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Pedagogical Machine

  1. Mar 21, 2015 #1
    1. The problem statement, all variables and given/known data

    A “pedagogical machine” is illustrated in the sketch (attached image). All surfaces are frictionless. What force ##F## must be applied to ##M_1## to keep ##M_3## from rising or falling?

    2. Relevant equations

    $$\sum_{}^{} \vec{F} = m \vec{a}$$

    3. The attempt at a solution

    I have chosen a coordinate system such that all particles have positive coordinates. My ##x##-axis points to the right and my ##y##-axis points upwards. The length of the rope remains constant, and it can be expressed as:
    $$l = x_p - x_2 + \frac{\pi R}{2} + y_p - y_3$$
    The pulley is located at ##(x_p,y_p)##.
    Differentiating twice with respect to time and rearranging:
    $$\ddot{x}_2 + \ddot{y}_3 = \ddot{x}_p$$
    Since the pulley and ##M_1## are fixed:
    $$\ddot{x}_2 + \ddot{y}_3 = \ddot{x}_1$$
    Since ##\dot{y}_3 = 0## for all ##t##, ##\ddot{y}_3 = 0##.
    In other words:
    $$\ddot{x}_1 = \ddot{x}_2$$
    Now, applying Newton's second law to each mass:
    $$T - M_3 g = 0 ⇒ T = M_3 g$$
    $$T = M_2 \ddot{x}_2$$
    $$F - T = M_1 \ddot{x}_1$$
    Where ##T## is the tension in the rope.
    $$M_3 g = M_2 \ddot{x}_2 ⇒ \ddot{x}_2 = \frac{M_3 g}{M_2} = \ddot{x}_1$$
    $$F = T + M_1 \ddot{x}_1 = M_3 g (1 + \frac{M_1}{M_2})$$
    $$F = M_3 g \frac{M_1 + M_2}{M_2}$$
    Which doesn't match the clue given in my book. According to my book, if all three masses are equal, then ##F = 3Mg##.
    Where did I go wrong?
     

    Attached Files:

  2. jcsd
  3. Mar 21, 2015 #2

    haruspex

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    How do you get your equation ##F-T=M_1\ddot x_1##? What are the forces acting on M1?
     
  4. Mar 21, 2015 #3
    In the ##x##-direction, there's ##F## and ##T##.
    ##T## acts on the pulley, which is a part of ##M_1##.
    All surfaces are frictionless.
     
  5. Mar 21, 2015 #4

    haruspex

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    Yes, but you've missed a force. What makes M3 accelerate?
     
  6. Mar 21, 2015 #5
    The net force caused by the difference between ##N## and ##N'## (different normal reactions)?
     
  7. Mar 21, 2015 #6
    $$\vec{F}_{31} = -\vec{F}_{13}$$
    The net force on ##M_3## is to the right, so the force ##M_3## exerts on ##M_1## must be to the left.
    I guess I missed an additional constraint:
    ##\ddot{x}_1 = \ddot{x}_3## since ##x_3## and ##x_1## differ by a constant.
    I got the right answer, thanks!
    I have a question though, the equation only holds if ##M_3## is always in direct contact with ##M_1##, right?
    I have not taken into account the fact that ##M_3## can swing left and right while deriving the first constraint equation.
     
  8. Mar 21, 2015 #7

    haruspex

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    Yes, if the two do not start quite in contact then there will be a different behaviour initially, but it wouldn't last long.
     
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