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Gravitational Acceleration Problem

  1. Apr 2, 2009 #1
    1. The problem statement, all variables and given/known data
    A 95 kg man jumps, therefore exerting a force of 1000 N on a bathroom scale. The scale is in a motionless elevator in Earth's gravity. What is the man's acceleration?


    2. Relevant equations
    Newton's 2nd Law: F=ma
    Possibly Fg=mg ?


    3. The attempt at a solution
    I tried just using F=ma.
    therefore, it'd be:
    1000 N=(95kg)(a)
    a=10 m/s^2

    But this does not take into consideration the earth's gravity.
     
  2. jcsd
  3. Apr 2, 2009 #2

    Redbelly98

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    Welcome to PF.

    "F=ma" uses the net force (vector sum of all forces acting on the man)

    Your F uses the 1000N force that the scale exerts on the man. As you suspect, you need to add the force due to Earth's gravity. Just take the vector sum of the scale force and the force due to gravity.
     
  4. Apr 2, 2009 #3
    So would it be:

    F(normal)-F(gravity)=ma
    F(normal) - mg=ma
    1000 N [up] - (95 kg)(9.8 N/kg [down])=(95 kg)(a)
    1000 N [up] - 931 N [down] = (95 kg)(a)
    69 N [up] = (95 kg)(a)
    0.73 m/s^2 = a

    This seems very low..
     
  5. Apr 2, 2009 #4
    wait!

    instead of:
    69 N [up] = (95 kg)(a)
    0.73 m/s^2 = a

    is it:
    1931 N [up] = (95 kg)(a)
    20. m/s^2 = a
     
  6. Apr 2, 2009 #5

    Redbelly98

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    Looks good to me.

    Note, the man's weight is about 930 N, so the scale would read 930 N when a=0.

    The scale reading of 1000N is not much more than this, so expect a to be considerably less than g=9.8 m/s^2.

    An acceleration of g=9.8 m/s^2 upward would require a scale reading of 2*930N, which would give Fnet = +930 N upward.

    EDIT:
    Nope. We have +1000N (upward) and -931N (downward), for Fnet=31N upward
     
  7. Apr 2, 2009 #6
    Alright.
    Thank you!!
     
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