13
FLUID MECHANICS
Answers to Multiple-Choice Problems
1. B
2. A 3. A, D 4. C
5. A 6. B
7. C
8. D 9. D 10. A 11. C
12. C
13. E 14. B
15. A
Solutions to Problems
13.1. Set Up: The density of gold is 19.3 3 103 kg / m3.
Solve: V 5 1 5.0 3 1023 m 2 1 15.0 3 1023 m 2 1 30.0 3 1023 m 2 5 2.25 3 1026 m3.
r5
0.0158 kg
m
5 7.02 3 103 kg / m3.
5
V
2.25 3 1026 m3
The metal is not pure gold.
13.2. Set Up: The density of platinum is 21.4 3 103 kg / m3.
Solve: r 5
40.0 kg
m
m
5 1.87 3 1023 m3. V 5 L 3 and L 5 0.123 m 5 12.3 cm.
so V 5 5
r
V
21.4 3 103 kg / m3
13.3. Set Up: The density of iron is 7.8 3 103 kg / m3. The volume of a cylinder of radius r and length L is pr 2L.
Solve: V 5 pr 2L 5 p 1 1.425 3 1022 m 2 2 1 0.858 m 2 5 5.47 3 1024 m3. r 5 m / V, so
m 5 rV 5 1 7.8 3 103 kg / m3 2 1 5.47 3 1024 m3 2 5 4.27 kg.
The weight is w 5 mg 5 41.8 N. This is about 9 lb; you will not need a cart.
1
2
13.4. Set Up: r 5 m / V so m 5 rV. w 5 mg. 1 gal 5 3.788 L 5 3.788 3 1023 m3.
Solve: V 5 1 11.5 gal 2
3.788 3 1023 m3
5 4.36 3 1022 m3
1 gal
m 5 rV 5 1 737 kg / m3 2 1 4.36 3 1022 m3 2 5 32.1 kg and w 5 mg 5 315 N
13.5. Set Up: r 5 m / V. For a sphere, V 5 43 pr 3.
Solve: (a) The mass of the nugget is
V5
r5
and its diameter is 2r 5 19.3 cm
1
2
31.1035 g
$1.00 3 106
5 7.29 3 104 g
5 1 2344 ounces 2
1 ounce
$426.6 / ounce
7.29 3 104 g
m
5
5 3.78 3 103 cm3
r
19.3 g / cm3
1 2
3 1 3.78 3 103 cm3 2 1/3
3V 1/3
5S
T 5 9.66 cm
4p
4p
13-1
13-2
Chapter 13
(b) The platinum nugget would have mass
m 5 rV 5 1 21.4 g / cm3 2 1 3.78 3 103 cm3 2 5 8.09 3 104 g 5 2.60 3 103 troy ounces
and would be worth 2.29 million dollars.
Reflect: The platinum nugget is worth more because a nugget of the same size has more mass, since platinum’s density is greater, and because it is worth more per ounce. The value if platinum is larger than the value if gold by
1 21.4 / 19.3 2 1 879.00 / 426.60 2 .
13.6. Set Up: The volume of a cube with side length L is L 3. The volume of a cylinder of radius r and length L is
pr 2L.
Solve: (a) The volume of the metal left after the hole is drilled is the volume of the solid cube minus the volume of
the cylindrical hole:
V 5 L 3 2 pr 2L 5 1 5.0 cm 2 3 2 p 1 1.0 cm 2 2 1 5.0 cm 2 5 109 cm3 5 1.09 3 1024 m3
The cube with the hole has mass
m5
7.50 N
w
5
5 0.765 kg
g
9.80 m / s2
and density
r5
0.765 kg
m
5
5 7.02 3 103 kg / m3
V
1.09 3 1024 m3
(b) The solid cube has volume V 5 L 3 5 125 cm3 5 1.25 3 1024 m3 and mass
m 5 rV 5 1 7.02 3 103 kg / m3 2 1 1.25 3 1024 m3 2 5 0.878 kg.
The original weight of the cube was w 5 mg 5 8.60 N.
1 2 1 /2
13.7. Set Up: r 5 m / V 5 m / L 3
L1
Solve: (a) m 5 rL 3 so r1L13 5 r2 L23 and r2 5 r1
1 2
1 2
3
5r
L1
3
5 8r
L2
L1 2
r 1 1/3
r 1 1/3
5L
5 L / 31/3
(b) L 2 5 L 1
r2
3r1
Reflect: If the volume is decreased while the mass is kept the same, then the density increases.
