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System of pulleys - find acceleration

  1. Jul 23, 2009 #1
    1. The problem statement, all variables and given/known data
    An object of mass m1 on a frictionless horizontal table is
    connected to an object of mass m2 through a very light pulley
    P1 and a light fixed pulley P2 as shown in Figure P5.34.
    (a) If a1 and a2 are the accelerations of m1 and m2, respectively,
    what is the relation between these accelerations? Express
    (b) the tensions in the strings and (c) the accelerations
    a1 and a2 in terms of the masses m1 and m2, and g.



    attachment.php?attachmentid=19690&d=1247441426.jpg


    2. Relevant equations



    3. The attempt at a solution
    How are these two pulleys connected? I don't think the string throught pulley 2 connects to m1 or else the tension and acceleration would be the same. I am having a hard time visualizing what is happening. Any help would be appreciated. Thank you.
     
  2. jcsd
  3. Jul 23, 2009 #2

    rl.bhat

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    Re: Pulleys...Confusing?

    In all such problems we have to follow five steps.
    Step 1: Find out the number of inextensible ropes. For every inextensible rope there will be one equation.
    Step 2: For every inextensible rope, find the bodies that are connected to the rope.
    Step 3: To relate the acceleration between the bodies, assume that the various bodies move by a distance x1, x2, ....and so on.Calculate the number of segments in the rope.
    Step 4: Relate the distance moved. (Total change in the length of the rope must be zero.) To do these calculate the change in length of each segment of the rope. Then add these changes to get the total change in the rope.
    Step 5: Once we get the relation between the distances moved, the acceleration relation will be the same
     
  4. Jul 24, 2009 #3

    jmb

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    Re: Pulleys...Confusing?

    As I understand the diagram, the string through pulley 2 connects to the center of pulley 1. That is, the extension of m2 controls the position of pulley 1. The two strings are not connected directly.

    Sorry if this is stating the obvious (!), but I thought from your wording it might be what was confusing you...
     
  5. Jul 24, 2009 #4
    Re: Pulleys...Confusing?

    You weren't stating the obvious, it was just what I needed to know. Ok so when m2 moves, it causes P1 to move. The net force acting on m2 is m2g-T2(Where T2 is the tension in that string). So m2g-T2=m2a2. T2=m2(g-a2). Let's say the tension in the string of P1 is T1. Because it is accelerating T2-T1=P1a=m1a. Simplifying I get T1=m2(g-a2)-m1a1. I try to get everything in terms of g m1 m2, but I keep going in circles.

    @rl.bhat- Can you show me what you mean? I don't really understand your post.
     
  6. Jul 24, 2009 #5

    rl.bhat

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    Re: Pulleys...Confusing?

    In the problems there are two ropes, two pulleys and two masses.
    The rope1, which passes through P1, fixed at one end and other end is free to move with m1
    The rope2, which passes through P2, connected to m2 at one end and other end is connected to P1.
    When m2 moves down due to gravity, segment 1 (P2m2) increases by x1 and segment 2 (P1P2) decreases by x2.
    Net change in the length of rope 1 = x1 - x2 = 0 or x1 = x2. Hence a1 = a2.
    The tension in each segment = T
    When the P1 moves towards right, segment 3, one end of which is fixed to the wall, increases by x2 and the segment 4, which is connected to m1, decreases by x2 - x3, because m1 is moving in the same direction as P1.
    Hence net change in rope 2 = x2 + x2 - x3 = 0 or 2x2 = x3 or 2a2 = 2a1 = a3.
    Tension in each segment = T/2.
    Now you can solve the rest of the problem.
     
  7. Jul 25, 2009 #6

    jmb

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    Re: Pulleys...Confusing?

    I just wanted to elaborate on rl.bhat's answer...

    In case it isn't clear, this last statement is due to force balance on the pulley P1 (it would be different if the pulley had non-zero mass).

    In fact another way to deal with these problems is to write down the force balance for every movable object (masses and pulleys) and then use inextensibility of the ropes.

    Let us know how you get on with the last part of the problem, or if you have trouble understanding anything already said.
     
  8. Jul 25, 2009 #7
    Re: Pulleys...Confusing?

    Well the thing I don't understand it how can you make the assumption that segment 2 decreases by x2 and segment 1 increases by x1. So if m2 moves 2m, then segment 1m? How did you get this? I am sorry I don't understand.
     
  9. Jul 26, 2009 #8

    rl.bhat

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    Re: Pulleys...Confusing?

    When m2 is not moving
    the distance between m2 and P2 = d1
    the distance between P2 and P1 = d2
    the distance between m1 and P1 = d3
    the distance between P1 and wall = d4
    When the mass m2 moves down through x1
    d1 becomes d1 + x1.....> change x1.
    d2 becomes d2 - x2...> change -x2
    So net change = x1 - x2 = 0. Or x1 = x2
    d3 becomes d3 + x2 -x3...> change x2- x3
    d4 becomes d4+ x2...>change ...x2
    Net change 2*x2 -x3 = 0 or 2x2= x3
     
  10. Jul 27, 2009 #9

    jmb

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    Re: Pulleys...Confusing?

    If you want to understand it physically rather than mathematically (although really you should make sure you are happy with both explanations) you can think of it like this...

    Questions:
    • What does moving m2 do to P1?
    • What does moving P1 do to the length of rope between P1 and the wall?
    • What does this do to the length of rope between P1 and m1?
    • So what is the net effect on m1?

    (Try and think through these yourself before checking the answers below...)

    Answers:
    • m2 is directly attached to P1 by an inextensible rope, so if m2 moves by a distance [tex]\Delta x[/tex] then P1 also moves by a distance [tex]\Delta x[/tex].
    • Since P1 moves by [tex]\Delta x[/tex] the length of rope between the wall and P1 also increases by [tex]\Delta x[/tex].
    • This rope is also inextensible so it means that length of rope between m1 and P1 (part of the same piece of rope) must decrease by [tex]\Delta x[/tex], bringing m1 closer to P1 by a distance [tex]\Delta x[/tex].
    • Since m1 is now closer to P1 by [tex]\Delta x[/tex] and P1 is itself [tex]\Delta x[/tex] further away from the wall, then m1 must have moved a total distance [tex]\Delta x + \Delta x = 2\Delta x[/tex] away from the wall.
     
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