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

  1. Apr 13, 2018 #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!
     
  2. jcsd
  3. Apr 13, 2018 #2

    russ_watters

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    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.
     
  4. Apr 13, 2018 #3

    navneet9431

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    In my textbook
     
  5. Apr 13, 2018 #4

    russ_watters

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    please provide more context than that. What does your textbook say about it? How do they use it.
     
  6. Apr 13, 2018 #5

    navneet9431

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    Check this link
     
  7. Apr 13, 2018 #6

    Dale

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    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.
     
    Last edited: Apr 13, 2018
  8. Apr 13, 2018 #7

    CWatters

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    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.
     
  9. Apr 14, 2018 #8

    navneet9431

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    Thank you, everyone, for a reply!
    Now, I just want to know why is it very much necessary to get k=1 anyhow?
     
  10. Apr 14, 2018 #9

    CWatters

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    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.
     
  11. Apr 14, 2018 #10

    Dale

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    It is not necessary at all, but it is convenient.
     
  12. Apr 14, 2018 #11

    navneet9431

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    Thanks!
    But,can you please explain how it is convenient?
     
  13. Apr 14, 2018 #12

    Dale

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    Multiplication by 1 is easy. And 1 is easy to remember.

    What positive number could be easier than 1?
     
    Last edited: Apr 14, 2018
  14. Apr 14, 2018 #13

    russ_watters

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    So easy you can ignore it and still be doing it correctly!
     
  15. Apr 14, 2018 #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.
     
  16. Apr 14, 2018 #15

    Merlin3189

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    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.
     
  17. Apr 14, 2018 #16
    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.
     
  18. Apr 14, 2018 #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.
     
  19. Aug 17, 2018 #18

    berkeman

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    Say what?
     
  20. Aug 17, 2018 #19
    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!
     
  21. Aug 17, 2018 #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.
     
    Last edited: Aug 17, 2018
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