Rotation of the Earth and Apparent Weight?

[wmrunner24]

Homework Statement

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.

Homework Equations

F_{centripetal}=ma_{centripetal}

F_{gravity}=mg

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.

 

[Xerxes1986]

Yes that is correct.