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How can you calc/estimate the gravitational force of a building on a person inside?

  1. Aug 2, 2011 #1
    1. The problem statement, all variables and given/known data
    I've posted this in the homework section but really, it was asked by a young kid who had just been learning about gravity in high school science. It made my realise that I wasn't able to provide a superb answer.

    So, you have a man standing inside a building, what is the gravitational force on him? Of course we are only talking about estimates and ball park figures here. If he were standing right in the middle of the building, and it had an evenly distributed mass, then it would be right to assume that the net gravitational force would be zero since it would all cancel?

    What about if the person were right on the left edge of the building? I have so little idea about weights of buildings that I'm struggling to even think of a ball park figure.


    2. Relevant equations

    I have a hunch that F = Gm1m2/r2 will come into it...

    3. The attempt at a solution

    Right now I am thinking that "on average, the net gravitational force will be zero because the forces in each direction will all cancel out" but that sounds a bit unsatisfactory?

    Cheers in advance for any help.
     
  2. jcsd
  3. Aug 2, 2011 #2

    tiny-tim

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    Hi Einstein2nd! :smile:
    Yes. :smile:
    A building weighs about the same as a ball park. :wink:
    You may as well assume that the whole mass of the huilding is concentrated at its centre.

    If the distance to the centre of the building, and to the centre of the Earth, are r and R,

    and if the mass of the building, and of the Earth, are m and M,

    then the extra gravitational acceleration from the building wil be g times m/M times R2/r2 :smile:
     
  4. Aug 2, 2011 #3
    Re: How can you calc/estimate the gravitational force of a building on a person insid

    Yeah that seems like sound logic. For my own curiosity, what about tidal forces? If you were standing in the middle of the building there could still be tidal forces?

    To make things simpler, assume it is a 1 storey building. Then perhaps divide the mass into 4 and assume that the forces are equal in all directions (as said before, so the net force on the person is zero).

    Take a person to be a 70kg, 1.7 metre sphere :p Consider the force in say the north direction, due to 1/4 of the building's mass, to be concentrated half way between the person (in the middle) and the edge of the building. I guess you could then calculate the difference in force on the near and far side of the sphere and that's the tidal force. This would be repeated 3 times and you would see that the person is actually being pulled apart! Maybe take into account only 60% of the human's body being water?
     
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