1. ## Normal Forces

While driving along a country lane with a constant speed of 17.0 m/s, you encounter a dip in the road. The dip can be approximated as a circular arc, with a radius of 65.0m. What is the normal force exerted by a car seat on a 80.0kg passenger when the car is at the bottom of the dip.

Do you have to do 2 equations here?? 1 will be centripetal force. Fc=mv^2/r?
The other v=d/t???
I am not quite sure on that one.
Any help to get me started would be great. Or do I have it wrong?
THanks again

2. Yes, you'll have centripetal force, but you'll also have gravity (pushes the passenger into the seat; hence, by Newton's 3rd law, the seat pushes back into the passenger). Just add them up to get the total.

3. The normal force provides the centripetal force.

So, find $\displaystyle F_c = \dfrac{80(17^2)}{65} = 355.7 N$

This is the resultant force. Hence the normal force = Resultant force - Weight = 355.7 - (-80g)

4. Ok so I find F=ma which is 80kg X 9.8 m/s.s gives me 784 added to 355 I get 1140 N correct??

5. Close enough, I think.

6. This is what I got