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Finding static friction of a block on a table on a pulley?

  1. Aug 21, 2016 #1
    1. The problem statement, all variables and given/known data
    "When the three blocks in Fig. 6-29 are released from rest, they accelerate with a magnitude of ##0.500 \frac{m}{s^2}##. Block 1 has mass M, Block 2 has mass 2M, and Block 3 has mass 3M. What is the coefficient of static friction between Block 2 and the table?"
    KyKIClY.png

    2. Relevant equations
    ##f_s=(μ_k)(F_N)##
    Answer from book: 0.37

    3. The attempt at a solution
    Okay, so I attempted this by beginning with the sum of all the forces at work, relative to the force on Block 3. There's the gravitational force on Block 1 negating half of Block 3's force, and then there's the static friction between Block 2 and the table also slowing the motion of Block 3. So:

    ##ΣF = ma = (-9.8\frac{m}{s^2})(M) + (μ_k)(-9.8\frac{m}{s^2})(2M) + (9.8\frac{m}{s^2})(2M)##
    ##ΣF = (μ_k)(-9.8\frac{m}{s^2})(2M) + (9.8\frac{m}{s^2})(M)##
    ##\frac{ΣF}{m} = a = (9.8\frac{m}{s^2})(-2μ_k + 1)##
    ##a = \frac{1}{2}\frac{m}{s^2}##
    ##\frac{1}{2}\frac{m}{s^2} = (9.8\frac{m}{s^2})(-2μ_k + 1)##
    ##\frac{1}{19.6}=-2μ_k + 1##
    ##-\frac{18.6}{19.6}=-2μ_k##
    ##μ_k=\frac{93}{196}##

    I have a feeling I messed up on the third step, in calculating acceleration and equating it to ##\frac{1}{2}##. Are there any other forces at work here, that are slowing Block 3's descent? Thank you for anyone who is willing to help me.
     
  2. jcsd
  3. Aug 21, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    Your problem statement gives block three a mass of 3M, but it looks like you've used 2M in your first step. Can you verify which is correct?

    What mass are you using for 'm' in F = ma? Hint: How much mass is accelerating?
     
  4. Aug 21, 2016 #3
    Right. I should be more mindful of reading the problem carefully; especially when doing physics homework.

    ##ΣF = -(μ_k)(9.8\frac{m}{s^2})(2M) + (9.8\frac{m}{s^2})(2M)##

    6M?

    ##\frac{ΣF}{6M} = -\frac{1}{3}(μ_k)(9.8\frac{m}{s^2}) + \frac{1}{3}(9.8\frac{m}{s^2})##
    ##\frac{ΣF}{6M} = \frac{1}{3}(9.8\frac{m}{s^2})(1-μ_k) = a##
    ##\frac{1}{2} = \frac{1}{3}(9.8\frac{m}{s^2})(1-μ_k)##
    ##\frac{3}{19.6} = 1 - μ_k##
    ##-μ_k=-\frac{16.6}{19.6}##
    ##μ_k=\frac{16.6}{19.6}##

    I wish I knew what I was doing wrong.
     
    Last edited: Aug 21, 2016
  5. Aug 22, 2016 #4

    gneill

    User Avatar

    Staff: Mentor

    You say the total amount of mass in the problem is 6M, so then the blocks have masses 1M, 2M, and 3M.

    When I do the math assuming those block masses I don't get the book answer. On the other hand, If the third block is 2M and the total mass is 5M, then I get the given answer.

    You might want to use a symbol for the gravitational acceleration (g) rather than lug around numbers and units all the way through the algebra. Only plug in numbers at the very end.
     
  6. Aug 22, 2016 #5
    Oops, I should learn to read more carefully.

    ##ΣF = (-μ_k)(g)(2M) + (g)(M)##
    ##a = \frac{1}{5}g(1-2μ_k)##
    ##\frac{1}{2} = \frac{1}{5}g(1-2μ_k)##
    ##\frac{5}{2g} = 1-2μ_k##
    ##\frac{5-2g}{2g} = -2μ_k##
    ##\frac{2g-5}{4g}=μ_k=0.37##

    Good to note. Thank you...
     
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