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

In summary: So "the force" in this equation is really just a word for "the force of gravity."Newtons laws lead to F=ma because he could not have known what people would use. He was working with imperial units, which are unit systems that were designed that way. 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.
  • #1
navneet9431
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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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  • #2
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.
 
  • #3
russ_watters said:
I've never heard of f=kma. where did you get it?
In my textbook
 
  • #4
navneet9431 said:
In my textbook
please provide more context than that. What does your textbook say about it? How do they use it.
 
  • #5
russ_watters said:
please provide more context than that. What does your textbook say about it? How do they use it.
Check this link
 
  • #6
navneet9431 said:
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.

navneet9431 said:
Check this link
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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  • #7
navneet9431 said:
Check this link
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.
 
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  • #8
Thank you, everyone, for a reply!
Now, I just want to know why is it very much necessary to get k=1 anyhow?
 
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  • #9
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.
 
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  • #10
navneet9431 said:
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.
 
  • #11
Dale said:
It is not necessary at all, but it is convenient.
Thanks!
But,can you please explain how it is convenient?
 
  • #12
navneet9431 said:
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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  • #13
Dale said:
Multiplication by 1 is easy.
So easy you can ignore it and still be doing it correctly!
 
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  • #14
Where did you find a "derivation" of F = m*a? I've never seen such, but rather understood it as more of a definition.
 
  • #15
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.
 
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  • #16
Merlin3189 said:
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.
 
  • #17
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.
 
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  • #18
hiffy said:
There is no experiment you can do test F=ma.
Say what?
 
  • #19
berkeman said:
Say what?

I mean simply that we conventionally take [itex] F=ma[/itex] 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 [itex] v=\frac{dx}{dt}[/itex]

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!
 
  • #20
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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  • #21
hiffy said:
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.
 
  • #22
hiffy said:
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).
 
  • #23
berkeman said:
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.
 
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  • #24
hiffy said:
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.
 
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  • #25
hiffy said:
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.
 
  • #26
e-pie said:
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.

I don't think you can really test a definition. I mean, with v=dx/dt, you could measure the distance your car travels in a certain time and compare that to what your speedometer says. But I think if you really analyze it, you'd find you're not really testing v=dx/dt. That statement is true by definition. You're really testing that whatever mechanism your car uses to output speed is consistent with v=dx/dt.

And similarly, I don't think you can test F=ma as a standalone law. That statement is true by definition. You're really testing Newton's Third Law. For instance, try analyzing the trolley experiment using just F=ma, without the Third Law. It can't be done because to analyze it, you need the concept of tension. You can try to graph "force" versus acceleration, but what is the "force" on the trolley? It's the force due to the tension in the cable, but the concept of tension only makes sense if forces come in equal and opposite pairs (the Third Law).

I guess this discussion just goes to show that there are even different interpretations of Newton's Laws, and that, as in quantum mechanics, for most practical purposes, physicists choose the "shut up and calculate" interpretation.
 
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  • #27
Put it in this way.

I am tabulating a velocity motion. Ideal experiment case.

Time Distance
0s 0 m
1s 10 m
2s 20 m
3s 30 m
...s ...m

Therefore from maths slope m=(d2-d1)/(t2-t1)=10 m/s.
... This formula gives a simple representation for the entire data range.

Upto this point I have not defined what m is physically. Now let's assign it a quantity based on dimensions. Say I name it for no particular reason "Velocity" and its dimension is m/s from slope. Then my slope formula gives that v=del x/del t. Thus I can define velocity to be the rate of change of position with respect to time.

Therefore I derived the equation and definition of velocity from experiments.
And I can test my equation/definitions in different inertial frames, different conditions, different sets of values. If they all match and produce expected results then my equations are correct through testing.
 
  • #28
e-pie said:
Therefore I derived the equation and definition of velocity from experiments.
Strictly speaking, that's not what is is usually meant by "derive". You are discovering, by analyzing your observational data, that a particular mathematical formula describes the behavior of the universe rather well.
 
  • #29
Nugatory said:
Strictly speaking, that's not what is is usually meant by "derive". You are discovering, by analyzing your observational data, that a particular mathematical formula describes the behavior of the universe rather well.

Indeed yes. "Derivation" is associated more commonly as producing an equation/proof from a given set of rules.

