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Two blocks and a spring

  1. Jul 2, 2008 #1
    1. The problem statement, all variables and given/known data

    20: Two boxes are sitting on top of one another. The top box has a mass of 5kg and is connected to the wall by an unstretched spring, k=2000N/m. The bottom box is 10kg. the static friction coefficient between the 2 boxes is 0.5, the kinetic friction coefficient between the 2 boxes is 0.25. the static friction coefficient between the bottom box and the floor is 0.4, and the kinetic friction coefficient is 0.2.
    a. Draw Free Body Diagrams
    b. What is the force F that causes the boxes to move relative to each other?
    c. How much does the spring stretch when force F is applied to the bottom box before the boxes move relative to each other?


    2. Relevant equations
    Force of a spring = -kx
    Force of friction = mu (static or kinetic) times normal force
    Sum of forces = ma

    3. The attempt at a solution

    I had no problem with part a

    For parts b and c I got about half way done and then found a problem.

    For part b, at first I assumed that for the two blocks to move relative to each other, the bottom block must already be moving so I should use the coefficient of kinetic friction. However, after I actually did the math out, I found that the force required to to overcome the first static force and make both blocks move together is greater than the sum of the two forces (kinetic friction of both, and static friction between the two) required to make them move relatively. So wouldn't that mean that the moment the lower block begins to move the two separate and so therefore the answer to part c is zero?

    Is this what occurs, or am I just misunderstanding the problem?

    I finished the problem ignoring that issue and got as answer for b and c 54 newtons and 1.2 centimeters. Are these right in that case?

    Thanks for the help?
     
  2. jcsd
  3. Jul 3, 2008 #2

    tiny-tim

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    Hi Piamedes! :smile:

    It's a lot easier for us to check if you show us your calculations. :smile:
     
  4. Jul 3, 2008 #3
    No problem, here they are:

    At first I started off by finding the minimum force necessary to make the two blocks move together.

    Sum of forces on the block system equals zero just before it starts to move because static friction is at a maximum. So:

    [tex]
    0 = F_(applied) - F_(friction)
    [/tex]

    [tex]
    F_(applied) = F_(friction)
    [/tex]
    [tex]
    F_(applied) = \mu_(static) [m_(top) + m_(bottom)] g
    [/tex]
    [tex]
    F_(applied) = (.4)(10 kg +5 kg) g
    [/tex]
    [tex]
    F_(applied) = 59 newtons
    [/tex]


    Then I calculated the forces for when the lower block is already moving and the two blocks are about the separate. Once again the sum of forces is zero because the static friction between the blocks is at a maximum.

    [tex]
    0 = F_(applied) - F_(fr kinetic) - F_(fr static)
    [/tex]
    [tex]
    F_(applied) = F_(fr kinetic) + F_(fr static)
    [/tex]
    [tex]
    F_(applied) = \mu_(kinetic) [m_(top) + m_(bottom)] g + \mu_(static)m_(top) g
    [/tex]
    [tex]
    F_(applied) = (.2)(10kg + 5 kg)g + (.5)(5 kg)g
    [/tex]
    [tex]
    F_(applied) = 54 newtons
    [/tex]

    After I calculated these two values I wondered how this would actually happen. Since you need to push with a greater force to make the blocks move at all, then wouldn't they immediately separate?

    I thought including the static friction between the blocks in the first step might solve this, but how can the top block resist movement when the two blocks haven't moved at all? Until there is some displacement, the spring connected to the top block will not cause a restorative force. Without that force, how can the top block cause friction?

    Thanks for the help.

    Sorry the latex looks so cramped; its my first time trying to use it. Any advice regarding that would also be greatly appreciated.
     
  5. Jul 4, 2008 #4

    tiny-tim

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    do the top block separately

    Hi Piamedes! :smile:

    You must consider the top block on its own.

    Do an FBD for the top block, showing the acceleration, the friction from the bottom block, the spring, and the weight. :smile:
     
  6. Jul 4, 2008 #5
    I drew a FBD diagram for the top block.

    [​IMG]
    [​IMG]

    When drawing it I questioned the direction of friction. I think that it should go to the right, along the blue line, otherwise there is no force accelerating the block in that direction.

    Apart from that I am still confused as to how the lower block will move out from under the top block. Will it occur immediately, or after the lower block is already moving?
     
  7. Jul 5, 2008 #6

    tiny-tim

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    Hi Piamedes! :smile:

    oooh, that is really pretty! :smile:
    The static coefficient of friction between the lower block and the floor is lower than between the two blocks, so the two blocks will move together first.

    The top block will stay on the lower block (with the same acceleration, of course) until the spring is strong enough to produce the extra force needed.
    The top block will want to be in uniform motion (good ol' Newton's first law), so the friction will be trying to push it forwards. :smile:
     
  8. Jul 5, 2008 #7
    Its amazing what you can do in MS paint.

    I think I understand it now. Since the only force between the two blocks is the static friction, regardless of the applied force on the lower block, the upper block will not slide off until the spring supplies the necessary force, because the only forces acting on the top block in the x direction are friction and the spring force. So the top block will slide off when:
    [tex]
    F_f = kx
    [/tex]

    Thanks a lot for the help
     
  9. Jul 5, 2008 #8

    tiny-tim

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    acceleration

    Yes, but you're ignoring the acceleration …

    the top block will slide if it cannot keep up with the acceleration of the lower block.
     
  10. Jul 5, 2008 #9
    So the two blocks will move uniformly until either the acceleration or spring separates them.

    So it would look like either:

    [tex]
    F_f = kx
    [/tex]

    or

    [tex]
    m_t a = F_f
    [/tex]

    [tex]
    m_t a = \mu m_t g
    [/tex]

    [tex]
    a = \mu g
    [/tex]

    So the top block will fall off once either the spring force supplies enough force to separate them, or the applied force is too large such that the acceleration of the top block equals or exceeds [tex] \mu g [/tex].

    Thank you for explaining this
     
  11. Jul 5, 2008 #10

    tiny-tim

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    No, everything acts together.

    Do Newton's second law for the top block:

    [tex]m_t a\,=\,\mu m_t g\,-\,kx[/tex]
     
  12. Jul 5, 2008 #11
    Sorry, mistyped that. The blocks move uniformly until the sum of acceleration and the spring force separates them.

    So in the end, if the initial acceleration is great enough, then the two block will separate at zero displacement because the force of static friction has been overcome. So in this situation, the force necessary to move the blocks will not immediately separate them, however, after a very small displacement the spring will supply enough restorative force to pull the top block off the bottom one.

    I think I have it now. Thanks for the help.
     
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