Is there any derivation of Newton's second law?

[Philip Wood]

considering the transfer of Δm into the system

As I said above, my system is the accreting body and the stuff it accretes in time delta t. So mass delta m isn't transferred into the system, but is transferred within the system. The overall mass of the system remains constant at (m + delta m).

Δp=p(t+Δt)−p(t)=(m+Δm)⋅(v+Δv)−m⋅v=F⋅Δt

In the last but one 'side' of your equation, it's not just mv that you should be subtracting but also the initial momentum of Delta m, which (see post 15) is (Delta m)w. That way we get the change in momentum of the system.

If we put F = 0 into your equation as quoted then we find that delta v is independent of w. This can't be right, as, with F = 0, we are dealing with a simple case of a two body collision. Note, too, that in terms of orders of magnitude (delta m) w is no more negligible than (delta m)v.

 

[DrStupid]

As I said above, my system is the accreting body and the stuff it accretes in time delta t. So mass delta m isn't transferred into the system, but is transferred within the system. The overall mass of the system remains constant at (m + delta m).

Than we do not talk about the force acting on a system with variable mass.

If we put F = 0 into your equation as quoted then we find that delta v is independent of w.

That's correct.

F_A = m_A \cdot \dot v_A + v_A \cdot \dot m_A = 0

results in

\dot v_A = - \frac{{v_A \cdot \dot m_A }}{{m_A }}

This can't be right, as, with F = 0, we are dealing with a simple case of a two body collision.

It's a special case of a two body collision.

 

[Philip Wood]

Than we do not talk about the force acting on a system with variable mass.

No, but I do talk about the force on a BODY of varying mass. My point is that in Galilean/Newtonian Physics a body can't acquire mass without the mass coming from somewhere outside the original body. And as the body acquires mass it also acquires momentum, dependent on the velocity of the acquired mass before it was acquired. You haven't (yet) convinced me that the equation I gave originally is wrong.

 

[DrStupid]

[

No, but I do talk about the force on a BODY of varying mass.

I still don't get it. What systems and what forced/interactions between which of the systems are you talking about exactly? I counted four systems

A: the body of initial mass m
B: the additional mass Δm
C: the common system of the body and the additional mass
D: an external system exerting a force the body

and two interactions

a: the collision of A and B
b: the force F between A and D

Is this correct?

If yes, than force F is not related to any kind of mass transfer. In that case your scenario is not suitable for the discussion of forces between open systems.

If not, please give a more detailed explanation of your scenario.

My point is that in Galilean/Newtonian Physics a body can't acquire mass without the mass coming from somewhere outside the original body. And as the body acquires mass it also acquires momentum, dependent on the velocity of the acquired mass before it was acquired.

That doesn't justify a modification of the second law.

You haven't (yet) convinced me that the equation I gave originally is wrong.

As it is your equation, you have to convince me that it is correct.

However, I can show you, that your force violates the second and the third law. The violation of the second law is obvious. Your force differs by the additional term $-w \cdot \dot m$ from the change of momentum. In order to show the violation of the second law, I repeat my calculation from #24 with your force. The forces between the interacting open systems are

F_A = m_A \cdot \dot v_A + \left( {v_A - v_B } \right) \cdot \dot m_A
F_B = m_B \cdot \dot v_B + \left( {v_B - v_A } \right) \cdot \dot m_B

with the third law and conservation of mass (you already agreed that the mass can change by mass transfer between the systems only) this results in

m_A \cdot \dot v_A = 2 \cdot \left( {v_B - v_A } \right) \cdot \dot m_A - m_B \cdot \dot v_B

This is obviously different from my result in #24. Thus at least one concept of force must be wrong. In order to find the correct result independent from the definition of force, I again repeat the calculation with momentum only. The changes of momentum are

\dot p_A = m_A \cdot \dot v_A + v_A \cdot \dot m_A
\dot p_B = m_B \cdot \dot v_B + v_B \cdot \dot m_B

with conservation of momentum

\dot p_A + \dot p_B = 0

and conservation of mass this results in.

m_A \cdot \dot v_A = \left( {v_B - v_A } \right) \cdot \dot m_A - m_B \cdot \dot v_B

That mean there is something wrong with your force.

 

[Philip Wood]

I counted four systems

A: the body of initial mass m
B: the additional mass Δm
C: the common system of the body and the additional mass
D: an external system exerting a force the body

and two interactions

a: the collision of A and B
b: the force F between A and D

Is this correct?

If yes, than force F is not related to any kind of mass transfer. In that case your scenario is not suitable for the discussion of forces between open systems.

I agree with your summary (A, B, C, D, a, b). I agree, too, that the force F_AD is not related to any kind of mass transfer, in that I'm trying to deal with cases where the force and the mass-accretion are independent, in the sense that we could conceive of either being turned off independently of the other e.g mass-accreting body experiencing a force (say from an electric field) unrelated to the mass accretion.

