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Laws of Motion - Basic Problem

  1. Jun 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Given in the pic .. .


    2. Relevant equations
    F=mg


    3. The attempt at a solution
    i think block 2 hitting wedge will take place 1st as the direction of displacement of B1 is perpendicular to 'g' .. .
    but i dont think thats a logical answer
    please answer and explain why
     

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  3. Jun 17, 2012 #2

    tiny-tim

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    hi draotic! :wink:

    call the displacement "x" and the angle "θ" …

    what equations do you get? :smile:
     
  4. Jun 18, 2012 #3
    sorry , i didnt quite get it .. .
    which angle are we concerned about?
     
  5. Jun 18, 2012 #4

    tiny-tim

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    the angle the right-hand part of the string makes with the horizontal at any particular time :wink:
     
  6. Jun 18, 2012 #5
    well the angle will vary from 0 to pi/2 , so are we going to involve integration ?
     
  7. Jun 18, 2012 #6
    Block#2 needs the highest force up at vertically down, weight plus centripetal force before continuing to hit the wedge.

    Thus block#1 must hit the pulley first before bigger block can continue the circular motion.
     
  8. Jun 19, 2012 #7

    tiny-tim

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    hi draotic! :smile:

    (just got up :zzz:)
    in problems like this, always try an energy equation first (which will avoid integration)

    if that doesn't work, use a force equation (which will) :smile:
     
  9. Jun 19, 2012 #8
    thanks tiny-tim and azizlwl for helping me out ..
    azizlwl's reason seems quite right to me..
    in B1 , only tension is involved ( ignoring kinetic friction ) and in B2 , we must have tension + centripetal force . .

    so B1 must hit pulley first..
     
  10. Jun 19, 2012 #9
    thanks tiny-tim and azizlwl for helping me out ..
    azizlwl's reason seems quite right to me..
    in B1 , only tension is involved ( ignoring kinetic friction ) and in B2 , we must have tension + centripetal force . .

    so B1 must hit pulley first..
     
  11. Jun 19, 2012 #10
    are all the parameters such as masses, distances from pulley variable?
     
  12. Jun 19, 2012 #11
    masses are same , no friction anywhere , no MI of pulley
     
  13. Jun 19, 2012 #12

    tiny-tim

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    you can assume the pulley is massless

    (alternatively, just add its "rolling mass" (I/r2) to the mass of the first block)
     
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