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A Spring Supporting a Box Containing a Block

  1. Jul 28, 2004 #1

    cj

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    A spring of stiffness k supports a box
    of mass M in which is placed a block of
    mass m.

    A) If the system is pulled downward
    a distance d from the equilibrium position
    and then released, find the force of reaction
    between the block and the bottom of the
    box as a funtion of time.

    B) For what value of d will the block just
    begin to leave the bottom of the box at the
    top of the vertical oscillations? Neglect any
    air resistance.
    -----
    I'm assuming undamped SHM:
    [tex](M+m)\frac{d^2x}{dt^2} + kx = 0 ??[/tex]

    Does the reaction force between the
    Block and the bottom of the Box simply
    equal the mass of Block times the
    acceleration of gravity + acceleration of
    the Box?
    [tex]F_{reaction} = m_{box} \cdot (g + \frac{d^2x}{dt^2}) ??[/tex]

    Is there any more development I could do
    to better represent the force of reaction
    between the block and the bottom of the
    box as a funtion of time?

    For Part B -- under what conditions would the Block
    leave the bottom of the Box? I'm perplexed here
    because it seems that the acceleration of the
    Block will always be the same as that of the
    Box, but experienced-based intuition tells
    me that, yes, the Block could at some point lose
    contact with the Box??
     
  2. jcsd
  3. Jul 28, 2004 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Notice that the acceleration of the box, block unit depends upon their total mass but the acceleration of the block alone depends on the mass of the block alone.
    As long as the block is in contact with the box, they have the same acceleration. The difference between the force necessary to give the block-box a given acceleration and the force necessary to give the block alone that acceleration is the force of the box on the block.

    What happens when the spring reaches its greatest length (i.e. its amplitude)? The spring is attached to the box and will start to pull it back but it is not attached to the block which has its own momentum.
     
  4. Jul 28, 2004 #3

    Doc Al

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    Staff: Mentor

    Just a couple of points to add to what HallsofIvy said.
    No. Three forces act on the box: its weight, the spring, the block-box contact force (the "reaction force").

    On the block, only two forces act: its weight and the block-box reaction force. So...
    [tex]F_{reaction} = m_{block} \cdot (g + a)[/tex]
    where a is the acceleration of the block.
    Note that the box is constrained by the spring. When it rises to a certain point, it will accelerate downward greater than g. Of course, the block is not so constrained: at that point the block-box force will equal zero and they will separate.
     
  5. Feb 7, 2008 #4
    I am trying to solve this problem as well, but I don't understand the block-box contact force that you say acts on the box. Can you further define this force?
     
    Last edited: Feb 7, 2008
  6. Feb 7, 2008 #5
    If the downward acceleration of the box is greater then g then the block will get thrown from the box.
     
  7. Feb 7, 2008 #6
    The force of reaction between the box and the block would simply be x(t) = d*cos{[k/(M+m)]^(1/2)*t}, and for the block to leave the bottom of the box: d > (m*g)/k. Am I thinking correctly?
     
    Last edited: Feb 7, 2008
  8. Feb 7, 2008 #7
    It seems obvious that that should be the case but it would still be interesting to write the problem out as a system of equations. The momentum of the small block would reduce the force which it applies on the larger block during negative acceleration.
     
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