13.8. Set Up: r 5 m / V. mmix 5 mg 1 mw, where mg and mw are the masses of the two volumes of liquid before
they are mixed. Vmix 5 Vg 1 Vw, where Vg and Vw are the volumes of the two liquids before they are mixed. From
Table 13.1, glycerin has density rg 5 1.26 g / cm3 and water has density r w 5 1.00 g / cm3.
Solve: (a) mg 5 r gVg 5 1 1.26 g / cm3 2 1 200 cm3 2 5 252 g and
mw 5 rwVw 5 1 1.00 g / cm3 2 1 300 cm3 2 5 300 g.
mtot 5 552 g and Vmix 5 500 cm3 so
r mix 5
(b) Vg 5
mg
rg
5
200 g
1.26 g / cm
3
552 g
m mix
5
5 1.10 g / cm3
Vmix
500 cm3
5 159 cm3 and Vw 5
300 g
mw
5
5 300 cm3
rw
1.00 g / cm3
mmix 5 500 g, Vmix 5 459 cm3 so
r mix 5
m mix
5 1.09 g / cm3
Vmix
Fluid Mechanics
13.9. Set Up: r 5 m / V. mmix 5 m1 1 m2 and Vmix 5 V1 1 V2
Solve: (a) m1 5 m 2 5 m so mmix 5 2m. V1 5
rmix 5
rav 5
m mix
5
Vmix
1
2
2r1r 2
2m
5
rav
r1 1 r2
r1 1 r2
m
r 1r 2
r1 1 r2
.
2
1
2 1
r1 1 r2
m
m
1
1
, V2 5
1
5m
so Vmix 5 m
r1
r2
r1
r2
r 1r 2
13-3
2
and
In general rmix is not equal to rav and this proves that the answer is no. For example, if r1 5 1.00 g / cm3 and
r2 5 2.00 g / cm3, then r mix 5 1.33 g / cm3 and r av 5 1.50 g / cm3, which is different,
(b) V1 5 V2 5 V so Vmix 5 2V. m1 5 r1V, m2 5 r 2V so mmix 5 1 r 1 1 r2 2 V
rmix 5
1 r1 1 r2 2 V r1 1 r2
mmix
5
,
5
Vmix
2V
2
which is the same as the average of the two densities. This proves that the answer is yes.
Reflect: If r1 5 r 2 5 r then in (a) both rmix and rav equal r.
13.10. Set Up: 1 mm Hg 5 1 torr. 1 atm 5 14.7 psi 5 1.103 3 105 Pa. 1 Pa 5 1 N / m2. r 5 13.6 3 103 kg / m3
(Table 13.1).
Solve: 1 mm Hg is the pressure p corresponding to h 5 1 mm and r 5 rHg in the equation p 5 rgh. So, 1 mm Hg
corresponds to p 5 1 13.6 3 103 kg / m3 2 1 9.80 m / s2 2 1 1.00 3 1023 m 2 5 133 Pa
(a) 1 120 mm Hg 2
be
0.158
.
0.105
1
1
21
2
133 Pa
1 atm
5 0.158 atm and 80 mm Hg 5 0.105 atm, so the expression would
1 mm Hg 1.013 3 105 Pa
2
(b) 1 mm Hg 5 1 torr so the expression is the same,
(c) 1 120 mm Hg 2
120
.
80
133 Pa
5 1.60 3 104 Pa and 80 mm Hg 5 1.06 3 104 Pa, so the expression would be
1 mm Hg
1.60 3 104
1.06 3 104
2
(d) 1 Pa 5 1 N / m so the expression would be the same as in (c).
(e) 120 mm Hg 5 0.158 atm 5 2.32 psi and 80 mm Hg 5 0.105 atm 5 1.54 psi, so the expression would be
2.32
.
1.54
13.11. Set Up: 5 L 5 5 3 1023 m3. r 5 m / V. F 5 pA. 1 cm2 5 1024 m2. A sphere of radius r has volume
V 5 43 pr 3. Specific gravity 5 5.0 means r 5 5.0 3 103 kg / m3.
