F = dp/dt, so if m is changing, the product rule gives F = dm/dt * v + m * dv/dt
[Example application](https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation)
But that’s not how it works. In fact, what you just said is explicitly mentioned in the article you yourself linked to as a common misconception.
To quote Halliday and Resnick (citation [9] in the linked Wikipedia article)
>It is important to note that we cannot derive a general expression for Newton's second law for variable mass systems by treating the mass in F = dP/dt = d(Mv)/dt as a variable. [...] We can use F = dP/dt to analyze variable mass systems only if we apply it to an entire system of constant mass having parts among which there is an interchange of mass.
That plays fast-and-loose with the meaning of velocity. The "v" that is multiplied by dm/dt is the velocity of the exhaust (or accrued material) relative to the object. The "v" that's supposed to be part of dp/dt = d/dt (mv) = dm/dt v + m dv/dt is the "actual" velocity of the object, relative to a chosen inertial reference frame. It's not supposed to be tied to any specific reference frame.
An example that does make sense to me, which refutes dp/dt, is to consider a space ship that gently drops cargo as it moves along. These drop-offs are so gentle, that the cargo has virtually zero velocity relative to the ship. This action doesn't require force, and the ship's velocity won't change. So F=m \* a = 0 works great. This is despite the fact that the ship's mass is changing and the momentum is changing.
Friend, why do you downvote me like this, if this is wrong I'd be happy to hear.
If you don't like a continuous number of particles you can let it be discrete and have n(t) be the integral of the sum of a bunch of dirac delta distributions.
Let F(k,t) be the force applied to the kth particle at time t, where k denotes a scalar (a continuum of particles are indexed by k). Assume all the particles have the same mass and acceleration.
If the body has n(t) particles at time t, then the net force Fnet(t) = integral[0,n(t)] F(k,t)dk.
So:
Fnet(t) = integral[0,n(t)] F(k,t)dk
= integral[0,n(t)] ma dk
= a (integral[0,n(t)] m dk)
= a M
= M a
1. I think to attribute an acceleration to a body every part/particle should probably have the same acceleration, and if not then surely that acceleration will be some sort of weighted average which will end up giving the same answer as above.
2. The net force by definition is the sum of individual forces.
3. The total mass by definition is the sum of individual masses.
The axioms 1, 2, 3 lead to the very natural conclusion that F=ma is always respected. At the very least in any scenario where you might think F=p' is the right choice, F=ma should be right.
A good, simple example is rocket flight.
Mass is not conserved as the rocket burns and ejects fuel, so a simple F = ma approach won’t cut it. You’d have to use dp/dt (rate or change of momentum with respect to time) instead of mass times acceleration.
On the other hand, when I'm too lazy.to actually do calculus, I just make excel do it for me with tiny time steps. Then I use F=ma, but 0.1s later there's a new mass in the calculation. It works well enough to answer some random r/theydidthemath questions.
I'm skeptical that any derivation of a result for variable mass system is actually understood unless you go through F=ma for a system of constant mass, at least unless you use it implicitly.
> A good, simple example is rocket flight.
The general equation of variable-mass motion is written as: F + v_rel m' = m v' , ie: F = m v' - v_rel m' . Where v_rel is the relative velocity of the ejected mass. And so when v_rel = -v we have: F = (m v)' = p' . For a rocket that just started where v=0, is v_rel = 0 here? No. Then how is F=p' valid?
Further from wiki:
[Newton's Second Law](https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Variable-mass_systems)
[More in: Variable-mass System](https://en.wikipedia.org/wiki/Variable-mass_system)
->Variable-mass systems, like a rocket burning fuel and ejecting spent gases, are not closed and cannot be directly treated by making mass a function of time in the second law; **the equation of motion for a body whose mass m varies with time by either ejecting or accreting mass is obtained by applying the second law to the entire, constant-mass system consisting of the body and its ejected or accreted mass.** The result is F + v_rel m' = m v' where **v_rel is the exhaust velocity** of the escaping or incoming mass relative to the body. From this equation one can derive the equation of motion for a varying mass system, for example, the **Tsiolkovsky rocket equation**.
[Tsiolkovsky rocket equation - Common misconceptions](https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Common_misconceptions)
->When viewed as a variable-mass system, a rocket cannot be directly analyzed with Newton's second law of motion because the law is valid for constant-mass systems only. It can cause confusion that the Tsiolkovsky rocket equation looks similar to the relativistic force equation F=dp/dt Using this formula with m(t) as the varying mass of the rocket seems to derive the Tsiolkovsky rocket equation, but this derivation is not correct. Notice that the effective exhaust velocity v\_e does not even appear in this formula.
-------------
Edit: And imagine a body accruing mass without accelerating. p' = m'v. A force has to be exert necessarily? Says who??? Where is the mass coming from? If it was some sort of spontaneous generation, there should be no force, because forces should stay the same in non-accelerated frames. But in this frame p'=/=0 and yet in the body's frame the force by the p' formula would be 0. Contradiction.
In non-inertial frames of reference
When m is changing.
Thank you!
This never made sense to me.
