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3 blocks dragging with friction, find tension

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Three blocks are located on a horizontal table. The coefficient of kinetic friction between the blocks and the table is 0.126. They are connected by a massless cord, as shown in the figure below, and pulled to the right. The masses of the three blocks are m1 = 3.0 kg, m2 = 5.0 kg, and m3 = 6.0 kg. The pulling force is equal to T3 = 82.0 N. What is the tension T2?

    2. Relevant equations


    3. The attempt at a solution

    I am just plain clueless
  2. jcsd
  3. Sep 20, 2010 #2


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    It's F_net = ma, per Newton's 2nd law. Look at all 3 blocks together (they move and accelerate together) , and apply Newton 2 to the system of blocks to calculate the acceleration. Then, to get T2, the cord tension pulling back on m3, you must learn how to draw free body diagrams. Isolate block m3 to denote the forces acting on it. Then apply Newton 2 again. Please show an attempt.
  4. Sep 21, 2010 #3
    Okay so to get acceleration for the whole system, I found all the frictional forces of the blocks and added them up (17.2872N) and subtracted them all by the pulling force (82.0N) and took the mass of all the blocks together and solved for a.



    Is this acceleration right?
  5. Sep 21, 2010 #4
    So i now have the a=4.62m/s^2 and the frictional force is 17.29N. I dont know if this is right but I used: T2-Fk=(M1+M2)(a) so T2=(M1+M2)(a)+Fk or with numbers T2=(3+5)(4.62)+17.29 and I got T2=54.25N is this right?
  6. Sep 21, 2010 #5


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    Looks good!
    Good effort, but not correct. It looks like you took a free body diagram (FBD)of the first 2 blocks, which is fine...but the friction force is the sum of the friction forces on M1 and M2. Do not include the friction force on M3; it doesn't show up in your FBD of the first 2 blocks.
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