Instead of saying "tends to" you can say "transfers". In static equilibrium the same amount that it transferred in is also transferred out. So that word covers the equilibrium case as well as the non-equilibrium case.

Good point. ___________

@Dale Such bi-directional balanced momentum transfer in static equilibrium is valid in kinetic molecular theory of matter, e.g. fast moving molecules in gas do carry and transfer momentum in both directions. But the concept of force in statics does not come from or rest on this concept, and it is anachronistic and misleading to explain the basic notion of force in terms of it. In statics, we already have forces and laws for how to calculate with them to determine loads and safety and there isn't any connection to momentum there.

This answer might be better if expanded. Something maybe clarifying how we might know there are forces acting on an object even when the net force is zero (e.g., a body in free fall at terminal velocity). I think a bit of discussion on that would clarify the phrase "tends to" also. Finally I think the other answer concerning devices that can measure force has some good insight. Perhaps mentioning spring displacement showing the existence of a force would clarify more.

@JánLalinský obviously I disagree. Any force is a transfer of momentum, just like any power is a transfer of energy. All locally conserved quantities can be traced as they flow through a system. That is the whole point of a local conservation law. Certainly, you can use shortcuts and simplifications, in specific applications, but for a general definition of force it is hard to beat one that has a direct connection to locally conserved quantities and therefore to symmetries.

@Dale When a body of charge $q$ and mass $m$ hovers in Newtonian static gravity field $\mathbf g$ and electrostatic field $\mathbf E$, due to electrostatic force balancing the gravity force, what is the value of transfer of momentum associated with electric force(in direction against gravity) to the body? I think it is zero, as there is no magnetic field, and gravity field carries no momentum. But the electric force isn't zero.

@JánLalinský see here for information on the electromagnetic stress tensor. en.m.wikipedia.org/wiki/Maxwell_stress_tensor Note the section devoted to electrostatics. The electromagnetic stress tensor is non-zero in the situation you described. The stress tensor also describes the flow of momentum in statics that you mentioned previously. It is also the spatial part of the stress energy tensor for gravity.

@Dale Maxwell's tensor components being non-zero does not mean there is a flow of momentum going on. In the example, momentum of both the electrostatic field and the gravity field is everywhere zero, so they can't transfer any momentum. A correct statement in that example is that no momentum transfer happens anywhere in the whole space. A statement that electrostatic field transfers non-zero momentum to the particle, and the particle transfers momentum of same magnitude and opposite direction to the field is incorrect.Momentum flow is local,and there is no point where you can see it non-zero.

@Dale what you're suggesting is correct in case of gas made of moving particles, or EM radiation; when there is non-zero pressure in the gas, there is momentum transfer going on in both directions. But this is not going on in static systems with zero motion.

@JánLalinský said “Maxwell's tensor components being non-zero does not mean there is a flow of momentum going on”. Yes, in fact it does. The ij component of any stress tensor is a flux of the i’th component of momentum in the j’th direction. This is well known and described in many places.

@Dale can you quote a reference to such a definition cuz I haven't seen such a definition anywhere... or is it just made up?

About stress tensors: any point of a bridge have a non zero Cauchy stress tensor in general. Even the divergence of the tensor equals the weight per volume in the point. But if there is no vibration there is no momentum or transfer of momentum.

@Dale No, that's true in some cases, like in microscopic model which has moving molecules and concentrated momentum transfers that average to zero. In contrast, in totally static model of solid (or liquid), momentum in direction of axis $x$ is zero everywhere and momentum transfer in direction of axis $x$ is zero everywhere, despite Cauchy's $\sigma_{xx}$ non-zero. It's force per unit area, not momentum transfer, these things differ.

@Dale In static electric field in vacuum, non-zero Maxwell's $T_{xx}$, due to non-zero electric field, does not even mean there is a force of one half of element on the other in direction $x$ - in fact, there is no matter, so there is no force. Maxwell's $T_{xx}$ is not a flux of EM $x$-momentum in direction $x$. EM momentum of any control volume is zero because magnetic field vanishes, and transfer of momentum in any direction (not just direction of $x$) is zero everywhere.

