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Help with Friction and Stopping Car

  1. Sep 24, 2008 #1
    1. The problem statement, all variables and given/known data
    A car is traveling at 50.0 mi/h on a horizontal highway. (a) If the coefficient of
    static friction between road and tires on a rainy day is .100, what is the minimum
    distance in which the car will stop? (b) What is the stopping distance when
    the surface is dry and coefficient of static friction is .600?


    2. Relevant equations[/]


    3. The attempt at a solution
    I'm not sure how to get started as I don't know the normal force, and I'm not
    sure what to solve for. If someone can help get the ball rolling I should be able
    to finish. Any help would be appreciated.

    Kevin
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 25, 2008 #2

    tiny-tim

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    Hi Husker70! :smile:

    Call the mass of the car m (it will cancel out later :wink:).

    Then the normal force (on a horizontal surface) is just mg.

    Now use F = ma to find the acceleration … which will be constant, so you can then use the usual uniform accleration kinetic equations. :smile:
     
  4. Sep 25, 2008 #3
    Thanks for the start and I figured that the mass shouldn't matter but in order to
    slove for the acceleration, I'm not sure how to find the force.

    Thanks,
    Kevin
     
  5. Sep 25, 2008 #4

    tiny-tim

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    Hi Kevin! :smile:

    Since it's a horizontal road, the only force slowing it down is the friction force …

    so multiply the normal force by µ. :smile:
     
  6. Sep 25, 2008 #5
    Does this seem right?
    50 mi/h = 80.4672 km/h = 22.352 m/s

    a= .100n
    ma = .100(g)
    a = .98

    solving for time
    v=vo + at
    22.352m/s=0+.98(t)
    t=22.81s

    solving for distance
    d=do+vot+at^2/2
    d= 0+0+.98(22.815)^2/2
    d= 255m

    Thanks for everyone's help
    Kevin
     
  7. Sep 26, 2008 #6

    tiny-tim

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    Hi Kevin! :smile:

    Right answer, long-winded method …
    a = .98 is right, but the other lines should be:

    ma= .100n
    a = .100(g) :wink:
    hmm … you have vf and vi and you need s but not t …

    so it would have been a lot quicker just to use vf2 = vi2 + 2as :smile:
     
  8. Sep 18, 2011 #7
    Hello,

    I am working on the same problem, but I do not comprehend the work that was done to solve this problem.
    Problem statement and variables given:
    Q: A car is traveling 50.0 mi/h on a horizontal highway. (a.) If the coefficient of static friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop? (b.) What is the stopping distance when the surface is dry and mu(s)=0.600? (mu(s) I think means static friction)
    Attempt:
    What I did was draw a free-body diagram where the weight force(earth is acting upon the car) is pointing down, normal force(road acting upon the car) is pointing up, and there is a friction force going to the left. I did convert 50.0mi/hr to 22.35m/s, but after that I am lost.

    Thank you,

    Joyce
     
  9. Sep 19, 2011 #8

    tiny-tim

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    Welcome to PF!

    Hello, Joyce! Welcome to PF! :smile:

    (have a mu: µ :wink:)
    i] First, find the friction force.

    ii] Then find the acceleration.

    What do you get? :smile:
     
  10. Sep 19, 2011 #9
    Hello Tiny-Tim,

    Would the friction force be: F=-w+N (friction equals negative weight force plus normal force)? I think I might be confused on how to obtain the friction force.

    Thank you,

    Joyce Kuang
     
  11. Sep 19, 2011 #10

    tiny-tim

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    Hello Joyce! :smile:
    That is a valid equation, but it's the equation for the net vertical force

    in this case the vertical acceleration is zero,

    so (from good ol' Newton's second law) mass times 0 = net force = -W + N.

    so W = N …

    in other words, when something stays on a flat surface (so that its vertical acceleration is zero), then the normal force equals the weight. :smile:

    (you can get a similar equation for the normal force on a slope, by doing 0 = net perpendicular force)

    To get the friction force, you don't use F = ma at all, you use F = µN,

    ie friction = µ times normal force

    This works either for kinetic friction (µk), or for maximum static friction (µs) …

    in this case, the static friction is maximum, since the question says so (because it asks for the minimum stopping distance). :wink:
     
  12. Sep 19, 2011 #11
    Hello,

    What I have so far is:
    Summation of F=m(0)=net force=-w+N
    In which you said w=N
    Then you said to use F=[itex]\mu[/itex] N
    I know that [itex]\mu[/itex] is 0.100
    so, F=(.100)N
    What is N? I know that it is equal to the weight force, but it does not give me a value for the weight or normal force. How can I obtain that value?

    And after finding the normal force, then I know the friction force, right?
    After finding the friction force why do I have to find the acceleration? Is it because we need acceleration to find the minimum distance?

    Thank you,

    Joyce
     
  13. Sep 19, 2011 #12

    tiny-tim

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    Hello Joyce! :smile:
    They don't give you the mass of the car, so call it "m" (it will cancel out later, when you find "a") …

    then W = mg. :wink:
    Yes, once you know "a", you can use one of the standard constant acceleration equations. :smile:
     
  14. Sep 19, 2011 #13
    I'm so sorry, but I am still a little confused.
    The normal force (N in newtons) is equal to the weight force.
    You said to use "m" to represent the mass, but what do I do after that?
    What do I do with my F=[itex]\mu[/itex]N equation? How does that relate to the mass?
    Maybe that is where my trouble lies. I do not see a connection between the mass and the equation, or I do not understand.

    Thank you,

    Joyce
     
  15. Sep 20, 2011 #14
    Hey, Joyce.

    You know that F=[itex]\mu[/itex]N. You also know that N = W = mg.

    Now, gravity is simply an acceleration in the y-direction, correct?

    So, now you know that N = ma[itex]_{y}[/itex]


    You already knew that F=[itex]\mu[/itex]N. Now F, in this situation, is friction force. So, it is a force in the x-direction. So, given that, you could say that.....

    ma[itex]_{x}[/itex]=[itex]\mu[/itex]ma[itex]_{y}[/itex]

    Since the mass is the same for both sides, the masses cancel out (since you would divide both sides by m). That leaves you with......

    a[itex]_{x}[/itex]=[itex]\mu[/itex]a[itex]_{y}[/itex]

    From that you can solve for a[itex]_{x}[/itex] and then use kinematics to solve for [itex]\Delta[/itex]x.


    Enjoy,
    Alex
     
  16. Sep 20, 2011 #15
    Aggression200,

    Thanks, I got it.
    What I did was:
    F=[itex]\mu[/itex]sN
    N=mg
    N-may (g is ay)
    max=[itex]\mu[/itex]s(may)

    a=(.100)(-9.8)= -0.98
    Vi=22.35 m/s
    Vf=0 m/s
    x=?

    So take the equation Vf2=Vi2 +2ax
    Solved for x, which gives you
    (Vf2-Vi2)/(2a)
    x=255m?

    Thank you very much,

    Joyce
     
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