Solve: (a) m 5 rV 5 1 1050 kg / m3 2 1 5 3 1023 m3 2 5 5.25 kg
(b) F 5 pA 5 1 13,000 Pa 2 1 1.0 3 1024 m2 2 5 1.3 N
(c) m 5 rV 5 1 5.0 3 103 kg / m3 2 A 43 Bp 1 3.75 3 1026 m 2 3 5 1.1 3 10212 kg 5 1.1 3 1029 g
13.12. Set Up: The surface area of a hemisphere is A 5 3pr 2. On earth, pair 5 1.01 3 105 Pa. F 5 pA.
Solve: (a) F 5 1 92 2 1 1.01 3 105 Pa 2 1 3p 2 1 1.25 m 2 2 5 1.4 3 108 N
1.4 3 108 N
(b) On earth the pressure is less by a factor of 92, so F 5
5 1.5 3 106 N
92
13-4
Chapter 13
13.13. Set Up: The density of seawater is 1.03 3 103 kg / m3.
Solve: (a) pgauge 5 p0 2 patm 5 rgh 5 1 1.03 3 103 kg / m3 2 1 9.80 m / s2 2 1 250 m 2 5 2.52 3 106 Pa.
(b) Fnet 5 pgauge A 5 1 2.52 3 106 Pa 2 p 1 0.150 m 2 2 5 1.78 3 105 N
Reflect: At the surface the gauge pressure is zero. At a depth of 250 m the force due to the water pressure is very
large.
1
2
13.14. Set Up: 1 mm Hg 5 133.3 Pa. F 5 pA. The surface area of a sphere of radius r is A 5 4pr 2.
Solve: DF 5 1 Dp 2 A 5 1 26 mm Hg 2
133.3 Pa
1 4p 2 1 1.25 3 1022 m 2 2 5 6.8 N
1 mm Hg
13.15. Set Up: Gauge pressure is p 2 patm. The density of seawater is 1.03 3 103 kg / m3.
Solve: p 2 patm 5 rgh.
At a depth of 1.5 km, p 2 patm 5 1 1.03 3 103 kg / m3 2 1 9.80 m / s2 2 1 1.5 3 103 m 2 5 1.5 3 107 Pa.
At a depth of 3.2 km, p 2 patm 5 3.2 3 107 Pa.
13.16. Set Up: 1 atm 5 1.013 3 105 Pa. The density of water is 1.00 3 103 kg / m3. The gauge pressure must
equal the pressure difference due to a column of water 1370 m 2 730 m 5 640 m tall.
Solve: pgauge 5 p0 2 patm 5 rgh 5 1 1.00 3 103 m3 2 1 9.80 m / s2 2 1 640 m 2 5 6.27 3 106 Pa 5 61.9 atm
13.17. Set Up: 1 mm Hg 5 133.3 Pa
Solve: p 5 rgh 5 1 1060 kg / m3 2 1 9.80 m / s2 2 1 1.20 m 2 5 1.25 3 104 Pa 5 93.5 mm Hg
13.18. Set Up: Gauge pressure is p 2 patm. The density of fresh water is 1.00 3 103 kg / m3 and the density of
seawater is 1.03 3 103 kg / m3.
Solve: p 2 patm 5 rgh
(a) p 2 patm 5 1 1.00 3 103 kg / m3 2 1 0.379 2 1 9.80 m / s2 2 1 0.50 3 103 m 2 5 1.9 3 106 Pa
(b) p 2 patm 5 1 1.03 3 103 kg / m3 2 1 9.80 m / s2 2 1 0.50 3 103 m 2 5 5.0 3 106 Pa
13.19. Set Up: 1 torr 5 1 mm Hg 5 133.3 Pa. 1 atm 5 1.013 3 105 Pa 5 760 torr.
p
133.3 Pa
5
5 1.36 cm
3
rg
1 1.00 3 10 kg / m3 2 1 9.80 m / s2 2
(b) 1 atm 5 760 torr so h 5 1 760 2 1 1.36 cm 2 5 10.3 m
The height of a column of mercury at p 5 1 atm would be only 0.76 m.
Reflect: If water is used, the height of the barometer for ordinary atmospheric pressures is impractical. Mercury is
used instead because its density is larger.
Solve: (a) p 5 rgh so h 5
13.20. Set Up: The added pressure is equal to rgh.
Solve: (a) rgh 5 1 1050 kg / m3 2 1 9.80 m / s2 2 1 1.85 m 2 5 1.90 3 104 Pa
(b) This additional pressure causes additional outward force on the walls of the blood vessels in your brain.
