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Simple friction problem?

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    a box having mass 100kg is dragged accross a level floor by means of a rope that makes a 30.0 degree angle with the floor. the coefficient of friction between the box and floor is .330. what force f just be applied to hte rope to move the box along at a constant velocity?
    meow

    2. Relevant equations

    joooooooooooooooooooooooooooooooooonwooooooooooooooooooooooooooo nhoooooooooooooooooooo

    3. The attempt at a solution

    i drew a body diagram
    x=zcos30-f
    y=n-980 +zsin30

    z is the force needed to pull the rope
     
    Last edited: Oct 18, 2009
  2. jcsd
  3. Oct 18, 2009 #2
    The box moves at a constant velocity when the force of kinetic friction equals the force of horizontal pull (net horizontal force = 0). You know for this problem that your net normal force is 980 - F*sin(30).

    so your force of friction in the negative horizontal direction, plus your Force of pull in the positive horizontal direction must be add to be zero.

    [980 - F*sin(30)]*(.330) + F*Cos(30) = 0

    Edit* where F = your force of pull
     
  4. Oct 18, 2009 #3
    Do you follow the reasoning?
     
  5. Oct 18, 2009 #4
    johnwoooooooooooooooo
     
    Last edited: Oct 18, 2009
  6. Oct 18, 2009 #5
    Also, i think i missed a negative sign


    [980 - F*sin(30)]*(.330) - F*Cos(30) = 0
     
  7. Oct 18, 2009 #6
    stupid typos heh
     
  8. Oct 18, 2009 #7
    yea i got it. my answer is 314nt...is that wat you got joonwoo??
     
    Last edited: Oct 18, 2009
  9. Oct 18, 2009 #8
    Ya, that is the answer I get as well. The trick to the problem was definitely remembering that Fnet = 0 at constant velocity. That would have probably given you the extra equation you would have needed to not have too many variables
     
  10. Oct 18, 2009 #9
    yea thanks for your help jooonwoo !~!@$!#%!
     
    Last edited: Oct 18, 2009
  11. Oct 18, 2009 #10
    No problem : )
     
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