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Free-body diagram to identify the forces acting on the car

  1. Oct 21, 2004 #1
    Where do I start?

    A curve of radius 56.5 m is banked so that
    a car traveling with uniform speed 59 km/hr
    can round the curve without relying on friction
    to keep it from slipping to its left or right.
    The acceleration of gravity is 9.8 m/s^2

    What is Q? Answer in units of degrees.
  2. jcsd
  3. Oct 21, 2004 #2
    Again, use a free-body diagram to identify the forces acting on the car. Remember, the car is travelling in circular motion; what does that tell you?
  4. Oct 23, 2004 #3
    Is this the right equation?

    Mg = 1.3
    g = 9.8 m/s^2
    Radius = 56.5 m
    Speed = 59km/hr
    m = 0

    Mg Tan = mV^2/R

    1.3 Mg Tan = 0 x (59)^2 / 56.5

    But wouldn't that just make it zero?? I'm lost.
  5. Oct 23, 2004 #4


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    What is M and what is m??
    and why is m=0???????????????????
  6. Oct 23, 2004 #5

    M x g tan = m (V^2/g)

    g tan = V^2 / g

    9.8 tan = [(16.2)^2 / 56.5]

    9.8 tan = 4.644955752 (Do I just type it in as that? or do I need to divide by 9.8)

    I am trying to find the degree of the angle.
  7. Oct 23, 2004 #6
    k...in this case....the force of gravity = centripital force....which I think you have figured out already....

    You are correct in stating that Centripital force= mv^2 / R

    Now draw a free body diagram to find the force of gravity.....its not mg tan (angle)....its mg * something......

    ...work from there.
  8. Oct 23, 2004 #7

    tan = V^2 / Rg ??

    tan = (59)^2 / (56.5) (9.8)

    tan = .62867905

    = .0109729727 ?

    that doesnt make sense
  9. Oct 23, 2004 #8


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    [tex] \tan \theta =\frac{v^2}{rg} [/tex]

    Only works without friction!, it's the formula to find the angle of a frictionless banked curve. To solve this problem do as the other says, do a freebody diagram and identify all the forces (components) pointing toward the center.
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