From my limited understanding, one newton is defined as the amount of force that gives a mass of 1 kilogram an acceleration of 1 meter per second squared. What I don't understand is why it corresponds to the mathematical definition of a newton, namely N = 1 kg · m/s^2. Maybe my question extends further. If we think about equations such as these mathematically they have no meaning. What does it truly mean to multiply mass with acceleration or displacement divided by time? If I say 3 ÷ 5, to me it means that I multiply the number 3 with the multiplicative inverse of 5. But if I say displacement ÷ time then it makes no sense to me, because it would imply that there's such a thing as the multiplicative inverse of time, which to me makes little sense. Of course, I am thinking about this narrowly. Mass, acceleration, time, displacement, etc., are all quantities in physics and not numbers. I guess my question is, why have we chosen to define it like this mathematically and not in some other way? Why multiplication and not some other operation?
Forces – Why is 1 Newton Defined as 1 kg·m/s²?
dimensional analysisforcesmetrologysi-units
Related Solutions
Lets see if we can make some sense of this confusion.
Start from what does a unit mean.
A unit length, for example, could be
1 meter
1 inch
1 foot
1 yard
1 kilometer
1 mile
1 stadium
1 parasangue
etc.
Units need definition and conversion factors from system to system, particularly if one wants to build a bridge or plan a road.
Ratio's of quantities where the units are eliminated are universal, whether you are talking of meters or parasangues ( an ancient persian equivalent of kilometer length)
Take the perimeter of a circle and divide it by its diameter. Whatever units you may have used to inscribe the circle, the ratio is pi, whether a kilometer diameter or an inch diameter.
Given the radius of a circle, whether it is small, in inches, or huge, in kilometers, one can find the perimeter in the appropriate units by multiplying by 2*pi.
This and similar quantities simplify the work not only for geometers, in map making, engineers and architects, but all scientists.
The same is true for units of weight ( don't let me make a long list of them). The ratio allows easy communication and calculations whether for tons or pounds.
etc.
Physics is independent of our choice of units
And for something like a length plus a time, there is no way to uniquely specify a result that does not depend on the units you choose for the length or for the time.
Any measurable quantity belongs to some set $\mathcal{M}$. Often, this measurable quantity comes with some notion of "addition" or "concatenation". For example, the length of a rod $L \in \mathcal{L}$ is a measurable quantity. You can define an addition operation $+$ on $\mathcal{L}$ by saying that $L_1 + L_2$ is the length of the rod formed by sticking rods 1 and 2 end-to-end.
The fact that we attach a real number to it means that we have an isomorphism $$ u_{\mathcal{M}} \colon \mathcal{M} \to \mathbb{R}, $$ in which $$ u_{\mathcal{M}}(L_1 + L_2) = u_{\mathcal{M}}(L_1) + u_{\mathcal{M}}(L_2). $$ A choice of units is essentially a choice of this isomorphism. Recall that an isomorphism is invertible, so for any real number $x$ you have a possible measurement $u_{\mathcal{M}}^{-1}(x)$. I'm being fuzzy about whether $\mathbb{R}$ is the set of real numbers or just the positive numbers; i.e. whether these are groups, monoids, or something else. I don't think it matters a lot for this post and, more importantly, I haven't figured it all out.
