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Homework Help: Coefficient of static friction with radius, speed, etc. given

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data

    To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 331 m. The car increases its speed at uniform rate of at ≡ d|v|dt = 3.68 m/s^2 until the tires start to skid. If the tires start to skid when the car reaches a speed of 27.8 m/s, what is the coefficient of static friction between the tires and the road? The acceleration of gravity is 9.8 m/s^2.

    2. Relevant equations
    F = ma(subscript c) = m(v^2/r)
    us = F/N

    3. The attempt at a solution
    I am Physics dummy, but I tried doing (27.8 m/s)^2 / 331 m. I tried to find the F of it. But I got all confused, I don't know what I am doing anymore. Can someone help? Please.
     
  2. jcsd
  3. Feb 20, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi MissPenguins! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Well, your approach looks ok …

    that is the radial acceleration (but don't forget there's also a tangential acceleration in this case :wink:) …

    where did you go from there? :smile:
     
  4. Feb 20, 2010 #3
    Re: Welcome to PF!

    I skipped to the next question because I got all confused and I don't know what else to do. :(
     
  5. Feb 20, 2010 #4

    tiny-tim

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    ok, next thing is to find the total acceleration, then find the friction force, then find the coefficient of friction. :smile:

    (to put it another way: you know what acceleration you are trying to produce … so how much force does that need, and where does the force come from?)
     
  6. Feb 21, 2010 #5
    But did I even start it right? I had ac = v2/r = (27.8)2/331 = 2.3348640....m/s2, and then I use u = a/g = 2.3348640..../9.8 = 0.238257.....and I got it wrong again :(.
     
  7. Feb 21, 2010 #6

    tiny-tim

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    (have a mu: µ :wink:)

    Yes, that method is correct (ma = Ffr, and |Ffr| = µmg), except that your a must be the total acceleration (ie the vector sum of the radial and tangential accelerations).
     
  8. Feb 21, 2010 #7
    Okie, I am confused when you use (ma = Ffr, and |Ffr| = µmg) since mine was ac = v2/r. But I think I got what u mean. So i redo it and have (2.3348640....+ 3.68 m/s2) / 9.8 = 0.61376163758555. Am I correct?
     
  9. Feb 21, 2010 #8

    tiny-tim

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    Nope :redface: … you need the magnitude of the vector sum of aradial and atangential, which is … ? :smile:
     
  10. Feb 21, 2010 #9
    Okie, I did sqrt of [((2.334864048...)2 + (3.68)2) / 9.8 ] = 0.44471525601966. correct? thanks
     
  11. Feb 21, 2010 #10

    tiny-tim

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    Should be! :smile:
     
  12. Feb 21, 2010 #11
    OMG, finally got it right. Thank you so much. :)
     
  13. Feb 23, 2010 #12
    I have the same exact problem and understand how to get to the end. But how do you continue the problem without a mass?
     
  14. Feb 24, 2010 #13

    tiny-tim

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    Welcome to PF!

    Hi ffejoriac! Welcome to PF! :smile:

    Call the mass m …

    you'll find it'll cancel in the end. :wink:
     
  15. Feb 24, 2010 #14
    what would be the radial force? is it the friction? thanks I know that the tangential component of the total force is from the car's acceleration.. you only have those 2 forces in the horizontal direction , so it must be the friction?
     
  16. Feb 24, 2010 #15

    tiny-tim

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    Hi holezch! :smile:

    Sorry, but you're putting the cart before the horse, and it's an imaginary cart.
    :redface:

    There aren't 2 forces in the horizontal direction

    For the motion given in the question, the acceleration is completely fixed. So first find that acceleration.

    Then find the force that will produce that acceleration …

    that's the friction force … one force.
     
  17. Feb 24, 2010 #16
    I thought that the car produces the fixed acceleration itself( from the engine or something), and then that would be a tangential force.. and there should be a radial component to the force as well, since the car is moving in circularly.. and you'll have an opposite frictional force along with it right? and that's what makes the car turn? thanks
     
  18. Feb 24, 2010 #17
    You need the magnitude of acceleration which you find from the tangential acceleration and centripetal acceleration
     
  19. Feb 24, 2010 #18

    tiny-tim

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    Nooo :redface:

    that's an internal force …

    only external forces can affect the movement of the car. :wink:

    The engine exerts a force on (ultimately) the tyres, but all this is internal. Then the tyres exert a friction force on the Earth, which makes the Earth rotate very slightly (the exact direction depends mostly on the ways the four wheels are pointing).

    The Earth in turn exerts an equal and opposite force on the tyres … that's the friction force that moves the car.​
     
  20. Feb 24, 2010 #19
    oh wow :| I didn't even know that haha.. That's interesting.. thanks a lot! so the friction creates both tangential and centripetal accelerations
     
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