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Homework Help: Question on pulleys and tension

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Two objects of masses m1 and m2 are connected via a massless string and ideal pulleys. The coefficients of friction between m1 and the horizontal surface are ms and mk
    a) what condition needs to be met to set the system in motion?
    b)assuming that this condition is (barely) met, describe the motion of the system.
    c) what is the tension in the string when the objects are in motion?

    2. Relevant equations

    3. The attempt at a solution
    a) m2 needs to be slightly heavier than m1
    b) m2 will be slightly heavier, the tension in the string increases causing m1 to move forward since they're not in equilibrium.
    c) ????

    This was on a test, but I got it wrong... he wanted actual equations proving everything and i had no idea how to do any of this. I was just really confused. Can anyone explain it to me?
  2. jcsd
  3. Dec 11, 2011 #2


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    Homework Helper

    m1 is on the horizontal surface. The Normal Reaction force = m1g so the maximum possible values of friction will be μsm1g when it is stationary, and μkm1g when it is moving [the s and k refer to static and kinetic friction]. Usually [always?] μk is less than μs.

    If the system is stationary, the tension in the string is m2g since the string is supporting the hanging mass m2.
    Up on the table, the friction force must be balancing that tension, so the tension must be no greater than the maximum possible friction; μsm1g

    If the mass m2 is big enough, the tension will exceed the possible friction, and the system will start to move. ie m2g > μsm1g

    Once the system starts to move, the friction force will drop to μkm1g.
    This means the masses will accelerate; m1 across the table and m2 down.

    Net force on the system [in the direction it can move] is m2g - μkm1g
    The total mass of the system is m1 + m2

    Using F = ma we get

    a = (m2g - μkm1g) / (m1 + m2)

    At this point it getting too hard to keep track of all those typing references, so lets pretend a has a value of 2, m2 = 3 kg and g = 10

    Net force on m2 = 3 * 2 = 6 N
    Weight force down = m2g = 3 * 10 = 30N
    SO the tension in the string must be 24N

    I will leave it to you to find the expression for tension in the real problem. It is an awful lot easier to write those subscripts with a pen on paper than it is to type them on this computer.
  4. Dec 11, 2011 #3


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    Since the masses were of a size that motion was only just achieved, m2g will be equal to the static friction μsm1g

    SO the acceleration can be expressed as:

    a = (μsm1g - μkm1g) / (m1 + m2)
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