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Topple and Centre of gravity

  1. Mar 7, 2008 #1
    Hi All,

    If I have block measuring 2m wide by 2m high and 3m long sat on the ground which has a centre of gravity 0.3m from the base how can I calculate the force required to cause the block to topple over. I am trying to calculate wind speed(force) required and how much effect Cog has

    Thanks

    Tom
     
  2. jcsd
  3. Mar 7, 2008 #2

    tiny-tim

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    Draw a picture!

    Hi 6060842! Welcome to PF! :smile:

    It often helps to draw a picture (just a rough one)!

    Hint: draw a picture of what the block will look like just as it's about to topple.

    What you need is the force to get it into that position! :smile:
     
  4. Mar 7, 2008 #3
    Tiny Tim thanks for the quick reply. Bit lost with your reply though.Not sure how to figure the force required
     
  5. Mar 7, 2008 #4

    tiny-tim

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    I assume you've drawn the picture.

    What do you notice? What's the difference between "before" and "after"? :smile:
     
  6. Mar 8, 2008 #5
    Okay

    Assumption 1000kg mass
    Assumption Force acting 1.5m from ground

    Weight Moment 2000kg * 1m

    Force Moment = F*1m

    Therefore when F>2000kg topple occurs (ie 20KN)

    (Does not matter about cog height)
     
    Last edited: Mar 8, 2008
  7. Mar 8, 2008 #6

    tiny-tim

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    6060842, you obviously haven't drawn any pictures. :frown:

    Look, to topple the block, you have to get it balanced on its edge with the cog above the line of contact.

    In that position it's balanced, and just breathing on it will push it one way or the other.

    And in that position, the cog is higher.

    So:
    (a) geometry: how high does the cog have to go?
    (b) physics: how much force is needes to get it there? :smile:
     
  8. Mar 8, 2008 #7

    russ_watters

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    Staff: Mentor

    Tim is right - once you've drawn a picture, the trigonometry is relatively straightforward.
     
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