\fdO'^ - Old Forge Coal Mines

HYDROMECHANICS. 193
(476) (a) 1 cu. in. of water weighs .03617 lb.
.03617 X 40 = 1.4468 lb. = weight of 40 cu. in. of water
= loss of weight of lead in water.
16. 4 — 1.447 = 14.953 lb. weight of lead in water. Ans.
(d) 16.4 -^ 40 = .41 — weight of 1 cu. in. of the lead.
16.4 — 2 = 14.4 lb. = weight of lead after cutting off 2 lb.
14.4 -i- .41 = 35.122 cu. in. = volume of lead after cutting
oflf 2 lb.
(477) (a) See Fig. 44. 13.5 X 9 X .7854 = 95.4261 sq.
in., area of base.
.03617 X 20 = .7234
due to the water only.
lb. per sq. in., pressure on the base
12 + .7234 = 12. 7234 lb.,
total pressure per square
inch on base.
12.7234 X 95.4261 =
1,214.144 1b. Ans.
{d) 47 X 12 = 564- lb.,
total upward pressure.
Ans.
(478) {a) See Fig. 44.
4 sin 53° = 3.195', nearly.
20 — 3.195 =16.805'=dis-
tance^ of center of gravity
of plate below the surface.
.03617 X 16.805 + 12 =
12. 60784 lb. per sq. in. = per-
pendicular pressure against the plate,
area of plate.
12.60784 X 40 = 504.314 lb. = perpendicular pressure on
plate.
Ans.
(d) 504.314 sin 53° = 402.76 lb. = horizontal pressure on
plate.
Ans.
(c) 504.314 cos 53° =303.5 lb. = vertical pressure on
plate.
Ans.
Fig. 44.
5 X 8 = 40 sq. in.