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Friction Clarification Question

  1. Jul 27, 2006 #1
    I think I'm just getting thrown off by this question because there's unnecessary information in it, but here it is:

    "A 1.0 kg brick is pushed against a vertical wall by a horizontal force of 24 N. If μs = 0.80 and μk = 0.70 what is the acceleration of the brick?"

    Is it just 0.0 m/s^2 because it's pushed up against the wall, or does it require some sort of calculation?
     
  2. jcsd
  3. Jul 27, 2006 #2

    Hootenanny

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    I would agree with you there, in my opinion the brick is stationary and not accelerating since the vertical wall will provide a normal force equal in magnitude but opposite in direction to the applied horizontal force leading to a net force of zero. The majority of the information is superfluous.
     
  4. Jul 27, 2006 #3

    Doc Al

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    They are asking about the vertical acceleration, not the horizontal. So you need a little calculation. :wink:

    (The information given is not superfluous.)
     
  5. Jul 27, 2006 #4

    Hootenanny

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    Ahhh! The block is held against the wall.. makes sense. In my mind I had the brick resting on the floor and being pushed against the wall. My bad :blushing:
     
  6. Jul 27, 2006 #5
    Thanks (again)! While I'm here, I have another more difficult (at least for me) problem which I'm having trouble phrasing my exact question to:

    "A block of mass m sits on top of a larger block of mass 2m, which in turn sits on a flat, frictionless table. The coefficient of static friction between the two blocks is μ_s.What is the largest possible horizontal acceleration you can give the bottom block without the top block slipping?"
    (a) μ_sg/2
    (b) μ_sg
    (c) 2μ_sg
    ---
    I'm having trouble combining Newton's second law and fs,max = μ_sF_N
     
  7. Jul 27, 2006 #6

    nrqed

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    You can't tell without doing a calculation. The brick *might* be at rest or it *might* be sliding down with some acceleration.

    Drwa a free-body diagram (*always* the first step in analyzing a problem like this). What you have to check is whether the static friction force can be strong enough to overcome the weight or not.
     
  8. Jul 27, 2006 #7

    Hootenanny

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    What happens if a horizontal force greater than the maximum frictional force (between the two blocks) to the larger block?

    Take note of Doc Al's comments above
     
  9. Jul 27, 2006 #8

    nrqed

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    Again, draw a free body diagram of the top mass (this is always the first step in such a problem. Without a clear FBD in you rmind, things will never be completely clear. and often, just drawing a FBD will make you realize how to solve a problem). Now, what is the *maximum* static friction force that the top mass may experience? Using this, what is the maximum acceleration it may have without slipping?
     
  10. Jul 27, 2006 #9

    Hootenanny

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    I'm gona duck out of this thread save all these double posts. :smile:
     
  11. Jul 27, 2006 #10
    For my first question:

    So if the brick's mass is 1.0 kg, then it's weight is 9.8 N
    The horizontal force is 24 N...and that's where I lose it
     
  12. Jul 27, 2006 #11
    Ohh...so using F_N as 24 N and multiplying it by μ_s (0.80) gives fs,max to be 19.2, which is enough to overcome the weight of 9.8 N.

    Right?
     
  13. Jul 27, 2006 #12

    Hootenanny

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    Okay I'm back in again :biggrin: . Yes that looks good to me, therefore the acceleration of the brick is ...
     
  14. Jul 27, 2006 #13
    It's 0.0 m/s^2 because it's not moving and its velocity is constant (0.0 m/s)

    Now, on to my second question...give me a minute :smile:
     
  15. Jul 27, 2006 #14
    Maxiumum static friction force:
    Okay, so in the FBD, the forces acting upon the top mass are: mg from the top and F_N from the bottom. F_N = mg, so:

    fs,max = μ_sF_N = (μ_s)(mg)

    ...right so far?
     
  16. Jul 27, 2006 #15

    Hootenanny

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    Looks good to me :smile:
     
  17. Jul 27, 2006 #16
    Okay, so for the maximum acceleration it can have without slipping...
    I honestly have no clue :grumpy:
     
  18. Jul 27, 2006 #17

    Hootenanny

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    HINT: [itex]F = ma \Rightarrow \mu_{s}mg = ma[/itex]
     
  19. Jul 27, 2006 #18
    Ohh, so if you solve for "a", the masses cancel out and the answer is (μ_s)(g). So nowhere does it take into account the differences in masses of the two blocks?
     
  20. Jul 27, 2006 #19

    Hootenanny

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    Well if [itex]\mu_{s}mg[/itex] is the maximum force that the bottom block (2m) can apply to the top block (m) via friction without it slipping. What is the maximum force one could apply to the bottom box without the top box slipping?
     
  21. Jul 27, 2006 #20
    Twice that, because it's twice as heavy? Honestly, I'm lost :frown: I really appreciate your persistent help, though!!!
     
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