# Why do we take k=1 in the derivation of F=k*ma?

• B
Gold Member

## Main Question or Discussion Point

In the derivation of F=ma, when we reach the point F=kma, we take k=1.
Why can't we take 'k' as some other value?
I will be thankful for help!

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russ_watters
Mentor
I've never heard of f=kma. where did you get it?

At most it could be a proportionality constant, which is 1 because unit systems are designed with it built in.

Gold Member
I've never heard of f=kma. where did you get it?
In my textbook

russ_watters
Mentor
In my textbook
please provide more context than that. What does your textbook say about it? How do they use it.

Gold Member
please provide more context than that. What does your textbook say about it? How do they use it.

Dale
Mentor
In the derivation of F=ma, when we reach the point F=kma, we take k=1.
Why can't we take 'k' as some other value?
I will be thankful for help!
You can and sometimes you need to. For example if you measure f in lbf, m in kg, and a in AU/day^2 then k would be 4.5 lbf/(kg*AU/day^2). We can only set it to 1 if you are using SI units or other unit systems that were designed that way, which are called consistent units.

As it says "The unit of force is so chosen that, k = 1, when m = 1 and a = 1." (emphasis added). We can do it because we defined the SI units that way.

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FactChecker, sophiecentaur and olivermsun
CWatters
Homework Helper
Gold Member
Ok so K is a constant of proportionality.

As Dale said, if you use SI units then K=1. If you use some other unit system then K has some other value. I believe this is entirely due to the way SI units are defined.

K is also 1 in imperial units but only if you use pounds force, slugs and feet per second^2. If you have the mass in pounds you have to convert them to slugs or K isn't 1.

Gold Member
Thank you, everyone, for a reply!
Now, I just want to know why is it very much necessary to get k=1 anyhow?

CWatters
Homework Helper
Gold Member
It's not essential that k=1 but it makes things easier to remember. The equations still work if you use obscure units, you just need the right value of k. Try working out what k would be in f=kma if the force was needed in Dyne, Mass was specified in Grains and the acceleration in furlongs per hour^2.

FactChecker
Dale
Mentor
Thank you, everyone, for a reply!
Now, I just want to know why is it very much necessary to get k=1 anyhow?
It is not necessary at all, but it is convenient.

Gold Member
It is not necessary at all, but it is convenient.
Thanks!
But,can you please explain how it is convenient?

Dale
Mentor
Thanks!
But,can you please explain how it is convenient?
Multiplication by 1 is easy. And 1 is easy to remember.

What positive number could be easier than 1?

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russ_watters
Mentor
Multiplication by 1 is easy.
So easy you can ignore it and still be doing it correctly!

Dale
Where did you find a "derivation" of F = m*a? I've never seen such, but rather understood it as more of a definition.

Merlin3189
Homework Helper
Gold Member
In the link of post #5.

Surely, as they say in that link, Newton's laws must lead to F=kma, because he could not have known what units people would use. I'm pretty sure he'd never heard of Newtons, nor kilograms nor metres for that matter.
If Newton had invented the Newton, presumably 1 N would equal 1 foot pounds per sec2 , again so that k was 1, then 32 N would equal 1 pound force.

PeroK and sophiecentaur
Surely, as they say in that link, Newton's laws must lead to F=kma, because he could not have known what units people would use. I'm pretty sure he'd never heard of Newtons, nor kilograms nor metres for that matter.
If Newton had invented the Newton, presumably 1 N would equal 1 foot pounds per sec2 , again so that k was 1, then 32 N would equal 1 pound force.
The metric system was a product of the French Revolution, so I'm sure you are correct that Newton never heard of a Newton. But I'm pretty sure that he did know about the foot, the pound, and the second. These are quite enough if mass is treated as a derived quantity.

I've actually seen a lot of students get confused on this point, and it doesn't help when a textbook asks you to "derive" ##F=ma##. ##F=ma## is not derived from any other principles or equations. It also does not come from experiment. There is no experiment you can do test ##F=ma##. ##F=ma## is a definition. You can define "force" any way you want. The question is, why is it useful to define "force" this way?

To see why, it might help to understand what Newton originally wrote, and why. Newton's second law was originally written as something like: "The force is proportional to the change in momentum over time." Back then, physicists didn't write equations the way we do today. They just wrote it out in plain language and wrote in terms of being proportional. When physicists did translate it into modern equations, since this is a definition, it made sense to write it as ##F=\frac{dp}{dt}##. Force is defined as the change in momentum over time. No sense putting in messy proportionality constants when it's simply a definition. At some point, physicists (I think Euler) decided to write it as ##F=ma##, which is equivalent and probably easier to work with, but loses something in the translation.

