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Centripetal Force - maxium downforce for no skidding

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data


    Downward force is used to keep a car in contact with the track. The motion through the air produces a force ppd to direction of travel pushing the car ONTO the track. A racing car of mass 720 kg takes an unbanked corner of radius 550 m at a speed of 50 m/s
    The max frictional force is 0.4*(reaction from track).

    Calculate the downward force necessary to prevent the car from skidding outward.

    Could someone please assist me in the drawing of a free body diagram
    2. Relevant equations

    3. The attempt at a solution
    So at the moment I know that the:
    R = mg + D (as the car does not rise above the track)

    thus the free body diagram would have R up and (mg + D) down

    assuming the track curves like this: ( (
    which direction will the frictional force act in, this is the final puzzle to the problem :S

    I would have said it acts towards the OUTER part of the track but the car needs to move in a circle and the only force providing centripetal acceleration appears to be friction

    any help is greatly appreciated
     
  2. jcsd
  3. Mar 28, 2012 #2
    http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html

    so by this it is towards the centre
    BUT HOW?? why is frictional force in the direction(kinda anyway)??
    so in this case it is the frictional force of the road on the car but why does it act towards the centre of circle not straight ahead?
     
    Last edited: Mar 28, 2012
  4. Mar 28, 2012 #3
    here seems to be a diagram (attachment)

    how does the force of friction on the tyres cause centripetal force, surely it would act in the opposite direction
     

    Attached Files:

  5. Mar 28, 2012 #4
    Friction is the only force that allows the car to turn a corner. Think about this: If you're sitting in the middle of the back seat of the car as it goes around a turn, and you must hold yourself steady in the seat with friction. Which direction would you feel the friction against the seat pulling you? This is the same direction that the friction against the road pulls the cars tires.
     
  6. Mar 28, 2012 #5
    Imagine a circle
    what ever point I am at my tyres will be at an angle (e.g: 30 degrees) BUT how does this provide the frictional force that provides the centripetal force 90 degrees to the direction of travel
     
  7. Mar 28, 2012 #6
    Friction acts in the direction opposite from where the car would accelerate otherwise. The friction is simply resisting your car's inertia wanting to go off track. Think about the situation I posed in my last post. When you're inside the car and the car turns, the inertia of your body wants to resist the car's turn and you feel accelerated directly away from the axis of your turn, but the friction of the seat pushes you towards the axis of the turn, keeping you on the seat.
    Think about the direction that you feel the seat pulling your pants. The friction from the seat is pulling on your pants in the direction of the axis of rotation.

    Assuming the car isn't skidding, it will go in the direction that the wheels are pointed. If the wheels are constantly turned so that the car goes in a circle, then the car would "feel" an acceleration away from the center of the circle, and so the friction will counter-act that acceleration, pointing in the opposite direction, which is toward the center.
     
  8. Mar 28, 2012 #7
    there is no such thing as a centrifugal force!! therefore this cannot be the cause of friction!!
     
  9. Mar 28, 2012 #8
    I never said that a centrifugal force would be the cause of friction. Not once did I mention a centrifugal force. I'm not sure where you got the idea that this would be the case.
     
  10. Mar 28, 2012 #9
    you said "then the car would "feel" an acceleration away from the center of the circle" this is centrifugal
     
  11. Mar 28, 2012 #10
    This feeling is the car's inertia. When the car accelerates towards the center of a circle, its inertia is going to resist this acceleration in the opposite direction. You can feel the same effect from your inertia every time you're in a car that makes a turn. I'm sure you can imagine this effect if you imagine you were in a car that's driving in circles. You'd feel like you were being pushed against the car door away from the center of the circle. This is not centrifugal force, this is your inertia resisting the car's turn.
    This resistance to the turn due to the car's inertia is what causes the friction to face the center of the circle. This friction is opposing the car's resistance to turning.
     
  12. Mar 29, 2012 #11
    would we have a situation like this:


    |/
    -----

    Where / is the fricitonal force and ------ is the horizontal compoenent and | the vertical componenet (all on a flat surface so | is straight ahead and ---- is to the side
    so the ---- is actually F cos(θ)
     
  13. Mar 29, 2012 #12

    tiny-tim

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    this is very confusing :confused:

    yes, if the car is turning left, the passenger is in a rotating frame, and feels a centrifugal force to the right, so the friction from the car is to the left, the same direction as the friction on the car that pulls the car in a circle

    but i don't see how it helps to compare two different frictions, one from the car, one on the car, in two different frames of reference :redface:

    the direction of friction from the road is obtained by F = ma
    the direction of the tyres is irrelevant …

    suppose the car had a jet engine, and blocks instead of tyres …

    point the jet at 30˚, and the car will slide in a circle …

    what provides the friction for it to do that? :smile:
     
  14. Mar 29, 2012 #13
    Man the thing is-
    It is the car that pushes against the friction and goes towards the center.
    Like you push an object and go away with the reaction force.
    And as long as tilting is concerned it automatically adjust itself such that it cause the force of gravity to aid(that is if u r tilted at an angle theta from the horizontal u will have a forve of mg sin theta inwards and so on.
     
  15. Mar 29, 2012 #14
    but I don't understand this
    If the car is moving forward, friction is backward
    if the car turns the wheels, the front wheels still move forward so friction should have a component providing centrpetal force not be the entire centrpietal force?
     
  16. Mar 29, 2012 #15

    tiny-tim

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    hi jsmith613! :smile:
    no, that's only for kinetic friction

    there's no kinetic friction between the tyres and the road (if there was, the tyres would slide, and leave tyre marks on the road :wink:)​

    you need to apply the rule for static friction, which is that the direction of friction is opposite to the relative motion that would occur if the road suddenly turned to ice :wink:
     
  17. Mar 29, 2012 #16
    http://en.wikipedia.org/wiki/Friction#Static_friction

    static friction is for: two or more solid objects that are not moving relative to each other
    here the road and the car are solid and are moving relative to each other

    so surely it is kinetic?
     
  18. Mar 29, 2012 #17

    tiny-tim

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    ah, but two paragraphs later, it says :wink:
    An example of static friction is the force that prevents a car wheel from slipping as it rolls on the ground. Even though the wheel is in motion, the patch of the tire in contact with the ground is stationary relative to the ground, so it is static rather than kinetic friction.​
     
  19. Mar 29, 2012 #18
    the question still is why does this act at right angles to direction of motion?
     
  20. Mar 30, 2012 #19
    As long as the wheels aren't skidding, the friction between the car's wheels and the road is static. The part of the wheel touching the ground doesn't slide against the ground; once it's pressed against the ground, it sticks to that spot on the ground (with static friction) until it's lifted back off.
     
  21. Mar 30, 2012 #20
    Static friction always opposes an acceleration. The car's inertia resists the circular path, causing the car to want to travel in a straight line, and not in a circle. Since the car's inertia would cause the car to go in a path that would cause the tires to skid, the rules for the direction of the static friction of the tires say that the direction must oppose the inertia. If you draw a force vector that would oppose the car's inertia and cause it to stay directly on the circle, this vector points directly at the center of the circle. Because of the rules for static friction (which say the tire's friction must oppose the car's inertia), this vector arrow is the direction that the tire's friction points.
     
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