Now, since physics should be independent of our choice of units, it should be independent of the particular isomorphisms $u_Q$, $u_R$, $u_S$, etc. that we use for our measurables $Q$, $R$, $S$, etc. A change of units is an automorphism of the real numbers; given two units $u_Q$ and $u'_Q$, the change of units is $$ \omega_{u,u'} \equiv u'_Q \circ u_Q^{-1}$$ or, equivalently, $$ \omega_{u,u'} \colon \mathbb{R} \to \mathbb{R} \ni \omega(x) = u'_Q(u_Q^{-1}(x)). $$ Therefore, \begin{align} \omega(x+y) &= u'_Q(u_Q^{-1}(x+y)) \\ &= u'_Q(u_Q^{-1}(x)+u_Q^{-1}(y)) \\ &= u'_Q(u_Q^{-1}(x)) + u'_Q(u_Q^{-1}(y)) \\ &= \omega(x) + \omega(y). \end{align}
So, since $\omega$ is an automorphism of the reals, it must be a rescaling $\omega(x) = \lambda x$ with some relative scale $\lambda$ (As pointed out by @SeleneRoutley, this requires the weak assumption that $\omega$ is a continuous function -- there are everywhere discontinuous solutions as well. Obviously units aren't useful if they are everywhere discontinuous; in particular, so that instrumental measurement error maps an allowed space of $u_\mathcal{M}$ into an interval of $\mathbb{R}$. If we allow the existence of an order operation on $\mathcal{M}$, or perhaps a unit-independent topology, this could be made more precise.).
Consider a typical physical formula, e.g., $$ F \colon Q \times R \to S \ni F(q,r) = s, $$ where $Q$, $R$, and $S$ are all additive measurable in the sense defined above. Give all three of these measurables units. Then there is a function $$ f \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R} $$ defined by $$ f(x,y) = u_S(F(u_Q^{-1}(x),u_R^{-1}(y)). $$
The requirement that physics must be independent of units means that if the units for $Q$ and $R$ are scaled by some amounts $\lambda_Q$ and $\lambda_R$, then there must be a rescaling of $S$, $\lambda_S$, such that
$$ f(\lambda_Q x, \lambda_R y) = \lambda_S f(x,y). $$
For example, imagine the momentum function taking a mass $m \in M$ and a velocity $v \in V$ to give a momentum $p \in P$. Choosing $\text{kg}$ for mass, $\text{m/s}$ for velocity, and $\text{kg}\,\text{m/s}$ for momentum, this equation is $$ p(m,v) = m*v. $$ Now, if the mass unit is changed to $\text{g}$, it is scaled by $1000$, and if the velocity is changed to $\text{cm/s}$, it is scaled by $100$. Unit dependence requires that there be a rescaling of momentum such that $$ p(1000m,100v) = \lambda p(m,v). $$ This is simple -- $10^5 mv = \lambda mv$ and so $\lambda = 10^5$. In other words, $$ p[\text{g} \, \text{cm/s}] = 10^5 p[\text{kg} \, \text{m/s}]. $$
Now, let's consider a hypothetical situation where we have a quantity called "length plus time", defined that when length is measured in meters and time in seconds, and "length plus time" in some hypothetical unit called "meter+second", the equation for "length plus time" is $$ f(l,t) = l + t. $$ This is what you've said - $10 \text{ m} + 5 \text{ s} = 15 \text{ “m+s”}$. Now, is this equation invariant under a change of units? Change the length scale by $\lambda_L$ and the time scale by $\lambda_T$. Is there a number $\Lambda$ such that $$ f(\lambda_L l, \lambda_T t) = \lambda_L l + \lambda_T t $$ is equal to $$ \Lambda f(l,t) = \Lambda(l+t) $$ for all lengths and times $l$ and $t$? No! Therefore, this equation $f = l + t$ cannot be a valid representation in real numbers of a physical formula.
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Best Answer
Don't overthink it. There is nothing deep going on.
Because $F=m a$, the units of force need to be the units of mass times the units of acceleration. In SI units, we measure mass in ${\rm kg}$ and acceleration in ${\rm m}/{\rm s^2}$, so the SI unit of force must by ${\rm kg\ m}/{\rm s^2}$.
The reason we keep track of units is because we can always choose to change units. Suppose we decide to measure mass in pounds instead of kilograms, distances in feet instead of meters, and times in years instead of seconds. The numerical value of $ma$ will be different in these new units. To compensate, the numerical value of $F$ must also change, in such a way that the equation $F=ma$ remains true. The units we assign $F$ are chosen so that it is easy to keep track of how to do this conversion consistently.