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How to remove mechanical advantage from a lever |
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| Aug12-08, 12:17 PM | #1 |
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How to remove mechanical advantage from a lever
I have a simple lever with two points of equal force, each at a different distance from the fulcrum.
--- What I want: As the arm turns, one point of force moves at a faster rate than the other. --- What I DO NOT want: One point of force having a mechanical advantage over the other. --- If I were to introduce two tracks in parallel, with each point of force on a track, but able to freely slide up and down the same lever arm as they moved across the tracks, would this remove the mechanical advantage from the force point farthest from the fulcrum? If not, is there another solution such as some lattice configuration, or even gears or something other than a lever? Again, all I want is two points of force moving together with one point of force moving at a faster rate than the other. Thanks. |
| Aug12-08, 02:34 PM | #2 |
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What I think you are describing violates the laws of thermodynamics. If you have two equal forces then they must have equal displacement. That is:
F1 x D1 = F2 x D2 where: F = force D = displacement The only way this relationship can not hold true is if there are losses in the lever or mechanical system you are using. There are two ways of accomplishing these losses. 1. A mechanical device with multiple degrees of freedom where one of those degrees is subjected to a damper, removing work from the system. 2. A hydraulic/pneumatic system where a working fluid is subject to some type of transformation that it absorbs external work in the working fluid itself, ie compression. Sorry if I misunderstood your question. If I have, can you draw a picture? |
| Aug12-08, 03:05 PM | #3 |
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Exactly - because if you turn it around, you are asking.
I want a lever where I can press down a short distance and have the other end move a large distance with the same force. It would be useful, but would have to create energy from nothing! |
| Aug12-08, 04:09 PM | #4 |
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How to remove mechanical advantage from a lever
It should be possible to build such a device as an active device. It certainly wouldn't be a simple lever. Probably the easiest way would be to build an electronic device to monitor the force and displacement on the control and then use something similar (in reverse) to produce the same force but a different displacement.
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| Aug12-08, 04:56 PM | #5 |
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Btw, you said you want to remove the mechanical advantage, but the mechanical advantage is simply a ratio and it manifests with both a change in displacement and a change in force. So really you are saying you want to keep the mechanical advantae but remove the mechanical advantage - an oxymoron.
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| Aug12-08, 06:37 PM | #6 |
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Thank you for your replies. You have in essence, answered my question. I was wanting to know if there was a sort of 'commonly known' way of removing the leverage 'gained' to the force position farthest from the fulcrum. I see now that there is not. Thank you.
Because of my limited knowledge of physics terminology (and principles unfortunately) it's hard to explain my dilemma. I am now thinking that the only way to achieve my goal of two points of force acting on the same arm at different distances from the fulcrum but without the farther position having a leverage advantage is to somehow change the angle of force on the farther position, so that it is not directed at a 90 degree angle to the arm, thus weakening the 'active' force. I've been reading the web about angular momentum. Times like this I really wish that I had a head for this sort of thing. If you have any thoughts on this 'angle of force' approach, please post them. Thanks again for your time and attention. FURTHER EXPLANATION: Just FYI, another way to explain what I am needing is this... I need a lever-arm with say 1 lb of force acting on the arm at 6 inches, and a second 1 lb force acting on the arm at 12 inches, but in the opposite direction. My problem is finding an arrangement in which these forces are balanced so that they cancel each other out and the arm doesn't move due to either force. Am I correct in thinking that I could achieve the scenario above if the force acting on the arm at 12 inches was arranged so as to point in a 45 degree angle to the arm, thus halving the 'active' force acting on the arm? |
| Aug12-08, 07:02 PM | #7 |
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He's looking for a practical solution, not a principle, so he has a braid range of ways he can get the dsired effect without violating any laws. There are lots of ways of balancing the forces. We'd need to have a better understanding though. How are the two levers connected? Where are they connected at the other end? |
| Aug12-08, 07:04 PM | #8 |
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Zeek, try doing some research on mechanical CCPM (continuously collective pitch mixing) for model helicopters. Some of the geometry gets a little complex but I believe accomplishes something similar to what you are after.
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| Aug12-08, 10:47 PM | #9 |
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Angling the forces doesn't matter - the vertical component is the only component that is used on the lever.
No, I suspect what you are looking to do isn't possible. |
| Aug13-08, 08:22 AM | #10 |
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| Aug13-08, 08:43 AM | #11 |
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I have whipped up a really simple diagram to ensure I've captured what's going on.
Have I missed anything? Input welcome. It does NOT contain the solution yet, it is merely the starting point. I'll modify it if anyone has any ideas. |
| Aug13-08, 09:42 AM | #12 |
| Aug13-08, 12:20 PM | #13 |
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| Aug13-08, 12:24 PM | #14 |
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| Aug13-08, 01:13 PM | #15 |
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Zeek, could we understand a little more about what's applying the force and the relation of the forces to the arm?
I get the feeling we're overthinking it. |
| Aug13-08, 01:25 PM | #16 |
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It seems to me there's no reason why it couldn't work. As simple a solution as a spring to dampen the force of F2 accomplishes what he wants in principle, would you agree? |
| Aug13-08, 07:08 PM | #17 |
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Perhaps I'm missing something. CS |
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