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Three-mass System with Friction

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data


    Block B, with mass 5.00 kg, rests on block A, with mass 8.00 kg, which in turn is on a horizontal tabletop. There is no friction between block A and the tabletop, but the coefficient of static friction between block A and block B is 0.750. A light string attached to block A passes over a frictionless, massless pulley, and block C is suspended from the other end of the string.

    What is the largest mass that block C can have so that blocks A and B still slide together when the system is released from rest?

    2. Relevant equations

    F=ma

    3. The attempt at a solution

    I can't figure this out for the life of me. I took blocks A and B as a system in themselves to find the acceleration for which the block B would not move.

    I set Fapplied-.75(Mb*9.81)=36.79N.

    I then took the force applied and set it as F=ma, 36.79=(Ma+Mb)a, a=2.83 m/s^2.

    This should be the most acceleration that block A can experience without having block B move. I used both Ma+Mb in the equation as it appears that the hanging block would need to move the two as a system.

    I then took the acceleration I had found and set a new force equation, utilizing the entire system:

    Fnet=Mc*g, then substitute in F=ma, Msys*a=Mc*g, (Ma+Mb+Mc)a=Mc*g.

    I solved for Mc and got a negative number, which obviously cannot be true.

    I'd appreciate any help on how I should go about solving this...
     
  2. jcsd
  3. Feb 6, 2014 #2

    BvU

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    The 36.79 N is the maximum force block A can exercise on block B. So a is a little bigger. To accelerate block B with 2.83 m/s2, a force of 14.15 N is sufficient.
    Better leave the g in there until you have a final expression. It cancels out.
     
  4. Feb 6, 2014 #3

    BvU

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    Also want to buy new solving equipment, because (Ma+Mb+Mc) * a=Mc*g solves easily : Mc = (Ma + Mb)*a/(g-a). Positive for all a up to g (Mc can't get higher then)
     
  5. Feb 6, 2014 #4
    First off, thanks for the assistance.

    So there is an error in the maximum acceleration I found? Should I have divided by only the mass of block A rather than the mass of A + B? If so, I get a=4.6m/s^2.

    I found an error I made when inputting the equation for Mc in the calculator, thanks for pointing out my mistake. However, after correcting, with the new acceleration of 4.6 m/s^2 I still receive an incorrect answer of 10.31 kg. The correct answer should be 39 kg. Which equation is incorrect?
     
  6. Feb 6, 2014 #5

    BvU

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    Eqns are OK. But filling in the mass of A is problematic. A can exercise a maximum force of 0.75 g on B, so B can be accelerated 0.75 g maximum by A. So A + B can be accelerated by 0.75 g maximum, etc.
    Remember my tip on leaving g in ? That way you get the exact answer (which, granted, you would also have gotten if you consistently use the same value for g...)
     
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