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Many people claim that force is defined by the second axiom of Newtonian Mechanics, which is:

"In an inertial reference frame, the rate of change of momentum of a particle equals the sum of the forces exerted on it"

However, force is a fundamental quantity in Classical Physics and therefore it shouldn't be defined in terms of other (fundamental) quantities. Thus, the second law cannot be a definition for the force.

Is my reasoning wrong? Am I stating something that is incorrect?

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marked as duplicate by Emilio Pisanty, Buzz, ZeroTheHero, John Rennie newtonian-mechanics Feb 9 at 10:08

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  • $\begingroup$ Newton's laws are terribly written. The 3 laws must be considered at the same time if you want them to be meaninful. $\endgroup$ – FGSUZ Feb 8 at 21:28
  • $\begingroup$ assume the physics is the same but the second law was different, then the definitions of the forces (say from a spring) would be different too $\endgroup$ – Wolphram jonny Feb 8 at 21:44
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    $\begingroup$ Why would a fundamental quantity be required to be defined in terms of other fundamental quantities? Wouldn't that make it non-fundamental, if it could be explained by other fundamental quantities? Or, if you are saying force is fundamental, why wouldn't you also say the momentum and inertia are equally "fundamental", therefore the definition does explain it through other equally "fundamental" quantities? $\endgroup$ – JMac Feb 8 at 21:45
  • $\begingroup$ Thanks for the replies. @JMac Just noticed, my bad. I meant 'shouldn't be defined' but i wrote instead 'should be'. Thanks for noticing. $\endgroup$ – Unstoppable Tachyon Feb 8 at 21:54
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    $\begingroup$ Possible duplicate of Are Newton's "laws" of motion laws or definitions of force and mass? $\endgroup$ – Emilio Pisanty Feb 8 at 23:57
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You should not think of $\mathbf{F}=m\mathbf{a}$ as defining force. You should think of it as telling you what the acceleration will be under the influence of a known force that depends on position, time, and velocity:

$$m\frac{d^2\mathbf{r}}{dt^2}=\mathbf{F}\left(\mathbf{r},t,\frac{d\mathbf{r}}{dt}\right)$$

It is the equation of motion, a second-order differential equation to be solved for the position $\mathbf{r}(t)$, when one knows the proper formula for the force.

If it were just a definition, it could have no predictive value. As an equation of motion expressing the dynamical evolution of a system, it predicts the future.

The causal relationship is force causes acceleration, not acceleration causes force.

Various kinds of interactions have specific formulas for the force as a function of position, time, and velocity. For example, for two point particles under Newtonian gravity, the force on one depends only on its position relative to the other, and not on time or velocity:

$$\mathbf{F}\left(\mathbf{r},t,\frac{d\mathbf{r}}{dt}\right)=\frac{GMm\,(\mathbf{r}-\mathbf{R})}{|\mathbf{r}-\mathbf{R}|^3}.$$

For a charged particle in an electromagnetic field, the force can depend on position, time, and velocity:

$$\mathbf{F}\left(\mathbf{r},t,\frac{d\mathbf{r}}{dt}\right)=q\left[\mathbf{E}(\mathbf{r},t)+\frac{1}{c}\frac{d\mathbf{r}}{dt}\times\mathbf{B}(\mathbf{r},t)\right].$$

These are the kinds of equations that define specific forces.

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This is just an extended comment, not an answer. You cannot replace the second law by $F=mv$ because it would violate the first law, but you could try $F=m(da/dt)$, where $a$ is the acceleration. In this case Hook's law will look like $F=-kv$. I am not sure if the third law will change though, but it does not seem to be.

In any case, notice that the form used by newton makes the gravitational and electric forces easy to express, with the modification here they will be a mess.

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  • $\begingroup$ No, it wouldn't. 2 and 3 are independent laws. $\endgroup$ – FGSUZ Feb 8 at 22:36

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