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Centripetal Force (Horizontally Banked Question)

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data
    If a curve with a radius of 60m is properly banked for a car travelling at 60km/h (16.66m/s) with no friction, what must be the coefficient of friction if the car is not to skid when travelling at 90km/h (25m/s)


    2. Relevant equations
    mac = mv^2/r

    3. The attempt at a solution
    The answer is 0.39

    First we need to find the angle, so since it said that when the speed was 60km/h, there is No friction, we can make this equation.. and do the following calculations
    Fgx = mac
    mgsinɵ = mv^2/r
    gsinɵ = v^2/r
    9.8sinɵ = 16.66^2/60
    sinɵ = 4.625/9.8
    ɵ = 28.16

    So now we have the angle..
    We can now use the speed 90km/h with our new angle.
    So the new equation is..
    Ff + Fgx = mv^2/r
    μFn + Fgx = mv^2/r
    μmgcos28 + mgsin28 = mv^2/r
    μ9.8cos28 + 9.8sin28 = (25)^2/60
    μ(8.65) + 4.6 = 10.41
    μ = 10.41 - 4.6/8.65
    μ = 0.67

    So I'm not sure where I went wrong..
    the answer is 0.39 :O
     
  2. jcsd
  3. Nov 2, 2011 #2

    PeterO

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    Homework Helper

    You will be familiar with the idea that when you drive through a dip in the road, you feel heavier. This is because when you move through the dip, the reaction force is larger than just mg.
    A similar thing happens when you drive round a banked turn. The reaction force is stronger than when the vehicle is parked on the banked turn.
    The net result is that your second line : mgsinɵ = mv2/r
    should actually have been : mgtanɵ = mv2/r
     
  4. Nov 2, 2011 #3
    Hmm I got angle as 25.24.
    And then the resulting coefficient of friction as 0.59 :S

    By the way thanks for the other thread, I got the right answer for it ^^
    Now..To finish this.. :(
     
  5. Nov 2, 2011 #4

    PeterO

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    Homework Helper

    By rounding off your 60 kph conversion you get angle 25.24. I didn't round, and got 25.28657. Only a slight difference, but it may make a difference.

    I hope you took into account the fact that at 90 kph, the Reaction force is even greater than it is at 60 kph.

    Draw a free body diagram and you will [hopefully] see how much bigger.
     
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