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Find Banking Angle and Normal/Friction Force on Curved Road

  1. Oct 27, 2014 #1
    1. The problem statement, all variables and given/known data
    A curved portion of highway has a radius of curvature of 65 m. As a highway engineer,
    you want to bank this curve at the proper angle for a steady speed of 22 m/s.
    (a) What banking angle should you specify for this curve?
    (b) At the proper banking angle, what normal force and what friction force does the
    highway exert on a 750-kg car going around the curve at the proper speed?

    [itex]r = 65m[/itex]
    [itex]v = 22 m/s[/itex]
    [itex]g = 9.8 m/s^2[/itex]
    [itex]m = 750 kg[/itex]

    2. Relevant equations
    [itex]F_{Ny} = F_N cos\theta = mg \Rightarrow F_N = \frac {mg}{cos\theta}[/itex]
    [itex]\vec F_c = \frac {mv^2} {r} = F_N sin\theta \Rightarrow (\frac {mg}{cos\theta}) sin\theta = \frac {mv^2} {r} \Rightarrow gtan\theta = \frac {v^2} {r} \Rightarrow \theta = tan^{-1}(\frac {v^2} {gr})[/itex]
    When the object is moving at constant velocity, the centripetal force will be the same as the force of friction.

    3. The attempt at a solution
    The wording "radius of curvature" is throwing me off, because we've never had it. I looked it up and pictures showed it as part of the circumference of the curved road, not the distance the to center. Nonetheless, I went ahead and treated "radius of curvature of 65 m" as simply r = 65m.

    (a) [itex]\theta = tan^{-1}(\frac {(22 m/s)^2} {(9.8 m/s^2)(65m)}) = 37.227^o \sim 37^o[/itex]
    (b) [itex]F_N = \frac {mg}{cos\theta} = \frac {(750 kg)(9.8 m/s^2)}{(cos 37.227^o)} = 9,230.949 N \sim 9,200N[/itex]
    [itex]\vec F_f = \vec F_c = F_N sin\theta \Rightarrow (9,230.949 N)(sin 37.227^o) = 5,584.48N \sim 5,600N[/itex]

    Thank-you
     
  2. jcsd
  3. Oct 27, 2014 #2

    haruspex

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    Good.
    That's only true on a horizontal surface. Here, those two forces act in different directions.
    Good move.
    Think about the logic/FBD behind your equation ##\vec F_c = F_N sin\theta##. What does that imply for [itex]\vec F_f[/itex]?
     
  4. Oct 28, 2014 #3
    I guess I'm a bit confused. Isn't there a different speed in order to prevent the car from sliding down the bank versus sliding up the bank? Where does the 22m/s fit in? Do I set the sliding down and up equations equal to each other and solve for the coefficient of friction, then use that to find the friction force?
     
  5. Oct 28, 2014 #4

    haruspex

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    For a given coefficient of friction, yes. But what is meant by 'proper angle' here? You solved that part ok, yet friction never appeared in the analysis. Why do you think that is?
     
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