0
$\begingroup$

I am wondering about what force really is after I have looked up wikipedia, where there is stated that $F=ma$ only holds to be true for a constant $m$. For inconstant mass force is:

$$F=\frac{dp}{dt}= \frac{d(m(t)v(t))}{dt} = \frac{dm(t)}{dt}v(t)+a(t)m(t)$$

I cannot imagine this, this breaks my understanding of force. As far as I know, in Newtons mechanics force describes how strong a mass accelerates. It is a variable of state. But if it is a variable of State, for any state in time $s_{T}$ the force should be defined as:

$$F=m(T)\cdot \frac{dv}{dt}$$

I really do not get, what the term: $\frac{dm(t)}{dt}\cdot v(t)$ means. I do not understand why suddenly force should be dependet of the amount of velocity. I hope someone can help me build a new understanding.

Edit: Wikipedia also says that my formula is just wrong, I do not fully understand the explanation given in "variable-mass"-system paragraph. Could you explain it more detailed?

$$\mathbf {F} +\mathbf {u} {\frac {\mathrm {d} m}{\mathrm {d} t}}=m{\mathrm {d} \mathbf {v} \over \mathrm {d} t}$$

I do not see why this holds to be true, or where this comes from. Makes no sense to me at all. Basically expelling mass should already be displayed by the force $F$ since newtons third law holds to be true.

Thanks

$\endgroup$
  • $\begingroup$ This is a widely held point of misunderstanding among physicists, but not so much among aeronautical engineers. It's also comes up here frequently, but the questions are always slightly different, so it's hard to say it's a duplicate. But in addition to the answers below, here's another $\endgroup$ – garyp Apr 28 at 0:00
1
$\begingroup$

The full equation of second law is inconsistent under transformations. Consider the a frame where $$F=m\frac{dv}{dt} + v\frac{dm}{dt}$$ Now consider another frame moving with a constant velocity $v'$. In that frame one can write $$F=m\frac{d(v-v')}{dt}+(v-v')\frac{dm}{dt}$$ There will be an additional factor of $-v'\frac{dm}{dt}$ in the expression. This means that in two different inertial frames there are different physical laws, which is obviously wrong. To keep the equations consistent with this transformation we write the equation as $$ F + u_{rel}\frac{dm}{dt} = m\frac{dv}{dt}$$ Where $u_{rel}$ is the relative velocity of the mass that is being added or subtracted to the system with respect to the system. If net force on a system with variable mass is zero then the change in mass accelerates the system.

The most general example of variable mass systems are rockets. Rockets are constantly releasing fuel, which reduces its mass. Another example can be machines guns firing bullets (though the bullets are generally much lighter than the gun).

In one of your comments I read that

Yes, but this spewing should already be in a or F. Since molecules are exerting pressure.

That is right the spewing force is included in $F$ but the variable mass is not. Since the spewing is constant the force of spewing is acting on different masses at different times. At time $t$ the constant force $F$ is acting on a body of mass $M(t)$. At time $t+\Delta t$ the same force is acting on a body of mass $M(t+\Delta t)$. That is what's missing from the equation.

Derivation of Variable Mass Equation

Consider a mass $dm$ that is added to a body of mass $M$ and velocity $v$. Let the velocity of $dm$ be $v_m$. Since the mass is being added the collision will be completely inelastic. Hence both $dm$ and $M$ will move with the same velocity, say $v+dv$. Since there is no net external force on the whole system we can conserve momentum. $$v_m dm +Mv = (dm +M)(v+dv)$$ or $$v_m dm + Mv =v dm + dmdv + Mv +Mdv$$ The term $dmdv$ is very small and hence can be are to zero. The equation becomes $$-(v-v_m)dm = Mdv$$ Dividing by a small time interval in which impulse occurs $$-u_{rel} \frac{dm}{dt} = Ma$$ The minus is because $dv$ is negative, i.e $v<v+dv$ (the mass was decelerated). If there is a net force $F$ then $$F = \frac{v dm + dmdv + Mv +Mdv - v_m dm -Mv}{dt}$$ Or $$F-u_{rel} \frac{dm}{dt} =Ma$$ Which is the desired result.

