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Max Speed on an Incline

  1. Mar 9, 2015 #1
    1. The problem statement, all variables and given/known data
    A concrete highway curve of radius 70 m is banked at a 15° angle. What is the maximum speed with which a 1500 kg rubber tired car can take this curve without sliding? The static friction coefficient is 1.

    2. Relevant equations
    Centripetal Force = ## (m v^2) / r ##
    Maximum Friction Force = ## μN ##

    3. The attempt at a solution
    ## (m v^2) / r ## = centripetal force

    Friction_max = 1 * normal force = ## 9.8 * 1500 * cos(15) = 14199.11 N ##

    Plug back into equation:

    ## (1500 v^2) / 70 = 14199.11 ##

    Which yields v = 25.7 m/s

    However, the answer is 34.5 m/s. What am I doing wrong? I don't understand where I am going wrong in my steps.
     
  2. jcsd
  3. Mar 9, 2015 #2
    Are you sure that the normal force is the only force you have to worry about? No other forces? None at all?
     
  4. Mar 9, 2015 #3
    I mean the frictional force
     
  5. Mar 9, 2015 #4
    I figured that when the centripetal force exceeds the maximum frictional force, the car would skid. Solving for the maximum frictional force only requires the normal force, so I didn't include any other forces. Where would I have to use the friction force?
     
  6. Mar 9, 2015 #5
    Look, I phrased that statement poorly. Reevaluate the forces acting on the car tire.
     
  7. Mar 9, 2015 #6
    Everything else is correct, your steps, your concept of friction, all correct.
     
  8. Mar 9, 2015 #7
    Forces on the tires: friction, normal, centripetal

    Am I missing anything?
     
  9. Mar 9, 2015 #8
    Yes. Yes, you are missing something very important. A force that is in every inclined plane problem you have ever seen...
     
  10. Mar 9, 2015 #9
    Ignore the centripetal forces. Just think about a tire on an incline. What forces act on it?
     
  11. Mar 9, 2015 #10
    OH, I'm missing gravity, aren't I?
     
  12. Mar 9, 2015 #11
    I do believe you are.
     
  13. Mar 9, 2015 #12
    Hmm, I still don't get the right answer. I tried adding the force of gravity in the direction of the incline to the frictional force (which would just double it), so I got:

    ## (m v^2) / r = 28398.22 ##
    ## (1500 v^2 / 70) = 28398.22 ##
    ## v = 36.4 ##
     
  14. Mar 9, 2015 #13
    Don't just give me numbers. Keep it symbolic. What are the two factors that make up the 28398.22N?
     
  15. Mar 9, 2015 #14
    The gravitational force directed down the incline would not be cosθmg
     
  16. Mar 9, 2015 #15
    I made an error, it should be 18003.75N

    Maximum Frictional Force: ## 9.8 * 1500 * cos(15) = 14199.11 ##
    Force due to gravity: ## 9.8 * 1500 * sin(15) = 3804.64 ##
    Added together: ## 14199.11 + 3804.64 = 18003.75 N ##

    ## (m v^2) / r = 18003.75 ##
    ## (1500 v^2 / 70) = 18003.75##
    ## v = 29.0 m/s ##
     
  17. Mar 9, 2015 #16
    I would wait until you have the solution to start plugging numbers in. You made a sign error.
     
  18. Mar 9, 2015 #17
    Should the friction force be negative?

    Even if I changed the sign on any of the numbers, my answer would still be incorrect. If I change the sign on either forces, I would get a smaller number and therefore a smaller velocity as an answer.
     
  19. Mar 9, 2015 #18
    You are right. Your solution above looks good. I think that you either made a calculation mistake, or that the answer you claim is correct, is not actually correct.
     
  20. Mar 9, 2015 #19
    Hmm, okay. Thank you for all the help! :) I really appreciate it.
     
  21. Mar 9, 2015 #20
    Any time.
     
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