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Question about the Normal Force

  1. May 27, 2012 #1
    Two buckets of nails are hung one above the other and are pulled up to a roof by a rope. Each bucket has a mass of 5.0kg. The tension in the rope connecting the buckets is 60 N. Calculate the acceleration of the buckets.

    So here's what I did:

    mt= ma + mb (m which is equal to mass)
    mt= 5.0kg + 5.0kg
    mt= 10.0 kg (total mass is 10.0kg)

    Fnet= Fn + Fg
    ma= Fn + mg

    if i isolate the a it would be:

    a= (Fn/m) + g

    How do I find the Normal force to solve the question?
  2. jcsd
  3. May 28, 2012 #2


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    Clue: forget about the upper bucket and just think about the forces and acceleration of the lower bucket.
  4. May 28, 2012 #3
    still dont get it please elaborate it
  5. May 28, 2012 #4
    Make a force equation of the lower bucket. Which are the forces acting on it? Due to the string(tension)? due to earth? how is the net force on it related to its acceleration, then?
  6. May 28, 2012 #5
    forcers that are acting on it are. Fn and Fg. I dont know what to do with the tension its a big question for me. Net force = ma
  7. May 28, 2012 #6
    Fn in your post #5, IS the tension in the string...You should probably call it FT

    Net force [tex]F_{net} = ma[/tex]

    But, what is [tex]F_{net}[/tex] in terms of Fn and Fg??

    remember, this is only for the lower bucket.
  8. May 28, 2012 #7
    Oh thanks for the tension thing.
    I think the Fnet is=T+mg ?
  9. May 29, 2012 #8
    But tension is acting upwards, mg is acting downwards, and Fnet, I'll leave to you. :wink:
  10. May 29, 2012 #9
    Oh ok now i got it, is this right? Fnet= T-mg
  11. May 29, 2012 #10
    Yep. :approve:

    Now use the relation with acceleration.
  12. May 29, 2012 #11
    a= Fnet/m Ok i used the 5 kg and I got the answer. But I have one quick question how come I didnt use the total mass?

    Thank You man ! :)

    But I could also do this

    when I isolate acceleration -->> a=Ft-(mg)/m Instead of a=Ft/m ( both correct)

    Thank You so much man :smile:
    Last edited: May 29, 2012
  13. May 29, 2012 #12
    Great! :biggrin:

    Why would you use something when you don't need it? :wink: Its simply not necessary to use the total mass for the situation you are given.

    That basically is using a = Fnet/m :wink: You just jumbled with terms to get there, in this answer.

    PS : Post such questions in homework section, next time!
    Last edited: May 29, 2012
  14. May 29, 2012 #13


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    The total mass is distributed across two buckets. If you want an equation involving the total mass then you'll need to treat the two buckets as a unit. So then you want the net force on that unit. That will be sum of the two gravitational forces, downwards, against the tension in the string above the top bucket acting upwards. But you are not told the tension in that part of the string, so it doesn't get you far.
    You can view the question as an exercise in figuring out which pieces of information are useful and which aren't.
  15. May 29, 2012 #14
    Thank You so much :smile:
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