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Force - rope pulling box

  1. Oct 7, 2003 #1
    Okay, a box of mass m is dragged across a floor by pulling a rope attached to the box and inclined at an angle @ above the horizontal.

    given that the coefficient of static friction is 0.05, what minimum force mag is required from the rope to start the crate moving?

    When it says it is inclined at an angle @, does that mean that the rope has an angle @ above the horizontal? If this is the case, could you let me know if this process is correct:

    Fx = Tcos@ - fs = Tcos@ - 0.5N = ma = 0
    Fy = N + Tsin@ - mg = ma = 0 ---> N = mg - Tsin@

    from Fx:
    Tcos@ - 0.5(mg - Tsin@) = 0

    Then I can solve for T, right?
  2. jcsd
  3. Oct 8, 2003 #2


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    Staff Emeritus
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    Yes, the "effective weight" is the actual weight minus the upward lift from the rope: mg- T sin θ. The force you must apply to overcome static friction is (0.5)(mg- T sin θ) and the horizontal force to do that is T cos θ : the equation is
    0.5(mg- T sin θ)= T cos θ as you have.
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