13.21. Set Up: 1 atm 5 1.013 3 105 Pa. The pressure difference is Dp 5 rgh.
Solve: (a) Dp 5 0.125 atm 5 1.27 3 104 Pa.
r5
Dp
1.27 3 104 Pa
5
5 864 kg / m3
gh
1 9.80 m / s2 2 1 1.50 m 2
(b) Dp 5 rgMarsh 5 1 864 kg / m3 2 1 0.379 2 1 9.80 m / s2 2 1 1.50 m 2 5 4.81 3 103 Pa 5 0.0475 atm
Reflect: The pressure difference on Mars is less than it is on earth by a factor of 0.379. Our calculated density of oil
is a reasonable value, somewhat less than the density of water.
Fluid Mechanics
13-5
13.22. Set Up: The density of seawater is 1.03 3 103 kg / m3. The area of the eardrum is A 5 pr 2, with
r 5 4.1 mm. The pressure increase with depth is Dp 5 rgh and F 5 pA.
Solve: DF 5 1 Dp 2 A 5 rghA
h5
DF
1.5 N
5 2.8 m
5
rgA
1 1.03 3 103 kg / m3 2 1 9.80 m / s2 2 p 1 4.1 3 1023 m 2 2
13.23. Set Up: p0 5 psurface 1 rgh where psurface is the pressure at the surface of a liquid and p0 is the pressure at a
depth h below the surface.
Solve: (a) For the oil layer, psurface 5 patm and p0 is the pressure at the oil-water interface.
p0 2 patm 5 pgauge 5 rgh 5 1 600 kg / m3 2 1 9.80 m / s2 2 1 0.120 m 2 5 706 Pa
(b) For the water layer, psurface 5 706 Pa 1 patm.
p0 2 patm 5 pgauge 5 706 Pa 1 rgh 5 706 Pa 1 1 1.00 3 103 kg / m3 2 1 9.80 m / s2 2 1 0.250 m 2 5 3.16 3 103 Pa
Reflect: The gauge pressure at the bottom of the barrel is due to the combined effects of the oil layer and water
layer. The pressure at the bottom of the oil layer is the pressure at the top of the water layer
13.24. Set Up: F 5 pA. 1 cm2 5 1 3 1024 m2. 1 torr 5 1 mm Hg 5 133.3 Pa. The pressure change with depth is
Dp 5 rgh. The arm is above the heart, so the pressure in the arm is less than the pressure at the heart.
Solve: (a) maximum F 5 1 120 torr 2 1 133.3 Pa / torr 2 1 1.0 3 1024 m2 2 5 1.6 N
minimum F 5 1 80 torr 2 1 133.3 Pa / torr 2 1 1.0 3 1024 m2 2 5 1.1 N
(b) The pressure in the arm is less by an amount
rgh 5 1 1050 kg / m3 2 1 9.80 m / s2 2 1 0.300 m 2 5 3.09 3 103 Pa 5 23 torr
The blood pressure reading should be
120 2 23
97
5 .
80 2 23
57
13.25. Set Up: 1 torr 5 1 mm Hg 5 133.3 Pa. The pressure change with depth is Dp 5 rgh. The arm is above the
heart, so the pressure in the arm is less than the pressure at the heart. On the moon, g 5 1.67 m / s2.
Solve: The pressure in the arm is less by an amount
rgh 5 1 1050 kg / m3 2 1 1.67 m / s2 2 1 0.25 m 2 5 438 Pa 5 3.3 torr
120 2 3
117
5
.
80 2 3
77
Reflect: The changes in blood pressure for points in the body above or below the heart is much less on the moon
than on Earth, since g is much less on the moon.
The blood pressure reading should be
13.26. Set Up: 1 atm 5 1.013 3 105 Pa. The force diagram for the piston is given in Figure 13.26. p is the
absolute pressure of the hydraulic fluid.
y
patm A
x
pA
w
Figure 13.26
Solve: pA 2 w 2 patm A 5 0 and
p 2 patm 5 pgauge 5
1 1200 kg 2 1 9.80 m / s2 2
mg
w
5
5
5 1.7 3 105 Pa 5 1.7 atm
A
pr 2
p 1 0.15 m 2 2
13-6
Chapter 13
13.27. Set Up: Pascal’s law says the pressure is the same everywhere in the hydraulic fluid, so F1 / A1 5 F2 / A2.
F1 5 100 N and F 5 1 3500 kg 2 1 9.80 m / s2 2 5 3.43 3 104 N, the weight of the car and platform that is being
lifted. The volume of fluid displaced at each piston when the pistons move distances d1 and d2 is the same, so
d1A1 5 d2A2.
F1
F2
F2
3.43 3 104 N
5 1 0.125 m 2
5 2.32 m and the diameter is 4.64 m.