I repeat my calculation from #24 with your force. The forces between the interacting open systems are

FA=mA⋅v˙A+(vA−vB)⋅m˙A

But my force wasn't the force of B on A, but was a force from outside (D in your notation).

At this point I have to leave this discussion for a few hours. I have to say that I am finding it very interesting indeed, and I must thank you for engaging in it. The difference of opinion clearly centres on the treatment of open systems.

May I please check that we're in agreement on the following two things…

(1) For a body of mass m moving at velocity v accreting mass at a rate dm/dt from a 'mist' of particles with (pre-capture) velocity w, then if no external force (FAD) acts, from momentum conservation we have

(m dv/dt + v dm/dt) = w dm/dt

(2) the force on an open system is equal to the limit as delta t approaches zero of (final p – initial p)/delta t.

 

[Jano L.]

No- Newton's second law is a definition.

... Newton's 2nd law is an empirical relationship that Newton formulated based on his laboratory experiments. The mass m was the proportionality constant between force and acceleration.

Chet

Since in science and mathematics being a definition excludes being a law, obviously there are at least two viewpoints that do not agree on what Newton's 2nd law is. Fortunately, this does not seem to lead to disagreements on the results of calculations. Still, it may be symptomatic of serious problems in understanding of physics.

I believe the misunderstanding is on the definitionists part.

Newton's II. law as he wrote it (translated to English):

The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.

This is a statement about what happens in a specific situation (one force, one body of constant mass); it clearly is not a definition (notice the word force occurs twice in the statement). There are other formulations of mechanics than Newton's, some of which are very different, and this may be the reason why people continue suggesting that Newton's laws are definitions, but: these other formulations are not Newton's, so even if those formulations are definitions, there is no justification to call his 2nd law a definition.

It is true that the equation
$\mathbf F=m\mathbf a$
is sometimes a good way to find out the vector of total experienced force. For example, the total force experienced by a car or artificial satellite can be found in practice only via this formula; the motion is measured with radar and the value of the force is derived.

However, above formula is generally not sufficient to find forces. For example, when normal force acting on a leaning ladder at rest is sought, we need other equations to find it (gravity, friction), or measure it with a scale (based on deformation). The normal force is obviously not found directly from the equation $\mathbf F = m\mathbf a$.

More generally, forces in statics are not sought through $\mathbf F = m\mathbf a$. It would not be possible, because all accelerations are zero but the partial forces do not generally vanish.

Moreover, all this was merely a value-finding, not concept-defining. The concept of force begins with feelings of effort, weight, hard-to obtain deformation of bodies and more abstractly, describes action of one body on another body, typically in mutual physical contact. The above formula cannot capture all that. Nor can Newton's II. law.

 

[Philip Wood]

Jano: You may have seen my fence-sitting post (12), but one argument in favour of your 'legalist' stance on N2L is that it accords better with the way we use language - which might be telling us something. Most of us, I think, wouldn't object to a sentence like "the force from the rope makes the sledge accelerate (if resistive forces are negligible)". A definitionist would have to allow this to be paraphrased something like "the rate of change of momentum of the sledge due to the rope makes the sledge accelerate". This is close to a tautology. Replacing force by rate of change of momentum has thrown away the idea of the rope pulling as something distinguishable from the effect of the pull.

I acknowledge, too, the pertinence of your distinction between individual forces and resultant force, only the latter giving rise to (or, according to the definitionists, being) the rate of change of momentum.

 

[Jano L.]

Jano: You may have seen my fence-sitting post (12), but one argument in favour of your 'legalist' stance on N2L is that it accords better with the way we use language - which might be telling us something. Most of us, I think, wouldn't object to a sentence like "the force from the rope makes the sledge accelerate (if resistive forces are negligible)". A definitionist would have to allow this to be paraphrased something like "the rate of change of momentum of the sledge due to the rope makes the sledge accelerate". This is close to a tautology. Replacing force by rate of change of momentum has thrown away the idea of the rope pulling as something distinguishable from the effect of the pull.

I acknowledge, too, the pertinence of your distinction between individual forces and resultant force, only the latter giving rise to (or, according to the definitionists, being) the rate of change of momentum.

Indeed, I have seen your fence-sitting post and I agree with much of it as well as with your last post. Except I am not sure what you mean by legalist - dictionary gives me this explanation:

legalist - strict adherence, or the principle of strict adherence, to law or prescription, especially to the letter rather than the spirit.

I would say I try to view Newton's laws in the way they were meant (the spirit), I do not adhere to strict formulations.

 

[Philip Wood]

Sorry: By legalist I meant taking N2L to be a law (with some experimentally testable content) rather than a definition. Legal might have been better.

