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Compute acceleration and tension in the cord

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data
    A cord passing over a frictionless, massless pulley has a 4.0 kg object tied to one end and a 12 kg object tied to the other. Compute the tension and acceleration in the cord.


    Right now, I'm just trying to calculate the acceleration. It is supposed to be 4.9 m/s^2

    2. Relevant equations

    F=ma
    T1=T2=0
    T1-mg=ma
    T2-mg=ma


    3. The attempt at a solution

    For the 4kg block-

    F=ma
    T1-mg = ma

    T1-39.2 = 4a
    Solving in terms of T1 to get
    T1= 4a+39.2

    For the 12kg block-
    F=ma
    T2-mg = ma

    T2-117.6= 12a
    Solving in terms of T2-
    T2= 12a+ 117.6

    Now, because T1=T2, I can say

    12a+ 117.6 = 4a+ 39.2
    But then I end up with a=9.8.

    The answer is 4.9, which, incidentally, is 9.8/2.

    Do I have to divide by 2 because I have two cords? Or is there some other reason I am missing?
     
  2. jcsd
  3. Dec 8, 2011 #2
    Assuming the following is correct can you solve for T and a?
     

    Attached Files:

  4. Dec 9, 2011 #3
    Yeah, you'd set T=4a + 4g , plug that into the other equation for T, to get

    12g+(-4a-4g) = 12a
    to get
    8g=16a
    8*9.8 = 16a

    a=4.9

    Ah. I must have screwed up somewhere in my algebra.

    Thank you!
     
  5. Dec 9, 2011 #4

    PeterO

    User Avatar
    Homework Helper

    co-incidence almost.

    One way to view this: the weight of a 4 kg mass is pulling one way, and the weight of a 12 kg mass is pulling the other way.

    If you consider the weight of the 4 kg mass cancells out [balances if you like] 4 of the other 12, we have a net force equal to the weight of an 8kg mass to accelerate this system - which has a total mass of 16 kg.

    Now the weight of a 16 kg mass will accelerate a 16kg mass at 9.8 ms-2, so we would expect that half that weight would achieve only half that acceleration → 4.9 ms-2
     
  6. Dec 9, 2011 #5
    Another way to solve this problem is to use just two directions for forces. It may sound crazy, but I have seen this procedure in two or more textbooks.
    The positive direction would the direction of motion (the direction of falling of the 12-kg block). This way, if m1 is the 12-kg block, and m2 is the 4-kg block, we would have the following system of equations (each equation for a separate free-body diagram):

    -T + m1g = m1a
    T - m2g = m2a

    I got 58.8 N for tension, and 4.9 m/s^2 for acceleration ;)
     
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