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dan667
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Homework Statement
An object weighs 100N at the South Pole. How much does it weigh at the equator?
Given: Earth's rotational spin speed 465m/s
Diameter of the Earth is 1.274 x 10^7 m
Homework Equations
Fc = m x v^2 / r
The Attempt at a Solution
Well I have this general understanding that the object should weigh less since it's experience uniform circular motion, and that it "wants" to fly off tangentially (Newton's first law) but is constrained by the force of friction, right? It seems I have calculated the answer of the centripetal force to be 0.3 N, but what I don't understand is that why I subtract it from 100N. The confusion lies in where the normal force and centripetal force vectors are located. It would be greatly appreciated if anyone could give a lucid answer to this question. The answer by the way, is 99.7N