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Center of mass of concrete wall

  1. Feb 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A cylin drical con crete sil o is 4m in diameter and 30m high. it consists of a 6000kg concreet base and a 38000kg concrete cylindrical concreete walls. locate the center of mass of the silo (A) when it's empty and (B) when it is two thrids full of silage whose density is 800kg/m^3. neglect the thickness of the silo walls and base.


    2. Relevant equations



    3. The attempt at a solution
    Part A

    6000kg*y = 38000kg (.5(30m) - y)
    y = 12.95 m
    or
    Center of mass is the y direction is 13m high and in the middle of the silo



    Part B

    Volume = 2m^2*PI*20m = 251.327 m^3

    Mass = 251.327 m^3 * 800 kg / m^3

    = 201062 kg

    h of grain = 20m

    201062kg*y = 38000kg*(20m -y)

    6.29*y = 20m

    y= 3.17m



    the real question is part B i dont think that makes sense...
     
  2. jcsd
  3. Feb 29, 2008 #2

    tiny-tim

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    Always draw a picture!

    This line is wrong.

    Always draw a picture so that you don't get confused. :smile:

    You'll see you have the (20m - y) in the wrong place.

    And you've left out the base, for which you need to use Part A's answer.
     
  4. Feb 29, 2008 #3
    i have a picture, i usualy upload a paint picture for you guys, but didnt this time, and the x direction is obvious, for x direction, Center of mass is the y direction is 13m high and in the x direction it is in the middle of the silo (1/2 the diameter)
     
  5. Feb 29, 2008 #4
    [​IMG]



    6000+(201062kg*(20*.5-y)) = 38000kg*(.5*30m - y)

    y= 8.87m



    i like that answer a lot

    ycm = 8.87 m
    xcm = 2m (or right in the middle however you look at it)
     
    Last edited: Feb 29, 2008
  6. Feb 29, 2008 #5
    anyone verify this?
     
  7. Mar 1, 2008 #6

    tiny-tim

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    don't do a balancing equation - choose a baseline

    Hi nightshade!

    (oops! I just meant draw a picture for yourself … there's no need to draw one for the forum if the question is clearly worded. :redface:)

    Your picture is very artistic, but you really need to spoil it (!) by marking the positions of the centres of gravity of the base the walls and the silage, with the weights of each next to them.

    The advantage in this case is that you would have seen that the c. of g. of the silage is at 10m, and (from Part A) that of the rest is at 12.95m - so the overall c. of g. must be somewhere in between!

    Have you been taught always to do a "balancing" equation about the expected position, y?

    If so, I suppose you'd better carry on doing what the teacher says, only more carefully.

    But if not, stop doing it, especially when you have more than two items to balance - there's too much risk of putting one on the wrong side - which is what you've done! :smile:

    Choose whatever baseline which makes the calculation easiest - in this case, choose the level of the base, for two reasons:
    (a) it makes one of the items zero, which always shortens the calculation!

    (b) height above the base is the answer you want anyway.

    For example, the calculation for Part A would be y = 38000*15/(38000 + 6000) = 12.95, as you said :smile:. Isn't that both easier and more intuitive? :cool:

    Now try it for Part B!
     
  8. Mar 1, 2008 #7
    38000*15 + 201062*10 / (3800+6000+201062)

    10.53m

    thankyou!
     
    Last edited: Mar 1, 2008
  9. Mar 1, 2008 #8

    tiny-tim

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    You're very welcome! :smile:

    [size=-2](don't forget to mark thread "solved"!)[/size]​
     
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