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Possible friction force needed for a car on a banked curve

  1. Aug 16, 2012 #1
    1. The problem statement, all variables and given/known data
    A 1200kg car rounds a dry curve (μ= .6)with a radius of 67 m banked at an angle of 12°. If the car is traveling @ 95 km/hr (26.4 m/s), will a friction force be required? If so how much and in what direction?


    2. Relevant equations
    Fn = mg cos 12°
    ƩFn sin 12° = m (v^2/r)
    Ffrmax = μFn


    3. The attempt at a solution
    Seen many different versions of this question asked so here is what I have attempted...

    Vertical Force on the car at that angle

    Fn = 1200 * 9.8 * cos 12°
    = 1.2 x10^4 N

    Horizontal force

    ƩFn sin 12° = m (v^2/r)
    ƩFn sin 12° = 1200 ( (26.4)^2/67)
    = 5.8 x 10^4 N

    max total static friction force of the
    Ffrmax = μFn
    = (.6) (1.2x10^4)
    = 7200 N

    Based on my answers I would say additional force would be needed (5.8 x 10^4 N-7200 N=5.1x10^4) perpendicular to the curve.

    Would you guys agree or where have I gone wrong?
     
  2. jcsd
  3. Aug 16, 2012 #2

    PeterO

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    The Normal force here is what you get if the car was parked on the banking. When driving in the banking, the normal force is bigger.

    It is the same effect when comparing parking at the bottom of a dip to driving through that dip.
     
  4. Aug 16, 2012 #3

    CAF123

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    Banking and turning the corner means you are under the influence of a centripetal force. It is this centripetal force that contributes to the normal force and that is why it is 'bigger'.
     
  5. Aug 16, 2012 #4

    PeterO

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    Not quite the way I would word it, but generally yes.
     
  6. Aug 16, 2012 #5

    CAF123

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    Just out of interest, how would you describe it?
     
  7. Aug 16, 2012 #6
    So what changes do I need to make to my equations so I address the problem correctly?
     
  8. Aug 16, 2012 #7

    CAF123

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    This is what I did:
    if you set up the coordinate system such that the positive [itex] x [/itex] direction is pointing radially inward and positive [itex] y [/itex] upwards, then I get that;

    [itex] \sum F_x = \frac{mv^2}{r}, [/itex] which gives [itex]\small Nsinθ = \frac{mv^2}{r}. [/itex]
    Similarly, [itex] \sum F_y = 0, [/itex] which gives [itex] \small Ncosθ = mg [/itex]

    The centripetal force required is [itex] F_c = \frac{mv^2}{r}, [/itex] so inputting numbers yields [itex] F_c ≈1.2 [/itex] x [itex] 10^{4} N [/itex].
    Now in this coordinate system, with the component of normal force contributing, we see that it provides [itex] ≈ 2.5 [/itex] x [itex] 10^{3} N [/itex], so the rest must come from friction, so that the car can successfully navigate the turn.
     
  9. Aug 16, 2012 #8
    I somewhat follow the coordinate system rewrite but am lost on where you got
    "the component of normal force contributing, we see that it provides ≈2.5 x 10 3 N"

    and where do I go from here???
     
  10. Aug 16, 2012 #9

    CAF123

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    I solved for [itex] N [/itex] first to give [tex] N = \frac{mg}{cosθ} [/tex] using the given values. Multiplying this by [itex] sinθ [/itex] gives the contribution of the normal force to the centripetal force.

    This means the contribution from friction must be ≈ 1 x 104 N, but according to your calculations the maximum friction force falls short of the required amount. Something is not adding up here.
    Do you have an answer from a book and/or does the question say anything sbout the car not making the corner?
     
  11. Aug 16, 2012 #10
    based on how the question is worded I would assume if the car is going to stay on the road is additional force needed, if it is needed how much and in what direction?
     
  12. Aug 16, 2012 #11

    PeterO

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    Since the car is on a banked turn it is undergoing centripetal acceleration, so the net force acting is horizontal.
    The weight force, mg, is unaltered, so the Normal force has to be bigger to give a net horizontal force.
    Since the angle between mg and N is fixed at 12o there may need to be a friction force parallel to the slope to make the vector sum correct [horizontal and the right size].

    It is that friction force [if any] you are after in this problem.
     
  13. Aug 16, 2012 #12
    and this is one of those where the example in the book is for a flat surface and it does not have a solution in the back.
     
  14. Aug 16, 2012 #13
    Okay so from all of the posts and help I just need some help making this unconfusing. I'm not looking for the answer outright but if you guys could reorder the correct equations so I know how to work through it, I know I can walk through it on my own. I'm feeling like I have a brain fart that is caught and won't get out over this.
     
  15. Aug 17, 2012 #14
    Sometimes it takes a day away to clear the head and get things lined up right. Thanks guys for the help.. Got it straigthened out now.
     
  16. Aug 18, 2012 #15

    CAF123

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    How did you get on in the end?
    Was a friction force required?

    I ask because I fear my method above may be incomplete - i may have not properly accounted for the friction or something.
    How did you solve it?
     
  17. Aug 18, 2012 #16
    Please correct me if it is wrong.

    FNCosθ=mg +FfSinθ
    FNSinθ +FfCosθ=mv2/r
     
  18. Aug 18, 2012 #17

    CAF123

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    Surely what you have written depends on what coordinate system you choose?
    How have you set yours up?
     
  19. Aug 18, 2012 #18

    CAF123

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    I think I realised your coordinate system.
    I agree with the first, but for the second, the Nsinθ and Ffcosθ components point rightwards, and we want a centripetal acceleration directed leftwards. I am not entirely sure if this changes things. Maybe another user could clarify things?
     
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