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A car travelling over a hill.

  1. Nov 21, 2006 #1
    Hi everyone,

    I'm having a bit of trouble with this homework problem. Here it is:

    A car is traveling at a constant speed along a hilly road, as the drawing shows. Point A is at the bottom of the dip, while point B is at the top of the hill. The radius of the curvature is 35.0 m at both points. The apparent weight of the car at point A exceeds the apparent weight at point B by a factor of four. What is the speed of the car?


    Fc=Centripetal force
    Fg=Force of gravity
    Fn=Normal force

    Fc = mv^2/r
    Fg = mg OR GMm/R (M = mass of larger mass, m = mass of smaller mass)

    At point A

    At point B

    Simultaneous Equations...


    I still have three unknown variables and I'm really stumped. My teacher never said if we need a to get a number answer, I can easily solve this for v in terms of Fn and m. The only other possibility is that at point B the car is going at critical speed, but the wording of the problem doesn't lead me to believe that this is true. Any help is appreciated.

    Edit: I made some progress with this problem.

    Solve the equation at point B for Fn.


    Substitute back into simultaneous equation...

    (3/2)g=(v^2/R) + (3v^2/2R)
    v=14.34 m/s
    Last edited: Nov 21, 2006
  2. jcsd
  3. Nov 21, 2006 #2
    Looks good. You didnt need to go to substitution if you had multiplied the point B equation by 4 before eliminating. Thats why it was wierd
  4. Nov 21, 2006 #3


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    Homework Helper
    Gold Member

    If you divide everything by m, the mass, you get:

    [tex]4a_n - g = \frac{v^2}{R} [/tex]

    [tex]g - a_n = \frac{v^2}{R}[/tex]

    where [tex] a_n [/tex] is the acceleration caused by the normal force. Get [tex]a_n[/tex] to cancel and you can solve for v. Try that it makes sense in my head anyway! Good Luck. (If I made a mistake, someone please correct me)

    (Actually, all u gotta do is get Fn to add out in the equations, then m cancels and you can solve for v)
    Last edited: Nov 21, 2006
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