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Pulley block system

  1. Feb 25, 2016 #1
    14564146227433.png


    1. The problem statement, all variables and given/known data
    FInd acceleration and tension. Take g=10m/s^2

    2. The attempt at a solution
    By drawing the free body diagrams of every block and simultaneously solving all the equations, I got the answer acceleration= 30/7 m/s^2, which is also correct.
    But when I try to treat all the pulleys and blocks as one system to get the acceleration, I get a different answer.
    f=ma


    ⇒ 40g+10g= 70 * a (since weight of 40 kg block and 10 kg block are the only forces acting downwards) ⇒ a=500/70 m/s^2
    Somebody told me that I should have rather subtracted 40g and 30 g, so that would be 40g-10g=70a ⇒a=30/7 which happens to be the right answer.
    But I don't understand why will we subtract the force vectors if they are all acting in the same direction ;ie, downwards. Where am I wrong with the second approach?
     
  2. jcsd
  3. Feb 25, 2016 #2

    BvU

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    Imagine one single pulley with two weights, 10 kg and 10 kg. Will the acceleration be (10+10) g / 20 or will it be (10-10) g / 20 ?
     
  4. Feb 25, 2016 #3
    20 g according to my logic. Since, weight( a force vector) of the bodies is in the downward direction, we will add the vectors together to get the total weight and hence the acceleration=mg=(10+10)g.
     
  5. Feb 25, 2016 #4
    Alright I get it now, when gravity acts on those blocks ,both will have acceleration in opposite directions; the 40 kg block will move downward being heavier and the lighter block will be forced to move up and hence we subtract their weight.(mg)
    Thank you so much.
     
    Last edited: Feb 25, 2016
  6. Feb 26, 2016 #5

    BvU

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    (Especially for later exercises) it is important to clearly identify the physical role of the rope: you should have seen that already when "drawing the free body diagrams of every block": the tensions at the two ends of a straight rope are equal and opposite forces. When the chord is run over a pulley, in the no friction case the magnitudes remain the same but the directions change.
    But for the straight sections - again - the tensions at each of the two ends are equal and opposite.

    As a simple trick you can (imagine to) cut each straight section halfway and apply two equal and opposite forces to keep the loose ends in place.
     
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