# Force Vectors and Circular Motion

1. Oct 11, 2006

### Ronnin

This is a case where I know the results but am fighting the math. At the top of a ferris wheel your acceleration in Y is 0. Ol' Newton says all all forces then must cancel, got that. I know my weight at this point will feel less at this point. Here is where I am troubled, vector for gravity points down so does the vector for the circle's accleration but my Normal points up and must have the same magnitude as Fg and Fc. The answer given subtracts Fc from Fg to get Fn (which feels right, but contradicts my vectors). Any conceptual help is appreciated.

2. Oct 11, 2006

### Chi Meson

Your acceleration at the top of a ferris wheel is NOT zero. Assuming you are undergoing perfect circular motion, you will have a centripetal force at the top. Gravity will supply the centripetal force, but all of the weight will be too much centripetal force, so therefore the normal force will balance some of the weight.

In perfect circular motion, the only unbalancd force will be centripetal and equal (mv^2)/r . At the top of the ferris wheel the unbalanced force is equal to mg-N.