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Acceleration of a stationary mass?

  1. Apr 14, 2015 #1
    so sitting in a chair gives me an acceleration of 1g. clearly I am not moving relative to the chair and Earth yet acceleration "a" is defined as the change in velocity ie; (Vfinal - Vinitial)/time

    in the chair Vfinal = Vinitial = 0 so by definition (and calculation) a = (Vfinal - Vinitial)/time = 0/t = 0

    so why am I accelerating at 1g when I just correctly calculated my acceleration to be 0m/s/s
     
  2. jcsd
  3. Apr 14, 2015 #2

    DrClaude

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    You're not accelerating at 1 g. You are feeling a force from gravity that is equal to your mass times g, and your chair is pushing you up with an equivalent force, such that the net force on you is zero, hence you are not accelerating.

    The fact that the force of gravity is ##F = mg## does not mean that you are being accelerated by ##g##, simply that you would be if no other forces were acting on you.
     
  4. Apr 14, 2015 #3
    Dr Claude given that F = ma or a = F/m, F is not zero, m is not zero so how do you get a to be zero?

    I thought F might be net force that cancels to zero ie gravity down reaction of chair up but F is a measurable quantity ie an accelerometer on the chair will give a non zero number.
     
  5. Apr 14, 2015 #4

    DaveC426913

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    Asked and answered:
     
  6. Apr 14, 2015 #5

    A.T.

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  7. Apr 14, 2015 #6

    DrClaude

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    Along the vertical direction:
    $$
    F_\mathrm{total} = F_\mathrm{gravity} + F_\mathrm{chair} = 0
    $$
    The force fro the chair exactly cancels out the force of gravity, even though ##F_\mathrm{gravity} = mg \neq 0##.


    How do you think that the accelerometer measures ##F_\mathrm{gravity}##? (Hint: think about the example of the chair.)
     
  8. Apr 14, 2015 #7
    i think i am getting it thanks, will work through the link (seems to be the ticket) to see the difference in defining accelerations.

    Geez that link goes into this in the explanation;

    4eea48349ad91c80da67232b83fdce7f.png
     
  9. Apr 14, 2015 #8
    OK so;

    Fg = - Fc giving Fg+Fc = 0 = Ftotal = 0

    I get that but it does raise the question if Ftotal = 0 why do I feel my own weight?

    I have no idea how an accelerometer works, I have a digital one, I doubt there is a spring and ball inside it.
     
  10. Apr 14, 2015 #9

    A.T.

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    You don't. You feel deformations in your body caused by the non-uniformly applied contact force form the chair.

    http://en.wikipedia.org/wiki/Accelerometer#Structure
     
    Last edited: Apr 14, 2015
  11. Apr 14, 2015 #10

    DrClaude

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    We'll get to that question when we have finished with the other.

    Let's stick with old-fashioned measuring apparatus. How does a dynamometer measure force?
    Dynamometer_small.jpg
     
  12. Apr 14, 2015 #11
    simple calibration of the spring under tension via Hooke's law restoring force, write a scale on the side in preferred units.

    the spring is not registering a force if it is unloaded but it still "feels" 1g force.
     
  13. Apr 14, 2015 #12
    going off line, not being rude if I do not respond to further posts for awhile.

    will check back here later.
     
  14. Apr 14, 2015 #13

    DrClaude

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    In other words, the dynamometer is measuring the force necessary for the spring to counter the force due to gravity, the equivalent of ##F_\mathrm{chair}## above.

    That's an arbitrary choice of the zero, since we want to measure the weight of a mass added to the spring. The spring can be slightly elongated compared to when it is at rest horizontally.
     
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