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Friction force

  1. Sep 26, 2007 #1
    1. The problem statement, all variables and given/known data

    A 5000 kg truck is parked on a 14 degree slope. How big is the friction force on the truck?

    2. Relevant equations

    friction = coefficient of friction X normal force

    3. The attempt at a solution

    I tried breaking the weight into components. I used 1.00 as the coefficient of friction (given in a table in the book...rubber on concrete). I got 60916.27 N, which of course was wrong.
  2. jcsd
  3. Sep 26, 2007 #2


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    You don't need no stinking tables. The force of friction acting on the truck is exactly equal and opposite to the force attempting to drag the truck down the slope. Since the truck is 'parked' and not moving, total force on it is zero.
  4. Sep 26, 2007 #3
    Zero isn't right.
  5. Sep 26, 2007 #4


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    Of course it isn't. Total force is tangential gravitational force plus frictional force. They act in opposite directions. What is the component of gravitational force acting down the incline? That's equal to the frictional force.
  6. Sep 26, 2007 #5
    The gravitational force down the incline is mgcos(theta), which equals 47593.005. I tried that as well.
  7. Sep 26, 2007 #6


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    I think the normal force is mg*cos(theta). I think the downward tangential force is mg*sin(theta).
    Last edited: Sep 26, 2007
  8. Sep 26, 2007 #7
    That was right. Thank you!
  9. Sep 26, 2007 #8


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    I hope you know why that was right. But you're welcome.
    Last edited: Sep 27, 2007
  10. Sep 27, 2007 #9
    Take a look at my thread for a similar problem called "Friction force problem".

    You are almost there to solving this. mgcos(14) gives you the normal force. which is perpendicular to the slope. It should also be in mega newtons.

    Look at your formula.

    The answer I am getting is 11.866 MegaN. I am studying the samething so. yea.

    EDIT: damn i was too late.
  11. Sep 27, 2007 #10


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    You are not only late, the answer you gave is pretty wrong. Suggest you figure out why.
  12. Sep 27, 2007 #11
    Ok. I re-did the problem and now am getting 11.8649 kN as the frictional force.

    For some reason I converted the 5000 kg to grams and then did multiplied by 9.81 m/s^2.

    Is that the correct answer?

    The static friction coeff = tan14 = 0.2493
    FN = 47593 N

    (0.2493 x FN)= FF

    = 11.8649 kN
  13. Sep 27, 2007 #12
    I got 11866.27 N
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