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Max. velocity with given coefficient of static friction (μ_s)

  1. Apr 24, 2007 #1
    1. The problem statement, all variables and given/known data
    A car goes around a curved stretch of flat roadway of radius R = 104.5 m. The magnitudes of the horizontal and vertical components of force the car exerts on a securely seated passenger are, respectively, X = 236.0 N and Y = 572.0 N.

    This stretch of highway is a notorious hazard during the winter months when it can be quite slippery. Accordingly the DMV decides to bank it at an angle φ = 23.5 ° to the horizontal.

    (1) What is the maximum speed at which the car could safely negotiate this banked curve during a wintry night when the coefficient of static friction between the road and tyres is μs = 0.24 ?

    2. Relevant equations

    F_s = (m*v^2)/r
    F_s = μ_s * n
    n = m*g
    where n is the normal force and μ_s is the coefficient static friction.

    3. The attempt at a solution

    v = ((F_s*r)/m)^1/2
    v = ((μ_s*n*r)/m)^1/2
    v = ((μ_s*m*g*r)/m)^1/2
    v = (μ_s*g*r)^1/2

    v = (0.24*9.81*104.5)^1/2 * 3.6 (to convert ms^-1 to kmh^-1)
    = 56.5 kmh^-1 (3sf)

    Actual answer is 100 kmh^-1 (to 3sf)
  2. jcsd
  3. Apr 24, 2007 #2


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    Homework Helper

    For one thing, you didn't take into account that the curve is now banked at an angle of 23.5 degrees. That's important.
  4. Apr 24, 2007 #3
    the equations should be:
    with the appropiate sketch it's also understood why.
    and ofcourse f=mu*N.
    Last edited: Apr 24, 2007
  5. Apr 24, 2007 #4
    I still don't quite understand why I can't use the v = (μ_s*g*r)^1/2 formula.
    For this question I used it:

    At what speed could the car now negotiate this curve without needing to rely on any frictional force to prevent it slipping upwards or downwards on the banked surface?

    v = (μ_s*g*r)^1/2
    v = (tanφ*g*r)^1/2
    v = (tan23.5*9.81*104.5)^1/2 * 3.6 (to convert ms^-1 to kmh^-1)
    = 76.0 kmh^-1 (3sf)
  6. Apr 25, 2007 #5
    you cant use this formula, cause the force diagram of yours is incorrect.
    i reackon you have a sketch of the problem, look at it and then draw the forces.
    this question is a question of statics, the maximum speed is reached when there it neither slips to the sideways or inwards of the road, this we get when the net force is zero.
  7. Apr 26, 2007 #6
    I did draw a diagram originally to get part a of the question and it was right. The only thing that has changed is that they gave me a coefficient of static friction. It should be the same diagram still since the angle of 23.5 degrees still applies...

    I got two questions from it (the x and y components):

    1) Sigma F_x = m.g.sinφ - F_s = m.a_x = 0
    2) Sigma F_y = n - m.g.cosφ = m.a_y = 0

    I have tried using your equations in every way but I get no where near the answer, would you be so kind and provide me with a helpful hint to steer me in the right direction? I'm really going around in circles at the moment :(...

    Last edited: Apr 26, 2007
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