Laws of Motion - Basic Problem

  • Thread starter draotic
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  • #1
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Homework Statement


Given in the pic .. .


Homework Equations


F=mg


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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Answers and Replies

  • #2
tiny-tim
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hi draotic! :wink:

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

what equations do you get? :smile:
 
  • #3
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sorry , i didnt quite get it .. .
which angle are we concerned about?
 
  • #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:
 
  • #5
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well the angle will vary from 0 to pi/2 , so are we going to involve integration ?
 
  • #6
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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.
 
  • #7
tiny-tim
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hi draotic! :smile:

(just got up :zzz:)
well the angle will vary from 0 to pi/2 , so are we going to involve integration ?

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:
 
  • #8
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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..
 
  • #9
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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
are all the parameters such as masses, distances from pulley variable?
 
  • #11
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are all the parameters such as masses, distances from pulley variable?

masses are same , no friction anywhere , no MI of pulley
 
  • #12
tiny-tim
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… no MI of pulley

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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