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Homework hellp. coefficent of friction.

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data
    An 18.0 kg box is released on a 37.0 degree incline and accelerates down the incline at 0.270 m/s/s. Find the friction force impeding it's motion. How large is the coefficient of friction?


    2. Relevant equations
    Ff=uFn


    3. The attempt at a solution
    I have no idea how to do this.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 6, 2009 #2
    Can you do the problem if there were no friction, that is just asking for acceleration down an inclined plane?
     
  4. Dec 6, 2009 #3
    ohh, maybe how can i do it without friction though?
     
  5. Dec 6, 2009 #4
    Well you can't but the principles used apply:

    In other words the acceleration from gravity along the surface of the ramp is sin(theta) * g. It is not 9.8. So if there were no friction a=9.8 sin(37). What is this come out to be?
     
  6. Dec 6, 2009 #5
    5.9?
     
  7. Dec 6, 2009 #6
    OK. Now we are told that the acceleration is not 5.9 but a measly 0.27.

    So lets write an eqn using Newtons laws.

    ma=sum of all forces ? is the frictional force

    18.0*0.27=18*5.9-? Lets not worry about the components that make up ? yet. Can you compute ?
     
  8. Dec 6, 2009 #7
    4.86=106.2-? so 101.34? or is there more than one other force?, Thank you for helping I'm sorry I'm really bad at physics.
     
  9. Dec 6, 2009 #8
    Should be 18(5.9-0.27)

    Ok now the tricky part. This frictional force is equal to the normal force * u where u is the coefficient of friction.

    On level ground normal force is just opposite the weight. But on a plane it is
    weight* cos(37)

    So normal force is 18*9.8 cos (37). Lets call this N.

    Then we have N*u=the same ? as you calculated before. can you get u?
     
  10. Dec 6, 2009 #9
    176.4 cos(37)=140.9
    N*u=101.34
    140.9u=101.34
    so u=.719
     
  11. Dec 6, 2009 #10
    By george I think you got it. It makes more sense with a free body diagram.
     
  12. Dec 6, 2009 #11
    Thank you so much!
     
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