Fluid Mechanics
1415
14.61.
I
DENTIFY
:
Apply Newton°s 2nd law to the car. The buoyancy force is given by Archimedes° principle.
(a)
S
ET
U
P
:
The freebody diagram for the floating car is given in Figure 14.61.
sub
(
V
is the volume that is
submerged.)
E
XECUTE
:
y
y
F
ma
=
∑
0
B
mg
−
=
w
sub
0
V
g
mg
ρ
−
=
Figure 14.61
3
3
sub
w
/
(900 kg)/(1000 kg/m
)
0.900 m
V
m
ρ
=
=
=
3
3
sub
obj
/
(0.900 m
)/(3.0 m
)
0.30
30%
V
V
=
=
=
E
VALUATE
:
The average density of the car is
3
3
(900 kg)/(3.0 m
)
300 kg/m
.
=
car
water
/
0.30;
ρ
ρ
=
this equals
sub
obj
/
.
V
V
(b)
S
ET
U
P
:
When the car starts to sink it is fully submerged and the buoyant force is equal to the weight of the
car plus the water that is inside it.
E
XECUTE
:
When the car is full submerged
sub
,
V
V
=
the volume of the car and
3
3
2
4
water
(1000 kg/m
)(3.0 m
)(9.80 m/s
)
2.94
10
N
B
Vg
ρ
=
=
=
×
The weight of the car is
2
(900 kg)(9.80 m/s
)
8820 N.
mg
=
=
Thus the weight of the water in the car when it sinks is the buoyant force minus the weight of the car itself:
4
2
3
water
(2.94
10
N
8820 N)/(9.80 m/s
)
2.10
10
kg
m
=
×
−
=
×
And
3
3
3
water
water
water
/
(2.10
10
kg)/(1000 kg/m
)
2.10 m
V
m
ρ
=
=
×
=
The fraction this is of the total interior volume is
3
3
(2.10 m
)/(3.00 m
)
0.70
70%
=
=
E
VALUATE
:
The average density of the car plus the water inside it is
3
3
(900 kg
2100 kg)/(3.0 m
)
1000 kg/m
,
+
=
so
car
water
ρ
ρ
=
when the car starts to sink.
14.62.
I
DENTIFY
:
For a floating object, the buoyant force equals the weight of the object.
fluid
submerged
B
V
g
ρ
=
.
S
ET
U
P
:
Water has density
3
1.00 g/cm
ρ
=
.
E
XECUTE
:
(a)
The volume displaced must be that which has the same weight and mass as the ice,
3
3
9.70 gm
9.70 cm
1.00 gm/cm
=
.
(b)
No; when melted, the cube produces the same volume of water as was displaced by the floating cube, and the
water level does not change.
(c)
3
3
9.70 gm
9.24 cm
1.05 gm/cm
=
(d)
The melted water takes up more volume than the salt water displaced, and so
3
0.46 cm
flows over.
E
VALUATE
:
The volume of water from the melted cube is less than the volume of the ice cube, but the cube
floats with only part of its volume submerged.
14.63.
I
DENTIFY
:
For a floating object the buoyant force equals the weight of the object. The buoyant force when the
wood sinks is
water
tot
B
V
g
ρ
=
, where
tot
V
is the volume of the wood plus the volume of the lead.
/
m V
ρ
=
.
S
ET
U
P
:
The density of lead is
3
3
11.3
10
kg/m
×
.
E
XECUTE
:
3
wood
(0.600 m)(0.250 m)(0.080 m)
0.0120 m
V
=
=
.
3
3
wood
wood
wood
(600 kg/m
)(0.0120 m
)
7.20 kg
m
V
ρ
=
=
=
.
wood
lead
(
)
B
m
m
g
=
+
. Using
water
tot
B
V
g
ρ
=
and
tot
wood
lead
V
V
V
=
+
gives
water
wood
lead
wood
lead
(
)
(
)
V
V
g
m
m
g
ρ
+
=
+
.
lead
lead
lead
m
V
ρ
=
then gives
water
wood
water
lead
wood
lead
lead
V
V
m
V
ρ
ρ
ρ
+
=
+
.
3
3
4
3
water
wood
wood
lead
3
3
3
lead
water
(1000 kg/m
)(0.0120 m
)
7.20 kg
4.66
10
m
11.3
10
kg/m
1000 kg/m
V
m
V
ρ
ρ
ρ
−
−
−
=
=
=
×
−
×
−
.
lead
lead
lead
5.27 kg
m
V
ρ
=
=
.
E
VALUATE
:
The volume of the lead is only 3.9% of the volume of the wood. If the contribution of the volume of
the lead to
B
F
is neglected, the calculation is simplified:
water
wood
wood
lead
(
)
V
g
m
m
g
ρ
=
+
and
lead
4.8 kg
m
=
. The
result of this calculation is in error by about 9%.
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1416
Chapter 14
14.64.
I
DENTIFY
:
The fraction
f
of the volume that floats above the fluid is
fluid
1
,
f
ρ
ρ
=
−
where
ρ
is the average density
of the hydrometer (see Problem 14.29). This gives
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 Spring '06
 Buchler
 Physics, Buoyancy, Force, Buoyant Force, kg

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