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Kinetic friction

  1. Oct 9, 2006 #1
    To move a large crate across a rough floor, you push down on it at an angle of 21°. Find the force necessary to start the crate moving, given that the mass of the crate is m = 32 kg and the coefficient of static friction between the crate and the floor is 0.57.

    N=mg=(32)(9.81)= 313.92 * coefficaint of mew sub s=.57=178.9
    Ok, I think I did that right....only I'm not sure what to do with the 21° angle. I did 178.9cos(21)=167.0 but that's not right.
  2. jcsd
  3. Oct 9, 2006 #2
    Did you draw a diagram with the applied force resolved into x and y components? (BTW, that's where the angle comes in)
    Because in this case, N does not equal mg.
  4. Oct 9, 2006 #3
    Like this?
    I did Fk=mewkN=.57*(mgsin(21)=.57*112.5=64.12 I did somthing wrong...again...but I don't know what...
  5. Oct 9, 2006 #4
    where's the force of the push? It has both an x and a y component.
    But yeah, that's the sort of chart I had in mind. :grin:
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