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Balance of a car on an inclined plane

  1. Feb 22, 2012 #1

    okh

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    1. The problem statement, all variables and given/known data
    A 900 kg car goes up along an inclined road by 30 degrees characterized by an acceleration of 0.25 m/s^2. If friction coefficient is 0.5, find the magnitude of the force that moves the car.
    Solution: F = 12285.6 N


    2. Relevant equations



    3. The attempt at a solution
    http://img259.imageshack.us/img259/5361/imagetcq.jpg [Broken]
    Sum of all forces is: m *a = 900 * 0.25 = 225
    Then Ffriction is = 0.5 * FwPerpendicular = 0.5 * sin(60) * 900 * 9.8 = 3819 N
    Then Fwparallel is = sin(60) * 900 * 9.8 = 4410
    F = 3819 + 4410 + 225 = 8454 N != solution
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 22, 2012 #2
    You are missing something :D
     
  4. Feb 22, 2012 #3

    okh

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    Do you mean in the problem statement ( :smile: ) or in my attempt?
     
  5. Feb 22, 2012 #4
    are you getting 8454 ?
     
  6. Feb 22, 2012 #5

    okh

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    Yes, exactly.
     
  7. Feb 22, 2012 #6
    From where did you get this question
    because as far as I know (:)) you are doing correctly (hopefully)
     
  8. Feb 22, 2012 #7

    okh

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    The teacher dictated us that problem. However I've just asked my schoolmates and the teacher said the solution was wrong and the right one was 8454!!!!!
     
  9. Feb 22, 2012 #8
    Then why you posted this things here ( : ) ) i am still trying make a smiley smile oh god
     
  10. Feb 22, 2012 #9

    ehild

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    Homework Helper
    Gold Member

    FwPerpendicular is mgcos(60).
    Figure out which force acts in the direction of acceleration, up along the incline and which one acts against it.

    ehild
     
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