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Solve for equilibrium state

  1. Jan 30, 2014 #1
    Two Blocks Connected by Pulley System

    1. The problem statement, all variables and given/known data

    [EDIT] sorry, I forgot to post the goal of this problem.
    [EDIT] I wrote the question wrong.
    At what point does the system come to rest?

    What is the maximum distance it will fall?

    Two blocks are connected by a rope and pulley system. The block on the right is released and is allowed to slide down the frictionless surface. The masses are specified, and their initial distance apart.

    2. Relevant equations
    This is where I have the question. But, here are some diagrams that I made showing what I mean.

    3. The attempt at a solution

    I have tried approaching it from two different methods.

    1: Force-work relationships and momentum
    2: Energy relationships

    I end up getting stuck in my attempts to solve this problem. It is deceptively simple, and I keep telling myself that it has a simple answer.

    Let me give you the energy example of what I tried to do.

    I said that the energy of the system must always be conserved, from the point of release to its energy equilibrium state. Thus, the sum of the potential and kinetic energies of both systems must equal.

    I denoted a new constant, k, representing the length of the diagonal of the wire. I also denoted an angle from k to the diagonal, Θ. I found the vertical displacement of the left mass as k-a. The right mass has a vertical displacement d, which also relates to Θ.

    So, my initial equation is:
    Initial energy = 0 (what I defined as my 0 energy state, since I only care about the change of potential energy of the two masses)

    [itex]0 = 2mg(k-a)-mgd-0.5(2m)\dot{k}^2+0.5m\dot{d}^2[/itex]

    This seems to get rather complicated fast. I can parameterize both k and d as:


    But, I am getting stuck from here. This seems to be exploding in complexity from something so simple. Is there a better, more elegant way to approach this?

    I might be coming back with further questions, because I still do not have a solid grasp of when itis best to use force/momentum, or potential/kinetic energy when I am approaching these problems. I do not have a good grasp of when it is conserved and when it isn't. For instance, in this problem, since there are no impacts, and the problem seems to lack a distinct before/after to it, the energy of the system is conserved, which is why I tried to approach it from the energy perspective.

    Should I use both methods to resolve some of the complexity? I do not have a good grasp of this. All of the class notes consist of very pure theory and vector analysis, so I have yet to strongly link the class notes to the actual applications.
    Last edited: Jan 30, 2014
  2. jcsd
  3. Jan 30, 2014 #2


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    Looks to me like a force problem. The blocks can be at rest when the tension in rope balances the force of gravity on the smaller block. If you just release the system in it's initial state and there is no friction at all, then it will never come to rest. It will just oscillate. So I think you need to assume there is some friction someplace (even if it's not on the surface). That would make assuming energy conservation a bad idea. Or am I making it too simple?
  4. Jan 30, 2014 #3
    That was my first assumption when looking at the problem, that the system would oscillate and never come to rest.
    Dang, you know what, I asked the question wrong, and I might have to redo the whole thread so other readers can see what I said.

    The actual question was "What is the maximum distance it will fall?" in which case it is asking for the lowest point the block will go.

    Alright, then, how does one set up the differential equation for this? I really am still at a loss, even with that information.
  5. Jan 30, 2014 #4


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    Think about energy. What is the KE at the lowest point?
  6. Jan 30, 2014 #5


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    You don't need a differential equation. Use energy. Define the initial PE to be zero. Since KE+PE is conserved and you start with KE=0, when they momentarily at rest at the end of the fall then KE=0 again. So PE must be zero again. Where can that happen?
  7. Jan 31, 2014 #6
    When the lighter block falls, it moves down a distance d, losing a potential energy mgd. The larger block must lift up by the same energy change, since this is a conservative system.

    Thus mgd=2mgx, then the larger block moves up half the distance that the smaller block moves down. However, I do not know how to find the magnitude of this movement.
  8. Jan 31, 2014 #7


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    That's just geometry. When the block has descended distance d, what is the distance from it to the pulley?
  9. Feb 2, 2014 #8
    There is probably a higher level conceptual understanding that I am missing here.

    The distance from the pulley is [itex]\sqrt{a^2+d^2}[/itex] But, the goal is to completely eliminate d altogether and express it fully in terms of the given information (the masses, a, gravity if applicable...) I am definitely not thinking this through properly.

    I'm thinking, maybe,

    [itex]\sqrt{a^2+d^2}=a+\frac{d}{2}[/itex] Then I can solve for d in this manner.

    This yields a d = 4/3 a

    Ah ha! Thanks for the help. You're right. It was just geometry. I was just getting stuck on how to see it.
    Last edited: Feb 2, 2014
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