But I would define "derive" as a verb as: to obtain logical results by performing a sequence of interconnected logical steps based on a given set of rules(axiom/proposition etc).
 
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  • #30
But can we not "derive" mathematical forms from experiments?
Like Millikan's oil drop experiment. q/m ratio for electron?
 
  • #31
hiffy said:
But I'm just stating what is, I think, the conventional way of teaching it.
I am not sure it is the conventional way, but I agree with you on the validity of this approach. And since most people think of physics in terms of SI units I think this is the best view.
 
  • #32
Dale said:
I think this is the best view.

Maybe but what is physics without units? What is physics without some particular case?

To quote Feynman on this in my own words: I need a formula for Euclidean space. Mathematicians will first derive the base case say for n=1,2,3 and further generalize it into n. But Physics don't always require particular cases, Physicists need the formula for n=3. Physics is an exact science whereas mathematics to some extent is not(as it always gives generalized versions of everything).

Mass of a bar is 1 or n does not make any sense if no unit is given. Atleast it is true for Physics.
 
  • #33
e-pie said:
Maybe but what is physics without units?
This is a weird response. Did I suggest doing physics without units?
 
  • #34
Let's clear up on what each other is suggesting. I rushed a bit:smile:.

You go first.
 
  • #35
e-pie said:
Let's clear up on what each other is suggesting. I rushed a bit:smile:.

You go first.
Newton’s 2nd law is a definition in SI units. It cannot be tested using SI units.
 
<h2>1. Why do we use k=1 in the derivation of F=k*ma?</h2><p>The value of k=1 is used in the derivation of F=k*ma because it simplifies the equation and makes it easier to understand. It also allows us to focus on the relationship between force, mass, and acceleration without any additional variables.</p><h2>2. Is k=1 a universal constant in the equation F=k*ma?</h2><p>No, k=1 is not a universal constant in the equation F=k*ma. The value of k can vary depending on the units used for force, mass, and acceleration. For example, if force is measured in newtons, mass in kilograms, and acceleration in meters per second squared, then k=1. However, if different units are used, the value of k will change accordingly.</p><h2>3. How does changing the value of k affect the equation F=k*ma?</h2><p>Changing the value of k will directly affect the equation F=k*ma. As k is a constant, any change in its value will result in a proportional change in the value of force. For example, if k is doubled, the force will also be doubled.</p><h2>4. Can we use a different variable instead of k in the equation F=k*ma?</h2><p>Yes, we can use a different variable instead of k in the equation F=k*ma. The variable k is commonly used to represent the constant of proportionality between force, mass, and acceleration. However, other variables such as c or G can also be used to represent this constant.</p><h2>5. What is the significance of k=1 in the equation F=k*ma?</h2><p>The significance of k=1 in the equation F=k*ma is that it simplifies the equation and allows us to focus on the relationship between force, mass, and acceleration without any additional variables. It also helps us to understand the concept of proportionality between these variables and how they are related to each other.</p>

1. Why do we use k=1 in the derivation of F=k*ma?

The value of k=1 is used in the derivation of F=k*ma because it simplifies the equation and makes it easier to understand. It also allows us to focus on the relationship between force, mass, and acceleration without any additional variables.

2. Is k=1 a universal constant in the equation F=k*ma?

No, k=1 is not a universal constant in the equation F=k*ma. The value of k can vary depending on the units used for force, mass, and acceleration. For example, if force is measured in newtons, mass in kilograms, and acceleration in meters per second squared, then k=1. However, if different units are used, the value of k will change accordingly.

3. How does changing the value of k affect the equation F=k*ma?

Changing the value of k will directly affect the equation F=k*ma. As k is a constant, any change in its value will result in a proportional change in the value of force. For example, if k is doubled, the force will also be doubled.

4. Can we use a different variable instead of k in the equation F=k*ma?

Yes, we can use a different variable instead of k in the equation F=k*ma. The variable k is commonly used to represent the constant of proportionality between force, mass, and acceleration. However, other variables such as c or G can also be used to represent this constant.

5. What is the significance of k=1 in the equation F=k*ma?

The significance of k=1 in the equation F=k*ma is that it simplifies the equation and allows us to focus on the relationship between force, mass, and acceleration without any additional variables. It also helps us to understand the concept of proportionality between these variables and how they are related to each other.

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