Tension practice problem

  1. I'm trying to improve on tension problems so I am practicing using this website: http://www.education.com/study-help/article/tension/

    I am on problem 6, but I keep coming up with the wrong answers. I attempted this by trying to show the net forces for each box.

    Box M net forces
    T upwards
    Mg downwards
    F = T - Mg = Ma

    Box 2M net forces
    T upwards
    T' downwards
    ***should I include the weight (2Mg) of this box???

    Box 4M net forces
    4Mg downwards
    T' upwards
    F = 4Mg - T' = 4Ma

    I have a feeling there is an error in box 2M that is fudging with my results. Lastly, does T = 6Mg? Help?? :confused:
     
    Last edited: Aug 5, 2011
  2. jcsd
  3. PhanthomJay

    PhanthomJay 6,228
    Science Advisor
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    Gold Member

    yes
    always include the weight in any free body diagram
    yes
    yes, correct your error for the net forces acting in the FBD of the 2M block.
     
  4. was my error that I didn't include the weight? or is it another entirely separate error in block 2m?
     
  5. Is the equation for block 2m:

    F = 2Mg + T' - T = 2Ma?
     
  6. PhanthomJay

    PhanthomJay 6,228
    Science Advisor
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    yes. Now solve for T, T', and a. (Note: T is not equal to 6Mg)
     
  7. I solved it and it still incorrect.

    Block M
    T - mg = ma <---plug in m = 1 Kg
    T - 9.8 = a
    T = 9.8 + a equation 1

    Block 2M
    T' + 2mg - T = 2ma <---Plug in m = 1 kg
    T' + 2(9.8) - T = 2a
    T' + 19.6 - T = 2a equation 2

    Block 4M
    4mg - T' = 4ma < --- plug in m = 1 kg
    4(9.8) - T' = 4a
    T' = 39.2 - 4a equation 3

    Substitute equation 1 and equation 3 into equation 2
    equation 2: T' + 19.6 - T = 2a
    (39.2 - 4a) + 19.6 - (9.8 + a) = 2a
    Solve for a.

    a = 7 m/s^2

    Correction answer is 8.6 m/s^2.
     
  8. Thank you phanthomjay *air hug*
     
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