Why Is Newton's second law $F=ma$, and how did Newton “Discover it”? [duplicate]

I have been thinking lately, why is it that Newton's second law of motion takes the form $F = ma$? It seems to me the idea of a force is a somewhat made up concept, at least the value we assign to it seems made up. Is the a reason that $F= ma$ instead of maybe something $F = kma$ where $k$ is some constant or $F = ma^2$ or $F = m^2a$. The fact that it relates mass and acceleration makes sense to me, but it seems as if the type of relation is somewhat arbitrary.

However upon reflection I have come up with a way that I guess you could establish this law by experiment.The basic idea of my experiment would be to get some sort of "pusher" mechanism that would "push", with constant strength, a variety of movable objects that are subject to no external interaction and then measure the acceleration of the objects while they are being pushed. I guess you could make the assumption that a falling object is being "pushed" towards earth somehow and assume that this push is constant.

Does anyone one know how Newton himself conducted the experiment? I can't seem to find an answer simply by googling it.

marked as duplicate by Yashas, Kyle Kanos, Jon Custer, Qmechanic♦Jul 18 '17 at 16:05

• Possible duplicates: physics.stackexchange.com/q/104101/2451 , physics.stackexchange.com/q/265362/2451 , physics.stackexchange.com/q/315032/2451 and links therein. – Qmechanic Jul 17 '17 at 22:55
• Something that helps when understanding these "discoveries:" it's not that he wrote down an equation and then designed experiments to show that it was right. He looked at the results of a bunch of experiments and found relationships that kept showing up in each experiment. – Cort Ammon Jul 18 '17 at 3:35
• Newton's law are experimentally proven, that's why they hold and some other laws don't. It might have been otherwise, but it isn't. As for the "intuition", since a free falling object falls proportionally to $t^2$ (in constant gravity) and this is possible only if the force is proportional to the second derivative of the position. – gented Jul 18 '17 at 7:57
• Possible duplicate of How did Newton discover his second law? – Yashas Jul 18 '17 at 9:45

it seems as if the type of relation is somewhat arbitrary

It's not arbitrary.

It's matched by experiment.

All the alternative formulas do not match experiment as well. That's why these laws were successful. This is the rule used in physics to choose which laws are the most successful - they have to be accurate and consistent with experiment to within the margin of error of the experiment.

So not arbitrary, but more survival of the fittest.

• "It's matched by experiment" completely ignores the OP's question, which was "What experiment?" – WillO Jul 18 '17 at 18:09
• @willO The OP edited the question after I answered, adding the "experiment" part. Feel free to be constructive and add an answer yourself. – StephenG Jul 18 '17 at 19:52
• Ah. I'd failed to recognize the timing of the edit. My apologies. – WillO Jul 18 '17 at 20:55

It is $F=kma$ with $k > 0$, in different units systems. For example, if you measure the mass of something in kg, but measure forces in pounds, you have to use a constant for converting the units.

As for why it's linear like that, it was just an observable fact that if you apply two forces it will accelerate under the vector sum of the forces. This is, of course, approximate. When you get in to special relativity the law gets adjusted to, $$\vec{F} = \frac{\operatorname{d} \vec{p}}{\operatorname{d} t},$$ where $\vec{p}$ is the relativistic momentum. For more, see Wikipedia's section on relativistic mechanics.

According to Newton's second law of motion Force is directly proportional to the rate of change of momentum and when we solve it for a body with fixed mass and travels at uniform acceleration we get $F = kma$ and after doing so many experiments Newton found that the value of k turns out to be 1 so we don't write it every time and the other formulas which you suggested like F=ma² or F=m²a are dimensionally incorrect