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Tension in Cord

  1. Jan 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Two masses M1 = .600kg and M2 = 2.2kg connected by a cord of negligible mass and passes over a frictionless pulley of negligible mass. Assuming that y axis has a positive upward direction, what is the tension in the cord?


    2. Relevant equations
    F = ma

    W = mg

    3. The attempt at a solution
    I first tried to add up the Fg for both masses under the assumption that the opposing force would be the tension.

    I then subtracted the larger Fg from the smaller Fg which was also wrong.

    I then came upon this and tried to follow the process discussed there. I tried to solve for a based on the posts discussed in the other thread but was lost on posts 9 and 10 (NEwayz, back to what I did).

    I found the equations of the two blocks to be T - m1g = m1a and m2g - T = m2a.
    this makes T = m1a + m1g = m2g - m2a.
    Plugging in the numbers gives 5.88N + .6a = 21.56N - 2.2a
    solving for a gives a = 5.6ms-2
    Plugging this back into the T equation gives a T of 9.24N
    I then double the T because there is T working on both ends of the rope and get 18.48N.
    I turns out that this is an incorrect answer and I am now out of ideas. Please help.
     
  2. jcsd
  3. Jan 25, 2009 #2

    hage567

    User Avatar
    Homework Helper

    You don't double T. The tension T is acting throughout the rope. It isn't T on one side and T on the other giving 2T total.
     
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