1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    so

    Fnet = Fpull - Fg - F1,2

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

    so

    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

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook