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Banked Curves

  1. Jun 28, 2005 #1
    I've been working on this problem for hours and I just can't figure out what to do next! Here it is-

    A car drives around a curve with radius 410 m at a speed of 32 m/s. The road is banked at 5.0 degrees. The mass of the car is 1400 kg. What is the frictional force on the car?

    I drew the FBD and got net force equations for the x and y directions:

    Fxnet: Fsin(theta) = m(vsquared)/r -Nsin(theta)
    Fynet: Fcos(theta) = Ncos(theta) - mg

    But now I don't know what to do. I know what the correct answer is (back of the book), but I can't figure out how to get there. When I plug in the numbers, it doesn't come out right. Are my net force equations wrong?

    Hints? Suggestions? Please!!

    Thanks, Heather
     
  2. jcsd
  3. Jun 28, 2005 #2
    What about the frictional force f? Resolve it into horizontal and vertical components and add it to your equations of motion.
     
  4. Jun 28, 2005 #3
    The frictional force is the Fsin(theta) for the x/horizontal and Fcos(theta) for the y/vertical. I think I did that part right, but I'm really not that great at physics, so it could be wrong.
     
  5. Jun 28, 2005 #4
    ok im defining x to be radially inwards/outwars and the y direction to be up and down. These directions are as I stated them and not oriented along the incline. It seems in your equations you had the wrong trig function associated with the friction and some of your directions were opposite.

    From the FBD in the Y direction we have
    Normal*cos(theta) + Friction*sin(theta) - Weight = 0

    From the FBD in the X direction we have
    Normal*sin(theta) - Friction*cos(theta) = m*v^2/r

    2eqns, 2 unknowns (Normal and Friction), the system is linear too!

    Sorry I really don't feel like solving this algebraiclly so im just gonna throw in the numbers

    This gives

    Normal = 13972.5 N (perpendicularily outwards of the incline)
    Friction = -2257.5 N (up the incline)

    Notice in the model I chose I had friction acting up the incline, because this value is negative, we know the actual frictional force is

    Friction = 2257.5 N (down the incline)

    This is feasable as Friction < Normal
    These values give u = 0.161 which is a reasonable value for the coefficient of friction
     
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