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Calculating the friction on a curved path

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data
    mi6OvdY.gif
    A car drives on a straight path and from point A it enters a curved path. It stops at rest on point B of the curved path.
    The curved path has the radius R and the length is given by the angle theta.
    The entire path has friction. The velocity of the car is known at point A as [itex]v_A[/itex]
    Calculate the friction force's work on the curved path from point A to B.


    2. Relevant equations
    Work on a straight line:
    [itex]W=F \cdot s \cdot \cos(\phi)[/itex]

    Work on a curved path:
    [itex]W=\int_{P_1}^{P_2}F \cdot cos(\phi) dl[/itex]

    Work done by friction on a straight path:
    [itex]W_{fric}=f_k \cdot s \cdot cos(\phi)[/itex]

    3. The attempt at a solution
    I drew a Free body diagram of the car in the curved path (is it correct?)
    fSgFuZ3.gif

    The force F is what's dragging it and then there is a friction force f that has a direction depending on where on the curved path that the car is which is the angle theta.

    But my problem is that I don't know the force F. I'm not sure what to do next.
     
    Last edited: Nov 13, 2013
  2. jcsd
  3. Nov 13, 2013 #2

    Mentz114

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    This is confusing because you say the car stops at point B, but it has velocity ##v_B## at B.

    Is the curved path circular ? If so then the frictional force on the curve can be found from the centripetal normal force.
     
  4. Nov 13, 2013 #3
    Ah, sorry! I meant that the car has a velocity at point A. I have edited it out now. :)
    Which formula are you thinking about?
    I know that [itex]f_k=\mu_k \cdot n[/itex] but how do I find the normal force?
     
  5. Nov 13, 2013 #4

    Mentz114

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    Travelling in a circle produces a force F = mω2r = m v2/r in the radial direction, where v is the tangential velocity. There is also the component of the gravitational force to add to this. So the frictional force on the curve is the sum of these multiplied by the CoF.
     
  6. Nov 13, 2013 #5
    You can use the work-energy theorem and avoid the integration.
    But you need the mass of the car, either way.
     
  7. Nov 13, 2013 #6
    What is the CoF?
     
  8. Nov 13, 2013 #7
    But how would I find the work of the friction through the work-energy theorem? Would my total energy just be the friction?
     
  9. Nov 13, 2013 #8

    haruspex

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    Is there anywhere else all the energy can have gone?
    As nasu wrote, you cannot answer the question without knowing the mass. Are you sure this was not given?
     
  10. Nov 13, 2013 #9

    Mentz114

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    CoF = coefficient of friction. You probably don't need it because

    ##\frac{mv^2_A}{2}=mgh + W_f## where ##W_f## is the work done by the firction, g is the acceleration of gravity and h is the height of B.
     
    Last edited: Nov 13, 2013
  11. Nov 13, 2013 #10
    Yes, the mass and kinetic coefficient of friction is known values. :)

    So my energy-work theorem states that:
    [itex]W_{fric}=ΔK[/itex]

    Since the car has the velocity [itex]v_A[/itex] at point A and 0 at point B since it stops there, then I have:
    [itex]W_{fric}=\frac{1}{2}\cdot m \cdot 0-\frac{1}{2}\cdot m \cdot v_A^2[/itex]

    That would mean the work done by the friction is:
    [itex]W_{fric}=-\frac{1}{2}\cdot m \cdot v_A^2[/itex]

    Is it seriously this simple? I was sure I had to use the integral formula because it was a curved path.
     
  12. Nov 13, 2013 #11
    So it would be easier to use the work energy theorem because it is not dependent of the path, right?
     
  13. Nov 13, 2013 #12

    haruspex

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    Yes. Don't complain :biggrin:.
     
  14. Nov 14, 2013 #13
    No, the variation of kinetic energy is due to the work done by all the forces acting on the object, not just friction. Here you have the work done by gravity, as Mentz's formula showed you already.

    An alternative is to write the variation in mechanical energy (kinetic and potential) equal the work of the dissipative forces (friction in this case).
     
  15. Nov 14, 2013 #14

    haruspex

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    Despite the way the car is shown in elevation, my interpretation of the wording is that the track is horizontal.
     
  16. Nov 14, 2013 #15
    Of course you can neglect whatever parts of the problem you want.

    A "curved path" makes sense either way. Maybe more sense in your interpretation.
    But the figure has more impact than the text. :smile: (a thousand words)
    And then some data become irrelevant (angle, radius). doesn't it?
    So we neglect the figure and most of the given data.
    Unless they really, really want you to calculate the work as an integral along that path.

    Maybe the context of the problem may help.
     
  17. Nov 14, 2013 #16

    haruspex

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    "A car drives on a straight path and from point A it enters a curved path" seems odd wording if this is in a vertical plane. I would expect e.g. "A car drives on a horizontal path and at point A it encounters a hill curving up."
    It looks to me like PhyIsOhSoHard drew the diagram, so I wouldn't trust the details of that where they conflict with the wording and PhyIsOhSoHard's equations.
     
  18. Nov 14, 2013 #17

    rcgldr

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    The problem statement isn't clear. What is slowing down the car?

    If the path is curving upwards, and if the angular momentum of the tires is ingored, then only gravity performs work in slowing down the car and friction does no work.

    If the straight and curved paths are horizontal, then the car must be braking as it moves through the turn in order to end up stopped. In this case the path doesn't really matter, since the work done is just the change in energy from 1/2 m Va^2 to zero. If this is the case, it seems like something is missing, since the path doesn't matter.
     
  19. Nov 14, 2013 #18

    haruspex

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    It implies friction is. OK, I understand that friction between the tyres and the ground does not slow a car on a road, but I'm reading it in a more general sense here. E.g. it could be lateral friction of a rail on the flanges of the wheels of a railroad car. You're right, it's not clear.
     
  20. Nov 14, 2013 #19

    rcgldr

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    In the other thread started by the same person, the car is gliding "down" a curved path with a similar picture, so if these are related problems, then the path is probably curving upwards, and as posted before, ignoring issues like angular momentum of the tires, then friction doesn't do any work.
     
  21. Nov 14, 2013 #20
    As I said already, the wording agrees more with your interpretation. But I thought that the figure is part of the original problem. I mean copied from the source and not the personal vision of the OP.
    Maybe he can clarify this.

    In the other problem (mentioned above by rcgldr) it says that is "gliding". If this is the actual wording, it may be that "car" is some generic object. It is the car actually rolling on its wheels?
    Maybe the problems are made-up by the OP.
    I find unusual to associate a car with "gliding". Especially if it's rolling on the wheels.
     
    Last edited: Nov 14, 2013
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