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Car on banked turn, with reversed banking angle

  • Thread starter pech0706
  • Start date
  • #1
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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


Ffriction=[tex]\mu[/tex]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!!!
 

Answers and Replies

  • #2
why have you not taken components of centrifugal force
 
  • #3
Doc Al
Mentor
44,912
1,170
why have you not taken components of centrifugal force
There's no need for centrifugal force.
 
  • #4
Doc Al
Mentor
44,912
1,170

Homework Equations


Ffriction=[tex]\mu[/tex]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.)
 
  • #5
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
 
  • #6
Doc Al
Mentor
44,912
1,170
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.
 
  • #7
I did not consider the course. I am more with the non-inertial frame.
I will consider it in future. thanks
 

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