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Can someone tell me how many forces of friction are in here?

  1. Apr 3, 2016 #1
    Ch085.gif

    Let's assume that the force of friction is only present between the two blocks and that the pulley is massless. On top of that, force F pulling the two blocks at a constant velocity; as a result, the acceleration = 0m/s2.
    May someone tell me how many forces of friction are going on between the "2kg" and "7kg" blocks? I guessed two since the top block is moving left and the bottom box is moving right. Am I right?

    And is it true that the force of gravity that applies on the "7kg" block is actually (2kg+7kg)*g = 9kg*g?

    Also may you please check if my sum of all forces are corrrect?:
    ΣFtop blockx: -T1 + μ*2kg*g = [2kg*0] -> T1 = μ*2kg*g
    ΣFtop blocky: Ntop = 2kg*g

    ΣFbottom blockx: T2 - μ*9kg*g - T1 + F = [9kg*0] -> T2 = μ*9kg*g + 1 - F
    ΣFbottom blocky: Ntop = 9kg*g

    note: sorry but I don't know if I should still multiply kg * gravity or just leave it as kg.
     
  2. jcsd
  3. Apr 3, 2016 #2

    haruspex

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    Action and reaction are equal and opposite. If the normal force experienced by the 2kg block from the 7kg block is 2kg * g then what is the normal force the 7kg block experiences from the 2kg block?
     
  4. Apr 3, 2016 #3
    Hello again :). The normal force the 7kg block experiences from the 2kg block is 2kg * g because of Newton's 3rd law as you mentioned above.
     
  5. Apr 3, 2016 #4

    haruspex

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    Right, so what is the frictional force on the 7kg block?
     
  6. Apr 3, 2016 #5
    So... therefore, "ΣFbottom block y" Nbot should be subtracted to Ntop as it is below where Nbot = 2kg*g ?
    ΣFtop blockx: -T1 + μ*2kg*g = [2kg*0] -> T1 = μ*2kg*g
    ΣFtop blocky: Ntop = 2kg*g

    ΣFbottom blockx: T2 - μ*9kg*g - T1 + F = [9kg*0] -> T2 = μ*9kg*g + 1 - F
    ΣFbottom blocky: Ntop - Nbot = 9kg*g
     
  7. Apr 3, 2016 #6

    haruspex

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    Where does that μ*9kg*g term come from?
     
  8. Apr 3, 2016 #7
    The force of friction on the 7kg block should feel μ*2kg*g.
     
  9. Apr 3, 2016 #8
    Well I thought that since the 2kg block is on top of the 7kg block, then the force of gravity for the 7kg block would be (2kg+7kg)*g "the sum of both masses times gravity".
     
  10. Apr 3, 2016 #9

    haruspex

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    Sure, but where will that total weight be felt? You already accepted in post #3 that the normal force at the boundary between the blocks is only 2kg * g.
     
  11. Apr 3, 2016 #10
    I figured the "9kg * g" force would be felt from block 7kg to the ground and back to up at it again through the Nbot force.
     
  12. Apr 3, 2016 #11

    haruspex

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    Sure, but there's no friction there.
     
  13. Apr 3, 2016 #12
    Hang on let me double check.
     
  14. Apr 3, 2016 #13
    ΣFbottom blockx: T2 - μ*2kg*g - T1 + F = [2kg*0] -> T2 = μ*2kg*g + 1 - F
    ΣFbottom blocky: Ntop - Nbot = 7kg*g

    I think I've fixed it.
     
  15. Apr 3, 2016 #14

    haruspex

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    You have T1 and T2. There's only one tension acting on the bottom block (other than F).
     
  16. Apr 3, 2016 #15
    So I don't need T2 since there's already a force F pulling on it?

    since T2 = F it will just cancel out?
     
  17. Apr 3, 2016 #16

    haruspex

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    It can't cancel. Either take F as acting directly on the block, or take T2 as acting on the block and set T2=F. It comes to the same thing.
     
  18. Apr 3, 2016 #17
    Alright. I just thought that force F was acting on the rope since it's pulling it but thank you, if the bottom equation looks correct. You've been a great help.


    ΣFbottom blockx: μ*2kg*g - T1 + F = [2kg*0] -> F = μ*2kg*g + 1 + T1
    ΣFbottom blocky: Ntop - Nbot = 7kg*g
     
  19. Apr 3, 2016 #18

    haruspex

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    Yes, that's right.
    If you want to think of F as a force someone is applying to a rope, the tension in the rope is F. The block feels the tension F, but it knows nothing about the person pulling on the other end. When constructing the FBD for a body, it's important to think in terms of the forces that act directly on the body, and not get mixed up with more distant origins of the forces.
     
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