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When discussing with my children Newton's laws (high school level), I told them (long story short) that one can imagine that since a body on which all forces balance move straight and at constant speed, one can imagine that not moving at constant speed implies (or is linked to) the fact that there is a net force.

And that this relationship is linear, and that the coefficient is the mass. And that this is magical.

I then started to wonder why

  • the relationship is linear
  • the coefficient is the mass of the object

Are there fundamental reasons for the relationship to be exactly $F = m \times \dot v$, with $m$ being simply the mass?

What I am trying to say that this could have been $F = m^{1.346} \times \dot v^{0.6547} + 10^{-11}$ (obviously made up example for a non proportional relation, with a tiny loose end, and a less nice coefficient)

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  • $\begingroup$ This isn't really answer quality, but I would imagine that this is simply a matter of definition. We know $F=\frac{dp}{dt}$, so that reduces to $F=ma$ in the case of constant mass - which covers most of the wide variety cases we deal with. $\endgroup$
    – zh1
    Commented May 7, 2020 at 20:09
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    $\begingroup$ Does this answer your question? Why does $F=ma$? Is there a straightforward reason? $\endgroup$
    – PhysicsMathsLove
    Commented May 7, 2020 at 20:11
  • $\begingroup$ @zhutchens1 In fact it covers all of the wide variety of cases. Newton's laws apply only to closed systems which means that the mass must be constant. (Any mass leaving the system would provide an impulse to the system. That is, an additional force.) $\endgroup$
    – garyp
    Commented May 7, 2020 at 20:42

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The second question is not so difficult - at least in the context of Newtonian mechanics, the (inertial) mass is defined to be the ratio of the applied force to the resulting acceleration, so it doesn't make much sense to have some $f(m)$ there instead of $m$. In your example, if $\mathbf F = m^{\alpha}\mathbf a$ for some $\alpha$, then the quantity which has all of the properties which we associate with inertial mass would be $m^\alpha$, not $m$.

It is a different question entirely to ask why the coefficient which appears in Newton's 2nd Law also appears in the equation for the gravitational force - in other words, why inertial mass and gravitational mass (both active and passive) are equal (or at least proportional) to one another. In Newtonian physics, this question is left unanswered, and can only be justified by the empirical observation that it is true. A satisfactory explanation from first principles requires general relativity and the equivalence principle.

The presence of an additional term (your $10^{-11}$) would spoil the observation that in the absence of external forces, particles do not accelerate. So if you are willing to build a theory based on that observation, that term must be zero.

That leaves the question of why the relationship between force and acceleration is linear, and this too can be motivated only by experiment. If you attach a spring to a cart and pull carefully (so that the length $\ell$ of the spring is constant), then you can measure the cart's acceleration; attaching two identical springs and pulling in such a way that they both have length $\ell$ corresponds to twice the force, and via measurement, twice the acceleration.

This need not be true in principle - pulling with twice the force could cause four times the acceleration, for instance - but experiment suggests very strongly that $\mathbf F \propto \mathbf a$, so that is what we use. In relativistic physics, this is actually no longer true, so our definitions need to be updated accordingly.

One could go into more sophisticated arguments for these things, perhaps based on relativity or quantum theory, but at the end of the day we will always end up justifying our models by their agreement with empirical observation. This is the nature of science - there is no ultimate "why" other than "because that's how it seems to be."

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  • $\begingroup$ Thank you for the excellent answer. I added the tiny $10^{-11}$ term to account for the fact that all of these discussions are around observations and experiments. It could be that there is an extra factor which we do not observe (because it was so small that undetectable at the time these laws were thought out) $\endgroup$
    – WoJ
    Commented May 8, 2020 at 10:33
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The notion of force by itself is purely intensive. While we feel doing more or less force on something, we can not say we are doubling the force for example without inventing a way to quantify it.

One old way to deal with it is a spring. Up to its elastic range, the deflection of a spring can be used to quantify forces.

After that, we can use it to measure the relation between force and acceleration, varying the volumes of the same material to get $\mathbf F = m\mathbf a$.

We can also use the spring to weight the masses and verify that the same proportion of deflections apply.

Because that empirical relation happens to be so good, we can say that $\mathbf F = m\mathbf a$.

That relation, unlike springs, where the elasticity modulus depends on the material and manufacturing, is more universal.

So it can now be taken as a definition of force, and if a spring shows some small non linearity measuring this relation, we can say that the spring has a deviation from linearity!

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