\fdO'^ - Old Forge Coal Mines

aOO HYDROMECHANICS.
For z;„ = 4,/=.0230; forz/„ = 6,/= .0214; .0230 -.0214
= .0016. 5.8-4=1.8. 6-4=2. Hence, 2 : 1.8 :: .0016 :
;r, or;ir = .0014. Then, .0230 - .0014 = .0216 =/for v,^ = 5.8.
Ans.
Forz/,„= 6,/= .0214; for z/^ = 8, /= .0205; .0214 -.0205
= .0009. 7.4 - 6 = 1.4. 8-6 = 2. Hence, 2 : 1.4 =.0009 : x,
or;i:=.0006. Then, .0214 - .0006 = .0208 =/ for ?/„. =
7.4. Ans.
Forz/„. = 8,/=.0205; forz/„= 12,/= .0193; .0205 -.0193
= .0012. 9.83-8 = 1.83. 12-8 = 4. Hence, 4 : 1.83::
.0012 : ;r, or ;r = .0005. Then, .0205 — .0005 = .02 =/ for
z/„=9.83. Ans.
For v^ - 8, /= .0205; for v^ = 12, /= .0193; .0205 -
.0193 = .0012. 11.5-8=3.5. 12-8 = 4. Hence, 4 : 3.5
::.0012 : .r, or;r = .0011. .0205 — .0011 = .0194 = /for z;,„ =
11.5. Ans.
(502) Specific gravity of sea-water is 1.026. Total area
of cube = 10.5' X 6 = 661.5 sq. in. 1 mile = 5,280 ft. Hence,
total pressure on the cube = 661.5 X 5,280 X 3.5 X .434 X
1.026 = 5,443,383 lb. Ans.
(503) 19' X .7854 X 80 = 22,682 lb. Ans.