Hammer Shank Ratio

Richard Brekne Richard.Brekne@grieg.uib.no
Wed, 19 Feb 2003 23:33:38 +0100



Phillip Ford wrote:

> I didn't just change the 10 pound force to a 14 pound force, I also changed its direction.  A 14 pound load at 45 degrees to vertical on the right side balances a 10 pound load vertically down on the left side.  I don't need to change the load on the left side to 14 pounds for the system to be in balance.

Which is the same as saying you didnt change anything at all. At least as far as the lever is concerned. Which was my point. The equation is an equality.  So... since it still takes 10 pounds at W1 to balance 10 pounds at W2, and the distances D1 and D2 are still the same... then you still have a 1 to 1 ratio. It doesnt matter what direction or how much weight it is..... all that matters is what the lever feels as "weight".

If the distance from the fulcrum for both input and output is the same.... then its a 1 to 1 ratio.... and no matter how what angle you apply the force, no matter if you introduce shear, torque, or anything else you want to thats not going to change the leverage ratio one iota.


> The component of the load in the direction of the movement balances the load on the other side in the direction of the movement 1:1.  That's the point.  If you just take the magnitudes of the loads and divide them you won't get a 1:1 ratio because you're not looking at the right numbers.

Yes you will.... and you just did.... 10 sin 45 = 10.  You keep mixing apples and bananas. The "load" doesnt change at all. The lever still sees exactly the same thing. If it didnt, you'd have to change the other side of the lever as well. What you are doing is changing the direction you apply leverage and mixing that up with the leverage itself. Thats something else that I am sure there is a very good name for... but it is not leverage.

Its really like weird too... because you seem to see that the hammer shank assembly effectively changes its  "weight" because of the changing direction of force applied between the contact points in the action... AND you seem to see that this change is weight is (naturally enough) reflected in a change in momentary DW or UW  needed at the key to balance (or move) that changing weight (of the hammer shank assembly).... so why you get mixed up and see a changing ratio due to force vectors is beyond me. Especially when the law of leverage simply defines that out of the question to begin with.

> This is the point that I'm trying to make about taking measurements on actions.  The forces measured have to be at the same points and in the same directions as the distances measured
> for there to be a comparison.

And the point I am making is that you dont. Force vectors influence a the ratio of a lever exactly to the degree of zip diddly.


> Try changing my 45 degree angle to 90 degrees and there's no leverage at all.

Nooooooo no no no no no. You are not able to use the leverage that is inherent in this fashion. Thats an entirely different matter. I keep making this point... how you apply leverage... how you use a lever.... has nothing to do with a given levers ratio. As to this example.... if you put a 90 degree force on the the input or outout of a lever there is no effective weight on that point at all. And the amount of force needed to balance whats on the other side is infinite... as in not attainable.

You keep bringing in other issues, other physics components that describe things other then the leverage. Leverage is described by the law of levers. And that is about weight and distance. You cant just decide to divide force vectors anyway you want to too describe leverage. You bring force vectors, torgue or any of the rest into it and you are describing something other then the ratio of levers.

--
Richard Brekne
RPT, N.P.T.F.
UiB, Bergen, Norway
mailto:rbrekne@broadpark.no
http://home.broadpark.no/~rbrekne/ricmain.html



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