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Centripetal force of a car on a curve

  1. Nov 9, 2007 #1
    1. The problem statement, all variables and given/known data
    A 600 kg car is going over a curve with radius 120m that is banked at angle of 25 degrees with speed of 30 m/s. The coefficient of static friction between car and road is 0.3. What is the normal force?


    2. Relevant equations
    F(centripetal)= (mv^2)/r
    F(static) = Mu(normal force)


    3. The attempt at a solution
    I think that the centripetal force is equal to the friction force. So I can just substitute centripetal force for friction. But my teacher also said that is a parallel force but I'm not sure it is. From the free-body diagram, I will get..
    F(c)= (Friction + Parallel force) Cos(angle)

    I'm actually quite confused ~_~
     
    Last edited: Nov 9, 2007
  2. jcsd
  3. Nov 9, 2007 #2

    Astronuc

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    Staff: Mentor

    See if this helps - http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html#c1

    The normal force means the normal force on the back which comes from the component of the weight point normal to the bank and the component of centrifugal force pointing normal to the bank.

    The weight points downard (vertically) while the centrifugal/centripetal force points horizontally, but both have components pointing normal to the bank.
     
  4. Nov 9, 2007 #3
    cool, thanks
    it makes more sense now..
    except how come the friction force is up there? usually my teacher draws it on the other side of the car, like on the same side as the centripetal force. does it matter? since its negative now instead of positive and etc?
     
  5. Nov 9, 2007 #4

    Astronuc

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    Staff: Mentor

    The friction force operates at the contact surface between tires and bank (road surface). It doesn't matter where one draws the arrow, although it should be correctly drawn at the tire/road interface, but the direction should point down the bank since the friction opposes the force which try to push the car up the bank.
     
  6. Nov 9, 2007 #5
    i got:


    x-component: Nsin25 + mu(N)(cos25) = (mv^2)/r
    y-component: Nsin25 - mg - mu(N)(sin25) = 0

    i'm not sure how to find N from that since, theres like 2 N's in each equation? and i cant seem to just factor it out...or did i just get the equation wrong entirely?
     
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