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Block on a pulley

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data

    At a given instant the 10-lb block A is moving downward with a speed of 6 ft/s. determine its speed 2s later. Block B has a weight of 4-lb, and the coefficient of kinetic friction between it and the horizontal plane is [tex]\o_{k} = 0.2[/tex]. Neglect the mass of the pulleys and cord.

    2uo0t91.png

    2. Relevant equations

    [tex]v = v_{0} + a_{A}t[/tex]

    3. The attempt at a solution

    FBD:
    2h6zltw.png

    Block A:

    [tex]+\downarrow \sum F_{y} = ma_{y}[/tex]

    [tex]10 - 2T = (0.311)(a_{A})[/tex]

    Block B:

    [tex]+\rightarrow \sum F_{x} = ma_{x}[/tex]

    [tex]T - (0.2)(4) = (0.124)(a_{B})[/tex]

    [tex]+\downarrow \sum F_{y} = ma_{y}[/tex]

    [tex]4 - N = 0[/tex]

    Kinematics:

    [tex]2S_{A} + S_{B} = l[/tex]

    [tex]2a_{A} + a_{B} = 0[/tex]

    [tex]2a_{A} = -a_{B}[/tex]

    Solving for T:

    [tex]10 - 2T = (0.311)(a_{A})[/tex]

    [tex]10 - 2T = -(0.311)(a_{B})[/tex]

    [tex]a_{B} = -\frac{10 - 2T}{0.311} = \frac{T - (0.2)(4)}{0.124}[/tex]

    [tex]T = -16lb/2 = -8lb[/tex]

    [tex]a_{B} = -70.84ft/s^{2}[/tex]

    [tex]a_{A} = 83.72ft/s^{2}[/tex]

    [tex]v = v_{0} + a_{A}t[/tex]

    [tex]v = 6 + (83.72)(2) = 173.44 ft/s[/tex] I am not sure if this is right because that seems fast.
     
  2. jcsd
  3. Oct 8, 2009 #2
    you forgot a factor (1/2) here.

    T = -8 lb is not a solution of the above equaton. Try to give some more detail when you
    try to solve for T again.
    The negative tension in a rope should have tipped you off that something was wrong.
     
  4. Oct 10, 2009 #3
    I didn't have a chance to get back to this but I finally solved it I think:

    [tex]2a_{A} = -a_{B}[/tex]

    [tex]a_{A} = -\frac{a_{B}}{2}[/tex]

    [tex]10 - 2T = (0.311)(a_{A})[/tex]

    [tex]10 - 2T = -(0.311)(\frac{a_{B}}{2})[/tex]

    [tex]a_{B} = -\frac{20 - 4T}{0.311} = \frac{T - (0.2)(4)}{0.124}[/tex]

    [tex]-2.48 + 0.50T = 0.31T - 0.25[/tex]

    [tex]T = 2.769 lb[/tex]

    [tex]a_{A} = 14.37ft/s^{2}[/tex]

    [tex]a_{B} = 15.85ft/s^{2}[/tex]

    [tex]v = v_{0} + a_{A}t[/tex]

    [tex]v = 6 + (14.37)(2) = 34.73 ft/s[/tex]
     
  5. Oct 10, 2009 #4
    I'm sorry, this was an error I missed. a_A and a_B should have the same sign.

    this T is not the solution of the last equation. I get 0.19 T = 2.23, so T = 11.7
    your also using only 2 digits of precision here, and then give the final answer with much
    more digits.
     
  6. Oct 10, 2009 #5

    rl.bhat

    User Avatar
    Homework Helper

    10 - 2T = 0.311(aA)
    T - 0.8 = 0.124(aB)
    aA = 2*aB
    So 10 - 2T = 0.622*aB
    T - 0.8 = 0.124*aB
    So ( 10 - 2T)/(T - 0.8) = 0.622/0.124.
    Now solve for T.
     
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