Page 209 - Schaum's Outline of Theory and Problems of Applied Physics

194 FLUIDS AT REST [CHAP. 16

SOLVED PROBLEM 16.11
How much force is needed to support a 100-kg iron anchor when it is immersed in seawater? The density
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of iron is 7.8 × 10 kg/m and that of seawater is 1.03 × 10 kg/m .
The volume of the anchor is
m 100 kg
V = = = 0.0128 m 3
d 7.8 × 10 kg/m 3
3
The weight of seawater displaced by the anchor is
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w = mg = dVg = (1.03 × 10 kg/m )(0.0128 m )(9.8 m/s ) = 129 N
Thus the buoyant force on the anchor is 129 N, and the net force needed to support it in seawater is
2
F net = mg − F buoyant = (100 kg)(9.8 m/s ) − 129 N = 851 N

SOLVED PROBLEM 16.12
A 70-kg person dives off a raft 2 m square moored in a freshwater lake. By how much does the raft rise?
The volume of water that must be displaced by the raft to support the diver is

m 70 kg 3
V = = 3 = 0.07 m
d 10 kg/m
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2
The area of the raft is A = (2m)(2m) = 4m . Since volume = (height)(area), the raft rises by
V 0.07 m 3
h = = = 0.018 m = 1.8cm
A 4m 2

SOLVED PROBLEM 16.13
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The density of ice is 920 kg/m and that of seawater is 1030 kg/m . What percentage of the volume of
an iceberg is submerged?
When an iceberg of volume V floats, its weight of d ice gV is balanced by the buoyant force on it, which is equal
to the weight of water displaced. If V sub is the volume of the iceberg that is submerged, the weight of water displaced
is d water gV sub . Hence
Weight of iceberg = weight of displaced water

d ice gV = d water gV sub
V sub d ice 920 kg/m 3
= = = 0.89 = 89%
V d water 1030 kg/m 3
Eighty-nine percent of the volume of an iceberg is under the water’s surface.

SOLVED PROBLEM 16.14
A 100-gal steel tank weighs 50 lb when empty. Will it float in seawater when it is filled with gasoline?
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The weight density of gasoline is 42 lb/ft , that of seawater is 64 lb/ft , and 1 gal = 0.134 ft .
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The volume of the tank is V = (100 gal)(0.134 ft /gal) = 13.4ft . The total weight of the tank when it is filled
with gasoline is
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w = 50 lb + (dg) gasoline V = 50 lb + (42 lb/ft )(13.4ft )
= 50 lb + 563 lb = 613 lb