Treat it like a cantilever with a back span. The -Ve will be the approximate weight you'll need to balance the thing then fine tune it from there.
Hello, all. I've got a device I'd like to build which will require a small panel (maybe 3 feet tall by 2 feet wide) be pivoted. The panel will have a pivot at the narrow edge. I'd like to assist the lifting of panel by placing a hanging weight on the opposite side of the panel. I intend for the counter weight to attach to the pivoting panel at the end farthest from the pivot. I want the panel to always be heavier than the force applied by the counter weight so that it will fall due to gravity when the panel is pushed out of the way. I'm sure there is a way to calculate how much weight needs to be used as the counter weight, given the panel weight, but I've no idea what that calculation might be.
Here's the gist of it: weight----*pivot-----------------panel. Assuming the panel is uniform in weight distribution and disregarding friction in the pivot, 1) how do I calculate how much weight I need to place on the other side of the pivot and how far from the pivot does it need to be (directly related items, I'm sure)? 2) assuming the panel has the pivot at the "top" (farthest from the ground) and is leaning 60 degrees up from the ground (mostly vertical), can I use a straight arm from the pivot to the counter weight attachment point or will I need a cam/egg shaped piece to carry a cable and effectively change the amount of force applied by the counter weight by moving it farther from the pivot point? Oh, and the panel will start at a 60 degree angle like this / and be raised only to the horizontal position like this ----, so 60 degrees of rotation.
Thanks for your time and help. I'd like to make calculations ahead of time instead of trial-and-error for days.
--HC
Last edited by hcb; 03-23-2016 at 11:41 PM.
Treat it like a cantilever with a back span. The -Ve will be the approximate weight you'll need to balance the thing then fine tune it from there.
See the following webpage, towards the bottom of the webpage:
Machine Design Applications, Equations and Calculators
Tell me and I forget. Teach me and I remember. Involve me and I learn.
Too many ways of interpreting your description. Post a sketch.
Thank you for your reply. I will have to go through the other replies and then, if I've not gotten the answer I will look up cantilever calculations. I'm sorry but I'm not an engineer so I'm not familiar with these calculations but you've given me information and keywords to help search for the answer. Thanks again.
--HC
Adding a sketch in PDF...hopefully...I've not done this before.
Maybe?
--HC
OK. I THINK I understand what you're trying to do. It all gets down to weights and moments. You're designing a see-saw. The basic equation for perfect balance is this: Weight of door x horizontal distance from pivot to door CG = weight of counterbalance x horizontal distance from pivot to counterbalance CG. As you adjust and play with those weights and distances you can probably achieve the final downward force you are after.
One comment - the counterbalance doesn't necessarily have to be hung from a cable. it could be permanently attached to the pivoting frame.
Thank you for your reply. I think I understand your answer; distance of CG of supported weight from pivot multiplied by the door's weight = weight of counterbalance weight multiplied by distance from pivot, assuming horizontal distance from pivot. Further references based on horizontal position.
To clean that up: CGpd is Center of Gravity pivot distance for supported weight. DW is door weight. CWpd is counterweight applied force distance to pivot. CW is counterweight.
Replace CW symbol with X just to help me, so: DW * CGpd = CWpd * x, solve for X is: (DW * CGpd) / CWpd = X. X is CW, so that's it. Assuming a 36" supported, symmetrically-weighted door, 18" is the CG from pivot. Assume door weighs 30 pounds. Assume counterweight is pulling down at 8" from pivot: (30# * 18) / 8 = CW. 540 / 8 = CW. 67.5# = CW. CW = 67.5#.
Correct?
I drew weight suspended like I did because I 1) had envisioned maybe using a cam-shaped wheel to vary the lifting force during the arc and a cable across a cam-shaped pulley could do that and 2) for the ease of assembly and adjustment for a homemade device.
I'm having fits with IE 11 jumping this textbox for responding so it takes me forever to type a reply. Any quick suggestions?
Thank you again.
--HC
For perfect balance: W1 x D1 = W2 x D2
Thank you for the formula and the diagram. That formula is dead simple and I probably learned something like that back in school but that was so many years ago.... I don't do this kind of work very often...now trig...until a few years ago when I learned AutoCAD, trig was my thing.
Anyway, thank you for your help, I feel confident this is what I needed.
(p.s. switching to Firefox seems to have alleviated my freezing/hesitating experience with the site)
--HC