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Force of Friction on object on an incline

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data
    A 3812.0 kg truck is parked on a 18.1° slope. What is the friction force on the truck?


    2. Relevant equations

    3. The attempt at a solution
    So i set up a triangle with and then used:
    weight of truck= (9.8 m/s^2 *3812 Kg) = 37367.6 N

    sin(18.1) = (37367.6 N) /(-Ff)
    Ff = -120246 N

    Could someone explain why this is wrong.. im thinking maybe i set up my triangle incorrectly, but i dont know how else i would draw it.. thanks in advance
     
  2. jcsd
  3. Feb 16, 2009 #2

    LowlyPion

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    Friction is based on the force normal to the incline, not normal to the horizontal.
     
  4. Feb 16, 2009 #3
    I thought friction went parallel to the incline?
     
  5. Feb 16, 2009 #4

    LowlyPion

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  6. Feb 16, 2009 #5
    sorry but im not following what your saying.. i dont know the materials so i cant find μ...
     
  7. Feb 16, 2009 #6

    LowlyPion

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    You don't need μ in this case. The Frictional Force is what it balancing the truck's component of gravity down the incline. They just want you to find it. Not calculate it from the Normal force.
     
  8. Feb 16, 2009 #7
    ok so Ff = mgcos(theta) .. and then is 18.1 my angle?
     
  9. Feb 16, 2009 #8

    LowlyPion

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    Is M*g*cosθ the component of gravity || to the incline?
     
  10. Feb 16, 2009 #9
    oh no then i would be mgsinθ = (3812 kg)(9.81 m/s^3)(sin18.1) = 1.162e4 N
     
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