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Homework Help: One chain-link pulls another

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A student tries to raise a chain consisting of three identical links. Each link has a mass of 200 g. The three-piece chain is connected to light string and then suspended vertically, with the student holding the upper end of the string and pulling upward. Because of the student's pull, an upward force 15.0 N is applied to the chain by the string. Find the force exerted by the top link on the middle link.
    A) 3.0 N
    B) 6.0 N
    C) 8.0 N
    D) 10.0 N
    E) None of the above

    2. Relevant equations

    3. The attempt at a solution
    Due to the force, the chain accelerates upwards at a=15/0.6=25 m/s^2.
    Let A, B, C be the top, middle and bottom link respectively.
    Apply Newton's second law twice to C and B, we have
    [tex]F_{BC} - mg = ma[/tex]
    hence [tex]F_{BC} = 7.0 N[/tex]

    [tex]F_{AB} - F_{CB} - mg = ma[/tex]
    but [tex]F_{CB} = F_{BC} [/tex]
    thus [tex]F_{AB} = 14.0 N[/tex]
    There are no keys available so I really need a clarification to my answer. Thanks!
  2. jcsd
  3. Feb 26, 2012 #2
    You have computed the acceleration correctly. But after that you went awry. At the bottom of the top link you have 0.4 kg suspended and accelerating upward. Draw a FBD of the bottom of the top link and look at the forces. There are three. One is the upward pull that you seek.
  4. Feb 26, 2012 #3
    No, the acceleration isn't correct.
    NET FORCE = ma
    You missed out the weight of the system.
  5. Feb 26, 2012 #4
    But I can only figure out 2 forces. One is the upward force exerted by the top link and the other is the weight of the two lower links. Am I missing something?
  6. Feb 26, 2012 #5
    Its own weight
  7. Feb 26, 2012 #6
    Then the required pull is (10)(0.6)+(25)(0.4)=16.0 N, right?
  8. Feb 26, 2012 #7
    The acceleration of the system isn't 25m/s^2 because you missed out the weight of the system which is 6N.
  9. Feb 26, 2012 #8
    Ah I see, it should have been 15 m/s^2, then F=(10)(0.6)+(15)(0.4)=12.0 N. Does it seem fine now?
  10. Feb 26, 2012 #9
    The weight of link A (2N) is still missing
  11. Feb 26, 2012 #10
    Whoops, it was a little too early in the morning when I typed this. My first sentence was incorrect.
  12. Feb 26, 2012 #11
    Do I need to include this in calculation? I am considering the two lower links (B and C) accelerating under the 3 forces.
    I'm getting confused. Is the force acting on (B+C) by A equals the force acting on B by A? What if the chain now consists of m identical links and we are to find the force acting on the nth link by the (n-1)th link (n<=m)?
  13. Feb 26, 2012 #12
    If you are considering B and C, then it should not be 15 N there.
  14. Feb 26, 2012 #13
    Nah, it's not 15 N, it's the acceleration of (B+C)
  15. Feb 26, 2012 #14
    Oh I see, sorry.
    The net force acting on B+C is the vector sum of the tension in the mid string (answer required) and the weight of B+C, and this net force = ma = (0.2+0.2)(15)
    You should get the correct answer by now. :wink:
  16. Feb 26, 2012 #15
    So if we consider the motion of (B+C), then there are only 2 forces acting on it,
    which result in an acceleration of 15 m/s^2.
    Thank you for clearing it up for me :D
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