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Solve the coefficient of friction in a problem involving circular motion

  1. Dec 17, 2007 #1
    1. The problem statement, all variables and given/known data
    If a curve with a radius of 88m is perfectly banked for a car traveling 75km/hr, what must be the coefficient of static friction for a car not to skid when traveling at 95km/hr?

    2. Relevant equations
    i guess relevant equations would be tan θ = v^2/rg, but its says next to friction not needed?
    f=ma or fr=mar
    and Ffr = µs x Fn

    3. The attempt at a solution

    I made a free body diagram and since it said it was banked i set that in the y axis theres no acceleration, so following the book (Giancoli), I got Fn=mg/cos θ. I solved for that and got 95^2/(88)(9.8) = .40 then inverse tan 10.46 or 84.5 for the angle

    then in the x axis i got Fn sin θ - Ffr = mar for the sum of the forces. I substituted Ffr for µs x Fn and Fn = mg cos θ. I
    I then had Fn sin θ - µs mg cos θ = mar
    I divided mass from both sides and divided both sides by m.
    I then for ar substituted v^2/r
    Then I isolated µs by dividing cos θ and subtracting Fn sin θ
    So I then had µs = V^2/r x cos θ - Fn sin θ
    I plugged everything in and I got 9015, I know this wrong just by looking at it, the back of the book says .22, I am completely clueless, so if anyone could help, I'd greatly appreciate it. Sorry if it is a little confusing.
  2. jcsd
  3. Dec 17, 2007 #2
    The centripetal force of the car is caused by friction

    The car is accelerating at [tex]a = \frac{v^2}{r}[/tex]

    Equating the force of friction and the centripetal force, we have

    [tex]F_f = \mu F_N = ma[/tex]

    [tex]\mu * mg = \frac{mv^2}{r}[/tex]

    [tex]\mu = \frac{v^2}{gr}[/tex]
  4. Dec 17, 2007 #3
    thank you, I forgot to convert 95km/hr to m/s too, anyways thanks again!
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