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Chain Link Problem? [Tension Forces; Newton's Third Law]

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data

    A chain of three links, each with a mass 0.2 kg, is being pulled up by a person lifting the top link with 8.88 N of force and the chain accelerates upward. Calculate three forces that are acting on the middle link while the chain is accelerating.

    2. Relevant equations

    ƩF = ma
    Fg (weight) = mg

    3. The attempt at a solution

    I drew a free body diagram showing tension force pulling up by 8.88 N and the weight, Fg, pulling the chain down. I calculated Fg = mTg (total mass of chain) = (0.6 kg)(9.8m/s^2) = 5.88 N.
    By the way, is it right for me to assume the mass of the middle link = total mass of the chain (0.2 kg*3 = 0.6 kg)?

    So, I have found two forces, I believe... and I'm not too sure on what the third force would be.
    I've learned about the internal forces of each object if they are acting on one another but I'm not sure if that's it, or how to calculate it.

    Oh, and I found acceleration by a = ƩF/m = 3 N/0.6 kg = 5 m/s^2, if what I have done above is correct.
     
    Last edited: Oct 27, 2013
  2. jcsd
  3. Oct 27, 2013 #2

    tiny-tim

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    hi harujina! :smile:
    no

    a free body diagram shows all the external forces acting (directly) on one body, and that body is the middle link only
    no, the weight (as an external force for the free body diagram) is the external force acting directly on the link, ie the weight of the link only
    internal forces on one body can be external forces on a smaller body

    a free body diagram for a large body will not show the forces between the parts, because they are internal

    a free body diagram for one part will show the forces between the parts, because they are external :wink:
     
  4. Oct 27, 2013 #3
    hi :)

    I remember my teacher taught us that you could draw a fbd of combined objects in order to determine acceleration, like what I have done above. Is this incorrect?

    Ok, so before I adjust anything I want to make sure I understand this because it's confusing me a lot. A fbd for a large body will not show forces between the parts, but in this case aren't the chains all equal in mass? So would the tension force just equally divide amongst the three links?
     
  5. Oct 27, 2013 #4

    haruspex

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    That's fine, and you got the right acceleration, but next you need an FBD for the middle link only.
    Did you leave something out there?
     
  6. Oct 27, 2013 #5
    Wait, would applied force on the middle link be 8.88 N/3 (the three links) = 2.96 N?
    and mass of the middle link would just be 0.2 kg?

    On another note, thank you so much for your constant help, haruspex! I really appreciate it.
     
  7. Oct 27, 2013 #6

    tiny-tim

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    hi harujina! :smile:
    where does the 3N come from? :confused:
    that's correct: you can draw a free body diagram for a large body (and it will not show t the internal forces)

    however, the question asks "Calculate three forces that are acting on the middle link while the chain is accelerating", so you need a free body diagram for the middle link on its own

    no

    if you draw a proper free body diagram for the middle link, you'll see why

    have you drawn it? what does it show?
     
  8. Oct 27, 2013 #7

    haruspex

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    You are asked for three forces. Draw the FBD. What three forces are there?
    (... and you're welcome :biggrin:)
     
  9. Oct 27, 2013 #8
    I calculated total force to be 3N by ƩF = Fa - Fg = 8.88 N - 5.88 N.

    Ok, to answer the question I will use link A = top link, link B = middle link, and link C = bottom link.
    As for the fbd of the link B, I don't know if this is right at all but would it be...
    Fa (applied force of 8.88 N; pulling up), aFb (force exerted by link A on link B; also pulling up) and cFb (force exerted by link C on link B; pushing down)?

    [EDIT]: Wait, but there's also Fg (weight; pulling down) isn't there :confused:
     
    Last edited: Oct 27, 2013
  10. Oct 27, 2013 #9

    haruspex

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    How is B aware of the 8.88N force?
     
  11. Oct 27, 2013 #10

    tiny-tim

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    oh of course … yes, that's the correct acceleration, then :smile:

    "Fa (applied force of 8.88 N; pulling up)" is wrong: it's not a direct force on the middle link, so it doesn't get mentioned at all :wink:

    Fa would be shown on an fbd for the top link, or for the whole body of three links, but not for the middle link (or the bottom link)

    and yes, an Fg should be on the fbd for the middle link …

    how much should it be?​
     
  12. Oct 27, 2013 #11
    Oh, I'm starting to get it now.
    Fg (of middle link)= mg; = (0.2 kg)(9.8m/s^2) = 1.96 N; correct?
    So now if I want to determine aFb and cFb, would I have to draw fbd's for the top and bottom link as well?
     
    Last edited: Oct 27, 2013
  13. Oct 27, 2013 #12

    tiny-tim

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    correct :smile:
    yup! :biggrin:

    hint: start with the bottom one … it's easier! :wink:
     
  14. Oct 27, 2013 #13
    Thank you so much! Can you just see if I'm on the right track?
    So, the forces acting on the bottom link are Fg and bFc (force exerted by middle link on bottom link; pulling up). The ƩF = (0.2 kg)(5 m/s^2) = 1N, so cFb (force exerted by bottom link on middle link; pulling down) = 1N?

    And this is what I got for the top link:
    ƩF = Fa (applied force of 8.88 N) - Fg (1.96 N) - bFa (force exerted by middle link on top link; pulling down)
    bFa = Fa - Fg - ƩF
    bFa = (8.88 N) - (1.96 N) - (1 N); = 5.92 N, so aFb = 5.92 N?
     
    Last edited: Oct 27, 2013
  15. Oct 27, 2013 #14

    haruspex

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    You forgot Fg.
    Yes, that looks right.
     
  16. Oct 28, 2013 #15
    Whoops...

    ƩF = (0.2 kg)(5 m/s^2) = 1N
    ƩF = bFc - Fg; = Fg + ƩF; = (1.96 N) + (1 N) = 2.96 N so cFb (force exerted by bottom link on middle link; pulling down) = 2.96 N
    Is this ok?
     
  17. Oct 28, 2013 #16

    haruspex

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    Yes.
     
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