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Rotation of the Earth and Apparent Weight?

  1. Oct 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Because of Earth’s rotation about its axis, a point on the Equator experiences a centripetal acceleration of 0.034 m/s2, while a point at the poles experiences no centripetal acceleration.
    What is the apparent weight at the equator of a person having a mass of 118.1 kg? The
    acceleration of gravity is 9.8 m/s^2 .

    Answer in units of N.

    2. Relevant equations
    F[tex]_{centripetal}[/tex]=ma[tex]_{centripetal}[/tex]

    F[tex]_{gravity}[/tex]=mg

    3. The attempt at a solution

    So, at first this problem greatly confused me. Now I think I have an idea of how to approach it.

    F[tex]_{centripetal}[/tex]=F[tex]_{gravity}[/tex]-F[tex]_{normal}[/tex]

    This equation describes the net force toward the center of the Earth. Gravity acts toward the center of the Earth, while the normal force will resist it. If a person were to be standing on a scale, it would be the normal force that would produce the "apparent" reading. So then:

    F[tex]_{normal}[/tex]=F[tex]_{gravity}[/tex]-F[tex]_{centripetal}[/tex]

    F[tex]_{normal}[/tex]=mg-ma[tex]_{centripetal}[/tex]

    F[tex]_{normal}[/tex]=1153.3646 N

    Is this line of thinking correct? Thanks in advance for any help provided.
     
  2. jcsd
  3. Oct 22, 2010 #2
    Yes that is correct.
     
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