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Geek Culture / A quick question about leverage and center of gravity

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Dark Java Dude 64
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Posted: 2nd Mar 2012 00:41 Edited at: 2nd Mar 2012 00:44
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Hi! Let's say you have a a large beam. We'll say it's ten feet long. One nine foot section weighs 100 pounds and the other one foot section also weighs 100 pounds. Would the beam be able to balance on the nine foot mark, or would the nine foot long section's greater leverage keep that from happening? Even though both side would be pushing down with 100 pounds of force, the one side has 9 times more leverage. However maybe it would be able to balance as there would be less weight per inch on the long side. Thanks for answering!

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Indicium
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Posted: 2nd Mar 2012 00:46
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It would tip because there is a greater moment on nine foot section due to the center of gravity not being the same distance from the pivot as the one foot section. ( Moment = Force x Perpendicular distance from the pivot )

Hope that made sense.

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Dark Java Dude 64
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Posted: 2nd Mar 2012 01:27
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Made perfect sense! Thanks!

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zenassem
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Posted: 2nd Mar 2012 02:50 Edited at: 2nd Mar 2012 02:56
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[edit] the one I posted is not good for your description sorry.

But I believe I've seen a Center of Mass simulator in MatLab that demonstrates your example. Searching...

I hate that I never have the bookmarks I should or thought I added!!!

.oO()Oo.oO (I'm not a real programmer,, I just play one on the Forums!!!) Oo.oO()Oo.
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Neuro Fuzzy
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Posted: 2nd Mar 2012 06:19
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I'm learning mechanics, so I thought I'd work out exactly how much torque a continuous beam exerts on something that's holding it.
http://mathbin.net/89910

I got that the torque is g/2*length*mass. It makes sense, and the explanation is more long-winded than it needs to be, but whatever.

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Dark Java Dude 64
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Posted: 2nd Mar 2012 06:34
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Ah! Very nice!

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Indicium
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Posted: 2nd Mar 2012 18:38
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In reality, it's ridiculously hard to get something perfectly balance on a pivot anyway, at college I have to do a practical investigations, one of them was to see how a mass affects the position of the pivot in equilibrium.

It's safe to say that my results weren't entirely accurate.

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29 games
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Posted: 2nd Mar 2012 21:04
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This is quite a fun way of finding the center of gravity for a long stick (works with short sticks to).



If you were to put a weight at one end your fingers will still end up in the center of gravity, which will now be shifted from the geometric center and closer to the weighted end.
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