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CG calculation - Help

  1. Oct 9, 2008 #1
    Hi guys,

    I've a question regarding about CG calculation. Any kind soul pls help to advise..

    Okay, I need to make a base to support a pole. Pole diameter is 4.5" with 1 metre high. Weighs abt 25-30kg. I need to make a steel base to hold the pole but not sure what size of the steel base will be stable enough so that this pole does not topple when accidentally pushed.

    Was thinking a 500mm x 500mm x 8mm thk steel base. Is it sufficient to prevent the system from toppling?

  2. jcsd
  3. Oct 9, 2008 #2
    I'm sure this is dependant on the toppling force.

    what forces are being applied to the pole?

    a perfect, circular cross sectional pole, will stand upright on it's own as it's cg sits on the vertical axis and through the centre of the cross section, under perfect conditons and with no outside forces. you need to give more info or make some assumptions...
  4. Oct 9, 2008 #3


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    You can think of it this way without doing calculations. Draw a vertical line on a piece of paper, this represents the bars original position. Draw a circle at the center of the bar indicating the center of mass. Note that any appreciable mass added for the base will tend to move it down. More on that later.

    Now, rotate the beam about the bottom point by the amount that you wish to prevent it from falling over. If an accidental nudge will move it 10°, then rotate it by that much. Rotate the center of mass as well.

    Now, draw a line vertically down from the new center of mass. Now draw a line perpendicular to new line at the bottom point. This point represents the radius of a support plate needed to keep the beam stable.

    Essentially looking at the new drawing, you can see that as long as the beam doesn't rotate past that point, then the center of gravity will tend to stabilize it, another rotationg past it will result in it falling.

    Now to the center of gravity. You can see that by lowering the center of gravity, you have the option of either a: making the support plate smaller, or b: increasing the allowable rotation.

    If your base is then circular , then the assembly is axisymmetric about the vertical axis. This means that the center of gravity of course will be in the center of the beam/bottom plate. The center of gravity will be at the point where there is the same amount of mass on both sides.

    So, you have the size of your beam, it's easy enough to get the mass. Area of the cross section times the length times density. Since the base is a flat cylinder as well, it's easy to get that mass. Divide the mass of the bottom plate by 2. Now move your center of gravity down the beam by that amount. In equation form, what you want is the mass of the beam to be 1/2 the mass of the volume, that way it evens itself out. So:
    [tex] \left( \pi r^2 \Delta h \rho \right)_{beam} = \frac{1}{2} \left( V \rho\right)_{support} [/tex]
    Where V is the volume of the support. I wrote it like this because we can now cancel out the densities assuming they are the same material and solve for the height.

    [tex] \Delta h = \frac{1}{2} \frac{V_{support}}{\pi r^2_{beam}} [/tex]

    ...i think
  5. Oct 10, 2008 #4
    Hi Red,

    Well. This pole will be screwed attached to the support base plate.

    As this system will be rested on the table. Just wanted to ensure that if it doesn't topple when people walk passed the table and accidentally shove the table.

    The forces applied on the pole will be the weight itself which is in the vertical axis.
  6. Oct 10, 2008 #5


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    The question of what forces are applied to the pole are not the weight. The external forces created when someone nudges the pole is what we need. This is a perfect example of sometimes having to take your best guess, using your judgement/experience, on what is expected.

    Honestly, I would follow Minger's advice. It's quick and easy.
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