# Force without acceleration

My doubt is very basic and fundamental, by Newton's second law we can say that $F=\frac{dp}{dt}$. Hence, there can also be possible cases when $F=\frac{dm}{dt}v$, when the body is moving with constant velocity in the presence of a force! Then what is the effect of that force as a whole, what is it doing? We have always thought of force as an agent of acceleration, something that provides acceleration, but here the body is under the influence of a net force and still possess a constant velocity!! This whole idea seems to be absurd and can anyone help me in absorbing this concept.

• I think that's what another Newton's law is for, the one on actio and reactio. If there is no net acceleration that shows that there is a force that counteracts acceleration. If you skip that hinderance you get acceleration which obeys all laws. Nov 28, 2022 at 12:07

Yes such a situation is possible, but you are no longer considering point mechanics (where $m$ is by definition constant), but the mechanics of a system consisting of multiple point particles. In other words: to arrive at such an equation with changing mass, you have to analyse a system of point masses, for each of which $F = m\dot v$ (in other words, it all depends on how the mass is gained).

A simple model leading to an equation such as the above is the following. Consider an object, let's say an asteroid, of mass $M$ that moves through space filled with small objects at rest of mass $m$, let's say dust. The small objects are at rest. We assume that if the large object hits a dust particle there will be a completely inelastic collision (idealized to occure instantaneously). In other words we can compute the velocity afterwards by momentum conservation (energy is not conserved, since the non-elastic deformation of the two colliding objects creates heat): $$p = Mv = (M+m)v'$$ so the velocity after such an event will be $$v' = \frac{M}{M+m} v.$$ Now we can say that $M$ depends on $t$ since the asteroid gains mass $m$ each time it hits a dust particle. Each of these events can be handled as above, the momentum is conserved but the mass of the asteroid changes, in other words, we arrive at the equation $$F = \dot p = \partial_t (M(t) v(t)) = \dot M(t) v(t) + M(t) \dot v(t).$$ The force $F$ is assumed to only apply to the asteroid, not the dust. So if there is a dust trail which the asteroid sweeps up the mass will rise, and it will slow down, unless an external force is applied.

• Point mechanics do not require constant mass. Point mechanics is an abstraction of non-rotating bodies. Mass can still vary, as can be seen in this question physics.stackexchange.com/q/216895 Aug 23, 2018 at 16:53
• Yes you can do that, but to understand the physical meaning of that construction, you have to do what this answer is doing. If the mass changes due to other mechanisms (e.g. dust particles with non-zero momentum) just using a changing mass will give wrong results. Aug 23, 2018 at 16:56
• I can agree with you in this specific example, however the dynamics of a point particle with varying mass is still point particle mechanics, which was what I wanted to notice. Aug 23, 2018 at 16:58
• Your last equation is missing something. The right side is a momentum, but the left and middle have momenutm per time. Aug 23, 2018 at 17:22
• yes, indeed it is wrong, I'll fix it. Aug 24, 2018 at 16:44

This is the idea behind a rocket. Very simplified, while the rocket looses fuel mass, the exhaust produces thrust

The answer of your question itself lies in it. You have written F to be equal to $F=\frac{dm}{dt}v$. It becomes a variable mass system just like a rocket!

A Special Relativistic view :

In the rest system $\:\mathcal{S}_{o}\:$ of a particle, see ($\alpha$), by a mechanism power is transferred to the particle with rate $\:\overset{\boldsymbol{\cdot}}{\mathrm{q}}_{o}\:$. This rate is with respect to the proper time $\:\tau\:$ and this power changes the rest mass $\:m_{o}\:$ of the particle: $$\overset{\boldsymbol{\cdot}}{\mathrm{q}}_{o}=\dfrac{\mathrm{d}\left(m_{o}c^{2}\right)}{\mathrm{d}\tau}=c^{2}\dfrac{\mathrm{d}m_{o}}{\mathrm{d}\tau} \tag{B-01}$$ In an other inertial system $\:\mathcal{S}\:$ moving with constant 3-velocity $\:\boldsymbol{-}\mathbf{w}\:$ with respect to $\:\mathcal{S}_{o}\:$, the particle is moving with constant velocity $\:\mathbf{w}\:$, see ($\beta$), under the influence of a 'force' $$\boldsymbol{\mathcal{h}}=\dfrac{\overset{\boldsymbol{\cdot}}{\mathrm{q}}_{o}}{c^{2}}\mathbf{w}=\dfrac{\mathrm{d}m_{o}}{\mathrm{d}\tau}\mathbf{w}=\gamma(w)\dfrac{\mathrm{d}m_{o}}{\mathrm{d} t}\mathbf{w} \tag{B-02}$$ This 'force' $\:\boldsymbol{\mathcal{h}}\:$, although acting on the particle, keeps its velocity $\:\mathbf{w}\:$ constant. So, its 3-acceleration is $\:\mathbf{a}=\mathrm{d}\mathbf{w}/\mathrm{d}t =\boldsymbol{0}\:$ and consequently its 4-acceleration $\:\mathbf{A}=\boldsymbol{0}$. This 'force' is defined as heatlike.

The force is the rate of change of momentum and if the mass changes and the velocity remains constant, there is still the force. If the mass and velocity both are constants there can not be any force. One should change and one should remain fixed for the force to exist.

One example is when you determine wave speed as described in this article here, in which the force is applied and the velocity remains constant.

• Are you associated with physicskey? If yes, it's perfectly acceptable to include links, but you could be a bit more specific that you're associated with the website. If you aren't associated with it, "One example is when I determine..." is a bit misleading.
– user191954
Sep 9, 2018 at 5:42
• I've made a small edit; I hope it's ok. Feel free to change things about.
– user191954
Sep 9, 2018 at 10:16