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Friction problem of a crate

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data
    A person is dragging a packing crate of mass 100kg across a rough floor where the coefficient of kinetic friction is 0.4. He exerts a force F just sufficient to keep the crate moving at a constant velocity. At what angle above the horizontal should his pulling force F be for it to be minimum?

    2. Relevant equations



    3. The attempt at a solution

    Kinetic friction force = 0.4 x 100g = 40g

    Fcos theta = 40g
    F sin theta = 100g

    tan theta = 10/4
    theta = 68.1 degree

    am i right?
     
  2. jcsd
  3. Sep 29, 2011 #2
    Draw a diagram and put in ALL the forces. You are leaving out the normal reaction force of floor on crate. This normal reaction force IS NOT EQUAL to the weight of the crate since the pulling force is not horizontal.
     
  4. Sep 29, 2011 #3
    Ok.

    Let the angle be theta,

    Friction force = 0.4 x 100g x cos theta

    F - 100g sin theta - 40g cos theta = 0

    F = 980 sin theta + 392 cos theta

    dF/dtheta = 980 cos theta - 392 sin theta

    Let derivative = 0

    tan theta = 2/5

    i got back the same ans! What is wrong?
     
  5. Sep 29, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    Why? What does cos([itex]\theta[/itex]) have to do with the weight of the crate?

    What's the normal force?
     
  6. Sep 30, 2011 #5
    The suggestion was to DRAW A DIAGRAM WITH ALL THE FORCES. Then you can get more help.
     
  7. Sep 30, 2011 #6
    The suggestion was to DRAW A DIAGRAM WITH ALL THE FORCES. Then you can get more help.
     
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