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Homework Help: 3 masses, 1 pully

  1. Sep 21, 2004 #1
    I am having quite a time with this problem, I would appreciate any assistance.

    Block B, of mass mB, rests on block A, of mass mA, which in turn is on a horizontal table top . The coefficient of kinetic friction between block A and the table top is uK and the coefficient of static friction between block A and block B is uS. 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.

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

    My assumption is we want acceleration to be at the maximum that keeps block B over A. The larger the mC, I am assuming the larger the a. (I don’t know how to express that thought mathematically, if it is right)

    This is what I did.

    A: T-uK*(mA+mB)=mA*a

    B: fs=uS*mB*g=mB*a…..fs=uS*g=a

    C: mC-T=mC*a

    FA-FB=T , I plug eq.A into FA, eq.B into FB, and solve for T in eq.3 to isolate the mC.

    My answer for mC, which I know is wrong was...uk*(mA+mB)-us*mB

    I am so confused. I have been struggling with this problem all day. I separated into components numerous times. (I don’t think that’s where my problem lays)
    Someone plx help me to understand why I don’t understand.


    Attached Files:

  2. jcsd
  3. Sep 21, 2004 #2
    Hi Elbhi,
    I got Mc= Uk(Ma+Mb)+UsMb

    If the above answer is right then here's my argument:
    For Mc to be maximum,block A should offer max resistance.This would occur when max static friction force acts between Ma and Mb.(as Static friction >= kinetic friction)Max static friction force will come into play when the blocks A and B are just about to slip relative to each other(but are not slipping).
    Last edited: Sep 21, 2004
  4. Sep 21, 2004 #3
    Hi, Elbhi.

    I get mC = [(uS+uK)(mA+mB)]/(1-uS)

    For B not to slide, max horizontal force forward is Fs=uS*mB*g=mB*a
    So, the system acceleration cannot exceed the max acceleration for mB

    a = uS*g

    For the system

    mC*g - uK(mA+mB)g = (mA+mB+mC)a = (mA+mB+mC)(uS*g)

    rearrange, and collect like terms in mC

    mC*g - uS*mC*g = uS(mA + mB)g + uK(mA + mB)g

    common factor of g leaves

    mC(1 - uS) = (uS + uK)(mA + mB)

    and, finally

    mC = [(uS+uK)(mA+mB)]/(1-uS)

    Or, so it seems to me

  5. Sep 21, 2004 #4
    Hello Rhia & Minstrel, thank you very much for your help both of you:). Rhia your answer was actually my first answer, however that wasnt right, which led to my fustration. Minstrel actually did it the right way. Thank you sooooooooooooooo much for helping me see the light. I thought it was immpossible, but through your reasoning it makes absolute sense.

    Any tips for first year physics students?????
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