CHAPTER 18
FLUID MOTION
Section 181:
Describing Fluids: Density and
Pressure
Problem
1. The density of molasses is 1600 kg
/
m
3
. Find the
mass of the molasses in a 0.75liter jar.
Solution
The mass of molasses, which occupies a volume equal
to the capacity of the jar, is Δ
m
=
ρ
Δ
V
=
(1600 kg
/
m
3
)(0
.
75
×
10

3
m
3
) = 1
.
2 kg
.
Problem
2. Salad dressing is made from one part (by volume)
of vinegar (density 1
.
0 g
/
cm
3
) to three parts olive
oil (density 0
.
92 g
/
cm
3
)
.
What is the average
density of the dressing?
Solution
The average density of the mixture is its total mass,
Δ
m
= Δ
m
vin
+ Δ
m
oil
,
divided by its volume Δ
V,
which is
1
4
vinegar and
3
4
oil. Therefore,
ρ
av
=
(Δ
m
vin
+ Δ
m
oil
)
/
Δ
V
= (
ρ
vin
1
4
Δ
V
+
ρ
oil
3
4
Δ
V
)
/
Δ
V
=
[
1
4
(1
.
0) +
3
4
(0
.
92)](g
/
cm
3
) = 0
.
94 g
/
cm
3
.
Problem
3. The density of atomic nuclei is about 10
17
kg
/
m
3
,
while the density of water is 10
3
kg
/
m
3
.
Roughly
what fraction of the volume of water is
not
empty
space?
Solution
The average density of a mixture of two substances,
with definite volume fractions, is
ρ
av
=
ρ
1
(
V
1
/V
) +
ρ
2
(
V
2
/V
)
,
where
V
1
+
V
2
=
V
is the total volume.
(Try this formula in the preceding problem.) The
density of water is approximately the average density
(
ρ
av
= 10
3
kg
/
m
3
) of the nuclei (
ρ
1
= 10
17
kg
/
m
3
) and
empty space (
ρ
2
= 0) provided we neglect the mass of
the atomic electrons, so the volume fraction of nuclei
in water is (
V
1
/V
) =
ρ
av
/ρ
1
= 10
3
/
10
17
= 10

14
.
Problem
4. Compressed air with mass 8.8 kg is stored in a gas
cylinder with a volume of 0
.
050 m
3
.
(a) What is the
density of the compressed air? (b) How large a
volume would the same gas occupy at typical
atmospheric density of 1
.
2 kg
/
m
3
?
Solution
(a) The density of the compressed air is Δ
m/
Δ
V
=
8
.
8 kg
/
0
.
050 m
3
= 167 kg
/
m
3
.
(b) The same mass of
air, at density 1
.
2 kg
/
m
3
would occupy a volume of
V
=
m/ρ
= 8
.
8 kg
/
(1
.
2 kg
/
m
3
) = 7
.
33 m
3
. (Note:
The volumes are small enough that any variation in
the density of the air due to gravity may be ignored.)
Problem
5. A plant hangs from a 3.2cm diameter suction cup
affixed to a smooth horizontal surface (Fig. 1842).
What is the maximum weight that can be
suspended (a) at sea level and (b) in Denver, where
atmospheric pressure is about 0.80 atm?
Suction
cup
figure
1842 Problem 5.
Solution
(a) The force exerted on the suction cup by the
atmosphere is
F
=
PA
=
P
atm
(
πd
2
/
4) = (1
.
013
×
10
5
Pa)
π
(0
.
016 m)
2
= 81
.
5 N (perfect vacuum inside
cup assumed). This is equal to the maximum weight.
(b) At Denver,
P
= 0
.
8
P
atm
,
so the maximum weight
is 80% of that in part (a), or 65.2 N (a slight variation
in
g
with altitude is neglected).
Problem
6. The pressure unit
torr
is defined as the pressure
that will support a column of mercury 1 mm high.
Meteorologists often give barometric pressure in
inches of mercury
, defined analogously. Express
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2
CHAPTER 18
each of these units in SI. The density of mercury is
1
.
36
×
10
4
kg
/
m
3
.
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 Spring '08
 WILKE
 Fluid Dynamics

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