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Finding the Tension of a cord between two masses

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data

    Assume the three blocks (m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 2.5 kg) portrayed in the figure below move on a frictionless surface and a force F = 48 N acts as shown on the 2.5-kg block.

    Click link for image:

    (a) Determine the acceleration given this system.

    (b) Determine the tension in the cord connecting the 2.5-kg and the 1.0-kg blocks.


    (c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block.

    2. Relevant equations

    ƩF = ma
    F-T = ma

    3. The attempt at a solution

    a) 48N = ma
    a = 48/5.5 <----- sum of all masses
    a = 8.73 m/s^2 (which was correct).

    b) I made two FBD's, one with the mass of 1kg and another with 2.5kg.
    for the first mass: T=ma for the second mass: F-T = ma substitute first equation into this one: 48-(ma)= ma
    48= 2(ma)
    a = 24/3.5 <--- 3.5 was the sum of the two masses
    a = 6.857 m/s^2
    Then substitute the new value of a into the first equation: T = 3.5(6.875)
    T = 24 N (This was wrong)

    c) ƩF= ma
    F= 1kg(8.73) <------- this acceleration was found at the beginning, and was correct
    F = 8.72N (which was wrong)

    Your help would be really appreciated guys.
  2. jcsd
  3. Oct 17, 2011 #2


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    Gold Member

    Does this help?

    Attached Files:

  4. Feb 6, 2012 #3
    Great diagram! Thx a lot; I had this same question.
  5. Sep 30, 2012 #4
    Thanks. I was making it harder than it was.
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