You see, when you look at Newton's third law, you can see the point of defining force as Newton did. The third law is that for every force, there is an equal and opposite force. But if "force" is just the change in momentum, then this law is saying that for every change in momentum, there is an equal and opposite change in momentum. In other words, momentum is conserved. Newton was working off a bunch of experiments that had been performed and interpreted by Wren, Wallis, and Huygens, showing that momentum was conserved, and his laws were essentially just one way of systematically breaking down that principle into bite-sized pieces. That's why it's useful to define force as he did.

navneet9431
berkeman
Mentor
There is no experiment you can do test F=ma.
Say what?

Say what?
I mean simply that we conventionally take $F=ma$ to be the definition of force, and insofar as it is a definition, it can't be experimentally tested on it's own. Not anymore than one could experimentally test $v=\frac{dx}{dt}$

That's not to say we can't test Newton's laws as a whole in the lab. We conventionally take the real physics content of his laws to be in the third law, and we certainly can and do test this in a lab!

No there are, like the trolley experiment. A graph of Force vs Acceleration can be drawn where m is constant. The slope is is 1/m if Y axis is acceleration. This experiment can be used to test the validity of Newton's equation.

But other points are correct. And F=dp/dt gives a more correct "definition" of force than F=ma.

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berkeman
Mentor
I mean simply that we conventionally take F=ma F=ma to be the definition of force, and insofar as it is a definition, it can't be experimentally tested on it's own. Not anymore than one could experimentally test v=dxdt v=\frac{dx}{dt}
Sorry, that still makes no sense to me. I certainly did test those relations in my undergraduate physics labs.

And F=ma is not a definition of force, it is a relationship between force, mass and acceleration.

as it is a definition, it can't be experimentally tested on it's own.
I disagree with this.

Experiments are conducted and the data is analysed to create some simple representation for the entire range of data ie a formula. A formula translates into a definition in such a way that it is independent of the physical quantity being defined and the definition can be reversed to the formula,vice-versa

Like Galileo rolled a ball on an inclined plane hundreds of times, tabulated and analysed the data to create equations and definitions.

Now it may seem intuitive that v=dx/dt but Newton derived these "simple" equations using "complex" Euclidean geometry. By complex I mean that it is not "everyday-geometry". Source:Gravity by GA Gamow where he uses the same calculations to derive gravitational equations Newton did(without Calculus).

And F=ma is not a definition of force, it is a relationship between force, mass and acceleration.
There are apparently different ways of interpreting Newton's laws (what's a definition, what's an axiom, etc.). But I'm just stating what is, I think, the conventional way of teaching it. Although as I alluded to earlier, I think there's a good deal of confusion about this even among folks trained in physics. Marion and Thornton have a nice discussion of the philosophical foundations of the laws (pgs. 49-50 in my edition):

These laws are so familiar that we sometimes tend to lose sight of their true significance (or lack of it) as physical laws. The First Law, for example, is meaningless without the concept of "force," a word Newton used in all three laws. In fact, standing alone, the First Law conveys a precise meaning only for zero force...

In pointing out the lack of content in Newton's First Law, Sir Arthur Eddington observed... that all the law actually says is that "every particle continues in its state of rest or uniform motion in a straight line except insofar as it doesn't." This is hardly fair to Newton, who meant something very definite by his statement. But it does emphasize that the First Law by itself provides us with only a qualitative notion regarding "force."

The Second Law provides an explicit statement... The definition of force becomes complete and precise only when "mass" is defined. Thus the First and Second Laws are not really "laws" in the usual sense, rather they may be considered definitions. Because length, time, and mass are concepts normally already understood, we use Newton's First and Second Laws as the operational definition of force. Newton's Third Law, however, is indeed a law. It is a statement concerning the real physical world and contains all the physics in Newton's laws of motion.

The reasoning presented here, viz., that the First and Second Laws are actually definitions and the Third Law contains the physics, is not the only possible interpretation. Lindsay and Margenau for example, present the first two Laws as physical laws and then derive the Third Law as a consequence.

Dale
ZapperZ
Staff Emeritus
Marion and Thornton have a nice discussion of the philosophical foundations of the laws (pgs. 49-50 in my edition):
I tend to agree with their view on this. To me, the 3rd law is a manifestation of a more underlying principle, which is the principle of conservation of linear momentum (which of course, via the Noether theorem, is connected to the linear translation symmetry of space).

So yes, to me, this is where all the physics is in Newton's laws.

Zz.

Dale
There are apparently different ways of interpreting Newton's laws (what's a definition, what's an axiom, etc.). But I'm just stating what is, I think, the conventional way of teaching it. Although as I alluded to earlier, I think there's a good deal of confusion about this even among folks trained in physics. Marion and Thornton have a nice discussion of the philosophical foundations of the laws (pgs. 49-50 in my edition):
Very true indeed. First and second law only gives the operational definition of Mass or Force.