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Is there any intuitive reasoning behind why the units of a "Newton" is $kg \frac{m}{s^{2}}$ and how it represents force? I always wanted to understand why objects of different mass fall at the same rate on earth and I found an explanation that uses Newtons Second Law of Motion and it made sense, but I'm confused as to why the units of force have this weird formula. Did Newton just use those units to make his second law work out mathematically via cancelling? Or is there some way to interpret $kg \frac{m}{s^{2}}$ as meaning force? Any insight would be appreciated.

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This is precisely the type of question I've started asking my introductory students. There's indeed an underlying intuition with units. Consider a free particle. It moves through space with constant momentum. If a force acts on that particle, its momentum will change. Remember that momentum is a vector and a change in a vector can involve a change in magnitude, a change in direction, or both (any arbitrary change in a vector can be decomposed into a change in magnitude parallel or antiparallel to the vector and a change in direction perpendicular to the vector). If you look at the units of force $\mathrm{kg}\cdot\mathrm{m}/\mathrm{s}^2$ and think of it as $\frac{\mathrm{kg}\cdot\mathrm{m}/\mathrm{s}}{\mathrm{s}}$ you can see it's the quotient of momentum to time (more correctly, duration because it's a time interval rather than a clock reading). So, think of a force of $X\;\mathrm{N}$ acting on a particle as being the how much momentum the particle gains or loses for every duration of one second that elapses. That's easiest to understand and visualize if the force is collinear with the particle's momentum. It's a bit more difficult to understand when only the direction of the particle's momentum changes, as in uniform circular motion.

So, a force of $12\;\mathrm{N}$ acting on a particle means that the particle's momentum changes by $12\;\mathrm{kg}\cdot\mathrm{m}/\mathrm{s}$ for every $1\;\mathrm{s}$ duration that elapses.

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This relationship is most easily seen through the relation $F = dp/dt$. –  Muphrid Dec 21 '12 at 2:22
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Yes, plus it's valid relativistically since $\vec{F}=m\vec{a}$ is a low speed approximation to $\vec{F}=\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}$. –  user11266 Dec 21 '12 at 2:26

The units of a new physical quantity will be defined when you define the quantity itself. So if you define force by $F = ma$, then it will have units of mass (e.g. $\mathrm{kg}$) times acceleration (velocity per unit time, e.g. meters per second squared). If instead you define force as the time-derivative of momentum, then it is the units of momentum (mass times velocity) divided by the units of time.

If you wanted to start with force as primitive, you could do that, calling the unit $\mathrm{N}$. Then perhaps mass would be the thing defined in $F = ma$, and the unit of mass would be that of force per unit acceleration, e.g. $\mathrm{N}/(\mathrm{m}/\mathrm{s}^2)$, which we could abbreviate as $\mathrm{kg}$, and this would serve as a definition of the kilogram.

It all depends on where you take your starting place to be. As a side note, the development of mechanics was muddled at first because people used the word "force" and similar words for a whole variety of related concepts without appreciating the differences. Today we have separated these cleanly into forces, momenta, accelerations, inertia, kinetic energy, potential energy, etc.

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So if I understand you correctly your saying that there is no intuitive meaning behind the units in this case, rather the units are generated as a result of creating a new physical quantity (in this case force). Is this correct? –  Ockham Dec 20 '12 at 23:52
    
I guess I'm saying don't look at all four parts (kg, m, two copies of seconds) together. Instead decompose it into a product or quotient of things you already understand. –  Chris White Dec 21 '12 at 0:25

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