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Friction Force on a System at Rest

  1. Oct 23, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    3. The attempt at a solution
    [tex]\sum F_{x} = w_{x} - T = m_{x}a[/tex]
    [tex]T = (6.9)g - (6.9)a[/tex]

    [tex]\sum F_{a} = T - w_{a} - f_{s} = m_{a}a[/tex]
    [tex]T - m_{a}g - m_{a}a = f_{s}[/tex]
    [tex](6.9g - 6.9a) - m_{a}g - m_{a}a = f_{s}[/tex]
    I don't see a 5kg block with friction, but I do see a 3kg block with friction:
    [tex]6.9g - 6.9a - 3g - 3a = f_{s}[/tex]
    [tex]3.9g - 9.9a = f_{s}[/tex]

    Since the system is at rest, [tex]a = 0[/tex]
    [tex]38N = f_{s}[/tex]
    So the answer is B

    But I have two problems with my answer:
    1) [tex]f_{s-max} = (0.40)(50) = 20N < 38N[/tex] in my answer
    2) Using the "5kg" block, [tex]f_{s} = 19N[/tex]

    Am I missing something?
     
    Last edited: Oct 23, 2007
  2. jcsd
  3. Oct 23, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks to me like the 3 kg was a typo and they meant to write 5 kg. (Otherwise the problem doesn't work, as you point out.)
     
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