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Two Blocks and a Pulley with Friction

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Blocks A and B are connected by a cable that passes over support C. Friction between the blocks and the inclined surfaces may be neglected. Knowing that motion of block B up the incline is impending when mB = 4.5 kg, determine:

    (a) the coefficient of static friction between the rope and the support
    (b) the largest value of mB for which equilibrium is maintained

    2. Relevant equations

    T2 = T1e^(μsβ) (1)
    β is in angle in radians
    W = mg

    3. The attempt at a solution

    My problem is that for part (b), my answer is wrong. I would like to know where I went wrong in my calculations. Thanks for the help in advance.

    I found the forces working on Block A
    ƩFx = 0
    → TA - WAsin(16) = 0
    TA = 24.336 N

    Then, Block B
    ƩFx = 0
    → WBsin(16) - TB = 0
    TB = 12.168 N

    Then, I looked at support C where I found the angle to be 32°, and I used (1) to find the static friction.
    β = (32)(∏/180) ≈ 0.559
    TA = TBe^(μsβ)
    24.336 N = (12.168 N)e^(μsβ)
    e^(μsβ) = (24.336 N)/(12.168 N)
    μsβ = ln [(24.336 N)/(12.168 N)]
    μs = (1/0.559)*ln [(24.336 N)/(12.168 N)]
    μs ≈ 1.240

    Then, since I know TB has to be greater than itself to upset the equilibrium, I did

    TB = TAe^(μsβ)
    TB = (24.336 N)e^[(1.240)(0.559)]
    TB ≈ 48.673 N

    Then, to get the mass I divided the TB by g = 9.81

    mB = (48.673)/(9.81) ≈ 4.962 kg

    That is wrong.
     

    Attached Files:

  2. jcsd
  3. May 30, 2013 #2

    haruspex

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    You forgot that the mass isn't hanging straight down.
     
  4. May 30, 2013 #3
    I just tried multiplying it by sin(16), which of course didn't work, but dividing by it sure did work. Thanks.
     
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