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Coefficients of Friction

  1. Sep 29, 2005 #1
    This is the question I am working on:

    An adult is pulling two small children in a sleigh over level snow. The sleigh and children have a total mass of 47 kg. The sleigh rope makes an angle of 23 degrees with the horizontal. The coeffcient of friction between the sleigh and the snow is 0.11. Calculate the magnitude of the tension in the rope needed to keep the sleigh moving at a constant velocity. (Hint: The normal force is not equal to the force of gravity.)

    I drew a FBD diagram and broke the information down;

    m = 47 kg
    a = 0
    uk = 0.11

    Now the way I was going to do it was find normal force, then use that in (FN)(uk ) = Fk to find the kinetic friction.

    Then I was going to put Fk = Ft (kinetic friction = tension)

    However, I am having difficulting finding the normal force, since it is not equal to force of gravity.

    Thanks in advance.

    O, and it's my first day here, hello all posters.
     
  2. jcsd
  3. Sep 29, 2005 #2

    hotvette

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    Go back to your FBD. The rope pulls on the sleigh at an angle. Therefore, there is a component of rope force in the positive vertical direction. That's why the normal force isn't equal to gravity (it's something less).
     
  4. Sep 29, 2005 #3
    You know what, I was pondering that and think it would be ___sin23 but am not sure what would go in the blank.
     
  5. Sep 29, 2005 #4

    hotvette

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    That would be T, the tension in the rope. All you have to do is sum the forces, write F = ma (in the horizontal direction), and go from there. Hint: what can you say about acceleraton if velocity is constant?
     
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