newton’s Law Question HELP 2!?

NetherCraft 0

A highway curve that has a radius of curvature of 100 meters is banked at an angle of 15° as shown above.

a. Determine the vehicle speed for which this curve is appropriate if there is no friction between the road and the tires of the vehicle.

On a dry day when friction is present, an automobile successfully negotiates the curve at a speed of 25 m/s.

b. Determine the minimum value of the coefficient of friction necessary to keep this automobile from sliding as it goes around the curve.

1 Answer

  • first case, no friction

    we use a standard x, y coordinate system even though the track is banked

    the forces on the car are the horizontal and vertical components of the normal force (N) and the weight

    in the vertical direction, the vertical component of N balances the weight since the car does not accelerate vertically, so

    N cos(theta) = mg (eq. 1) draw a diagram showing the normal force (which is perpendicular to the track) and be sure you see what the vertical component is N cos(theta) where theta is the bank angle)

    the horizontal component of the normal force must match the centripetal force, so we have

    N sin(theta)=mv^/r (eq.2)

    now, divide eq. 2 by e1. 1:



    for r=100m and theta =15 deg, we have that v=sqrt[ r g tan(theta)]

    v=16.2 m/s

    second case, with friction

    to keep the car from sliding up the track at 25m/s, there must be friction opposing the motion

    there is both a horizontal component of friction and a vertical component of friction; the horizontal component of friction acts toward the center of the curve and has magnitude f cos(theta) where f is the total frictional force, the vertical component acts down with magnitude f sin(theta)

    remembering that f= u N where u is the coefficient of friction, our force laws become:

    in the x direction: N sin(theta) + f cos(theta) = mv^2/r

                          Nsin(theta)-+u N cos(theta)=mv^2/r
                          N[sin(theta)+u cos(theta)]=mv^2/r (eq.3)

    in the y direction: N cos(theta)-mg – fsin(theta)=0

                          N[cos(theta)- u sin(theta)]=mg  (eq.4)

    divide eq. 3 by e1 4

    N[sin (theta) + u cos(theta)]/N[cos(theta)- u sin(theta)]=mv^2/r/(mg)

    [sin(theta)+u cos(theta)]/[cos(theta)-u sin(theta)=v^2/rg

    knowing theta =15 deg, v=25m/s, r=100 m and g=9.8m/s/s, you can solve for u

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