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Case when there is no normal force in spite of contact

  1. Apr 14, 2015 #1
    1. The problem statement, all variables and given/known data

    "There is a pulley with 2 blocks on the table.Force "F" is acting on it ,as it can be seen here
    TOMORROW.png
    A=10 kg and B=20 kg
    F=294 N
    Now,my exact problem is my teacher says "there is no contact force between A and the table.."

    Why block A is not pushing against table while block B is?

    2. Relevant equations:

    F=2T
    T=147 N

    3. The attempt at a solution
    As it is not a numerical problem instead conceptual doubt,I can not attempt anything.But I will try some in order to obey the guidelines of filling the template.Condition for normal force is there should be deformation.The objects should press against each other.I don't know why there is no deformation.Evertytime when something is on the table
    there is normal force equal to weight.
    Why not in this case?Why block A is not pushing against table while block B is?
     
    Last edited: Apr 14, 2015
  2. jcsd
  3. Apr 14, 2015 #2

    berkeman

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    What is the normal force down on the table for the larger block?

    If you moved the table 1cm down and released the blocks, what would happen?
     
  4. Apr 14, 2015 #3
    200N (Taking g=10 m/s^2)
     
  5. Apr 14, 2015 #4

    ehild

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    Touch the table with your hand and move your hand on the surface. You feel the surface of the table, but your hand does not push it. there is contact, but the weight of your hand is opposed by your muscles, the normal force between the table and your hand is zero.

    The situation is the same here. What should be the tension so as it keeps block A in rest just touching the table?
    What should be the normal force acting on block B?
     
  6. Apr 14, 2015 #5

    SammyS

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    It seems to me that there is a problem here with the problem as it's stated.

    In order that block A not accelerate upward, the most that T can be is 10g = 98 N ,(assuming g = 9.8 m/s2, which seems to be what's assumed here.)

    Therefore, the upper limit for F is 196 N.

    In that case, (F = 196 N) we have T =98 N so the net force on block A is:
    T - mAg = 98 - 10⋅9.8 = 0​

    So, block A is resting on the table. Does the table exert any force on the block?
     
    Last edited: Apr 14, 2015
  7. Apr 14, 2015 #6
    Hi Gracy,

    Let's consider a free body diagram on mass A, and the forces acting on mass A. If mass A is not glued to the table, what is the least that the normal force can be (and still remain in contact with the table)? With this value of the normal force, from a force balance on mass A, will it be accelerating upwards or not? (Note that, if it is accelerating upwards, it is not in contact with the table).

    Chet
     
  8. Apr 15, 2015 #7

    ehild

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    @gracy, what is the exact wording of the problem? Are the blocks in rest? Is the force F given as F=294 N? Do you need to find the forces between the blocks and the ground? Is the pulley massless?Is the rope taut?
    If the blocks are in rest, you can find the tension in the rope and the normal forces by considering the free body diagrams at each block.
     
  9. Apr 15, 2015 #8
    Yes.
     
  10. Apr 15, 2015 #9
    Yes.
     
  11. Apr 15, 2015 #10

    ehild

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    And what is the question?
     
  12. Apr 15, 2015 #11
    There is a massless pulley on which force "F" is applied.The pulley is connected to block A and block B.Both of which were resting on a ground when force "F" was being applied.A=10 kg and B=20 kg
    Question was find acceleration of A and B.
    But I have a solution for that.My teacher has given me.But in that solution it is said that "there is no contact force between A and the table"
    That's my exact problem...
     
  13. Apr 15, 2015 #12

    haruspex

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    Quite so. So consider starting with each block resting happily on the table, spring slack. Each weight deforms the table slightly. Now gradually raise the pulley. The tension increases in the string as the deformations decrease. At some point, the deformation from the lighter block becomes zero. What is the normal force there now?
     
  14. Apr 15, 2015 #13
    Zero .(Normal force on lighter block.)
     
  15. Apr 15, 2015 #14

    haruspex

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    Right. Does that answer your question?
     
  16. Apr 15, 2015 #15
    Yes.But one thing I would like to ask is
    Where is it mentioned in my problem as stated in #11 post.
     
  17. Apr 15, 2015 #16

    haruspex

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    It doesn't need to be. You are given that there is no normal force. How that came to be is irrelevant. You asked how it is possible, and now you know how it is possible.
     
  18. Apr 15, 2015 #17

    ehild

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    There is a massless pulley on which force "F" is applied.The pulley is connected to block A and block B.Both of which were resting on a ground when force "F" was being applied.
    Initially the blocks were in rest on the table. Then the force was applied. That force accelerated the pulley. The raising of the pulley made the rope taut. Block A lost contact with the ground.
    The net force acting on the whole system became F-WA-WB+NB=NB. This is still an upward net force, so the CM accelerates upward. At the end, B also loses contact with the ground. With no interaction between the blocks and the ground, the net force is just zero. The CM does not accelerate any more, but the pulley does, and so are the blocks.

    The blocks accelerate with respect to the accelerating pulley, and the relative accelerations have the same magnitude.
     
    Last edited: Apr 15, 2015
  19. Apr 15, 2015 #18
    What does it denote?
     
  20. Apr 15, 2015 #19

    ehild

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    NB is the normal force on B from the ground.
     
  21. Apr 15, 2015 #20
    Gracy,
    Are you saying that the force F is unspecified, or are you saying that it is 294 N?

    Chet
     
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