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Math Help - Mechanical question

  1. #1
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    Mechanical question

    Any help would be great thanks

    A light inextensible string passes over a fixed smooth pulley. Particles A and B, of o.o3kg and o.o5kg respectively, are attached to the ends of the string. The system is held at rest with A and B at the same horizontal level and the string taut. The two parts of the string not in contact with the pulley are vertical. The system is released at time t=0, where t is measured in seconds, the particle B moves downwards for 2s before it hits the floor. The string then becomes slack and B remains a rest. Neglecting air resistance, show that the string becomes taut again when t=3.
    thanks
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  2. #2
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    Quote Originally Posted by gracey View Post
    Any help would be great thanks

    A light inextensible string passes over a fixed smooth pulley. Particles A and B, of o.o3kg and o.o5kg respectively, are attached to the ends of the string. The system is held at rest with A and B at the same horizontal level and the string taut. The two parts of the string not in contact with the pulley are vertical. The system is released at time t=0, where t is measured in seconds, the particle B moves downwards for 2s before it hits the floor. The string then becomes slack and B remains a rest. Neglecting air resistance, show that the string becomes taut again when t=3.
    thanks
    When released, there is a difference in forces, so the system moves down.
    This is F = 0.05*9.8 -0.03*9.8 = 0.196 newton

    The system will accelerate downward
    F = m*a
    0.196 = (0.05 +0.03)a
    a = 0.196 /0.08 = 2.45 m/sec/sec

    When B hit the floor, A continued moving up because of inertia.
    A will stop moving after somer time, t1 second
    V = Vo -g(t1) -------------(i)
    where
    V = final velocity of A after t1......which is zero
    Vo = velocity of A at t = 2 sec

    Vo = (initial velocity) +a*t
    Vo = 0 +2.45(2) = 4.9 m/sec

    So, substitute those into (i),
    0 = 4.9 -9.8(t1)
    t1 = 4.9 /9.8 = 1/2 sec

    Total time now is 2 +1/2 = 2.5 sec after the release.

    In that 0.5 sec it moved upward some more before stopping, A traveled:
    h1 = Vo*t -(1/2)g(t^2)
    h1 = 4.9(0.5) -(1/2)(9.8)(0.5)^2
    h1 = 1.225 meters

    Then, from rest, A will fall down.
    When it falls down, it will travel 1.225 m and then the string will become taut again because by then, A will start pulling up B.
    h1 = (initial velocity)*t2 +(1/2)g(t2)^2 ....."+" because going down is positive here
    1.225 = (0)(t2) +(1/2)(9.8)(t2)^2
    (t2)^2 = 1.225 / 4.9 = 0.25
    t2 = 0.5 sec

    So, total time now after release is 2.5 +0.5 = 3 seconds.

    Shown.
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