As part of the construction of new third-rail beams for the 36, several new metal brackets are needed to hold the copper buss fuses in place, in case they blow out. These are attached to the wooden beam with a couple of nails. (I keep meaning to take a picture of what they look like when installed.) But you can see how rusty the one spare original is.
So the Hicks Iron Works had its first new order in several years, after Rod ordered the metal blanks we needed. 
I also was able to borrow a small sledge hammer for flattening the steel, having had no use for such a device as a homeowner. The 180º bend is the only real challenge in this process. The work went quickly after a brief learning curve, and we now have all ten done, except for drilling the holes for the nails, which I'll do on the Museum's drill press.
I also was able to borrow a small sledge hammer for flattening the steel, having had no use for such a device as a homeowner. The 180º bend is the only real challenge in this process. The work went quickly after a brief learning curve, and we now have all ten done, except for drilling the holes for the nails, which I'll do on the Museum's drill press. 
But say, speaking of homework, that made me think of an interesting physics problem. In the figure, a uniform bar is bent 180º. What is the maximum ratio of L/R such that the bar will balance as shown? I want an exact expression for the answer. The first person to answer correctly gets fame, adulation, and a chance to help install them.
6 comments:
At the risk of being labeled one who didn't stay awake in physics class, I think that L=2R would be stable, and the exact tipping point would be L=2R+H (where H is the height of the bend)
Sorry, that's not correct. Also, you need to put your name on your paper in order to get credit.
You may assume that the bend is negligibly small compared to the horizontal dimensions, by the way.
This may be a duplicate of sorts
if L weighs R, an their are 2 R's - the total weight is 3L. Dividing by 2 gives 1.5L as the center point but since L is doubled.. to get 1/2 - each side must be .25... so the center is 1.25L (giving each side a weight of 1.5) (the R side being 2x .75.
Long way from a physics class about 43 years ago at the UofI in which I got a "D". So this is continuation of my giving incorrect answers to physics problems. It did convince me that I had no aptitude for science - even though I was as chem major ..so I went into law and that worked out too..
Bob
All right, I'll stick my neck out. The tipping point is L = sqrt(2) * R, right? The trick is to realize that the lever-arm can be simplified to a force of the entire weight on each side of the fulcrum applied at half the total length of that side.
R. W. Schauer
(Drum roll, please...) Rich is exactly right. And I appreciate that he shows his work, so to speak. Good work!
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