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Inertial frame of reference question for stacked boxes

  1. Dec 14, 2014 #1

    I was wondering about a question and how it would be reconciled within Newton's laws of motion. Take a case where two boxes are stacked on top of each other, and the bottom box rests on a frictionless surface. Now, imagine a rope is attached to the top box, and tension is applied to the rope. Also, imagine that there is static friction on the surface where the top box meets the bottom box.

    Imagine the tension applied is less than the max value of static friction that can be achieved at that interface. From this case, it would seem that static friction would match tension when viewed at rest from the ground (the inertial reference frame) but the system of boxes would still be seen to accelerate from the ground since the bottom box is on a frictionless surface and is pushed by the top boxes static frictional force.

    It seems that one from the ground would conclude that the forces on the top box were balanced despite the fact that both of the boxes as a system would accelerate. I feel my misunderstanding has something to do with reference frames, but am not sure how to answer this myself.

    Any help is appreciated.

  2. jcsd
  3. Dec 14, 2014 #2


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  4. Dec 14, 2014 #3
    That is the nature of how static friction at that interface would work. As tension was applied to the top box, the static friction would match the tension as long as tension was at or below the max static friction.
  5. Dec 14, 2014 #4


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    Says who?
  6. Dec 14, 2014 #5
    This diagram is from hyper physics. Read the boxed information in the middle of the graph.
  7. Dec 14, 2014 #6


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    Static friction does not prevent motion, it only prevents slipping. It assumes any value necessary to prevent slipping up to the max.
  8. Dec 14, 2014 #7

    This assumes that the acceleration of the body is zero. This is not the case here.
    But indeed this is a case worth considering.
    In the ground reference frame the difference between T and the friction force will be equal to the mass of the top body times its acceleration.
  9. Dec 15, 2014 #8


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    What nasu said.
  10. Dec 16, 2014 #9

    So, what you are saying is that static friction must actually be less than tension in that case, and that the amount less depends on the acceleration?

    I think the reason this explanation gives me a problem is because if this is the case where tension is greater than static friction, it would seem that the top block should begin to move relative to the bottom block.

    I think what does make this correct in my head though, is despite the fact that it would seem the top block would begin to move relative to the lower block, the lower block would be moving relative to the frictionless ground at the same rate as the top block is moving relative to the bottom block therefore causing both blocks to remain static relative to each other. Does this seem like a proper interpretation, or have I missed something?

  11. Dec 16, 2014 #10


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    This can happen, if the static friction is not sufficient to accelerate the lower block at the same rate as the upper one accelerates.
  12. Dec 16, 2014 #11
    Yes. This is a correct interpretation. But you can get a much better understanding if you use free body diagrams. Draw a free body diagram for each of the boxes, showing the external forces acting on each. Now write a force balance on each of the boxes, based on the free body diagram. In these force balances, set the acceleration of each of the boxes to the same acceleration a. What do you get?

  13. Dec 17, 2014 #12
    Both situations can happen. They move together (same acceleration) for small values of tension, up to a threshold value.
    After that, the top body starts to slide over the bottom one (which moves too, but with a different acceleration).
    To see all these, do the math, starting with the free body diagrams, as was suggested already.
  14. Dec 18, 2014 #13
    When I do that, it shows that the tension is the net force for the system of boxes, and it shows that when the system is accelerating uniformly, the tension varies from greater than 1 to less than or equal to 2 times the value of static friction currently acting depending on the ratio of masses between the boxes. As the mass of the top box gets larger in relation to the mass of the bottom box, the tension can increase. Therefore, it appears that if they are boxes of equal mass, the max tension can be twice that of max static friction and still cause uniform acceleration for the boxes as a system.

    This appears to mean that uniform acceleration of the system will occur until the Tension applied is greater than twice that of the max static friction possible. It also appears to mean that in this case, static friction will assume its value between 1/2 and up to 1 (depending on mass) times the amount of tension. When the system would accelerate uniformly, it would seem that static friction would assume half of tension's value when the masses were equal, and could approach but not be equal to tension as the ratio of masses got smaller.

    This is an interesting result to me. Have I made some mistake in this interpretation?

    Thank you very much for your time,
  15. Dec 18, 2014 #14
    It's hard to tell without our seeing your actual equations. I don't want to take the time to do the problem myself, but I can tell from seeing your equations if what you say is correct.

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