Posted by Matthew Goldberg on .
Driving in your car with a constant speed of v = 11.7 m/s, you encounter a bump in the road that has a circular cross section, as indicated in the figure below. If the radius of curvature of the bump is r = 34.4 m, calculate the apparent weight of a 65.9 kg person in your car as you pass over the top of the bump.
I honestly don't even know where to start with this question. I am having a very hard time with apparent weight in general so please don't criticize me. Thank you for any and all help!

PHYSICS 
drwls,
Assume that car tires and shock absorbers do not "cushion" the bump, so that the passenger's center of mass follows the profile of the bump.
The apparent weight is the force that the seat applies to the person to keep him or her in place. At the top of the bump, less force is required because there is a downward centripetal acceleration.
The person's actual weight is M g = 645.8 N
That gets reduced by
M V^2/R = 262.2 N
temporarily, due to the bump
The apparent weight is then 383.6 N 
PHYSICS 
Damon,
The centripetal acceleration is down, toward the center of the circle
Ac = v^2/R = 11.7^2/34.4 = 3.98 m/s^2
The gravity force down on driver = m g = 65.9*9.8 = 646 Newtons
The force up on the driver from the car is F, the weight a scale would measure if placed on the seat.
so
total force down on driver = m (Ac)
646  F = 65.9 (3.98)
F = 646  262
F = 384 Newtons