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The normal reaction between blocks A and B is?

  1. Oct 3, 2016 #1
    1. The problem statement, all variables and given/known data

    Capture.PNG

    2. The attempt at a solution

    Let rightward acceleration of the block (A+B) be $$a$$.
    Let downward acceleration of block (A) be $$a'$$ down.

    For block A:

    $$mg-T=ma'$$

    $$N+ma=0$$

    For block B:

    $$T+N=Ma$$

    On solving $$N=-\dfrac{m^2g}{M+2m}$$ by taking $$a=a'$$.But that is wrong answer.

    Where did I go wrong?
     
  2. jcsd
  3. Oct 4, 2016 #2

    andrewkirk

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    Why do you think there will be any normal reaction? For there to be a normal reaction they need to remain in contact. Do they?
     
  4. Oct 4, 2016 #3
    Because the blocks are touching...
     
  5. Oct 4, 2016 #4

    andrewkirk

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    Do they remain touching, as B accelerates rightwards? if so why?
     
  6. Oct 4, 2016 #5
    Because the pulley moves when the bigger block moves right.In turn the amount of string (horizontal) released moves to vertical position and the smaller block moves further down.Now since the released horizontal part of string becomes vertical naturally the smaller block will keep touching the bigger block...
     
  7. Oct 4, 2016 #6

    haruspex

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    You found that the normal force is negative. What does that tell you?
     
  8. Oct 4, 2016 #7
    Truly speaking i'm confused by the fact that it is negative...Can you provide an explanation?
     
  9. Oct 4, 2016 #8

    haruspex

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    Unless they are glued together it cannot be negative.
    You assumed that the blocks remained in contact and arrived at a contradiction, so....
     
  10. Oct 4, 2016 #9
    Well that means they are not in contact.But it is difficult to visualize such a thing.Can you visualize that happening?
     
  11. Oct 4, 2016 #10

    andrewkirk

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    Sure it can happen. B accelerates rightward and A accelerates downwards but not rightward (not at first, anyway), which means it swings to the left relative to B. It's probably easiest to first work out what it is that makes B accelerate rightward. What is it?
     
  12. Oct 4, 2016 #11
    I dont think anything accelerates B rightwards.I guess I got your point!
     
  13. Oct 4, 2016 #12

    andrewkirk

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    Sure it does. How does the pulley stay at the corner of block B? What must it be attached to?
     
  14. Oct 4, 2016 #13
    Sorry.I meant no force accelerates the smaller block right.Of course tension pulls the bigger block.
     
  15. Oct 4, 2016 #14
    Hello Andrew ,

    Would it be correct to say that the string connecting block A would not be vertical , and the horizontal component of tension is the force responsible for accelerating A rightwards ?
     
  16. Oct 4, 2016 #15

    haruspex

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    Yes, but a full description of the motion might be rather complicated.
     
  17. Oct 4, 2016 #16

    andrewkirk

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    Yes (as haruspex said). I think, because it would then get so complicated, the problem setter wants the reader to focus on with the situation immediately after block A is released, while the angle of its cord to the vertical is still small enough for horizontal components of force to be ignored on the grounds of immateriality.
    Correct. The pulley is subject to a downwards force from cord because of the weight of A, and a rightwards force from the cord connecting it to the wall. These add to give a force that points diagonally downwards to the right, which is transmitted to B by the strut that holds the pulley in place. The vertical component of that force is cancelled out by the normal force on B from the floor, leaving a rightwards force that accelerates B (but not A).
     
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