Why $k$ is chosen to be unitless?

Question

In $F = kma$, why $k$ is taken to be unitless? If $k$ is unitless and 1, then we have $F=ma$. This means (I guess) the physical quantity Force is product of different (from Force) physical quantities mass and acceleration. But doesn't that mean we are equating entities that are physically different? What is the basis for the choice of $k$ to be unitless.

Answer 1

In an equation, units on both sides must equate to the same unit - only then you compare identical things.

But of course you can Multiply or divide things of different units:

An example: a velocity (or speed, call it v) is always a certain distance (d) traveled over some time t, as such $v = d / t$ or in units 10 km/h = 5 km / 0.5 h = 5/0.5 km/h = 10 km/h.

Now, consider you are not traveling on a flat surface but through a rough landscape and climbing a mountain which has an incline of angle $\alpha$. But you are interested in the velocity on the map, thus ignoring height differences.

so the velocity as projected on a map then is $v_{projected} = k * d / t$ with the "projection factor" $k = \cos(\alpha)$. This projection factor is unitless.

Generally, when you add or subtract, all parts must be of equal unit; you cannot add kilometres and seconds. If you divide or multiply... not an issue.

So, no, we are not equating properties which are different as we are only equating the whole left side with the whole right side of the equation. On the right side we are calculating the velocity from distance and time (and possibly an angle).

Answer 2

The fact that k is taken to be 1 is that till that time (when this relation was established) there was no formal definition of a unit force. So we had the freedom to choose any constant and it was chosen to be 1. But, you will see that the value of G is not 1, this is because, we had established the definition of a unit force (i.e., the amount of force to be provided to a body of mass 1 kg to produce an acceleration of 1 m.s^-2 in it ) so we no longer had that liberty to choose its value and that value had to be calculated with respect to the quantity force.