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Two blocks on a tabletop

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Block A (mass 2.25 kg) rests on a tabletop. It is connected by a horizontal cord passing over a frictionless pulley to a hanging block B (mass 1.30 kg.) The coefficient of kinetic friction between block A and the table is 0.450. After the blocks are released from rest, find a. the speed of each block after moving 3.00 cm and b. the tension of the cord.


    2. Relevant equations



    3. The attempt at a solution
    Block A:
    Fgravity: (2.25kg)(9.8m/s2) = 22.05N
    Fnormal: 22.05N
    Ffriction: (22.05N)(0.450) = 9.92N
    Resultant force: 22.05N - 9.92N = 12.13N

    Block B:
    Fgravity: (1.30kg)(9.8m/s2) = 12.74N

    I have no idea what to do after this. I found the exact same problem through a search but I don't understand the explanation.
     
  2. jcsd
  3. Feb 15, 2010 #2

    tiny-tim

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    Hi cdotter! :smile:
    Sorry, but this is completely wrong :redface:

    you haven't used the tension, T, at all;

    and the 22.05 and the 9.92 are perpendicular, so you can't add (or subtract) them anyway.

    Try again. :smile:
     
  4. Feb 15, 2010 #3
    I now understand that the forces are perpendicular and that I can't subtract them, but how are my normal force and frictional force calculations incorrect?
     
  5. Feb 16, 2010 #4

    tiny-tim

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    Your normal force and frictional force were correct, but your "resultant force" equation was completely wrong.

    Try again, using T. :smile:
     
  6. Feb 17, 2010 #5
    Block B pulls down with 12.74N of force. Block A has a frictional force of 9.92N. Shouldn't the tension then be 12.74N+9.92N = 22.66N? My book says it's 11.7N. I have absolutely no idea where 11.7N comes from.
     
  7. Feb 17, 2010 #6

    tiny-tim

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    No, everything is accelerating, and you've left "a" out of your F = ma. :wink:
     
  8. Feb 17, 2010 #7
    Oh, but I don't understand how to put what I have into the equation? :confused:

    Block A: [itex]F_{fric}-T_{rope}=ma[/itex]
    Block B: [itex]F_{gravity}-T_{rope}=ma[/itex]

    Right?

    edit: I guess not. I'm so lost haha.
     
    Last edited: Feb 17, 2010
  9. Feb 17, 2010 #8

    tiny-tim

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    Start with the lower block …

    there are only two forces on it, and their total has to equal its mass times its acceleration. :smile:
     
  10. Feb 17, 2010 #9
    Could you show me what you mean because I have absolutely no idea what to do. It has two forces acting on it: weight pulling down and the tension of the rope pulling up. I don't know how to calculate those or put them into the equation.
     
  11. Feb 17, 2010 #10
    I figured it out by reading another post, thank you for the hints tiny-tim.
     
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