# Rotation of the Earth and Apparent Weight?

by wmrunner24
Tags: centripetal, force, rotation, weight
 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$$_{centripetal}$$=ma$$_{centripetal}$$ F$$_{gravity}$$=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$$_{centripetal}$$=F$$_{gravity}$$-F$$_{normal}$$ 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$$_{normal}$$=F$$_{gravity}$$-F$$_{centripetal}$$ F$$_{normal}$$=mg-ma$$_{centripetal}$$ F$$_{normal}$$=1153.3646 N Is this line of thinking correct? Thanks in advance for any help provided.
 P: 50 Yes that is correct.

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