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Finding the coefficient of Kinetic force with an angle

  1. Oct 14, 2015 #1
    1. The problem statement, all variables and given/known data
    A 1000N crate is being pushes across a level floor at a constant speed by a force (F) of 300N at an angle of 20 degrees below the horizontal, what is the coefficient of kinetic force?

    2. Relevant equations
    Fk= uk*N
    uk= Fk/N

    Fk=F2

    3. The attempt at a solution
    in the back of the textbook it comes up as 0.256, but when I try I do:
    Fk = F2
    F2 = 300cos20 = 281.9
    uk= F2/n
    uk= 281.9/1000
    uk= 0.281

    Can someone tell me where I've gone wrong? It seemed like it should be a simple question
     
  2. jcsd
  3. Oct 14, 2015 #2

    Nathanael

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    The normal force from the ground is not always the weight of the object.
    The normal force is what ever it needs to be to keep the block from accelerating vertically.
     
  4. Oct 14, 2015 #3
    okay, so should I try and use that to try and find the weight? Because that was all the question gave me, it didn't mention any weights.
     
  5. Oct 14, 2015 #4

    Nathanael

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    The weight of the crate is given, it is 1000N.
    In your equations you seem to be saying that the normal force is equal to the weight of the crate (1000N).

    If I push down on a piece of paper, the normal force on the paper (from the table or whatever) is going to be much more than the weight of the paper, right?
     
  6. Oct 15, 2015 #5

    haruspex

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    To be precise and general, the normal force from a surface is the force of least magnitude required to prevent the object penetrating the surface. Necessarily, that will be at right angles to the surface.

    @KellyTarzuoty, draw a free body diagram of the block. Three forces act on it. What are they? Use the usual statics equations to find the normal force.
     
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