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Acceleration of the pulley

  1. Aug 14, 2006 #1

    danago

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    Hey. IN class, my teacher went through a question, which i did not understand at all. Could someone please maybe help me?

    What i have to do, is determine the acceleration of this pulley system:
    [​IMG]

    ANy help greatly appreciated,
    Thanks,
    Dan.
     
  2. jcsd
  3. Aug 14, 2006 #2
    The teacher was most likely asking "What is the acceleration of the blocks that are falling?" There are three different accelerations for which the teacher could have asked: the acceleration of the blocks that are going up, the acceleration of the blocks that are going down, and the angular acceleration of the pulley. I'm guessing that the pulley is of negligible size, so finding the other two accelerations is simpler - they are the opposites of each other.

    Draw a forces diagram, and use Newton's second law. Figure out which weights are acting in which directions.
     
  4. Aug 15, 2006 #3

    daniel_i_l

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    You have to find the acceleration of the system = the acceleration of the blocks. one side will be positive and the other negative depending on what direction you define positive. Keep in mind that the tension of the string is the same everywere.
     
  5. Aug 15, 2006 #4

    danago

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    Im still really stuck :(
     
  6. Aug 15, 2006 #5

    danago

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    ok. Ive looked over the working my teacher has given me. Im not sure if im understanding it properly though.

    Ill label the diagram like below, with the tensions of the strings labeled as [tex]T_x[/tex].

    [​IMG]

    Ill start with the 10kg block. The force that it is accelerating upward with is equal to the force of [tex]T_1[/tex] pulling up on it, plus the force of gravity pulling down on it. So i can get:

    [tex]10a = T_1 - 10g[/tex]

    Now. With the 5kg block, the overall force acting on it will be the force of T2 pulling up on it, plus the force of T1 pulling down, plus the force of gravity pulling down. So:

    [tex]
    5a = T_2 - T_1 - 5g
    [/tex]

    And then if i do the same with the 20kg block. The overall force pulling it down is equal to the force of gravity pulling it down, plus the tension pulling it back up. So:

    [tex]
    20a = 20g - T_3
    [/tex]

    Now that i have the three equations, i can solve simultaneously for a (keeping in mind all tensions are the same), which gives me a=1.4. Sound right?
     
    Last edited: Aug 15, 2006
  7. Aug 15, 2006 #6

    danago

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    ok. Instead of what he showed me, i put my own method to the test. The 20kb block has a weight force of 196N, and the other side has a weight force of 147N. Since the 20kg side is applying a greater force, it will move downwards, with an overall force of 49N. Since 49N is being used to pull a mass of 25kg in total, i can then use the F=ma formula and get 1.4m/s/s.

    Is the theory behind that correct? Cos it seems MUCH simpler than that my teacher showed me.
     
  8. Aug 15, 2006 #7
    ur method sounds good
     
  9. Aug 15, 2006 #8

    danago

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    ok. Now ive gotta figure out why my teacher is showing us the long drawn out way :P

    Thanks for the help everyone :)
     
  10. Aug 15, 2006 #9
    ur most welcome :smile:
     
  11. Aug 15, 2006 #10

    Office_Shredder

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    danago, your teacher's method is more general. While your method works for simplified cases like this, once you have systems with 6 or 7 different blocks wrapped around pulleys, spheres that have mass, lying on tables, etc. you need a consistent, mathematically rigorous way to sort out all the forces
     
  12. Aug 15, 2006 #11
    The difference between your method and the teacher's method is that you intuitively canceled out the tension forces, while the teacher's method requires you to map out the forces and cancel out tensions manually.

    My advice to you: Learn the teacher's method for solving the problem, even though it confuses you right now. As Office_Shredder said, more complex problems become simpler when you have a consistent way of dealing with them. If you see a problem like this one again, you can do it your way, but many problems require the use of Newton's second law and free-body diagrams.
     
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