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Homework Help: Car on the banked turn with friction

  1. May 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Car traveling along the "reverse" banked turn (road tilted away from the curvature). Radius of curvature is 15 m, static friction between road and tires is 0.5, and banking angle is 10 degrees. Find the maximum velocity possible on this reverse banked turn.

    frictionbank1.gif
    2. Relevant equations

    v=sqrt(Usgr)

    3. The attempt at a solution
    sqrt(0.5 x 9.8 x 15)= 8.57 m/s

    Its wrong as actual answer is 7.2m/s
     

    Attached Files:

    Last edited: May 30, 2010
  2. jcsd
  3. May 30, 2010 #2
    i get 7.8m/s..
     
  4. May 30, 2010 #3
    how?
     
  5. May 31, 2010 #4
    I have tried and got answer using this equation:
    v=sqrt(((MUcostheta-sintheta) x g x r)/(costheta-MUsintheta))

    I derived it by:
    sum of x forces: Fx=Fcostheta-MU(Nsintheta)=(mv)^2/r
    sum of y forces: Fy=0=Ncostheta-MU(Nsintheta)-mg
    mg=N(costheta-MU(sintheta)
    N=(mg)/(costheta-MU(sintheta))

    Netforce: -Fnet(opposite direction)=Nsintheta-fcostheta(Opposite direction)
    Fnet=-Nsintheta+MU(Ncostheta)
    =N(-sintheta+MU(costheta))
    =((mg)/(costheta-MU(sintheta)) x (-sintheta+MU(costheta)
    Fnet=Fcentripetal

    (mv^2)/r=((mg)/(costheta-MU(sintheta)) x (-sintheta+MU(costheta)
    v=sqrt((MUcostheta-sintheta) x g x r)/(costheta-MUsintheta))
     
    Last edited: May 31, 2010
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