Fluid Mechanics
145
(c)
58
92(1.013 10 Pa) (1.72 m)(11.5 m)
5.79 10 N
Fp
A
π
⊥
==
×
=
×
EVALUATE:
Most of the force in part (a) is due to the 92 atm of air pressure above the surface of the benzene and
the net force on the bottom of the tank is much less than the inward and outward forces.
14.23.
IDENTIFY:
The gauge pressure at the top of the oil column must produce a force on the disk that is equal to its weight.
SET UP:
The area of the bottom of the disk is
22
2
(0.150 m)
0.0707 m
Ar
ππ
=
.
EXECUTE:
(a)
0
2
45.0 N
636 Pa
0.0707 m
w
pp
A
−==
=
.
(b)
The increase in pressure produces a force on the disk equal to the increase in weight. By Pascal’s law the
increase in pressure is transmitted to all points in the oil.
(i)
2
83.0 N
1170 Pa
0.0707 m
p
Δ=
=
. (ii) 1170 Pa
EVALUATE:
The absolute pressure at the top of the oil produces an upward force on the disk but this force is
partially balanced by the force due to the air pressure at the top of the disk.
14.24.
IDENTIFY:
2
21
1
A
FF
A
=
.
2
F
must equal the weight
wm
g
=
of the car.
SET UP:
2
/4
AD
=
.
1
D
is the diameter of the vessel at the piston where
1
F
is applied and
2
D
of the diameter at
the car.
EXECUTE:
2
2
1
2
1
D
mg
F
D
=
.
2
2
11
(1520 kg)(9.80 m/s )
10.9
125 N
Dm
g
DF
=
EVALUATE:
The diameter is smaller where the force is smaller, so the pressure will be the same at both pistons.
14.25.
IDENTIFY:
Apply
y
y
Fm
a
=
∑
to the piston, with
y
+
upward.
A
=
.
SET UP:
5
1 atm
1.013 10 Pa
=×
. The force diagram for the piston is given in Figure 14.25.
p
is the absolute
pressure of the hydraulic fluid.
EXECUTE:
atm
0
pA w
p
A
−−
=
and
2
5
atm
gauge
(1200 kg)(9.80 m/s )
1.7 10 Pa
1.7 atm
(0.15 m)
g
p
−=
=
=
×
=
EVALUATE:
The larger the diameter of the piston, the smaller the gauge pressure required to lift the car.
Figure 14.25
14.26.
IDENTIFY:
Apply Newton&s 2nd law to the woman plus slab. The buoyancy force exerted by the water is upward
and given by
water
displ
,
B
Vg
ρ
=
where
displ
V
is the volume of water displaced.
SET UP:
The floating object is the slab of ice plus the woman; the buoyant force must support both. The volume
of water displaced equals the volume
ice
V
of the ice. The freebody diagram is given in Figure 14.26.
EXECUTE:
y
y
a
=
∑
tot
0
Bmg
−
=
water
ice
ice
(45.0 kg
)
m g
=+
But
/
mV
=
so
ice
ice ice
mp
V
=
Figure 14.26
3
ice
33
water
ice
45.0 kg
45.0 kg
0.562 m .
1000 kg/m
920 kg/m
V
ρρ
=
EVALUATE:
The mass of ice is
ice
ice ice
517 kg.
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Chapter 14
14.27.
IDENTIFY:
Apply
y
y
Fm
a
=
∑
to the sample, with
y
+
upward.
water
obj
B
Vg
ρ
=
.
SET UP:
17.50 N
wm
g
==
and
1.79 kg
m
=
.
EXECUTE:
0
TBm
g
+− =
.
17.50 N 11.20 N
6.30 N
Bm
gT
=−
=
−
=
.
43
obj
33
2
water
6.30 N
6.43 10 m
(1.00 10 kg/m )(9.80 m/s )
B
V
g
−
=
×
×
.
1.79 kg
2.78 10 kg/m
m
V
−
= ×
×
.
EVALUATE:
The density of the sample is greater than that of water and it doesn&t float.
14.28.
IDENTIFY:
The upward buoyant force
B
exerted by the liquid equals the weight of the fluid displaced by the
object. Since the object floats the buoyant force equals its weight.
SET UP:
Glycerin has density
gly
1.26 10 kg/m
=×
and seawater has density
sw
1.03 10 kg/m
. Let
obj
V
be
the volume of the apparatus.
2
E
9.80 m/s
g
=
;
2
C
4.15 m/s
g
=
. Let
sub
V
be the volume submerged on Caasi.
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 Spring '06
 Buchler
 Physics, Buoyancy, Force, m/s, volume flow rate

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