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Centripital force

  1. Feb 24, 2008 #1
    1. The problem statement, all variables and given/known data

    A banked circular highway curve is designed for traffic moving at 70 km/h. The radius of the curve is 193 m. Traffic is moving along the highway at 40 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding off the road?

    2. Relevant equations

    coefficient of static friction = Fstaticmax / normal force
    Fnet = ma
    for circular motion: Fnet = m (v^2 / radius)

    3. The attempt at a solution

    y is up and down, x is left and right (up and righ are positive)
    N = normal force
    fstat = static friction
    @ = angle

    So because the bottom of the tire is always instintaneously static, I solved for the coefficient of static friction:
    1. Find the angle the road is tilted at using the speed it was designed for:

    70km/h = 19.44 m/s
    Fnety = 0 = Ny - mg = Ncos(@) - mg

    Fnetx = m(v^2/r) = Nx = Nsin(@)
    solve for N... N = (mv^2) / rsin(@)

    Fnety = 0 = (mv^2) / rsin(@) * cos(@) - mg
    factor out mass... plug in values...

    0 = [(19.44m/s^2)^2 * cos(@)] / [193m*sin(@)] - g

    g = 1.958m/s^2 * [cos(@) / sin(@)]

    sin(@) / cos(@) = 1.958m/s^2 / g = tan(@)

    inverse tan to get @, which = 11.299 degrees

    2. Find the normal force and the frictional force on the car going 40 km/h
    40 km/h = 11.11 m/s
    Vertical forces: weight, Ny,
    Horizontal forces: fstat, Nx

    sin(11.299) = Nx / N,
    N = Nx / sin(11.299)

    Fnetx = 0 because the minimum amount of friction should cancel out the centripital force (Nx) so the car does not slide. fstat = fstatmax in this case

    Fnetx = 0 = Nsin(11.299) - fstat
    fstat = Nsin(11.299)

    coefficient of static friction = fstat / N
    = Nsin(11.299) / N = sin(11.299) = .196

    but this is not the right answer...
    thanks for the help!
  2. jcsd
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