lever calculations

[Kaizen]

This should be simple but all of the on line calculators seem to be for a fulcrum not a lever.....am I right a lever is like a crane but no winch?
So i'm trying to see what it would take to lift a 200 pound weight 16 feet in the air. so I need to know what the center point is, arm length on both sides and weight to add. I guess too how tall of a tower would be needed to support the center point. say with two 2x6x16...probably need to sister a few together to get the complete length needed. no turning just a simple machine. anyone have any references?

 

[IlliniJeeper]

The point of a lever is to provide mechanical advantage. What's your goal? Is it to be able to lift the 200 pounds with 50 pounds of force? Is it to counterbalance the weight?

Do you need to lift 16' exactly or can you overshoot? Your max height should be a bit more than 16' to account for deflection of your lever.

 

[Kaizen]

ok so i'm talking about lifting a long arm of say 16 feet by putting x amount of weight on a 4 foot arm as a counterbalance= total beam would be 20 feet. 16 feet high should be fine.
The end of the long arm will have a 200 pound weight.
with the above guesstimates i'm looking at about 800 pounds plus the weight of the long arm right?

 

[larry_g]

I think your going to have to draw a picture for me. I don't understand a lever without a fulcrum. Am I correct that you do not want a teeter/totter like lever with a fulcrum in the middle?
Any lever is basic trig.

lg
no neat sig line

 

[UpNorther]

ok so i'm talking about lifting a long arm of say 16 feet by putting x amount of weight on a 4 foot arm as a counterbalance= total beam would be 20 feet. 16 feet high should be fine.
The end of the long arm will have a 200 pound weight.
with the above guesstimates i'm looking at about 800 pounds plus the weight of the long arm right?

^ this
4:1 ratio = 800 lbs + arm

 

[IlliniJeeper]

800 lbs, plus friction in the fulcrum, plus the weight of the 16' length multiplied by the distance to its center of mass (~8') minus the weight of the 4' length multiplied the distance to its center of mass (~2') times your mechanical advantage (1:4 or 0.25).

 

[Kaizen]

thanks. guess I just had to talk it out.