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Frictionless pulley system: same tension changing masses

  1. Apr 8, 2012 #1
    1. The problem statement, all variables and given/known data

    An astronaut lands on a distant planet with an unknown gravitational field g’. On the bottom of his spaceship, he attaches an ideal frictionless pulley supported overhead with two light buckets hanging downwards, attached by a light rope. He places 24 kg of the same rock type in both buckets and removes his hand, the rope is stressed to its limit; any increase in tension would break the rope. He now removes 6kg from one bucket, but holds on to prevent motion. What maximum amount may he add to the other bucket so that the same limiting tension will be present in the rope when he removes his hand?



    2. Relevant equations
    Fnet = ma
    fg = mg


    3. The attempt at a solution

    So, this question has given me much grief, and this is pretty much the progress I made.

    In the first case, when there is 24 kg in both buckets,
    Fnet = 0 because the two buckets would not move so,
    Fg= T

    In the second case, with the 18kg and 24 kg buckets
    18kg: Fnet = T-Fg
    24kg: Fnet = Fg- T

    I know the fg = (g’)(mass)

    18kg: Fnet = T-18g’
    24kg: Fnet = 24g’ –T

    This is where my answer gets a bit weird. I know that by adding mass to the 24kg bucket, it should accelerate downwards.

    A= (T-18g’)/18
    A=(24g’-T)/24

    (T-18g’)/18 = (24g’-T)/24
    24T-432g’ = 432g’-18T
    42T = 864g’
    T= 20.6 g’

    So then do I plug 20.6g’ into the acceleration formula for T and use the calculated acceleration and Fnet=ma to get mass?

    If I do that my acceleration is 0.14g’ and subbing into Fnet gives me a mass of 3.26g’. but how do I calculate g’?
     
  2. jcsd
  3. Apr 10, 2012 #2
    i know that I can't actually include 24 in my problem as the mass is changing so do I replace 24 with m? and then have my g' cancel at some point?
     
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