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Homework Help: Forces involved in spring-pulley system

  1. Mar 14, 2004 #1
    I am trying to answer the following question:

    Two equal masses are constrained by the spring-and-pulley system shown (the pulley has no mass and the surface is frictionless). Determine the equation of motion for the system in terms of x, the extension of the spring from its unstretched length. Solve for x as a function of time with the boundary conditions x = dx/dt = 0 at t = 0.

    I have attached a word document with the diagram for the system. All that I've added to the diagram so far is T1, T2 and mg for the mass dangling over the edge.

    Now, the last part of the question (solving for x with the boundary conditions) I can solve easily once I've actually figured out the equation of motion.

    The problem I'm having is getting to the point where I have a second-order differential that I can solve. I'm not entirely sure how to choose my coordinate system (should I have an x and y coordinate system?) and I'm also not sure about the forces involved.

    Are the tensions (T1 and T2) that I've drawn in correct, and what do they equal?

    Is it just T1 = T2 = -kx + mg , or am I missing something?

    I'd really appreciate it if someone can point me in the right direction here.
    Last edited: Mar 14, 2004
  2. jcsd
  3. Mar 14, 2004 #2

    Doc Al

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    Staff: Mentor

    Your attachment didn't make it; try again.
  4. Mar 14, 2004 #3
    Okay, hopefully it'll work this time...

    Attached Files:

  5. Mar 15, 2004 #4
    Question: Is the mass on the table subject to friction, or are we working in super-perfect imaginary physics world?

  6. Mar 15, 2004 #5
    We're working in a super-perfect imaginary physics world :)

    No friction forces involved.
  7. Mar 15, 2004 #6
    Well, that makes life a bit easier!

    We don't really need to consider tension, do we? All it's doing is communicating force between the blocks and the spring.

    We got two forces, right? The restoring force of the spring and the force of gravity acting on the hanging block. So,

    F = mg - kx

    (I took gravitational force to be positive by assuming that the positive x direction was to the right, and you can see that gravity will always be pulling to the right)

    And then we got

    [tex]F = m_\textrm{total}\ddot{x}[/tex]

    Now we have to address something. What mass are these forces acting upon, i.e. what is [itex]m_\textrm{total}[/itex]? The forces are trying to move both blocks, right? So [itex]m_\textrm{total}[/itex] should equal 2m. That leaves us with

    [tex](2m)\ddot{x} = mg - kx[/tex]

    which you'll need to solve.

    Does that help?

  8. Mar 15, 2004 #7
    It certainly does. Thanks a lot!
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