Bathroom scale on one of the Earth's poles

[psruler]

I need help on this problem:

Suppose you stand on a bathroom scale on one of the Earth's poles, and it reads 700.0 N. If you stand on the same scale on the Earth's equator, and your mass is unchanged, what will be the scale reading on the equator? (Remember that the scale tells you the normal force of the ground on you.) Assume the Earth is a perfect sphere of constant radius.

 

[HallsofIvy]

The point of this problem is that the scale reads the "net" force on you: the force of gravity minus the "centrifugal force" due to rotation.

(I just know I'm going to get into trouble for saying that! People are going to complain that "centrifugal force" is a fictitious force. Yes, that's why I put it in quotes! It would be more correct to say that part of the gravitational force goes to keep you rotating with the Earth and the rest to pressing you against your scale.)

At the pole you are not "rotating" (you are kind of slowly pivoting in place). The scale reads the gravitational force.

At the equator you are rotating. You are going in a circle equal to the circumference of the Earth every 24 hours. Calculate the (centripetal) force necessary to do that and subtract from your weight at the pole.

 

[M98Ranger]

The scale reading on the equator will be slightly less than 700.0 N due to the centrifugal force caused by the Earth's rotation. This force is strongest at the equator and decreases towards the poles. Therefore, the normal force of the ground on you will be slightly less on the equator compared to the poles. However, the difference in scale readings will be very small and likely not noticeable on a regular bathroom scale. Additionally, the scale reading may also be affected by the slight difference in gravitational pull at the equator compared to the poles, but again, this difference would be very small. Overall, the scale reading on the equator would be very close to 700.0 N, but slightly less due to the effects of the Earth's rotation.