Uniform Circular Motion Free Body Diagrams

[archelon]

Homework Statement

A circular-motion addict of mass 82.0 kg rides a Ferris wheel around in a vertical circle of radius 14.0 m at a constant speed of 7.10 m/s. (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?

Homework Equations

a = v^2/r = 3.6 m/s
ma = -295.2
mg = -804.42

The Attempt at a Solution

I already know the answer to this problem (12.39 s, 509.22 N, and 1099.62 N respectively) but I'm having trouble understanding the FBD that leads to the answers for (b) and (c).
For example, for the FBD at the time of the Ferris wheel, I have the force from centrifugal acceleration and the force from gravity pointing towards the center and the normal force pointing opposite from those two. Then I solve for the normal force and get -Fn=Fa+Fg, or in other words -Fn = ma + mg, since the normal force has to cancel out the two forces pointing downwards. So the answer I keep getting is -Fn = -295.2 + -804.42. The same thing (just the opposite) happens to me at the bottom.
Please show me where I am mistaken and help me understand the correct FBD's.

 

[PhanthomJay]

You should avoid the conecept of centrifugal force caused by centripetal acceleration, because it is a ficticious 'pseudo' force. Instead, there are only 2 forces acting in your FBD, the normal force and the weight (gravity) force. At the top of the ferris wheel, the normal force acts up and the weight acts down. The sum of these 2 forces, that is, the net force, is equal to ma, per Newton 2,where a is the centripetal acceleration acting inward (down) toward the center of the circle. The centripetal acceleration is always inward toward the center of the circle, and always in the direction of the net force.