sand constitutes a transfer of momentum out of the system. And "force", in a more general sense, can be seen as the rate of momentum transfer.

I do not think I understand, could you please elaborate?

 

[jbriggs444]

I do not think I understand, could you please elaborate?

If we write Newton's second law as $F=\frac{\mathrm{d} p}{\mathrm{d t}}$ then a "force" is a rate of momentum change with respect to time.

Momentum conservation assures us that any time there is a change of momentum in one system, there must be an equal and opposite change in some other system. So a force can be regarded as not just a rate of change in momentum but a rate of transfer of momentum.

Usually we consider "forces" that act on immutable objects. But we can widen our viewpoint and consider "forces" that act on abstract systems instead. With an abstract system instead of an immutable object, there is the possibility of a mass flow in or out of the system. [Mass is conserved, so it will be a mass transfer].

If we allow ourselves to contemplate situations where there is a mass flow across an interface between systems and if momentum is transferred along with the mass, then this can be regarded as a "force" across that interface.

The rate of momentum transfer due to a mass flow is given by $F=\vec{v}\ \mathrm{d}m$ where $\vec{v}$ is the average velocity of the arriving (or departing in the case of negative $\mathrm{d}m$) mass.

 

[brochesspro]

So a force can be regarded as not just a rate of change in momentum but a rate of transfer of momentum.

I do not understand how this statement can be inferred from your previous statement.

If we allow ourselves to contemplate situations where there is a mass flow across an interface between systems and if momentum is transferred along with the mass, then this can be regarded as a "force" across that interface.

The rate of momentum transfer due to a mass flow is given by F=v→ dm where v→ is the average velocity of the arriving (or departing in the case of negative dm) mass.

This part is too complicated for me to understand, but let me see what I can infer from here. Did you not just give an example when momentum was imparted but it was still not a force? And why is v the average velocity? Is it not the relative velocity in this case? And it is still unclear to me why I could not use the equation $F_{net} = m\dot v + v\dot m$.

 

[jbriggs444]

I do not understand how this statement can be inferred from your previous statement.

Without conservation of momentum, a force could simply be a change in momentum. With conservation of momentum, it is not just a change. It has to be a transfer.

 

[brochesspro]

It is the fallacy of equivocation. You use v twice in the same formula. One (the v in v˙) is the velocity of the object and the other (the plain old v) should be the velocity of the ejected mass flow. Two different velocities. Should be two different variables.

But do we not get this from $F_{net} = \dot P$? What is the mistake in the derivation, if any? And by the way, I noticed that a part of your reply disappeared when I changed my post a bit, I did not know that was a thing, can I undo it?

Without conservation of momentum, a force could simply be a change in momentum. With conservation of momentum, it is not just a change. It has to be a transfer.

I still do not understand, is there an article or something that explains it?

 

[A.T.]

I do not think I understand, could you please elaborate?

"Force" can mean different things in physics. It can mean a force, that can be measured using stain gauges or accelerometers. Or it can be any term in the equations, that has the same dimension as force.

Here the change of momentum is due to definition of the control volume. Another example are inertial forces due to the definition of the reference frame. Both are mere artifacts of the chosen analysis method.

 

[brochesspro]

"Force" can mean different things in physics. It can mean a force, that can be measured using stain gauges or accelerometers. Or it can be any term in the equations, that has the same dimension as force.

Here the change of momentum is due to definition of the control volume. Another example are inertial forces due to the definition of the reference frame. Both are mere artifacts of the chosen analysis method.

I understand, I guess.

 

[PeroK]

And it is still unclear to me why I could not use the equation $F_{net} = m\dot v + v\dot m$.

This equation is simply wrong. Mass cannot be created or destroyed. The only way that the mass of a system can change is by changing the definition of the system (adding or subtracting particles). That equation is incomplete, as it does not take account of the momentum of the incoming or outgoing particles.

There are, however, numerous sources online that give that equation as a generalisation of Newton's second law. That is wrong. Newton's second law applies to a well-defined system - in the simplest terms a particle of fixed mass.

 

[PeroK]

I understand, I guess.

Wikipedia gets it right:

https://en.wikipedia.org/wiki/Variable-mass_system

 

[PeroK]

Whereas, this site gets it wrong:

https://www.nagwa.com/en/explainers/163185454392/

There are lots of those all the Internet.

 

[brochesspro]

Hey, sorry for the late reply. I think I am a bit biased towards the "wrong" explanation of the second law since I have been using that all this time. I think I will take some time to see for myself what is what. Thank you for all your help.