# Car on banked turn, with reversed banking angle

## 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 $$\mu$$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

Ffriction=$$\mu$$s[(mgcos10-mv2/r)
Fparallel=mgsin10
N=mgcos10-(mgcos10-mv2/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-v2/15)=1.703(m)
divide out the m to cancel mass giving you:
(0.5)(9.81cos10-v2/15)=1.703
divide out 0.5
9.8cos10-v2/15=3.407
subtract 9.8 cos10
-v2/15=-6.244138
multiply both sides by -15
v2=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!!!

Related Introductory Physics Homework Help News on Phys.org
why have you not taken components of centrifugal force

Doc Al
Mentor
why have you not taken components of centrifugal force
There's no need for centrifugal force.

Doc Al
Mentor

## Homework Equations

Ffriction=$$\mu$$s[(mgcos10-mv2/r)
Fparallel=mgsin10
N=mgcos10-(mgcos10-mv2/r)
Show how you arrived at these equations. (Always start with a free body diagram and then apply Newton's 2nd law.)

in the car's frame mv^2/r is the centrifugal force. pech0706 has used it in his equations. how can there be no need for it as the car is moving in a circular track

Doc Al
Mentor
in the car's frame mv^2/r is the centrifugal force. pech0706 has used it in his equations. how can there be no need for it as the car is moving in a circular track
Centrifugal force is only used when viewing things from an non-inertial frame. There is no need to use anything other than the usual inertial frame for this problem. You could use such a frame, but you don't need to. (And unless non-inertial frames have been explicitly covered in the course, it will just add to the confusion.)

Note that v^2/r is the magnitude of the centripetal acceleration.

I did not consider the course. I am more with the non-inertial frame.
I will consider it in future. thanks