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Mass falling and pulling others on a rough surface

  1. Sep 8, 2016 #1
    1. The problem statement, all variables and given/known data
    Snap1.jpg 4 m masses, μ is the coefficient of friction. what is the tension and what should the maximum μ be to allow acceleration.

    2. Relevant equations
    Mass-acceleration: F=ma

    3. The attempt at a solution
    $$\left\{\begin{array}{l} mg-T=ma \\ T-3mg\mu=3ma \end{array}\right.~~\rightarrow~~a=\frac{1-\mu}{4}g$$
    So, from here, μ≤1 in order for a to be positive. but logically:
    $$3mg\mu\leq mg~~\rightarrow~~\mu\leq \frac{1}{3}$$
     
  2. jcsd
  3. Sep 8, 2016 #2

    NascentOxygen

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    Check your working for solving simultaneous equations.
     
  4. Sep 9, 2016 #3

    rude man

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    You're mixing up static & dynamic friction coefficients. Your "logically" expression is correct if μ = μs.
     
  5. Sep 9, 2016 #4
    I add the equations:
    $$+\left\{\begin{array}{l} mg-T=ma \\ T-3mg\mu=3ma \end{array}\right.$$
    $$mg-3mg\mu=4ma~~\rightarrow~~a=\frac{1-\mu}{4}g$$
    The question didn't distinct between the coefficients, it mentions only one.
     
  6. Sep 9, 2016 #5
    I believe that should give you a = g(1 - 3μ)/4

    Then you need to ensure that a is positive, g and 1/4 are positive constants so you have the restriction that: 1-3μ ≥ 0
     
  7. Sep 9, 2016 #6
    $$\left\{\begin{array}{l} mg-T=ma \\ T-3mg\mu=3ma \end{array}\right.~~\rightarrow~~a=\frac{1-3\mu}{4}g$$
    Thanks HoodedFreak
     
  8. Sep 9, 2016 #7

    PeroK

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    The first part of the question asks you to find ##T##. I assume this means ##T## in terms of ##m, g## and ##\mu##. But, you've eliminated ##T## from your equations.
     
  9. Sep 9, 2016 #8
    That's simple:
    $$T=(g-a)m=...=\frac{3+3\mu}{4}mg$$
     
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