Rotation of the Earth and Apparent Weight?

by wmrunner24
Tags: centripetal, force, rotation, weight
wmrunner24 is offline
Oct22-10, 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


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.


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]=1153.3646 N

Is this line of thinking correct? Thanks in advance for any help provided.
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Xerxes1986 is offline
Oct22-10, 11:55 AM
P: 50
Yes that is correct.

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