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Does friction work as an action reaction pair with 2 obects on top of eachother?

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data
    the 1.0kg block is tied to the wall with a rope. it sits on top of the 2.9kg block. the lower block si pulled to the right with a tension force of 20N. the coefficient of kinetic friction at both the lower and upper surfaces of the 2kg block is mk=.40. what is the tension in the rope holding the 1kg block to the wall?


    2. Relevant equations
    newtons third law, perhaps the second law. ffk=fn *mewk fnet=ma fg=mg


    3. The attempt at a solution
    when i tried thisi said that the backward friction in the 2kg block was equal to the forward fricttion in the 1kg block (action reaction pair. this also makes sense i think because the 1kg is tending backwards "slipping" as the 2.0kg object accelerates forward). i noticed that the fg of the 2kg box was actually 3*9.8 because it has the 1kg box on top of it. this would also be the Fn of the 2kg box. so ff = 11.76 subbing in fN and mewk. so there is a forward force on box 1kg. then we also have to take into account another action reaction pair. that is, the forward force 20N on box 2kg has a reaction pair on the 1kg block of 20N back. then we have yet another force acting on the 1kg, and that is its own Ffk which is Fk=9.8*1 *.4 = 3.92 newtons forward. so the tension in the rope would be T = 3.92 + 11.76 - 20 = 4.32 N (back). There are no solutions for these questions and i just want to see if i did it right. i have a feeling its perfection but like in the title i infered that i wasnt sure about the action reaction forces i assumed (through newtons second law)
     
  2. jcsd
  3. Oct 5, 2008 #2

    HallsofIvy

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    Much of the information given here is irrelevant. The friction force is just the coefficient of kinetic friction multiplied by the weight of the top block. As long as the lower block is moving, so that there IS friction, how fast it is moving is irrelevant. Then tension in the line holding the upper block in place is equal to the friction force.
     
  4. Oct 5, 2008 #3
    i think you're wrong sorry. any other people think they know how to do it?
     
  5. Oct 5, 2008 #4

    Vanadium 50

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    I don't think he's wrong. What horizontal forces are on the 1 kg block?
     
  6. Oct 5, 2008 #5

    Doc Al

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    For once, HallsofIvy is not wrong. :rofl: (Sorry Halls... couldn't resist.)
     
  7. Oct 5, 2008 #6

    HallsofIvy

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    Well, at least you didn't do what mrjoe2 did- he sent me a pm telling me that if I don't understand a question, I shouldn't respond to it! I was tempted to delete the entire thread.
     
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