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Friction Force Question

  1. Jan 23, 2017 #1
    1. The problem statement, all variables and given/known data
    Two blocks,in contact,slide down an inclined plane ##AC## of inclination ##30^°##,The coefficient of kinetic friction between the ##2.0 kg## block and the incline is ##μ_1=0.20## and that between the ##4.0kg## block and the incline is ##μ_2=0.30##.

    Find the magnitude of the acceleration.
    2. Relevant equations
    ##\vec {F_t }=m\vec a##
    ##F_f=μN##
    ##w=mg##

    3. The attempt at a solution
    Ok I need to draw free body diagram,after that I can solve the question I guess.I stucked in the free body diagram,I dont know its true or not.Theres pic in the attached files.

    Is there will be any action-reaction force pairs ?
     

    Attached Files:

  2. jcsd
  3. Jan 23, 2017 #2

    Doc Al

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    Staff: Mentor

    If you draw individual free body diagrams, don't forget the forces the blocks exert on each other. (Those would be 3rd law pairs.)

    But why not treat the two blocks as a single "system"?
     
  4. Jan 23, 2017 #3
    When will there be action-reaction pairs? What do you think?
     
  5. Jan 23, 2017 #4
    They have differenet friction constant and ##N## are different This 2 body diagrams seem simpler ( maybe not but for me yes )

    If an object exerts a force on something.That something should exert a force on that object

    Well,If ##2.0 kg## object exerts a force on ##4.0kg## object. which its I guess, cause theres ##mgsin30## (component of ##2.0kg## block) exerts a force on 4 kg block so there will be action-reaction force pair ? Is it true ? the magnitude of it ? Or the idea ?
     
  6. Jan 23, 2017 #5

    Doc Al

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    Sure. So what? (You can include the two friction forces acting on the blocks.)

    Using separate diagrams is perfectly fine! But then you must include the unknown contact force that the blocks exert on each other.
     
  7. Jan 23, 2017 #6
    I dont know how to think them as a single object..

    I wanna write down my solution ( I ll add in a minute )
     
  8. Jan 23, 2017 #7

    Doc Al

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    Staff: Mentor

    Since they move together, it's easy to treat them as a single system. But you don't have to.

    (Think about it: If a single block were sliding down the incline you could -- mentally -- imagine the block divided into two pieces and treat each piece separately. You'll get the same answer as usual, though.)
     
  9. Jan 23, 2017 #8
    Parallel to inclined plane is x and and the perpendicular to that axis is y.Down is +x and up is +y

    ##m_2a=m_2gsin(30^°)-μ_2m_2gcos(30^°)+F_{12}##
    ##m_1a=m_1gsin(30^°)-μ_1m_2gcos(30^°)-F_{21}##

    Since
    ##F_{12}=-F_{21}##

    lets subtract them.

    ##a(m_2-m_1)=gsin(30^°)(m_2-m_1)+gcos(30^°)(m_1μ_!-μ_2m_2)##
     
  10. Jan 23, 2017 #9
    I guess I understand
     
  11. Jan 23, 2017 #10

    Doc Al

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    Careful with signs.

    Let's let ##F_{12}## stand for the magnitude of the contact force. In the first equation, it will appear as ##+ F_{12}##; in the second, as ##- F_{12}##.
     
  12. Jan 23, 2017 #11
    But ##F_{12}## in the +x direction and ##F_{21}## in the -x direction. thats why theres "-" sign
     
  13. Jan 23, 2017 #12
    This approach though only works for this question. I think we should check first whether they do move together?
     
  14. Jan 23, 2017 #13
    I thought that too but the question asks one acceleration so they must have same acceleration I guess.
     
  15. Jan 23, 2017 #14

    Doc Al

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    You are counting the sign twice, which is why your final equation is incorrect.

    What you meant to do was use ##+ F_{12}## in the first equation and ##+ F_{21}## in the second equation. Then, using ##F_{12} = - F_{21}## would work out fine.
     
  16. Jan 23, 2017 #15
    What you're doing now (along with Doc Al's correction) is correct. For a different question, I think it's better to check first.
     
  17. Jan 23, 2017 #16
    I am writing ##\vec F_{21}=F_{21} (-i)## but actually it is ##\vec F_{21}=F_{12} (-i)## my mistake I couldnt notice this.I solved and answer is correct thanks
     
  18. Jan 23, 2017 #17
    So is it always ##\vec F_{21}=F_{12} (-i)## like this right but always (of course if ##\vec F_{12}=F_{12} (i)##
     
  19. Jan 23, 2017 #18

    Doc Al

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    I would say it this way: ##\vec{F}_{12} = - \vec{F}_{21}##. The directions are opposite but the magnitudes are equal.

    And note that when you combine the two equations correctly, you'll get an equation for the two-block system. (The masses will add.)
     
  20. Jan 23, 2017 #19
    Newtons Third Law yeah...

    I noticed two-block system equation when you say right know.Thanks again

    Problem solved.I have one more question and I ll open another thread soon If you also help me in that I ll happy
     
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