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Minimum amount of friction in a turn

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data
    What minimum coefficient of friction between the tires and the road will allow a 3700 kg truck to navigate an unbanked curve of radius 25 m at a speed of 45 km/h?



    2. Relevant equations



    3. The attempt at a solution
    The minimum coefficient is 7.62. To arrive at this i used the formula k(min)=(V2/(gr)). This will be substituted to 252/(3700/45).
     
  2. jcsd
  3. Mar 18, 2012 #2

    Doc Al

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    Staff: Mentor

    Your formula is correct, but your substitution is not. Hint: Convert the speed to standard units of m/s.
     
  4. Mar 18, 2012 #3
    so instead put in 12.5, and then 625/296 = 2.11 ???
     
    Last edited: Mar 18, 2012
  5. Mar 18, 2012 #4

    Doc Al

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    No. Where do the numbers 625 and 296 come from?
     
  6. Mar 18, 2012 #5
    well 25^2 is 625 and 3700/12.5 is 296
     
  7. Mar 18, 2012 #6

    Doc Al

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    You need to evaluate: (V2)/(gr)

    No need to square 25 or involve 3700.
     
  8. Mar 18, 2012 #7
    you lost me now, why would i square V^2? :(
     
  9. Mar 18, 2012 #8

    Doc Al

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    See your first post. (You need to evaluate V^2, which is V squared.)
     
  10. Mar 19, 2012 #9
    but i thought it would only SQRT if both sides have an exponent of 2 to get rid of the 2, because what i do to one side i have to do to the other do i not?
     
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