@ojasdessai certainly, this is common and well known. Here are several such definitions in a variety of contexts. "For example, Txx is the flux in the x direction of x-momentum" phys.libretexts.org/Bookshelves/Relativity/… "In general, the stress energy tensor is the flux of momentum p^\mu over the surface x^\nu." theoretical-physics.com/dev/fluid-dynamics/general.html

"The element ij of the Maxwell stress tensor has units of momentum per unit of area per unit time and gives the flux of momentum parallel to the ith axis crossing a surface normal to the jth axis" en.wikipedia.org/wiki/Maxwell_stress_tensor "shear stress in a fluid can also be interpreted as the flux of momentum" en.wikipedia.org/wiki/Momentum_diffusion Anyway, I think that suffices. If you or anyone else (i.e. Jan or Claudio) have more detailed questions, please post a full question. This is standard known physics, although clearly there is opportunity for confusion

@Dale In electric field vacuum,there is no force, thus no impulse to talk about; there is no momentum, and no EM momentum. Yet you believe there is flux of EM momentum. That is indeed a confusion. You're interpreting Maxwell's tensor like people did in aether theory. But it's only a math concept that we can use to calculate force on material body, when we integrate over closed surface. Values of the tensor components at single point of space are arbitrary, and do not correspond to any real force or transfer of momentum there. Cf. journals.aps.org/pre/abstract/10.1103/PhysRevE.65.036615

@Dale I now tend to almost agree with you in case of stress tensor in solid, non-zero component $\sigma_{xx}$ means there definitely is $x$-impulse transfer between two parts across the interface normal to axis $x$, because there is force component parallel to $x$ acting on both parts in time. But impulse is not change of momentum, only sum of all impulses on a body equals change of momentum, so I still think calling these components flux of momentum or momentum transfer is wrong and is due to confusing momentum with impulse.

@JánLalinský it is not wrong. It is well known and commonly used. It is self-consistent, and it is also consistent with experiment. That said, it is not a mandatory interpretation, but it is completely valid and one that I like due to its consistency. If you want details, then ask a question, otherwise the fact that it is valid I think is already well supported above.

@Dale Well-known and used is not the same as correct, this is also one the messages of the paper I linked above. The interpretation of Maxwell's tensor you propagate (as real stress or even transfer of momentum) is a real and well-known issue, and leads to wrong conclusions, as explained in the paper. I do not really "want details", that's a bit arrogant of you to say in reaction to my substantive comments. I do not want you to defend your position by posting links to Wikipedia. I want you to think about my objections and react to them in good faith when contributing here.

@JánLalinský I am reacting in good faith here. Comments are not the location for this. It should be as a formal question and answer. Hence my repeated suggestion to do it properly. Reacting in good faith does not imply that I should try to treat a complex topic where it doesn’t fit and doesn’t belong. If you don’t want details that is fine, but don’t disparage my intentions

@Dale comments are the location for pointing out problems with an answer, or comments, which I did. If you want to expand your comments into a consistent convincing exposition, indeed please do it elsewhere, do not post links to Wikipedia, which is not a reliable source on physics. You can write both the question which would be appropriate for explaining your view and the answer. I'll have a look then.

@JánLalinský I posted two links to Wikipedia, and two to online textbooks. Wikipedia is, on average, reliable and particularly on specific points like these where it agrees with other sources. Your link is behind a paywall so while it is valid the links to Wikipedia are more useful.

@Dale You said this is a complex topic, and I think it is a subtle topic, and in my experience, Wikipedia is not reliable even on much more straigthforward things. But let's not devolve into comparing sources. Please do make a write up on this issue here or elsewhere, I think that would be useful.

@JánLalinský do you have a non-paywalled link to that article. I couldn’t find one. I suspect that I know the general argument, but I am curious about their specific examples.

@Dale I don't have a direct link, but you can get it from sci-hub. If you can't access that, the paper has also a webpage here, where one can request it from the author: researchgate.net/publication/…