- #1

pech0706

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## Homework Statement

Consider the problem of a car traveling along a banked turn. Sometimes roads have a "reversed" banking angle. That is the road is tilted "away" from the center of curvature of the road. The coefficient of static friction between the tires and the road is [tex]\mu[/tex]

_{s}=0.50, the radius of curvature is 15m, and the banking angle is 10 degrees. What is the max speed at which a car can safely navigate such a turn?

## Homework Equations

F

_{friction}=[tex]\mu[/tex]

_{s}[(mgcos10-mv

^{2}/r)

Fparallel=mgsin10

N=mgcos10-(mgcos10-mv

^{2}/r)

## The Attempt at a Solution

I know the answer is supposed to be 7.2 m/s, however I keep getting in the whereabouts of 10m/s.

here is what I did: (0.5)(m)(9.81cos10-v

^{2}/15)=1.703(m)

divide out the m to cancel mass giving you:

(0.5)(9.81cos10-v

^{2}/15)=1.703

divide out 0.5

9.8cos10-v

^{2}/15=3.407

subtract 9.8 cos10

-v

^{2}/15=-6.244138

multiply both sides by -15

v

^{2}=93.662085

take the square root of 93.662085

v=9.6779m/s

v=10m/s

I don't know what I'm doing wrong, but from what I understand the normal force is reduced due to the reversed banking angle. Other than that I'm not real sure where I'm going wrong, I don't know if the equations I've figured out are not correct. I could really use some help, my homework is due on Wednesday and this is the only one I can't figure out. Thanks!!!