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Homework Help: Chain with 2 links lifted vertically - Force

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data

    A chain consisting of two links, each of mass 0.5 kg, is lifted vertically with an acceleration of 3.0 m/s2 upward.

    The magnitude of the downward force exerted on the top link by the bottom link is?

    2. Relevant equations

    F = ma

    3. The attempt at a solution

    1 is the top link
    2 is the bottom link
    F1,2 = The force on 1 by 2

    The Fnet on either of the two links is (0.5kg)(3m/s2) = 1.5N


    Fnet = Fpull - Fg - F1,2

    Fpull = 1kg(3m/s2) = 3N


    1.5N = 3N - (.5kg)(9.8m/s2) - F1,2

    F1,2 = 3.4 N downward

    The answer is 6.4N but i dont understand why? where did i go wrong?

    this is exactly what i did on a homework assignment and got it right but this is what i did on the test and i got it wrong since the answer is 6.4N not 3.4N like i got.

    Thanks for any help
  2. jcsd
  3. May 8, 2010 #2


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    You didn't calculate F_pull correctly. Where's the weight force? You might want to check your new result by drawing a FBD of the bottom link.
  4. May 8, 2010 #3
    I dont see how F_pull is wrong?

    The only other thing i can think of doing to calculate this is to factor in the the acceleration of gravity?

    and by weight force do you mean Fg cause i have that in there....
  5. May 8, 2010 #4
    What i am really confused about is i used this same method on a mastering physics problem and got all of the answers right so i guess i got lucky on that. I really need to figure out how to do this right because im sure something similar to it is going to be on my final.
  6. May 8, 2010 #5


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    You have got to draw free body diagrams to solve these problems. You noted that
    But F-pull is the pulling force which must be greater than the weight of the links in order to accelerate them upward at 3m/s/s. Since the links weigh about 9.8 N, the pull force can't be just 3 N.
    Again, you can also calculate F_1,2 directy by drawing a FBD of the botton link alone.
  7. May 8, 2010 #6
    oh that makes sense.

    ok so using my fbd of the botton link i have the force pulling it up which i believe is (3+9.8)(1) = 12.8 N in the upward direction

    there is also the F_2,1 force in the upward direction (this is equal to F_1,2)

    then there is the force of gravity of link 2 going downwards.

    Does that seem right?
  8. May 8, 2010 #7


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    No. If you draw a FBD of the bottom link, there are only two forces acting on it, its weight down, and the F-2,1 force acting up. Use newton 2 to solve for F_2,1. Don't put in forces that don't exist when you draw your free body diagram.
  9. May 8, 2010 #8
    Oh jeez i feel stupid. That was a lot easier then i was making it.

    so sum of the forces in the y = ma

    Fy = 0.5kg(9.8+3)

    Fy = 6.4 N

    Thank you Jay
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