What is wrong with $F = m^2a$?

Question

Newton's second law states $\vec F \propto \vec a$. The proportionality constant is defined as the mass ($m$) of the particle on which the force ($\vec F$) is acting and whose acceleration ($\vec a$) is being measured. Therefore, a definition such as $\vec F = m^2 \vec a$ would be consistent with Newton's second law.

what problem will "$\vec F = m^2 \vec a$" cause in the formulation of mechanics. What about a multi-particle system.

possible duplicate: How do we know that $F = ma$, not $F = k \cdot ma$ and links therein. — AccidentalFourierTransform, Aug 10 '17 at 10:27
This would guarantee the mass is positive, having $ m = \mu^2$ where $\mu$ is some other quantity. The same can be done with (absolute) temperature $T = \tau^2$ which forces $T$ to be always positive. — John Alexiou, Aug 10 '17 at 15:54
What is right about $F=m^2a$? This definition is not consistent with Newton's 2nd Law because Newton said that change in momentum is proportional to impulse, and momentum is proportional to $m$ not $m^2$. — sammy gerbil, Aug 10 '17 at 17:28

Razor · Answer 1 · 2017-09-03T08:59:59.540

Imagine two identical point objects of masses $m$ and at a height $h$ above the ground, initially at rest. If $\vec F = m^2 \vec a$ , then the force on each of them would be $m^2 \vec a$ with $\vec a$ pointing downward. The combined force on the balls is $\vec F_{net} = 2m^2 \vec a$ (we want the forces to add up). Now join the particles with a massless rod. Imagine this rod to be very thin. Then the two particles join to form one object. The force on this object would then be $(2m)^2 \vec a=4m^2 \vec a$ with does not match with $\vec F_{net}$.

Since the particles always move downward, we did not add any constraint or change anything in the system by joining them by a massless rod. Yet, the forces do not match. Hence the only way out is by having $\vec F$ to be linearly dependent on mass.

What an intuitive answer utilizing linearity! – Superfast Jellyfish Mar 27 '19 at 17:45 — Superfast Jellyfish, Mar 27 '19 at 17:45

score 3 · Answer 2 · answered Aug 10 '17 at 10:33

Firstly, it would be dimensionally inconsistent, without changing the dimensions of force. Answer continues, ignoring this.

If you want $m$ to be the same mass as we are used to using, then it won't work, simply because it doesn't describe the world we live in anymore. Simple example, 1d gravitational free-fall would become: $$m^2 a = \frac{G M m}{r^2},$$ meaning the acceleration would be: $$a = \frac{G M }{m r^2}.$$

That is the acceleration depends on mass, that is simply not what we observe (cit. Galileo!)

If we want to redefine your $m^2$ as $m^2= m'$, where $m'$ is the mass we are used to using, then of course the equation is unchanged, and will still describe physics accurately. Just now by having $m^2$, it would make it explicit that mass is positive.

score 0 · Answer 3 · answered Aug 10 '17 at 15:11

All the above answers are sufficient and I am giving my approach supporting the above (obviously I have faith in Newton ). You can make a graph too of force as a function of mass with constant acceleration , and you ll get a straight line not a curved path which supports that force is linearly related to mass.

What is wrong with $F = m^2a$?

3 Answers3