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Calculating coefficient of kinetic friction

  1. Apr 30, 2008 #1
    [SOLVED] Calculating coefficient of kinetic friction

    1. The problem statement, all variables and given/known data

    An applied force of 450 N {forward} is needed to drag a 1000-kg crate at constant speed across a horizontal, rough floor. Calculate the coefficient of kinetic friction for the crate on the floor.

    Fapp=450 N {forward}
    m=1000 kg
    a=0
    g=9.81 m/s^2

    2. Relevant equations
    Fnet=F(f(kinetic)) + F(app) + F(g)
    Fnet=ma
    Ffkinetic=-(mu)kFn
    Fg=-mg


    3. The attempt at a solution

    Fnet=-(mu)kFn + Fapp + Fg
    Since Fnet = ma
    ma=-(mu)kFn + Fapp + Fg
    0=-(mu)k1000 + 450 - 1000(9.81)
    (mu)k1000=-9360
    (mu)k=-9.36

    The answer is supposed to be 4.59 x 10^-2, which makes a whole lot more sense than -9.36. I'm completely lost. The text only has one example and it's completely different from this question. tia!
     
  2. jcsd
  3. Apr 30, 2008 #2

    nrqed

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    You are mixing x and y components!!! You must apply Newton's second law separately to the x and y components! So you have two equations :

    [tex] \sum F_x = m a_x [/tex]

    and

    [tex] \sum F_y = m a_y [/tex]

    Draw a FBD, identify the forces along x and along y, write down their x and y components (including the correct signs) and plug in the above equations.
     
  4. Apr 30, 2008 #3
    Thank you. I looked at the text again and retried writing my information making sure I was dealing with just the horizontal components (I was looking at the equations for slopes and combining the wrong things), using from the text this formula:

    F(net_h) = F(app) + F(f_kinetic)

    Since there is no acceleration Fnet is 0:

    0 = F(app) + F(f_kinetic)
    F(f_kinetic) = -F(app)

    Since F(f_kinetic) = -(mu)k_F(N)

    -(mu)k_F(n) = -F(app)
    (mu)k = F(app)/F(N)

    Since F(N) is mg:

    (mu)k = 450/(1000*9.81)
    = 4.59 * 10^-2
     
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