1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Car Driving over a hill

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill the normal force on the driver from the car seat is 0. The driver's mass is 70.0 kg. What is the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley??


    2. Relevant equations
    Fg - Fn = 0 ?
    F = mv^2/r


    3. The attempt at a solution
    Stumped. I looked at this question and was like... wtf?

    Only thing I can think of is Fg - Fn cannot equal zero because there is always Fg and if Fn is 0, then that equation doesn't make sense. But I don't know any better than this equation. I'm so confused!! =S
     
  2. jcsd
  3. Feb 22, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    At the top of the hill if his normal force is 0, then what must his centripetal acceleration be?
     
  4. Feb 22, 2009 #3
    Fg = 9.8 * 70 = 686
     
  5. Feb 22, 2009 #4

    LowlyPion

    User Avatar
    Homework Helper

    How is the direction of the centripetal acceleration different at the top as opposed to the valley?
     
  6. Feb 22, 2009 #5
    At the top, it is directed downwards toward the centre, at the bottom is directed up toward the centre...
     
  7. Feb 22, 2009 #6

    LowlyPion

    User Avatar
    Homework Helper

    ... and you still can't answer the original question?
     
  8. Feb 22, 2009 #7
    Fn = Fg .. which would be al;so 686 at the bottom of the hill.

    The answer should be 1.37 x 10^3 N

    =P

    No I still can't answer the original question.

    Edit: and yes I see the answer is 2* 686... I have no idea why.
     
  9. Feb 22, 2009 #8

    LowlyPion

    User Avatar
    Homework Helper

    In the first case Fn = 0 = mg - mv2/r

    That means that mv2/r = mg.

    In the valley

    Fn = mg + mv2/r

    But you just figured out that mg = mv2/r so

    Fn = 2*mg
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Car Driving over a hill
  1. Driving a car (Replies: 3)

Loading...