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Homework Help: Find force link 3 exerts on link 2

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A chain consisting of five links, each of mass 0.105 kg, is lifted vertically with a constant acceleration of a = 3.5 m/s2. Find the magnitude of the force that link 3 exerts on link 2.

    2. Relevant equations
    F = ma

    3. The attempt at a solution
    I did:
    F = ma
    FU - (0.105 * 3)(9.8) = (0.105) * 3.5
    FU = 4.19 N

    where FU is the force pulling up, (0.105*3) is the mass of the link's 5, 4, and 3 weighing down on link 2 (mg=weight) and then the mass*accel on the other side.

    This is wrong. why? and how do i get the right answer?
  2. jcsd
  3. Oct 1, 2008 #2
    does no one know how to do it? ...if it helps. i found the force required to move the top link at that acceleration to be 6.98 N....at it was the right answer...
  4. Oct 2, 2008 #3


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    Science Advisor
    Homework Helper
    Gold Member

    It looks like your calculation and answer is correct, although you left out a 3 in an apparent typo omission (see above in red). Unless, the links are numbered 1 thru 5 bottom to top, rather than 1 thru 5 top to bottom??
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