Car on the banked turn with friction

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Homework Help Overview

The problem involves a car traveling on a reverse banked turn, where the road is tilted away from the curvature. The parameters include a radius of curvature of 15 m, a static friction coefficient of 0.5, and a banking angle of 10 degrees. Participants are tasked with finding the maximum velocity possible on this turn.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the maximum velocity using a formula derived from the static friction and gravitational forces. Some participants provide their own calculated velocities, which differ from the original poster's result. Others express confusion about the discrepancies in their results and seek clarification on the calculations.

Discussion Status

Participants are actively discussing their different approaches to the problem, with some providing equations and derivations. There is no explicit consensus on the correct method or answer, as various interpretations and calculations are being explored.

Contextual Notes

Participants are working under the constraints of the problem statement, including the specific values for radius, friction, and banking angle. The discussion reflects uncertainty regarding the correct application of the equations involved.

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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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i get 7.8m/s..
 
pat666 said:
i get 7.8m/s..

how?
 
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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