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2 Blocks Stacked, Pullied, Bottom Accelerating, Newton's 3rd

  1. Mar 20, 2009 #1
    Unfortunately this is due today, I will go work on another problem and watch these boards.
    And its not a last minute thing, I have been working on this since it was assigned... :cry:

    1. The problem statement, all variables and given/known data

    The lower block in Figure P8.28 is pulled on by a rope with a tension force of F = 28 N. The coefficient of kinetic friction between the lower block and the surface is 0.34. The coefficient of kinetic friction between the lower block and the upper block is also 0.34. What is the acceleration of the 2.0 kg block?

    http://img27.imageshack.us/img27/6212/abaabab.gif [Broken]



    2. Relevant equations

    F=ma
    fkkn

    3. The attempt at a solution

    I drew out my free-body diagrams. Here are the formulas I gained from them (if they were right).

    m1 (Top Mass)

    Fx1=fk2 on 1-T2 on 1

    Fy1=n1-m1g

    m2 (Bottom Mass)

    Fx2=F-T1 on 2-fk1 on 2-fk2

    Fy2=n2-mg-n1 on 2

    etc etc etc...

    Question is, after I have formulas

    T2 on 1k(m2g+m1g)

    T1 on 2=F-μkm1g-μk(m2g+m1g)

    T2 on 1 is coming out to be 9.996N

    while

    T1 on 2 is coming out 14.672N

    Am I missing something? They are connected so would the tensions not be the same? I am thinking I have a concept mixed up or are they not supposed to be the same?

    Anyway I ask because I tried to use formula Fx2=m2ax2 and it gave me the wrong acceleration, so I am curious if anyone can see my problem?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 20, 2009 #2

    Doc Al

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    Staff: Mentor

    There's only one tension in the rope, of course. You need that assumption to solve for the acceleration and the tension, so I don't know how you managed to solve for the tensions.

    I suspect a sign error. Show your Newton's 2nd law equations for both masses. What is the relationship between the accelerations of the two masses?
     
  4. Mar 20, 2009 #3
    You have confirmed what I thought, I knew my tensions were wrong.. I'm going to look over things.

    Sorry, I am unsure of which part of the equation you are asking for, I don't typically hear parts of my problem described that way.

    I thought I included almost all of my work other than the free-bodies above?

    The two masses accelerations of course are bound to each other, if the bottom block moves at 1m/s^2, the top block will as well.
     
    Last edited: Mar 20, 2009
  5. Mar 20, 2009 #4
    AHHHHH!


    I was setting the F=ma=0

    As if my acceleration was zero to solve for the tensions..


    stupid stupid stupid stupid stupid!

    ok, i'm going to lay out my a=F/m and see what it looks like, i have a feeling its going to just confuse me even more since i see that the tensions are going to be in the middle of the acceleration equation and i'm still not going to know them
     
    Last edited: Mar 20, 2009
  6. Mar 20, 2009 #5

    Doc Al

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    D'oh!


    Write your equations in the form:
    ∑F = ma

    You'll have two unknowns and two equations--which works out fine. (Hint: Combine them so as to eliminate the tension.)

    Be careful with signs. Hint: Assume the bottom mass has an acceleration of a to the right.
     
  7. Mar 20, 2009 #6
    Ok my equations for acceleration are.


    ax1=(fk21-T21)/m1

    ax2=(F-T12-fk12-fk2)/m2

    Yet I am still unsure of the values of T12 and 21.

    I know that the accelerations both have to be the same.. I know that both tensions have to be the same.. could I create some sort of equation with (m1+m2)

    Sec, I'll try it and see if I come up with anything.. assuming it can be done? Idk. Just thinking..
     
  8. Mar 20, 2009 #7

    Doc Al

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    Please rewrite your equations in the form I suggested. Call the acceleration of the bottom block a. (Don't clutter your equations with uneeded subscripts.)

    For one thing, call the single tension by a single symbol: T.
     
  9. Mar 20, 2009 #8
    a=(fk21-T)/m1

    a=(F-T-fk12-fk2)/m2
     
  10. Mar 20, 2009 #9

    Doc Al

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    Write your equations in the form:
    ∑F = ma

    (not a = ∑F/m)

    And express those friction forces in terms of μ, m, and g.
     
  11. Mar 20, 2009 #10
    Fx1km1g-T

    Fx2=F-T-μk(m2g+m1g)-μk(m2g+m1g)
     
  12. Mar 20, 2009 #11

    Doc Al

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    What happened to the "ma" part?
     
  13. Mar 20, 2009 #12
    Fx1km1g-T=m1a

    Fx2=F-T-μk(m2g+m1g)-μk(m2g+m1g)=m2a

    Like that? I just left it as F before since they mean the same thing.
     
  14. Mar 20, 2009 #13

    Doc Al

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    Two comments:

    (1) Sign problem in your first equation. You are using a convention of right = positive, left = negative. Good! But if the acceleration of the bottom block is +a, what's the acceleration of the top block?

    (2) One of your friction forces in your second is incorrect. (Too many masses.)
     
  15. Mar 20, 2009 #14


    Fx1km1g-T=m1(-a)

    Fx2=F-T-μk(m1g)-μk(m2g+m1g)=m2a

    I see what you are saying, how is this?
     
  16. Mar 20, 2009 #15

    Doc Al

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    Perfect. So how can you combine those equations to eliminate T?
     
  17. Mar 20, 2009 #16
    I'll work this out really fast, thanks for your quick responses! I'll edit this post and add my formula once I get it.

    edit: Okay so T=am1km1g

    I plug this into the larger equation and solve for a?


    another edit: I tried what I mentioned above and came up with. Oh wait I think I may have messed up a sign..

    a=(F-μk(m2g+m1g))/(m2+m1m2)

    maybe it should be

    a=(F-2(μkm1g)-μk(m2g+m1g))/(m2+m1m2)

    2.268m/s2, that doesn't seem too outrageous.
     
    Last edited: Mar 20, 2009
  18. Mar 20, 2009 #17

    Doc Al

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    That will work. Another way to eliminate T is to subtract one equation from the other.


    Both are incorrect, but the second one is almost there. Your mistake is in the denominator.

    Also, simplify that expression a bit. Combine like terms.
     
  19. Mar 20, 2009 #18
    I had kept refreshing that other page and didn't see that it had scrolled over to the second page.. give me a minute to get it back out.

    edit: new equation, starting over i have

    a=((F-μkm1g-am1)/m2) + μkg

    edit 2:then solving for "a" i get..

    a=((F-μkm1g)/(m2(1+m1)) - μkg

    This equation gives 2.835 m/s2
     
    Last edited: Mar 20, 2009
  20. Mar 20, 2009 #19

    Doc Al

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    Still not quite right. When you get terms like 1 + m or m + m², you know something must be wrong since the units don't match.

    Try my idea: Subtract your initial two equations to eliminate T, then solve the resulting equation for a.
     
  21. Mar 20, 2009 #20
    This was a big mess, deleted and continued below.
     
    Last edited: Mar 20, 2009
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