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Speed at which a car on a banked curve will slide.

  1. Apr 29, 2008 #1
    1. The problem statement, all variables and given/known data
    If a car goes around a banked curve too fast, the car will slide out of the curve. Find an expression for the car speed [tex] v_{max} [/tex] that puts the car on the verge of sliding out. What is the value for R=200, [tex] \theta = 10 [/tex], [tex] \mu_s = 0.60 [/tex]?

    2. Relevant equations

    a = v^2/r, F = ma.

    3. The attempt at a solution
    Place the x axis along the bank, and write the for equations for x and y:

    [tex] \sum F_x = \frac{mv^2}{R}\cos\theta = mg\sin\theta + f_s [/tex]
    [tex] \sum F_y = 0 = F_n - \frac{mv^2}{R}\sin\theta - mg\cos\theta [/tex]

    It seems that the car should start to slide when
    [tex] \frac{mv^2}{R}\cos\theta > mg\sin\theta + f_s[/tex]

    Using [tex] fs = \mu_s F_n [/tex]

    [tex] \frac{mv^2}{R}\cos\theta > mg\sin\theta + \mu_s\left(\frac{mv^2}{R}\sin\theta + mg\cos\theta\right) [/tex]

    Solving for v:

    [tex] v > \sqrt{\frac{gR(\tan\theta + \mu_s/\tan\theta)}{1-\mu_s\tan\theta} [/tex]

    Unfortunately, the figure I get for v using R = 200, theta = 10, coefficient of static friction = 0.60 is much too high (88.6 m/s).

    Could someone give me a hint about where I went wrong?

    Thank you,
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 29, 2008 #2


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    Staff Emeritus
    Science Advisor

    In this term [tex]\mu_s mg\cos\theta[/tex], if one divides by cos[itex]\theta[/itex], then one does not get a tan function, but simply

    [tex]\mu_s mg[/tex]

    So one would obtain

    [tex] v > \sqrt{\frac{gR(\tan\theta + \mu_s)}{1-\mu_s\tan\theta} [/tex]
    Last edited: Apr 29, 2008
  4. Apr 29, 2008 #3
    Thank you, Astronuc. I am in your debt. What a stupid error :-)

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