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.

Does this answer your question? How is it possible that SI units allow us to write $F=ma$ without an extra constant? — Superfast Jellyfish, Feb 26 '20 at 12:24
Does this answer your question? How do we know that $F = ma$, not $F = k \cdot ma$ — fqq, Feb 26 '20 at 12:40

planetmaker · Answer 1 · 2020-02-26T12:32:08.160

0

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).

edited Feb 26 '20 at 12:32

answered Feb 26 '20 at 12:26

planetmaker

1,029

1

The question is essentially asking about defining the units of force. This does not seem to have much to do with that. – fqq Feb 26 '20 at 12:37
To me the question reads as about equating different properties - with force as an example. And I chose the much more hands-on property of velocity, distance and time to illustrate. The argument would be the same with force, mass and acceleration (and a fudge factor to accomodate for friction or whatever) – planetmaker Feb 26 '20 at 12:40
Yes. That's what I meant in the question. We see that Area = k.l.b. k is chosen to be unitless and 1. If that's so, we have unit of area equal to unit of length squared. But, area is physically different from length squared right? I don't think that length squared physically mean anything at all. This leaves me a question that what actually equality (=) mean? – Rajendra Reddy Malluri Feb 26 '20 at 13:08
The area of a rectangle is the length times the width. So yes, an area is the product of two lengths. If I take a 10m long stick and drag it 10m vertically, I covered an area. Or an area is the ratio of a volume by a length... – planetmaker Feb 26 '20 at 13:19

Abhirup Adhikary · Answer 2 · 2020-02-26T12:55:30.263

-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.

edited Feb 26 '20 at 12:55

answered Feb 26 '20 at 12:32

Abhirup Adhikary

108

I’m sorry but $k$ is unitless unless you redefine the unit of force, mass or acceleration (which would mean redefining length and time). This is NOT the same as $G$ since the expression for the gravitational force already contains masses and distances. – ZeroTheHero Feb 26 '20 at 12:41
I am sorry but could you please elucidate what you said – Abhirup Adhikary Feb 26 '20 at 12:45
There’s nothing more to say... in $F=G m_1m_2/r^2$ you need the units in $G$ to make the right hand side have units of forces. In $F=k ma $ the units of $m a$ are already those of force, so $k$ must be unitless. – ZeroTheHero Feb 26 '20 at 12:47
So you are saying we could write G as m^3.s^2.kg^-1 Right? – Abhirup Adhikary Feb 26 '20 at 12:52
I understand now extremely sorry for sharing wrong information but the edited answer is correct now – Abhirup Adhikary Feb 26 '20 at 12:56

Why $k$ is chosen to be unitless?

2 Answers2