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Rotation of the Earth and Apparent Weight? 
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#1
Oct2210, 11:43 AM

P: 57

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]=mgma[tex]_{centripetal}[/tex] F[tex]_{normal}[/tex]=1153.3646 N Is this line of thinking correct? Thanks in advance for any help provided. 


#2
Oct2210, 11:55 AM

P: 50

Yes that is correct.



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