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Banked curve

  1. Mar 28, 2005 #1
    A highway curve of radius 30 m is banked so that a car travelling at 40 km/hr can travel around it without slipping even if there is no friction between the car's tires and the road surface. Without friction, a car travelling faster than this will slide up the curve, while a car travelling slower than this will slide down the curve. Find the angle of elevation of the banked highway curve.

    Ok... What am i doing wrong here, i don't seem to be able to find the right answer to this problem.

    After i drew a free-body diagram, the centripetal force, a component of the Normal force is what is letting the car turn successfully.

    so, cos theta *g*m = the Normal force.

    The x component of the Normal force is providing the centripetal acceleration.

    the line of the x component of the Normal force and the horizontal line of the ground is parallel so i figured out the x component as sin theta * Normal force

    so sin theta * cos theta *g*m = m * 11.111m/s^2/30

    using the trig identity sin2theta = 2sintheta*cos theta

    sin 2 theta/2 * g = 4.11m/s^2

    theta = 29 degrees

    This is not the correct answer. The correct answer is 23 degrees. Can some one please explain my error. Thanks in advance...
  2. jcsd
  3. Mar 28, 2005 #2
    At the given speed this car does not slide up or down the curve. So the net force in the direction of the slope is zero. Normal force does not come in to play in this direction.

    m v^2/r * cos (theta) = mg sin (theta)


    theta = 23 degrees.
  4. Mar 29, 2005 #3


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    Wrong. You do not know the normal force. All what you know about it that it is normal to the highway. It makes an angle theta with the vertical. You can decompose the normal force into vertical and horizontal components. The vertical component is of the same magnitude as gravity, mg, and opposite to it, so all the vertical forces cancel. from that you get the magnitude of the normal force:

    [tex] F_N \cos(\theta) = mg \rightarrow F_N= \frac{mg}{\cos(\theta)} [/tex]

    The centripetal force is equal to the horizontal component of the normal force:

    [tex]\frac{ mv^2}{R}=F_N\sin(\theta)\rightarrow \frac {mv^2}{R}=\frac{mg}{\cos(\theta)}\sin(\theta)\rightarrow \tan(\theta)= \frac{v^2}{gR}[/tex]

  5. Mar 29, 2005 #4
    What have you done differently except to find the net force on a different direction?
  6. Mar 29, 2005 #5


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    I answered apchemstudent, and corrected him. He wrote that the magnitude of the normal force was equal to the normal component of gravity, i.e. mgcos(theta).

    As for your method, it is correct, but some of the students tend to include the centripetal force among the "real" ones. "The forces acting on the body are gravity and the normal force and the centripetal force" they say. So it must be emphasized that the centripetal force is the resultant of the "real" forces. Here they are gravity and the normal force. The car not only does not slide up or down, but moves along a horizontal circle, with constant speed. From its motion, we know that the resultant of all the forces acting on it is horizontal, perpendicular to the velocity, pointing towards the center of the circle, and the magnitude is mv^2/R.
    Now we can choose a cordinate system as we like. You are right, you can choose the x axis of your coordinate system parallel with the road, and then the normal force does not come into play. The parallel component of gravity equals the parallel component of the net force (and it is the centripetal force).

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