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Friction force in Newton's law

  1. Oct 7, 2011 #1
    A block of mass m1 = 4 kg is put on top of a block of mass m2 = 5 kg. To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table. Find the magnitude of
    a. The max. horizontal force F that can be applied to the lower block so that the block will move together
    b. The resulting acceleration of the blocks.

    Ok, so I'm not here to ask for how to do it, not yet ;). I'm here to ask if any one here could explain the problem to me, especially the one that is bold. I have no idea what they are talking about.....Thanks
     
  2. jcsd
  3. Oct 7, 2011 #2
    If I understood it corretly, it basically means this:

    the minimum force applied to the block in order to "win" against static friction - which is holding the block in its place - is 12N.
    To help visualize things: just forget about the bottom block.
    Pretend the block you want to move is on the ground. Can you understand why the minimum force is 12N?
     
  4. Oct 7, 2011 #3
    Yes, now I got it, so basically if it's put on the ground, 12N is the minimum force that you need to apply in order to move it. So fs maximum will be 12 N, right ? and since we have fs = us * mg = 12N, we could find coefficient of static friction....uhm, or coefficient of kinetic friction ???? wasn't it called uk when its moving and us when its not ??? I got confused again.... :(((
     
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