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Incline block

  1. Dec 24, 2007 #1
    1. The problem statement, all variables and given/known data
    A block weighing 75 N rests on a plane inclined at 25 degrees. A force F is applied to the object at 40 degrees to the horizontal, pushing it upward on the plane. Coefficient of static friction between block and plane is .363

    EDIT: What is the minimum value of F that will prevent the block from slipping down the plane?

    2. Relevant equations
    Ff = uFn. F = ma, etc.

    3. The attempt at a solution

    So the free-body diagram has F and Ff in the same direction (diff angles) and F(parallel to surface) is in the other way.

    Ff + Fcos(15) = Fgsin(25) (15 degrees because 45-20 = 15)
    .363*75cos(25) + Fcos(15) = 75sin(25)

    I solved for F and got 7.3 N while the answer is 8.05 N. Thank you for any hints/help
    Last edited: Dec 25, 2007
  2. jcsd
  3. Dec 25, 2007 #2
    If you have a block moving up a plane then it's kinetic friction (which will most likely be <= static friction) for starters... Also the friction force opposes the movement of the object you it should be

    Fcos(15) = Fgsin(25) + Ff
  4. Dec 25, 2007 #3
    Wow I forgot to put the actual question down, they're actually asking

    What is the minimum value of F that will prevent the block from slipping down the plane?
  5. Dec 25, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Hint: What's the normal force? How does F affect the normal force?
  6. Dec 25, 2007 #5
    got it thank you.
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