Car on hill centripetal force One of the exercises in my textbook (Giancoli) deals with a car going up the slope of a hill and then decending into a valley. Curvature/radius is $R$ in both situation, and I'm supposed to answer the question at which point (top of hill/lowest point in valley)
the normal force is smallest and largest respectively. Intuitively I'm ok answering the question, in that I feel lightest at the top of the hill (lowest normal force), and heavier at the bottom of the valley (largest normal force), but I'm having a hard
time figuring out the equations. In the answers section it says, at the crest of the hill the centripal force is directed downwards, so the normal force  must be less than the weight. But force implies acceleration in the same direction, does it not? But if there was acceleration in the y direction, at that very instant, would the car not "sink" into the ground?
edit: I'm sorry this is most certainly not a homework assignment, I'm 45 and I'm pursuing physics out of interest in astronomy.
 A: If the car were at rest at the top of the hill, then you'd be right:  a non-zero downwards acceleration would mean that it would be gaining downwards velocity, which would mean that it would be sinking into the road.  Obviously that would be bad.
But the car is moving past the peak of a curved road.  An instant before it gets to the peak, its velocity vector has an upwards component, because it's still going uphill.  An instant after, its velocity vector has a downwards component, because now it's going downhill.  Since its vertical velocity is changing (downwards) in this small span of time, the car must have a downwards acceleration at the peak of the road.
Another way of looking at it is that the car does move downwards after it gets to the peak of the road.  If the car had no horizontal velocity, it would sink into the road.  But because the car has a horizontal velocity as well, it doesn't sink into the road;  it moves further along the road, to a point that is vertically below the peak but still on the road surface.
