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3 Blocks Question

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Blocks with masses of 1 kg, 3 kg, and 6 kg are lined up in a row on a frictionless table. All three are pushed forward by a 17 N force applied to the 1 kg block.
    (a) How much force does the 3 kg block exert on the 6 kg block?
    (b) How much force does the 3 kg block exert on the 1 kg block?

    2. Relevant equations
    F=MA


    3. The attempt at a solution
    (a) 17N=1kg(a)
    17=a

    3kg+1kg=4kg 1+3+6=10 170-68=102
    F=4kg(17) F=10kg(17)
    68=F 170=F

    Completely missing this one I need some direction please!!!
     
  2. jcsd
  3. Sep 29, 2009 #2

    Delphi51

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    This isn't correct because the 1 kg mass pushes against the 3 kg mass, and by Newton's 3rd law, it pushes BACK on the 1 kg mass. So the acceleration is reduced.

    You really have to consider all the masses together. You could write an F = ma equation for each mass. The first one would be
    17 - F1 = 1a
    where F1 is the force that the 2nd mass pushes on the first, and the first pushes on the second. Keep going on this and you'll have a set of three F = ma equations with unknowns a, F1 and F2. The two forces are the answers to the two parts of the question.

    If you just want to find the acceleration, you could think of the 17 N being applied to the combined mass of all three blocks.
     
  4. Sep 30, 2009 #3
    thanks for the help, i am having trouble following what you are saying could u give me a starting point so I can go from there and finish the problem. Physics is not my cup of tea I can't visualize this in my head at all!
     
  5. Sep 30, 2009 #4

    Delphi51

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    Start with the 17 N acting on a mass of 1+3+6 and find the acceleration.

    Next, consider the 1 kg block pushing on the 3 + 6 kg blocks.
    What force must be acting on the 3+6 to make it accelerate with the value you found first?
     
  6. Sep 30, 2009 #5
    ok, so I did 17N = 10kg(a). Therefor, a must equal 1.7 correct?

    Then, I plugged 1.7 into F=4kg(1.7) which gave me 6.8. I thought this would be correct for portion (a) but it wasn't. I did not attempt the second part of the question because this was wrong. Does this mean that only the 3kg force is the answer, bc that would be 5.1.

    That does not make sense to me though because wouldn't the 3kg force also include the force that the 1kg block is acting on that? Im really sorry but I just cannot understand why Im not getting the right answer physics is my weakness and I really am trying.
     
  7. Sep 30, 2009 #6

    Delphi51

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    The 1.7 m/s^2 is correct - good start!

    Next, consider the 1 kg block pushing on the 3 + 6 kg blocks.
    What force must be acting on the 3+6 to make it accelerate with 1.7 m/s^2?
    This is the force of the 1 kg on the 3 kg; and of course the force of the 3 on the 1 which is question (b).
     
  8. Sep 30, 2009 #7
    OK, so that would mean F=9kg(1.7m/s2)... this gives an answer of 15.3 N. So if block 1 must be using this force, how do you take that force and figure out what the 3 must be using on the 6?
     
  9. Sep 30, 2009 #8
    Would it be F=6kg(1.7), because that would give you the force that must be used in order to get the 6kg block to accelerate at 1.7 thus what the 3 kg block must be pushing with?
     
  10. Sep 30, 2009 #9

    Delphi51

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    I think you've got it!
     
  11. Sep 30, 2009 #10
    Awesome, thank you so much delphi you are awesome!!
     
  12. Sep 30, 2009 #11

    Delphi51

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    Most welcome!
     
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