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: simple leverage variables
: simple leverage variables -- landscaper	Post Reply	Top of thread	Engineering Forum

Posted by: jboggs ® 02/03/2011, 14:54:25 Author Profile eMail author Edit

Thinking more about your problem. First, your question wasn't clear. Length and weight are two different things. They cannot be equal. In leverage calculations the effects of length and force (weight) combine to form a "moment". M = L x F. A moment is like torque, a "rotary force". "Length" is usually the horizontal distance from the fulcrum to the centerline of the force. So it doesn't matter if the weight is above or below the fulcrum since it is the horizontal distance. The lever will rotate toward whichever direction produces the greater moment. When the moments in both directions are equal it stops.

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: simple leverage variables
: simple leverage variables -- landscaper	Post Reply	Top of thread	Engineering Forum

Posted by: jboggs ® 02/02/2011, 18:09:31 Author Profile eMail author Edit

Welcome to the forum. If you look around you will see instructions on how to post graphical images. If you could create a sketch of what you are trying to do, with all the relevant points labeled, we could probably help you a little better.

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: : simple leverage variables
: : simple leverage variables -- jboggs	Post Reply	Top of thread	Engineering Forum

Posted by: RWOLFEJR ® 02/03/2011, 14:31:30 Author Profile eMail author Edit

Your first question that ended something like... the weight equaling the length to fulcrum... I'm not understanding what your after there... The load sitting on your beam or hanging below your beam won't make a difference far as your leverage goes. It's all about the distance from the fulcrum for leverage or torque calculations. But... Load on the beam would mean you'd need to have a way to get the rock on and off of your beam. Load under the beam you can lift the rock and set it. There are a bunch of ways to lift rocks without having a strap under it so you can place them pretty much exactly where you need them. There are toggle types that sort of pinch the piece when you lift. (Google jersey barrier lifting or chinese finger trap) If the rocks are round-ish, you can hang three or four straps from a hook with loops at the bottom of the straps. Then run a fourth ratchet strap through those loops and cinch it up at a point a fair amount under the center of the rock. Follow me? Like pulling the string on an onion bag. A drawback to the load under your beam is you might need a higher fulcrum point to get the required lift? Far as keeping one end of your beam up goes... That means you'll have to have more weight on one end to make it want to settle that way. You can do that by just making the side you want to rise up with no load be a little bit shorter. If you want your lever or the "X" and "L" in the fulcrum equation to be equal length then put your hook for your load the same distance from fulcrum as the length of your shorter leg. The weight of stone is approximately 150 pounds per cubic foot... more or less depending on type of stone. When you are figuring out how big a beam you'll need remember to double the weight of your rock for the calculation. Say you have a ten ton stone on one end of your beam... you'll need a ten ton force on the other side to lift it if levers are equal length. So you now have twenty tons trying to bend your beam over x+y length with the load at center. (Center being your fulcrum pushing up while the "supports" push down.) If you're going after some mechanical advantage then you will either need to have a moveable counter balance or a longer leg on the non boulder side. This is probably where you were going with your first question about weight equalling length? Same ten ton rock... 5 foot of beam on the rock side of fulcrum and 15 feet on the work side of the fulcrum. Now you only need a third of the force to lift the rock. But now you'll need a counter weight to hold the 5 foot long side down until you can get a strap on the rock and then remove it when you're ready to lift. The counterweights will need to be slightly more than the weight of ten feet of your beam. (That's the extra length from fulcrum) So if your beam is 50 pounds per foot you'd need 550-ish lbs. of counter weight to hold the rock side of the beam down while hooking up. (I DIDN'T CALCULATE FOR REQUIRED BEAM WEIGHT... JUST PULLED A NUMBER OUT OF THE AIR... MIGHT TAKE A MUCH HEAVIER BEAM!!) If you want to keep the counter weight on there permanently then your beam size will need to deal with the added load and you'll require a little more work or force to lift the rock plus the counterweight. So same 10 ton rock with 5 foot and 15 foot on fulcrum and the counter weight you'll need 3.3 tons to lift rock plus an additional third of the counter weight or 183 lbs. So that's 6,783 lbs. to move a boulder weighing 20,000 lbs. Like the saying goes... give me a long enough lever and I can move the world. Hope this helps you out some... Good luck and be carefull...!! Bob

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: : : simple leverage variables
: : : simple leverage variables -- RWOLFEJR	Post Reply	Top of thread	Engineering Forum

Posted by: landscaper ® 02/03/2011, 16:20:49 Author Profile eMail author Edit

Thank you. I am sorry i was not very clear with my questions. I tried to scan a sketch but am having coputer difficulties.This specific problem is in designing a art piece in a landscape project but uses the same principals as lifting boulders or objects with a lever and fulcrum. Bob your last paragraph answered my most important question. If i can get my computer issues worked out i will get the sketch up so it will explain things better Jackie

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