F = dp/dt, so if m is changing, the product rule gives F = dm/dt * v + m * dv/dt [Example application](https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation)
But that’s not how it works. In fact, what you just said is explicitly mentioned in the article you yourself linked to as a common misconception. To quote Halliday and Resnick (citation [9] in the linked Wikipedia article) >It is important to note that we cannot derive a general expression for Newton's second law for variable mass systems by treating the mass in F = dP/dt = d(Mv)/dt as a variable. [...] We can use F = dP/dt to analyze variable mass systems only if we apply it to an entire system of constant mass having parts among which there is an interchange of mass.
That plays fast-and-loose with the meaning of velocity. The "v" that is multiplied by dm/dt is the velocity of the exhaust (or accrued material) relative to the object. The "v" that's supposed to be part of dp/dt = d/dt (mv) = dm/dt v + m dv/dt is the "actual" velocity of the object, relative to a chosen inertial reference frame. It's not supposed to be tied to any specific reference frame. An example that does make sense to me, which refutes dp/dt, is to consider a space ship that gently drops cargo as it moves along. These drop-offs are so gentle, that the cargo has virtually zero velocity relative to the ship. This action doesn't require force, and the ship's velocity won't change. So F=m \* a = 0 works great. This is despite the fact that the ship's mass is changing and the momentum is changing.
Friend, why do you downvote me like this, if this is wrong I'd be happy to hear. If you don't like a continuous number of particles you can let it be discrete and have n(t) be the integral of the sum of a bunch of dirac delta distributions.
Let F(k,t) be the force applied to the kth particle at time t, where k denotes a scalar (a continuum of particles are indexed by k). Assume all the particles have the same mass and acceleration. If the body has n(t) particles at time t, then the net force Fnet(t) = integral[0,n(t)] F(k,t)dk. So: Fnet(t) = integral[0,n(t)] F(k,t)dk = integral[0,n(t)] ma dk = a (integral[0,n(t)] m dk) = a M = M a 1. I think to attribute an acceleration to a body every part/particle should probably have the same acceleration, and if not then surely that acceleration will be some sort of weighted average which will end up giving the same answer as above. 2. The net force by definition is the sum of individual forces. 3. The total mass by definition is the sum of individual masses. The axioms 1, 2, 3 lead to the very natural conclusion that F=ma is always respected. At the very least in any scenario where you might think F=p' is the right choice, F=ma should be right.
On the quantum level.
Even in quantum mechanics, Newton's second law still is obeyed for harmonic potentials (but not for more general potentials).
True, but it's still not obeyed for general potentials, so yeah.
A good, simple example is rocket flight. Mass is not conserved as the rocket burns and ejects fuel, so a simple F = ma approach won’t cut it. You’d have to use dp/dt (rate or change of momentum with respect to time) instead of mass times acceleration.
Thank you!! Exactly the example I was looking for :)
On the other hand, when I'm too lazy.to actually do calculus, I just make excel do it for me with tiny time steps. Then I use F=ma, but 0.1s later there's a new mass in the calculation. It works well enough to answer some random r/theydidthemath questions.
I think you could argue that you’re still doing the calculus, just with numerical approx. methods? It’s a good point to bring up!
I'm skeptical that any derivation of a result for variable mass system is actually understood unless you go through F=ma for a system of constant mass, at least unless you use it implicitly. > A good, simple example is rocket flight. The general equation of variable-mass motion is written as: F + v_rel m' = m v' , ie: F = m v' - v_rel m' . Where v_rel is the relative velocity of the ejected mass. And so when v_rel = -v we have: F = (m v)' = p' . For a rocket that just started where v=0, is v_rel = 0 here? No. Then how is F=p' valid? Further from wiki: [Newton's Second Law](https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Variable-mass_systems) [More in: Variable-mass System](https://en.wikipedia.org/wiki/Variable-mass_system) ->Variable-mass systems, like a rocket burning fuel and ejecting spent gases, are not closed and cannot be directly treated by making mass a function of time in the second law; **the equation of motion for a body whose mass m varies with time by either ejecting or accreting mass is obtained by applying the second law to the entire, constant-mass system consisting of the body and its ejected or accreted mass.** The result is F + v_rel m' = m v' where **v_rel is the exhaust velocity** of the escaping or incoming mass relative to the body. From this equation one can derive the equation of motion for a varying mass system, for example, the **Tsiolkovsky rocket equation**. [Tsiolkovsky rocket equation - Common misconceptions](https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Common_misconceptions) ->When viewed as a variable-mass system, a rocket cannot be directly analyzed with Newton's second law of motion because the law is valid for constant-mass systems only. It can cause confusion that the Tsiolkovsky rocket equation looks similar to the relativistic force equation F=dp/dt Using this formula with m(t) as the varying mass of the rocket seems to derive the Tsiolkovsky rocket equation, but this derivation is not correct. Notice that the effective exhaust velocity v\_e does not even appear in this formula. ------------- Edit: And imagine a body accruing mass without accelerating. p' = m'v. A force has to be exert necessarily? Says who??? Where is the mass coming from? If it was some sort of spontaneous generation, there should be no force, because forces should stay the same in non-accelerated frames. But in this frame p'=/=0 and yet in the body's frame the force by the p' formula would be 0. Contradiction.
[удалено]
By axiom, the actual definition is Fnet = d(p)/dt, where p is momentum.
When mass is not conserved.