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Pulley problem involves Friction static and Kinetic problem

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data

    M1 has a mass of 6.51 kg. It is on a horizontal surface, connected by a light string to a hook. Mass M2 can be increased smoothly by adding masses little at a time. The pulley has a negligible mass and no friction.

    When M2 is 3.12 kg it begins to accelerate downwards at a rate of 1.52 m/s2. Calculate mu static -mu kinetic for M1 on the surface.

    I was also given this hint, First calculate ms (from knowing the force it takes to begin moving) and then mk (from the given acceleration) and subtract

    2. Relevant equations

    F= Ma


    Mass2
    Fnety = T-Mg


    Mass1

    Fnetx= T-Fstatic

    Fs= Mus*Fn

    Fk= Muk* FN

    3. The attempt at a solution

    Mass2
    Fnety = T-Mg
    -4.74N= T-30.576N
    T= 25.83

    Mass1

    Fnetx= T-Fkinetic
    9.89N= 25.83-15.934
    Fk= Muk*Fn
    15.9348/63.798= 25.83

    Muk= .24976

    Fnet=0
    T-Fs=
    25.83

    Fs= Mus* FN

    25.836= Mus * FN
    25.836/63.798

    Mus=.405

    Mus-Muk= .155
     

    Attached Files:

  2. jcsd
  3. Oct 3, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    There is something odd about this solution and I can't quite pin it down. I think you are supposed to write F = ma for each mass as your starting point. This would give you 2 equations:
    [1] T - u*m1*g = m1*a and [2] m2*g - T = m2*a
    where T is the tension in the string between the two masses.
    When a = 0, you get T = u*m1*g from [1] and when you put it into [2] you have m2*g = u*m1*g which gives u = m2/m1 = 0.479 for the static coefficient.
     
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