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Forces and Pulleys problem! Help! due in the morning!

  1. Dec 14, 2004 #1
    Construction worker Fred, who has a mass of 96 kg, stands on a girder. He sees a sandbag of mass 33 kg sitting on the ground, a distance 12 m below. A massless rope tied to the sandbag runs up and over a frictionless pulley connected to the girder. "Aha!", says Fred, "I'll lower myself to the ground gently by grabbing the rope. I'll fall slowly to the ground while the sandbag rises up to the girder. Brilliant!"

    How fast will Fred be moving when he reaches the ground?

    Ok. Now, I assume velocity is negative, since the man is "falling"/moving down.

    To find acceleration down I thought it would be Force down over the summation of weight: i.e.
    (9.8 m/s^2 * 96kg) / (96 + 63) = 5.9169 m/s^2

    Then, I need his velocity when he reaches the ground, i.e.: V final.

    Vf^2 = Vi^2 + 2ad

    Vf^2 = 0^2 + 2 (5.9169 m/s^2)(12)
    Vf^2 = 142.0075
    Vf = 11.9166 m/s

    Since the man is using the pulley to move DOWN, the position is going negative and thus velocity must be negative (I think? :( )

    Vf = -11.9166 m/s

    The "online-homework grader" is saying my answer is wrong, can anyone tell me where I messed up and how to fix it?
  2. jcsd
  3. Dec 14, 2004 #2
    well, just be consistent in how you set up your axes. If you would like your velocity to be negative downwards, then so is gravity (which acts downwards too)

    Now, the other thing.. is that.. you said the sandbag is 33kg, but you wrote 63 in your equation.. did you make an arithmetic mistake, or is there more?
  4. Dec 15, 2004 #3
    oh my bad, the sandbag is 63 not 33, thats a typo :(

    I don't know what I did wrong or what to do though, do you know how this is suppose to work?
  5. Dec 15, 2004 #4
    Well, did you draw your Force diagram? Can you show it to us? Then maybe we can deal with the mathematics...
  6. Dec 15, 2004 #5
    uhm, thats hard to do with ASCII.....
    a pulley.

    | | 96kg man going DOWN, Gravity pushing both objects DOWN (obviously)
    |63kg sandbag going UP

    Height = 12m

    Thats about the best I can show
  7. Dec 15, 2004 #6
    I know how it looks like. I am asking you to draw a free-body diagram so you can use newton's second law to find the net force. You can draw it/scan it in and then post a link or use an attachment or something.
  8. Dec 15, 2004 #7
    ok is this what you meant?

    Attached Files:

  9. Dec 15, 2004 #8
    Ok, good. Now, we see that the net force is (mass of man - mass of sandbag) * gravity. Note that I subtracted the masses because the forces are in opposite directions.

    Now try again and see what you get :-)
  10. Dec 15, 2004 #9
    If I do that I get ( 96 kg - 63kg ) = 33 kg

    33 kg * 9.8m/s^2 = 323.4 N

    In order to solve for a velocity I need an acceleration don't I? so how can I go farther if I do it like this?
  11. Dec 15, 2004 #10
    You have a net force, and so you will have a net acceleration on the system as well.
  12. Dec 15, 2004 #11
    So net acceleration would be:

    NetForce = NetMass * NetAccel
    323.4 N = 159 kg * NetAccel
    324.4N/159kg = netaccel
    2.03396226415 m/s^2 = netaccel

    Vf^2 = vi^2 + 2ad

    Vf^2 = 0^2 + 2(netaccel)(12)

    Vf^2 = 48.8150943396
    Vf = 6.98677996932 m/s when hitting ground

    Velocity must be negative!

    So final answer is -6.98677996932 m/s, right? or did I do something wrong again?
  13. Dec 15, 2004 #12
    Again, please be consistent. If you have a positive Net Force, and it is pointing downwards, your velocity and acceleration vectors should also point downwards. (You used a positive net force downwards)...

    So you should have a positive velocity downwards. I won't check your numbers, but your concepts appear sound.
  14. Dec 15, 2004 #13
    Can anyone please confirm if my answer is correct? thanks!
  15. Dec 16, 2004 #14

    Andrew Mason

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    Science Advisor
    Homework Helper

    When Fred leaves the girder he has initial KE before the rope has any tension.
    It is much easier to do this by analysing the energy.

    [tex]\Delta PE = \Delta KE[/tex]

    [tex](M-m)gh = 1/2 (m+M) v^2[/tex]

    [tex]v^2 = 2(M-m)gh/(m+M)[/tex]

    h represents the change in height of Fred's center of mass (assuming he does not bend his knees when landing) which is the same as that of the sandbag.

    Last edited: Dec 16, 2004
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