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Force Formula - Question

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Does the force formula ##F=m χ 10\frac{N}{kg}## which means ##F=G##, apply for bodies on rest or motion?


    2. Relevant equations
    ##F=G=m χ 10\frac{N}{kg}##

    3. The attempt at a solution
    I think it only applies for bodies in rest.

    I'm waiting an answer , thanks!
     
  2. jcsd
  3. Feb 13, 2012 #2

    Doc Al

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    Staff: Mentor

    The formula W = mg gives you the weight of a mass m, regardless of whether it's at rest or in motion.
     
  4. Feb 13, 2012 #3
    Weight is G, not W.., anyways I wasn't asking about other formulas, i was asking about my formula ##F=mχ10\frac{N}{kg}##
     
  5. Feb 13, 2012 #4

    Doc Al

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    Staff: Mentor

    :rolleyes:
    Where did you get that formula? It sure looks like 'mg' to me, which is weight. Seems like you're using g = 10 m/s^2.
     
  6. Feb 13, 2012 #5
    I see you don't know that formula,we use that to find the Pushing force or weight of a body using the mass, which i think that the body is in rest. For example if we have a body with mass 2g, then the weight would be:

    ##G=mχ10\frac{N}{kg}=\frac{2}{1000}kg χ 10\frac{N}{kg} = \frac{200}{1000}N##(btw. kilogram and kilogram get simplified )

    What I wanted to ask is if that formula applies also if the body is in motion
     
  7. Feb 13, 2012 #6
    No ! Doc Al got you correct. Your formula is nothing but of weight which is mass times gravity.

    F=mg

    Newton per kg in your unit gives you gravitational pull..

    1 N/kg = 1 m/s2

    You are assuming gravity in your formula to be 10 m/s2.

    So you state that weight = mg = m*10 N/kg
    OR
    weight = m*10 m/s2
    Of course , I guess everyone here know that kg and kg gets simplified..:rolleyes:

    Your formula is nothing but simple mg where you substituted for g by 10 metre per second squared.

    Edit : Of course this formula applies when body is in motion. Doc Al already stated that in previous posts.
     
  8. Feb 13, 2012 #7
    So, the gravity is not always ##10\frac{N}{kg}##, then why is the Weight/Force formula
    ##F=mχ10\frac{N}{kg}##
     
  9. Feb 13, 2012 #8

    Doc Al

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    Staff: Mentor

    That formula applies near the earth's surface.
     
  10. Feb 13, 2012 #9
    Thank you very much, I am new on physics, I am only on the first grade so , I need to learn more :) Thanks@!
     
  11. Feb 13, 2012 #10
    Anyways, can the body near the earth surface be in motion and in rest aswell, or only in rest or only in motion?
    ----------------------
    Would...:
    ##1\frac{N}{kg}=1\frac{m}{s^2}## because :
    ##1\frac{N}{kg}=1\frac{m}{s^2}=1\frac{kg\frac{m}{s^2}}{kg}=1\frac{m}{s^2}##kg and kg get simplified ##=1\frac{m}{s^2}=1\frac{m}{s^2}##
     
  12. Feb 13, 2012 #11

    Doc Al

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    Staff: Mentor

    I don't know what you mean. A rock (for example) can be in motion or at rest. (Not at the same time within the same reference frame, of course.)

    Sure, those units are equivalent.
     
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