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Homework Help: Car travelling around a banked corner with friction

  1. Oct 27, 2011 #1
    The problem:
    A car is travelling along a curve having a radius of 74.0m, banked at an angle of theta = 23 deg. The coefficient of static friction is 0.09. What is the slowest speed the car can negotiate the curve?

    Relevant equations:
    Fc = Fnet
    Fc = mv^2/r
    Fnet = [(tan(theta) + u)/(1-utan(theta))]mg

    The attempt at a solution
    v^2 = (74)[(tan23 + 0.09)/(1-0.09tan23)](9.8)
    v = (387.92)^0.5
    v= 19.70 m/s

    This answer isn't being accepted by the online program we use in class and I can't figure out what I'm doing wrong. Any help would be much appreciate
  2. jcsd
  3. Oct 27, 2011 #2
    Seems rather unusual for the question to be asking for the slowest speed.
  4. Oct 27, 2011 #3
    I think it means the slowest the car will be able to travel without sliding into the centre - it'll have to be moving fast enough to offset the other forces. That's how I took it at least. Part 2 of the question asks for the fastest speed, I figured I'd just try to figure out part 1 first:)
  5. Oct 27, 2011 #4
    Okay so i just realized that the speed I had found, 19.7 m/s, is for the fastest speed.

    Now I have no idea how to solve for the slowest speed
  6. Oct 27, 2011 #5
    You calculated the maximum velocity. You should have calculated the minimum.

    [tex]v_{min}=\sqrt{\frac{rg(sin\theta -\mu cos\theta)}{cos\theta +\mu sin\theta}}[/tex]
  7. Oct 27, 2011 #6
    To find your the slowest speed, your force of friction will be pointed the other direction. Redraw your forces. Think of any object (toy car) rolling on a notebook at a certain angle. If it is going really slow, it will start to 'slide' downwards right? It does not have enough speed to keep it at a constant radius. Since it's sliding downward force of friction is pointing up, away from the center.

    Solve the same way you did with the maximum speed, where you had force of friction pointing downward (thus keeping the object from flying outwards).
  8. Oct 27, 2011 #7
    The equation worked, which is great, and the explanation was really helpful - I was just going to ask where the equation came from when i saw it - so thank you all!
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