<!doctype html public "-//w3c//dtd html 4.0 transitional//en"> <html>   Richard Brekne wrote: Kept thinking about where you were at with this segment... <blockquote TYPE=CITE>Phillip Ford wrote: <blockquote TYPE=CITE>Well, I would say that it depends on the way you look at it.  You have to compare apples to apples.  Let's take an example.  A see-saw or teeter-totter.  A simple lever.  Let's say it is horizontal.  Put a 1 N force vertically down at 1 m to the left of the fulcrum.  Now put a (Balancing) force  vertically down at a point 1 m to the right of the fulcrum.  What will that force be?  1 N.  The 'leverage' is 1.  Your 'ratio', obtained by dividing one force by the other is 1.  1 mm of downward movement on the left side will result in 1 mm of upward movement on the right side.</blockquote> <blockquote TYPE=CITE>Now, on the right side, rather than putting the force vertically down, angle it at 45 degrees.  Since the points of force application and distance measurement have not changed then the 'leverage' should still be 1.</blockquote> And so it should be.... if thats what you really do in the continuence. <blockquote TYPE=CITE>A 1 mm downward movement on the left side will still result in a 1 mm upward movement on the right side.</blockquote> ok..still fine. <blockquote TYPE=CITE>But what will the force be?  1.414 N.  Your 'ratio' obtained by simply dividing one force by the other would be 1.414.  How did that happen?  The effective ratio for speed, weight, and distance are supposed to be the same any way you look at it right?</blockquote> </blockquote> Been looking at this wondering how you got where you did and I think I now see. To begin with, I want to restate that the only thing the affects the ratio of a lever is the distance from the fulcrum of the input and output points. Secondly, let me restate that the law of leverages states that for the lever to be in balance, the products of the weight at either of these points, times their repective distances from the fulcrum are equal. This law is also expressed in terms of force F1 * d1 = F2 * d2. Now lets take your teeter totter which has points 1 meter on either side of a fulcrum and starts off with 10 pounds on each side. You want to put a  rougly 14 pound force angled at 45 degrees on one of these same points and say the result is that there is still movement at a 1 to 1 ratio, still speed at a 1 to 1 ratio, but suddenly there is a 1.4 to 1 weight ratio. The problem here is that at the exact point on the lever where there is a 1 to 1 motion and speed ratio... there is still 10 pounds of weight on each side. You put it like this..     "The vertical component of the load on the right side (14 sin 45 = 10)" But then you turn around and seem to want to treat this force like it was a dead weight of 14 pounds that is sitting (all 14 pounds of it) exactly on that same point. Its not. Where ever that 14 pounds is in reality... it is not on that point.. its somewhere off on your 45 degree angle line. And if you want to find out exactly how much that moves for 1 mm of downward movement on the otherside of the lever... then you have to measure to that point... find the relevant ratio.  Point is,  if you change the ratio of a lever... then you change it for all three things (weight, distance, and speed) and equally so... and the ONLY way to make this change is by changing the length of one or both arms.       -- Richard Brekne RPT, N.P.T.F. UiB, Bergen, Norway <a href="mailto:rbrekne@broadpark.no">mailto:rbrekne@broadpark.no</a> <a href="http://home.broadpark.no/~rbrekne/ricmain.html">http://home.broadpark.no/~rbrekne/ricmain.html</a>  </html>