Hope this helps.

$\endgroup$
2
$\begingroup$

I cannot imagine this, this breaks my understanding of force.

I think you're overlooking two commonly understood examples of systems in motion, with changing mass:

  1. Classical propulsion of a rocket:

The rocket spews out large amounts of burnt fuel. It relies on conservation of momentum but the rocket's mass drops during flight. See the rocket equation.

  1. The relativistic rocket:

Increase in mass due to Relativity.

$\endgroup$
  • $\begingroup$ Yes, but this spewing should already be in a or F. Since molecules are exerting pressure. $\endgroup$ – TVSuchty Apr 27 at 20:04
  • $\begingroup$ @TVSuchty: the simple derivations for these systems don't lie and have been confirmed time and time again. Don't "overthink" things: it leads to nowhere. Best regs. $\endgroup$ – Gert Apr 27 at 20:12
  • $\begingroup$ I never said that. $\endgroup$ – TVSuchty Apr 27 at 20:15
1
$\begingroup$

First let me assume we are arguing on constant mass.

As far as I know, in Newtons mechanics force describes how strong a mass accelerates.

The mass, in Newtonian Mechanics, is a property of an object that tells us the resistance it exhibits to be accelerated. Force causes a change in object's motion, so the magnitude of such quantity can be understood as how much the motion of an object has changed. To sum up, I'd say 'force describes how much a mass has been accelerated' instead of 'force describes how strong a mass accelerates'.

I do not understand why suddenly force should be dependent on the amount of velocity.

Note that the force is not characterised by the velocity of the object but by the change of it. I always like to think of the Newton's First Law; note that an object can move at constant velocity in the absence of forces

How to understand force when the mass of an object changes

Think of the classic propulsion of a rocket. The spacecraft must lose mass in the form of gases to be accelerated. This change on mass means a change on rocket's momentum, that equals the force exerted by the gases on the rocket (which is also called thrust or impulse).

Actually, you can model a system rocket + gases whose momentum and mass are conserved.

$\endgroup$
  • $\begingroup$ To sum up, I'd say 'force describes how much a mass has been accelerated' instead of 'force describes how strong a mass accelerates'. What is the difference, only the tense? Surely not? $\endgroup$ – TVSuchty Apr 28 at 10:08
  • $\begingroup$ @TVSuchty I think using the word strong is not accurate enough. As force is a quantity, I'd rather use how much. It is not a big deal though. $\endgroup$ – JD_PM Apr 28 at 10:48
-2
$\begingroup$

Force is defined as $F=dp/dt$, where $p=mv$. Since both $m$ and $v$ are varying with time, you must use the product rule to arrive at a proper derivative. From calculus, the product rule is $d(uv)=udv + vdu$. Following this rule, and taking the derivative of both $m$ and $v$ with respect to time, you get:

$d(mv)/dt = mdv/dt + vdm/dt$

$dv/dt=a$, so the derivative simplifies to

$d(mv)/dt = ma+vdm/dt$

$\endgroup$
  • $\begingroup$ The equation you have given is wrong. The last equation does not follow Galilean transformation, and hence is wrong. Your answer does not address the question as well. $\endgroup$ – Manvendra Somvanshi Apr 27 at 23:18
  • $\begingroup$ @ManvendraSomvanshi, the math behind physics still has to be correct, and the derivation that I showed stays. And note - my interpretation of the OP's misunderstanding was that he didn't see where the extra term was coming from in the force equation. I DIDN'T interpret his question as requiring a long and drawn-out explanation of Galilean transformations. $\endgroup$ – David White Apr 28 at 0:37
  • $\begingroup$ the OP has clearly asked an explanation for why the equation derived by you is wrong. The OP clearly states in his edit “I do not see why this holds to be true, or come from?” The reason for that is Galilean transformation. The OP had asked the physical interpretation of the extra term not the derivation, which you have not provided. $\endgroup$ – Manvendra Somvanshi Apr 28 at 8:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.