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How is it possible for a car to skid away from the center at a curve?

  1. Aug 15, 2012 #1
    When a car turns too fast, it skids away from the center and I don't understand how that's possible in terms of forces.

    Background idea: The confusion came about when I was approaching a question where the net force of a car on a curve was towards the center of the curve as static friction. The question was what is the maximum speed before skidding. Easy. But wait. What happens if it does exceed that speed? Where does the skidding come from? The only force that's acting on the car is toward the center. Even if you do speed up and exceed the static friction so that you go into kinetic friction that is STILL towards the center. And the velocity of the car was tangential so there was no outwards component (by outward, i mean radially away from the center). So in order to skid, there HAD to be a force on the car away from the center. What is this force???
     
  2. jcsd
  3. Aug 15, 2012 #2
    The three laws of Newton may be useful perhaps? The force on the car (by the road) is a reaction to the force by the car on the road. In detail:

    1. Newton's first law: The car tries to continue in a straight line, unless forced to change that state of motion. This tendency as expressed by the first law is also called "the force of inertia". That is no "force" in the common, active sense of the word, but yields a force as described next.

    2. Newton's second law: The car's change of motion in order to stay in the curve (its acceleration) is proportional to the force with which the road keeps the tires sticking to it.

    And of course there is the force by the car as expressed with Newton's third law:

    3. The mutual actions of the road and the car upon each other are always equal, and directed contrary.
     
  4. Aug 15, 2012 #3

    CAF123

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    Gold Member

    I am wondering if this is similar to the reason of getting jolted leftwards or rightwards in a car when it turns a corner suddenly. This could however just be due to insufficient friction from the seat, yes? Or perhaps it is the effect of the centrifugal force?
     
  5. Aug 15, 2012 #4
    It's exactly the same effect, and Newton called the true centrifugal force due to inertia (third law, see above) "centrifugal force" (What else! :rolleyes:). However, later other people called a fictitious (imaginary) force "centrifugal force", and perhaps that's what you mean. IMHO, bringing that up only leads to confusion.
     
  6. Aug 15, 2012 #5
    Yes I understand all that. But still where is the force that is directed ON the car that is pointing AWAY from the curve that is required for the car to skid?


    PS.
    There is no centrifugal force on the car since the car is the object that is doing the spinning. A centrifugal force is only an inertial thing. ex) a spinning liquid bottle has the liquid lined up on the wall because the bottle is spinning while the liquid tries to be stationary. But the bottle itself does not have any "centrifugal force" applied. (well "centrifugal force" is not even a force) But in this case, the car IS the object that is spinning

    SO DON'T SAY THIS IS CENTRIFUGAL FORCE. Just putting it out there so that the conversation doesn't go tangent
     
    Last edited: Aug 15, 2012
  7. Aug 15, 2012 #6
    The only thing causing your car to turn when you steer the front wheels is the static friction between the tire and the road that gives the motion only one way to go: in circular motion. Notice that when the tire is turned the force is pointing towards the center of circular motion. The resultant acceleration on the car from the force causes circular motion and this acceleration times the mass is usually called the "centripetal force" no matter the context (that's why it's always [itex]\frac{v^2}{r}[/itex] no matter the situation and thus can be set equal to some sum of the forces on the object). Note this is the alternative force to the centrifugal force but not in the non-inertial frame (so for the inertial frame).

    This is really no different than planetary rotation. What happens to the moon when you remove the earth (static friction) suddenly? It's really the removal of a force that causes the change in motion from circular to linear motion. You are not looking for a force that causes the car to break its circular motion - you are looking for a breaking of the static friction force, and thus a resultant no-force situation in which linear motion would again return to the car.

    Tires are important. Don't neglect replacing them..
     
  8. Aug 15, 2012 #7
    That is impossible, for then you would not state:
    If you understood "all that", then you would state that there is no force that is directed ON the car that is pointing AWAY from the curve that is required for the car to skid. As I tried to explain, there is instead a centripetal force on the car.
    Exactly! :smile: "Pointing AWAY from the curve" = centrifugal; there is NO force that is directed ON the car that is pointing AWAY from the curve.

    But why did you first contradict that? :bugeye: Perhaps (or probably!) you got bugged by the fictitious force concept. Let's avoid that here.

    If I read it correctly, your starting phrase "the net force of a car on a curve was towards the center of the curve as static friction", was a complete mix-up.
    Thus it might be useful if you analyze the force balance when the car is not skidding, based on my first post (#2). In this case, which are the two forces of Newton's third law? You will see that the force by the car, is pointing AWAY from the curve.
     
    Last edited: Aug 15, 2012
  9. Aug 15, 2012 #8
    There is NO force required for the car to skid, just as there is no force required for the car to continue going in a straight line. That's simply what things do: they go in a straight line, unless acted upon by a force. In the case of a car skidding, the "skid" = going in a straight line.
     
  10. Aug 15, 2012 #9

    Dale

    Staff: Mentor

    All of this is correct.

    Here is not correct. For concreteness, let's say that you are executing a left turn and begin to skid. When you skid in a left turn you do not actually accelerate towards the right. You either go straight or you turn left at a larger radius; either of these increase your distance from the center and, if the road is curved, run you off the road. There is no force to the right.
     
  11. Aug 15, 2012 #10
    It's so much simpler than all this. The friction defines a certain maximum radius at which you can make your circle. If you try to circle too quickly, exceeding what friction allows, you simply have too little frictional force in to keep your inertia from forcing you into a wider circle. The car wants to move in a straight line, so to accelerate it in a circle requires or force or else you'll make a larger circle (be pulled outward) to a circle that better approximates a straight line.
     
  12. Aug 15, 2012 #11

    rcgldr

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    The issue is that the transition into kinetic friction and lower rate of centripetal acceleration results in the car's path circling about a different center, one that is further away (larger radius) than the previous center. The two paths are both circles, one larger than the other, that touch at the point of transition (assume the transition is instantaneous).
     
  13. Aug 15, 2012 #12
    Correct if you mean a minimum radius. And we all tried to explain Newton's first law. Maybe all explanations together work. :rolleyes:
     
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