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Fletcher's trolley apparatus - maximum value of coefficient

  1. Jun 17, 2011 #1
    1. The problem statement, all variables and given/known data
    A Fletcher's trolley apparatus is set up with a block of wood of mass 4.0kg and a suspended mass of 2.0kg.

    3. The attempt at a solution
    a) Calculate the acceleration of the sys. and the tension in the string when the mass is released.

    m1
    FnetX = [T - Ff = m1(a)]
    FnetY = [Fn - Fg = 0]

    m2
    No FnetX
    FnetY = [T - Fg = -m2(a)]

    Equation to find the acceleration in this situation:
    [tex]a = \frac{Fg - Ff}{mass1 + mass2}[/tex]

    The second part of the question is:
    b) What is the max. value of the coefficient of friction that will allow the sys. to move?

    I am stuck as to how to find this out.
    I know:

    [tex]Ff = μFn[/tex]
    [tex]Ff = μ(mass1)(g)[/tex]
     
  2. jcsd
  3. Jun 17, 2011 #2

    Doc Al

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    Staff: Mentor

    Hint: Express the acceleration in terms of μ. (Replace Fg and Ff with what they equal.)
     
  4. Jun 17, 2011 #3
    That will give me the following:
    [tex]a = [\frac{(mass2)(g) - μ(mass1)(g)}{(mass1 + mass2)}][/tex]
    But then I have a, which I have nothing to sub in for.
     
  5. Jun 17, 2011 #4

    Doc Al

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    Staff: Mentor

    That's good.
    As μ increases, what happens to a?
     
  6. Jun 17, 2011 #5
    a increases, but will become negative.. a is proportional to μ ....
     
  7. Jun 18, 2011 #6

    Doc Al

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    Staff: Mentor

    Have another look at the formula. (And if something increases, how can it become negative?)
    Not exactly.
     
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