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  • WoAiMeiMei
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    146 posts since Jul '07
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    eagle
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    14,473 posts since Aug '01
    • 08 May `08, 7:00PM

      first question is JC physics with friction added only

      2nd question need to consider force

      In short, you need to draw out all the forces to analyse.

       

      Try first... Cannot do tmr I do... my last paper tmr... after wait for my first class cert only :D

  • WoAiMeiMei
    WoAiMeiMei's Avatar
    146 posts since Jul '07
  • WoAiMeiMei
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    146 posts since Jul '07
    • 09 May `08, 9:12PM

      okay.. i lost some marks for the first question.. still wrong :x

      im not sure about the second question

  • WoAiMeiMei
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    146 posts since Jul '07
    • 10 May `08, 7:42PM

      mass times friction coefficient to test slipping... or is it force. it should be mass :x ahh wth

  • WoAiMeiMei
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    146 posts since Jul '07
    • 12 May `08, 2:30PM

      help D:?

      the force pulling the block up is the same as the one going down?

      Edited by WoAiMeiMei 12 May `08, 2:31PM
  • WoAiMeiMei
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    146 posts since Jul '07
    • 13 May `08, 7:37PM

      far out im tank.

       

      derived
      acceleration of the block = a = (mg + uk Mgcos # - Mgsin# )/(M+m) and the T = mg - ma

      # = the given angle
      M = the greater mass
      m = the smaller mass
      uk = the given kinetic coefficient of friction

       

      so answer for first bit should be 281.6126N

       

      okay second question...

  • WoAiMeiMei
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    146 posts since Jul '07
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    eagle
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    14,473 posts since Aug '01
    • 14 May `08, 11:36AM

      Q1

      Assume it is moving already... If it is not, we have to redo with μs... So now we do with μk

      Let the tension in the string be T

      For the 54kg block,
      Reaction force is 54g cos 42
      Downward force is 54g sin 42

      Taking moving downwards left as positive,
      54 * a = 54g sin42 - T - μk 54g cos42
      54a = 315.0979 - T -- (1)

      For the 27kg block,
      upward force = T
      downward force = 27g

      Taking upwards as positive,
      T - 27g = 27a
      27a = T - 264.87
      54a = 2T - 529.74  -- (2)

      (2) - (1): 3T - 844.8779 = 0
      T = 281.626N

  • Moderator
    eagle
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    14,473 posts since Aug '01
    • 14 May `08, 11:56AM

      Q2

      Reaction force = mg cos 54 + mrω^2 = m * (5.766 + 18.28224) = 24.0484 m

      let coefficient of static friction be μs

      μs *  24.0484 m = mg sin 54
      μs = 0.33

  • WoAiMeiMei
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    146 posts since Jul '07
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