Car on the banked turn with friction

  • Thread starter Mehtab
  • Start date
  • #1
Mehtab
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Homework Statement



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

Homework Equations



v=sqrt(Usgr)

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
 

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    frictionbank1.gif
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Answers and Replies

  • #2
pat666
709
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i get 7.8m/s..
 
  • #3
Mehtab
3
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i get 7.8m/s..

how?
 
  • #4
Mehtab
3
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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:

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