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Simple Forces Question

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A car of mass 500kg climbs a 15 deg. hill. If the car's engine provides a forward thrust of 8000 N and the drag and friction on the car add up to 800 N calculate the car's rate of acceleration


    2. Relevant equations
    F=ma
    Normal F= mg x 15 deg



    3. The attempt at a solution

    Normal force = mg x cos 15

    NF = 500 x 10 x 0.9

    NF = 4500 N

    Backward Force = 800 + mg x sin15

    BF = 800 +5000 x 0.2

    BF=1800 N

    Fwd Force = 8000 N

    Resultant force = 8000 N - 1800 = 7200 N


    F=ma

    7200 = 500 x a
    500a=7200
    a= 14 ms

    I have been killing myself and I still can't find the mistake !

    Help me! please !

    Thanks,

    dave
     
  2. jcsd
  3. Nov 9, 2009 #2
    Can someone please help me ?
     
  4. Nov 9, 2009 #3
    What add up to 800 mean?I don't think it means that friction is 800N
     
  5. Nov 9, 2009 #4
    Maybe that is the error.
     
  6. Nov 9, 2009 #5
    No mate I just checked twice with the textbook thats exactly ow the question looks like.
     
  7. Nov 9, 2009 #6
    I don't said that the error was what the textbook said.i said that maybe you have misunderstood it.maybe it doesnot mean that T=800N
     
  8. Nov 9, 2009 #7
    I think the mistake is somewhere in the resultant force and then there is a follow through :(
     
  9. Nov 9, 2009 #8
    It says drag nad friction add up to 800
     
  10. Nov 9, 2009 #9
    so backward force is just 800 ?
     
  11. Nov 9, 2009 #10
    It cannot because only backward from weight is above 1000.But it may mean that friction is 800 less or more.I donnot know but the mistake must be there
     
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