In my quest to find interesting problems for my students, I have often gone back in time to articles from the past. I've just found one, and it has me a bit flummoxed…
"If a glass tube, 36 inches long, closed at top, be sunk perpendicularly into water, till its lower, or open end be 30 inches below the surface of the water,; how high will the water rise within the tube[,] the quicksilver in the common barometer at the same time standing at 29 1/2 inches?" (1798)
I realize that for such problems, the 'norm' in freshman physics is for a vacuum to be at the top of the tube -- now we have air in the upper portion exerting a pressure downward… In any event, I am struggling to arrive at the stated formulation, given in 1799:
"Let l=36 inches the length of the tube, b=30 inches the part immersed, x=the height of the water in the tube, and f=413 inches, the height of a column of water equal to the pressure of the atmosphere, when the quicksilver stands at 29 1/2 inches. Then, since the spaces occupied by the same quantity of air, are reciprocally as the compressing forces… (lf)/(l-x) +x = b + f, and x = 2.2654115 inches."
1. What is the "modern" law/principle that gives: "...the spaces occupied by the same quantity of air, are reciprocally as the compressing forces…"
2. Is there a way to formulate this problem in terms of hydrostatics for freshman physics -- I think it would be cool to show it to them, and maybe do the demo.