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Car moving along a banked curve with friction

  1. Nov 5, 2005 #1
    car moving along a banked curve with friction (solved)

    first off, thanks to whoever helps me. you're a life saver. i could just copy it from a friend, but i want to be skilled enough for the AP Physics B exam. so yeah... i just want to understand what i have to do. the logic with this is killing me.
    ok here's the problem:
    A 1200-kg car rounds a curve of radius 67m banked at an angle of 12 degrees. If the car is traveling at 95 km/h, will a friction force be required? If so, how much and in what direction?
    ok, here's what i got so far. first, i converted 95 km/h to 26.388... m/s. then, i found a formula in the book for things with no friction required: tan (theta) = (v^2)/(rg). so i plugged in the angle, radius, and gravity constant to see what speed would be necessary for no friction required. i got 11.8 m/s. nowhere near. so i now have that friction is required. simple enough.
    so now i need to find out how much friction force is needed. i drew a diagram to help me figure out how weight, normal force, friction force, and centripetal force are related as vectors. i understand centripetal force and weight are constant in this case. i know i have to find the friction force as a vector since the coefficient of friction isn't provided, so i can't find it directly from normal force. thing is though, normal force and friction force affect each other, and i have no clue how to figure out normal force in this case. i tried messing around with the vectors and solving for friction force by pythagorean theorem, sin, and cos, but i ended up getting three different answers... i am massively confused. assistance would be GREATLY appreciated.
     
    Last edited: Nov 5, 2005
  2. jcsd
  3. Nov 5, 2005 #2

    rho

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    This is a circular motion question, the car is moving in a circle so is therefore accelerating so there must be a resultant (centripetal) force directed inwards (due to Newton's second law) which is equal to the frictional force.

    work out the horizontal component of v then use :

    F = mv^2/r
     
  4. Nov 5, 2005 #3
    i wish it were that easy. that would be true if the curve was flat, but the curve is banked. the friction force is at an angle.

    [​IMG]
     
    Last edited: Nov 5, 2005
  5. Nov 5, 2005 #4
    There is no friction force pointing out of the circle. The only frictional force is pointing in because it keeps the car in the circle. The normal forces and friction forces are vectors and can be broken up into horizonal and vertical components.

    Then you can say that the sum of all vertical forces(weight, y component of normal force, and y component of frictional force) = 0 because there is no vertical acceleration.

    Sum of all horizonal forces = mv^2/r and go from there.
     
  6. Nov 5, 2005 #5

    rho

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    Your diagram contains a 'centrifugal' force!

    The method I said should work, just convert everything to ms^-1
    Then work out the centripetal force, and resolve it down the slope to get the frictional force.
     
  7. Nov 5, 2005 #6
    very, very good!!!!
    ...but how do i go about splitting normal force and friction force into horizontal and vertical vectors??? :confused:
     
  8. Nov 5, 2005 #7

    rho

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    You can't say the sum of all forces = 0 because there is a resultant force on the body as it is accelerating.
     
    Last edited: Nov 5, 2005
  9. Nov 5, 2005 #8
    centrifugal force is force pointing out from the circle. the inertia of the car provides that. it's not a factor.

    converting all the forces into acceleration won't do anything. i'll just have accelerations in all different directions instead of forces. that won't solve for any of the values.

    unless i'm misunderstanding what you're saying?
     
  10. Nov 5, 2005 #9
    the sum of all vertical forces = 0. this is because the car is not accelerating up or down. this means the net force is zero vertically.

    however, the car is accelerating horizontally by going in a circle (centripetal acceleration). that's why the horizontal accelerations and forces equal the centripetal acceleration and forces because that is the net acceleration and force.
     
  11. Nov 5, 2005 #10

    rho

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    The answer should just be: F(friction)= F(centripetal)/ cos 12

    unless i misunderstand this question.
     
    Last edited: Nov 5, 2005
  12. Nov 5, 2005 #11
    ...wow. i feel like a complete moron. thanks rho!!! ^.^
     
  13. Nov 6, 2005 #12

    Fermat

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    There is also the slope component of the mass of the car providing some of the centripetal force. That should also be taken into account.
     
  14. Nov 6, 2005 #13
    Isn't that considered by the fact that we're taking the friction force in consideration with the angle (cos 12)?
     
  15. Nov 6, 2005 #14

    Fermat

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    The forces acting down the slope are: Friction and the slope component of the mass of the car, mg.sinα.
    This is equal to the slope component of the centripetal force, Fc. viz.

    Fr + mgsinα = Fc.cosα
    Fc = (Fr/cosα + mg.tanα)
    mv²/r = (Fr/cosα + mg.tanα)
    v² = (Fr.r/m.cosα + (g/r).tanα)
    ========================

    and if there is no friction, then v_crit is given by,
    v²_crit = (g/r).tanα
    ===============

    our standard result.
     
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