 

[f95toli]

There is a actually very famous book that tries to answer that question. The Science of Mechanics by Ernst Mach. It was written late 19th century,
Mach is nowadays most famous for the Mach number, as in supersonic jets can fly faster than Mach 1; but he was also a philosopher,

Btw, the answer is no; but the reasons why it is no can -if you want- be made to be pretty complicated.

 

[Philip Wood]

In case anyone (other than the long-suffering DrS and I) was following the sub-plot on F = ma + v dm/dt, introduced by H Smith in post 4, challenged by me in post 15 and debated between DrS and me in posts 20-35, I'd like to show a bit of humility.

(1) The accreting body (like a hailstone falling through a mist) still seems to me a good example of a body of changing mass. In Galilean/Newtonian Physics any mass that a body acquires has to come from somewhere else; mass is conserved. The acquired mass may bring momentum with it.

(2) If the accreting body and the 'pool' of mass from which it accretes has no external forces (e.g.pull of gravity) acting on it, then (see thumbnail in post 23, omitting F):

m dv/dt + v dm/dt = w dm/dt.

v is the velocity of the accreting body and w is the velocity of the pool of mass from which it is accreting. I hope the equation is uncontroversial.

(3) The left hand side of the equation is the rate of change of momentum of the accreting body. So it is equal to the resultant force on the accreting body. This was H Smith's contention in post 4, which I challenged in post 15. I shouldn't have done so.

(4) The right hand side is the force on the accreting body given in terms relating to the pool of (possibly moving) mass. [This is analogous to our writing, for a fixed mass m on a spring: m d²x/dt² = –kx in that the right hand side is the force on the mass given in terms relating to the spring.]

(5) Suppose now that we switch on a force, F^ext, such as a gravitational pull from outside the body and the pool from which it accretes. We must surely add this to the rhs of the equation, giving…

m dv/dt + v dm/dt = w dm/dt + F^ext.

(6) Re-arranged, this is the equation that I put forward in post 15 and derived in post 23. But it was misleading of me to have put –w dm/dt on the lhs rather than w dm/dt on the rhs. As I now see it, there are two forces on the accreting body: w dm/dt from the pool, and F^ext from some agency external to the accreting body and the pool. The resultant of these forces is equal to the rate of change of the accreting body's momentum.

(7) When w = 0, that is the pool is stationary, then we have simply m dv/dt + v dm/dt = F^ext.
My sole original intention was to point out that this last equation applies only in the special case of w = 0. In general the velocity of the accreted mass before accretion has to be taken into account. My exposition was faulty because I failed to distinguish clearly between the resultant force on the accreting body, which is equal to m dv/dt + v dm/dt, and F^ext, which, in general, isn't.

 

[DrStupid]

May I please check that we're in agreement on the following two things…

(1) For a body of mass m moving at velocity v accreting mass at a rate dm/dt from a 'mist' of particles with (pre-capture) velocity w, then if no external force (FAD) acts, from momentum conservation we have

(m dv/dt + v dm/dt) = w dm/dt

(2) the force on an open system is equal to the limit as delta t approaches zero of (final p – initial p)/delta t.

Yes, I agree with both. (1) result from conservation of momentum and (2) from the second law.

[edit]

(5) Suppose now that we switch on a force, F^ext, such as a gravitational pull from outside the body and the pool from which it accretes. We must surely add this to the rhs of the equation, giving…

m dv/dt + v dm/dt = w dm/dt + F^ext.

The general equation is (neglecting interactions between the other systems)

F = m \cdot \dot v + v \cdot \dot m = \sum\limits_i {F_i } = -\sum\limits_i {m_i \cdot \dot v_i } - \sum\limits_i {v_i \cdot \dot m_i }

You equation applies to the special case that the sum of the Forces F_i acting on the system consists of two forces only, where one (the internal force) is the result of mass transfer only and the other (the external force) is not related to any mass transfer.
[/edit]

This is a statement about what happens in a specific situation (one force, one body of constant mass); it clearly is not a definition (notice the word force occurs twice in the statement).

Newton's laws of motion define what he means if he use the term "force":
The first law is a qualitative definition: Forces are the only reason for a change of the state of motion.
The second law is a qualitative definition: Forces are proportional to the change of momentum (as given in definition 2).
The third law distinguishes forces from fictitious forces.

This definition was necessary because the term "force" has often been used different before. According to Aristoteles force was required to keep the state of motion (and not to change it).

 

[Philip Wood]

You equation applies to the special case that the sum of the Forces Fi acting on the system consists of two forces only, where one (the internal force) is the result of mass transfer only and the other (the external force) is not related to any mass transfer.

That is indeed all I set out to consider. Have always lacked ambition.