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Dynamic Pulley System

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img53.imageshack.us/img53/6141/41622697.png [Broken]


    2. Relevant equations
    F = ma, all that basic stuff.


    3. The attempt at a solution
    Alright
    g = 32.2
    Ma = 150/g = 4.66lb
    Mb = 75/g = 2.33lb

    so Box A
    __
    |....}==== 2T
    |__|

    box B

    3T
    -----|....|______ 60lb
    ===={__|

    Sum of Fx = ma

    so for A
    2T = 4.66aA
    for B
    60 - 3T = 2.33aB

    then for the lengths of the wire I did
    _________
    O________
    ........n__O
    |-sa-||sb|

    so L = 2sa + 3sb
    0 = 2aA + 3aB
    but clearly aA and aB must have opposite signs so
    2aA = 3aB

    using simultaneous equations with the following you get
    60 = 2.33aB +3T
    0 = 4.66(3/2)aB - 2T
    which makes aB = 7.22ft s-2 (ANS)
    and T = 14.4 lb

    then i figured there were two ways to go about work done

    KEi(A+B) + Wd = KEf(A+B)
    or Wd = 2T*sA + (60-3T)*sB

    Which is great cause you get to doublecheck your result

    the problem is that for method 1 I get 334.1J
    and for method 2 I get 216.6J

    So, yeah, any help is appreciated.

    EDIT1: Formatting Errors
    EDIT2: I found my mistake. My simultaneous eqs were wrong, it was 6.99, not 4.99.
    Now the answers are aA = 7.02 ft s-2 and W = 140.4J
    I'd still appreciate some double-checking on my method since I'm not really sure if I'm doing this right.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
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