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Question on banked curve, Application of Newtons Laws

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data

    A turn in the road with a radius of 30m is banked at an angle of 40 degrees. If [itex]\mu[/itex]=.75, what is the minimum and max speed a 1200 kg car can travel without sliding off the turn.

    2. Relevant equations

    F=ma, Mv^2/R, Fw = mg

    3. The attempt at a solution

    Alright, so using the information given i made a force diagram. Basically what i did is make the car, then a vector going to the right as Fc, another vector which makes the hypotenuse coming from the car to North-East, as Fn since it is perpendicular to the curved road, and Since Fw is down, upwards it would fit in the triangle to make a triangle with the legs of Fw and Fc, and a hypotenuse of Fn.

    Since the velocity isn't given I solved for the velocity by replacing Fc with Mv^2/r.

    So

    (tan(40)*m*v^2)/r = mg

    Plugging in all the values I got a Velocity of 18.7 m/s, but this is without friction so its better not to use this, and to find the Fn and Ff.

    So mg/sin(40)=Fn. Fn works out to be 18295N, so using [itex]\mu[/itex]*Fn = Ff, the Ff is 13722N.

    I'm honestly very confused on where to go from here, not even sure if what i've done so far is correct.. If anyone could help it'd be greatly appreciated..
     
  2. jcsd
  3. Oct 26, 2011 #2
    Thats not entirely correct what you did. First of all There are three forces acting on the car plus the centripetal force. tht is 4. The first is the normal force which is perpendicular to the plane on which the car is travelling. The friction force is in the plane of the road acting either "up" or "down" depending on the velocity. The third is the gravity force which acts downwards and is perpendicular to the centripetal force.
    The car doesnt move "up or down" but stays on the bank. Hence the gravity force keeps an equilibrium with the vertical components of the friction force and the normal force. for e.g. When we are searching for the minimum velocity i.e. the car shouldnt slide down the bank so the friction force is "up". Hence this equilibrium condition is: mu*N*sinTheta+N*cosTheta = mg.

    Similarly you can write up the horizontal components of the normal and friction force and this should be equal with the centripetal force and from these two equation you can eliminate the N normal force.
    For the maximum velocity take the friction force in the other direction.
     
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