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How to remove mechanical advantage from a lever

 
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Aug13-08, 09:27 PM   #18
 

How to remove mechanical advantage from a lever


Quote by stewartcs View Post
How exactly would this not move? If F1 = F2 then the torque at the fulcrum from one direction would be double that from the other direction since the distance from the fulcrum and the points of applied force is double. The rotation would be counter-clockwise.

Perhaps I'm missing something.

CS
The spring on F2 would absorb half of the force, balancing it with the force from F1. It's not meant to be the solution, merely to demonstrate that, in principle, there are ways to modify the forces.
 
Aug14-08, 05:28 AM   #19
 
Mentor
But that's the thing: in principle the internal forces are irrelevant. It is the extrenal forces on the system that make the lever arm move or not move. You cannot absorb/redirect forces like this, it's a violation of conservation law. This is how perpetual motion crackpots are made - they look for ways to do exactly what is being looked at in this thread.
 
Aug14-08, 06:06 AM   #20
 
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Quote by zeek View Post
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?
OK, if you want to keep a lever arm from rotating you need to balance torques (r x f = |r| |f| sin(a)). As you said, if you angle the force with the longer r you can get a smaller torque, but you need a sharp 30º angle, not just 45º. The other thing you could do is add a third torque, e.g. add an electric motor at the fulcrum to provide the extra torque you need. Since you are keeping everything stationary there are no concerns about violating conservation laws, no work is being done.
 
Aug14-08, 07:29 AM   #21
 
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Quote by DaveC426913 View Post
The spring on F2 would absorb half of the force, balancing it with the force from F1. It's not meant to be the solution, merely to demonstrate that, in principle, there are ways to modify the forces.
The spring still has an equal reaction force against the lever, regardless of the spring's displacement. For example, place a scale on a table top, then place a spring on the scale, then place a 1 lbf weight on top of the spring. The scale will still read 1 lbf (ignoring the spring's weight in this case). The normal force on the scale is unchanged (with or without the spring) even though the spring "absorbs" the force.

CS
 
Aug14-08, 07:36 AM   #22
 
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Quote by DaleSpam View Post
OK, if you want to keep a lever arm from rotating you need to balance torques (r x f = |r| |f| sin(a)). As you said, if you angle the force with the longer r you can get a smaller torque, but you need a sharp 30º angle, not just 45º. The other thing you could do is add a third torque, e.g. add an electric motor at the fulcrum to provide the extra torque you need. Since you are keeping everything stationary there are no concerns about violating conservation laws, no work is being done.
Exactly, as long as the system stays in equilibrium, it won't move!

That idea sounds feasible though.

CS
 
Aug14-08, 08:55 AM   #23
 
Quote by russ_watters View Post
But that's the thing: in principle the internal forces are irrelevant. It is the extrenal forces on the system that make the lever arm move or not move. You cannot absorb/redirect forces like this, it's a violation of conservation law. This is how perpetual motion crackpots are made - they look for ways to do exactly what is being looked at in this thread.
So where does the spring idea fail?
 
Aug14-08, 12:55 PM   #24
 
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Quote by DaveC426913 View Post
So where does the spring idea fail?
Take a look at post #21. When the spring has a force applied to it, the opposite end of the spring (where the lever is in your drawing) has a reaction force equal to the applied force. The spring's displacement (length) is just reduced. The magnitude of the applied force is still seen at the lever (i.e. the other side of spring). So nothing is gained from it.

CS
 
Aug14-08, 02:43 PM   #25
 
Take a look at post #21...
Sorry. Somehow I got behind a post or two. I wouldn't have asked if I'd read that.
 
Aug14-08, 08:43 PM   #26
 
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After discussion in the mentor's forum, we've decided this thread doesn't go anywhere good. The concept is just plain wrong and the flaw has been explained, so there is nothing to be gained by pursuing it further.
 
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