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How to find tension of a single object in a pulley system?

  1. Nov 19, 2015 #1
    1. The problem statement, all variables and given/known data

    LETTER d only

    phsy.PNG

    2. Relevant equations
    Fnet = ma

    3. The attempt at a solution
    So I considered 2m independently and set its net force equal to its gravity minus its tension.

    (2m)a = Fg - Ft = (2m)g - Ft

    I then rearranged the equation so that Ft = (2m)g - (2m)a = 2m(g-a)

    There's no answer key, so I'm not sure if this is correct. I appreciate any assistance with this problem! Thanks!
     
  2. jcsd
  3. Nov 19, 2015 #2
    The problem is that your answer contains a. The directions say to express your answers in terms of m and g.

    So, you need to find an expression for a in terms of m and g.
     
  4. Nov 19, 2015 #3
    Ah, ok. Since all the blocks are moving in a system, I'm going to assume that the acceleration calculated in letter b will be used.

    Fnet = (5m)a = 5mg - 4mg (I hope I'm calculating this correctly)
    (5m)a = mg
    a = g/5

    So time to plug back in: Ft = 2m(g-a) = 2m(g-(g/5))
    Ft = 2m(4g/5)
    Ft = 8mg/5

    It seems like that's it. Was my original methodology correct though?
     
  5. Nov 19, 2015 #4

    ehild

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    Gold Member

    What do you mean? The blocks are connected, and the magnitude of their acceleration is the same.

    The net force is correct, mg, but the whole mass is not 5m.
     
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