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Crate being pulled

  1. Dec 19, 2004 #1
    Good evening.

    I would appreciate if someone could verify the following please.

    The question:

    A 100kg crate is pulled across a horizontal floor by a force P that makes an angle of 30 degrees above the horizontal. The coefficient of kinetic friction between the crate and floor is 0.200. What is P if the net work done is zero.

    My solution:

    First determine Ff = umg. This would give me 196N. My next step is to determine P. I use 196N and put that over cos30. My answer is 226.3N. Does this seem reasonable.

    thanks for your time
  2. jcsd
  3. Dec 19, 2004 #2

    Doc Al

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    Staff: Mentor

    No, [itex]F_f = \mu N[/itex], where N is the normal force. To solve this problem, recognize that the crate is in equilibrium. Set up equations for the vertical and horizontal forces. (Net force in each direction equals zero.)
  4. Dec 19, 2004 #3
    Thanks for the direction.

    Is this what you meant?
    mgsin30 = f + Fcos30 = uN + Fcos30
    N = Fsin30 = mgcos30

    To resolve = Add N into N in equation for Horizontal
  5. Dec 19, 2004 #4

    Doc Al

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    Staff: Mentor

    I don't understand how you got your equations. Here's what I get:
    Horizontal: [itex]P cos(30) = \mu N[/itex]
    Vertical: [itex]N + P sin(30) = mg[/itex]
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