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Applications of Newton's Laws

  1. Aug 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A meteor of mass .25 kg is falling vertically through Earth's atmosphere with an acceleration of 9.2 m/s. In addition to gravity, a vertical retarding force (due to the frictional drag of the atmosphere) acts on the meteor. What is the magnitude of this retarding force?


    2. Relevant equations
    F = ma

    but I think, because of "in addition to gravity"

    F = ma + mg


    3. The attempt at a solution

    Fm-Ffr= ma + mg

    Fm = (.25)(9.2) = 2.3 N


    -Ffr= ma + mg - Fm

    -Ffr = mg

    = -2.45 N
    the friction force greater than the weight? Doesnt sound right
     
  2. jcsd
  3. Aug 18, 2012 #2

    Doc Al

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    Staff: Mentor

    No, you were right the first time:
    ƩF = ma

    Fix this.
     
  4. Aug 18, 2012 #3
    So, Fm-Ffr= ma

    -Ffr = ma - Fm

    But that equals 0. What other forces are in act here?
     
  5. Aug 18, 2012 #4

    Doc Al

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    Why do you think it equals 0? (I assume Fm is the object's weight, right?)
     
  6. Aug 18, 2012 #5
    Fm is the meteor's weight, yes.

    So gravity plays no part in this?

    Now I am thinking this
    (im switching to Y+ up, i forgot to consider that i was making Y+ down)
    ƩF = ma

    Ffr - Fm - w = may

    Ffr = may + Fm + W

    I think this is correct because there is 1 force pulling the meteor back, the Force of friction, and 2 forces acting down, the weight of the meteor(with earths gravity) and the meteor moving with the 9.2 m/s2 acceleration.

    Or am I just adding the weight of the meteor twice?
    I think I am overthinking this.
     
  7. Aug 18, 2012 #6

    Doc Al

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    What do you think gravity is?
     
  8. Aug 18, 2012 #7
    Okay, I'm settling with this.

    Ffr - Fm = may

    Ffr = may + Fm

    Ffr = (.25kg)(9.2m/s2) + (.25kg)(9.8m/s2)

    Ffr = 4.75 N
     
  9. Aug 18, 2012 #8

    Doc Al

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    Almost. What's the sign of the acceleration?
     
  10. Aug 18, 2012 #9
    negative

    Ffr = (.25)(-9.2) + (.25)(-9.8)

    Ffr = -4.75 N

    Thanks a bunch!
     
  11. Aug 18, 2012 #10

    Doc Al

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    The acceleration of the object is -9.2; the weight is just mg, not -mg. (You already included the direction of the weight--downward--in your first equation.)

    One more time!
     
  12. Aug 18, 2012 #11
    Ffr = (.25)(-9.2) + (.25)(9.8) = .15 N

    finally, 5th times the try.

    Thank you very much for the help
     
  13. Aug 18, 2012 #12

    Doc Al

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    Good! And you're welcome.

    It's always useful to do a 'sanity check' of your answers. In this case, you know the acceleration is just a little bit less than free fall acceleration. That should tell you that the resistance is small compared to the weight.
     
  14. Jan 5, 2014 #13
    I 'am curious would the total force be: Fm+Ffr with Fm=(m*-g)+(m*-a)+Ffr. Therefore Ft=-4.6N
     
  15. Jan 5, 2014 #14

    Doc Al

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    No. Only two forces act: gravity and the retarding force. You can find the net force directly using ∑F = ma.
     
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