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Force of Fs

  1. Dec 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A 2000kg car makes a left run of radius 6.5 m. If this turn is made in 2 seconds, what isthe force of static friction of the car during the turn?

    2. Relevant equations

    a = v^2/r, F = (m*v^2)/r ; d = v*t;

    3. The attempt at a solution

    d/t = v = 3.25 m/s

    Fc = (m*v^2)/r = 3250

    Fs(static friction) <= Fc
  2. jcsd
  3. Dec 18, 2009 #2

    Doc Al

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    Staff: Mentor

    What distance is traveled in making the turn? How would you calculate it?
  4. Dec 18, 2009 #3
    Oh whoops the distance would be 2*pi*r/4 since it goes a quarter of a circle.

    That would make d/t = 5.1 m/s

    Fc = 3.2 kN. Still stuck from there
  5. Dec 18, 2009 #4

    Doc Al

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    Staff: Mentor

    Don't forget to recalculate Fc using the corrected speed.
  6. Dec 18, 2009 #5
    Yea I did but I forgot to square! :tongue2: Got it now thanks
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