Solve: (a) A 5 pr 2 so 2 5 2 . r2 5 r1
Å F1
Å
100 N
r1
r2
(b) d1A1 5 d2A2.
d2 5 d1
1
A1
pr12
0.125 m
5 d1 2 5 1 50 cm 2
A2
2.32 m
pr2
2
2
5 1.45 mm
Reflect: The work done by the man is F1d1 5 1 100 N 2 1 0.50 m 2 5 50 J. The work done on the car is
F2d2 5 1 3.43 3 104 N 2 1 1.43 3 1023 m 2 5 50 J. Energy conservation requires that these two quantities be equal,
and they are equal.
13.28. Set Up: After the barrier is removed the top of the water moves downward a distance x and the top of the oil
moves up a distance x, as shown in Figure 13.28. After the heights have changed, the gauge pressure at the bottom of
each of the tubes is the same.
x
25.0 cm
x
25.0 cm 2 x
x
Figure 13.28
Solve: The gauge pressure at the bottom of arm A of the tube is p 2 patm 5 r wg 1 25.0 cm 2 x 2 . The gauge pressure at the bottom of arm B of the tube is p 2 patm 5 roilg 1 25.0 cm 2 1 r wgx. The gauge pressures must be
equal, so rwg 1 25.0 cm 2 x 2 5 roilg 1 25.0 cm 2 1 rwgx. g divides out and roil 5 0.80r w, so r w 1 25.0 cm 2 x 2 5
0.80rw 1 25.0 cm 2 1 rw x. r w divides out and leaves 25.0 cm 2 x 5 20.0 cm 1 x, so x 5 2.5 cm
The height of fluid in arm A is 25.0 cm 2 x 5 22.5 cm and the height in arm B is 25.0 cm 1 x 5 27.5 cm.
(b) (i) If the densities were the same there would be no reason for a difference in height and the height would be
25.0 cm on each side. (ii) The pressure exerted by the column of oil would be very small and the water would
divide equally on both sides. The height in arm A would be 12.5 cm and the height in arm B would be
25.0 cm 1 12.5 cm 5 37.5 cm.
13.29. Set Up: The density of aluminum is 2.7 3 103 kg / m3. The density of water is 1.00 3 103 kg / m3. r 5 m / V.
The buoyant force is FB 5 rwaterVobjg.
9.08 kg
m
5
5 3.36 3 1023 m3 5 3.4 L.
r
2.7 3 103 kg / m3
(b) When the ingot is totally immersed in the water while suspended, T 1 FB 2 mg 5 0.
Solve: (a) T 5 mg 5 89 N so m 5 9.08 kg. V 5
FB 5 rwaterVobjg 5 1 1.00 3 103 kg / m3 2 1 3.36 3 1023 m3 2 1 9.80 m / s2 2 5 32.9 N.
T 5 mg 2 FB 5 89 N 2 32.9 N 5 56 N.
Reflect: The buoyant force is equal to the difference between the apparent weight when the object is submerged in
the fluid and the actual gravity force on the object.
Fluid Mechanics
13-7
13.30. Set Up: The buoyant force exerted by the water is FB 5 rwgVfish. When the fish is fully submerged the
buoyant force on it must equal its weight.
Solve: (a) The average density of the fish is very close to the density of water.
(b) Before inflation, FB 5 w 5 1 2.75 kg 2 1 9.80 m / s2 2 5 27.0 N. When the volume increases by a factor of 1.10, the
buoyant force also increases by a factor of 1.10 and becomes 1 1.10 2 1 27.0 N 2 5 29.7 N.
(c) The water exerts an upward force 29.7 N and gravity exerts a downward force of 27.0 N so there is a net upward
force of 2.7 N; the fish moves upward.
13.31. Set Up: FB 5 rfluidgVsub, where Vsub is the volume of the object that is below the fluid’s surface.
Solve: (a) Floats, so the buoyant force equals the weight of the object: FB 5 mg. Using Archimedes’s principle
gives rwgV 5 mg and
5750 kg
m
5
5 5.75 m3.
V5
rw
1.00 3 103 kg / m3
(b) FB 5 mg and rwgVsub 5 mg. Vsub 5 0.80V 5 4.60 m3, so the mass of the floating object is
m 5 r wVsub 5 1 1.00 3 103 kg / m3 2 1 4.60 m3 2 5 4600 kg.
He must throw out 5750 kg 2 4600 kg 5 1150 kg.
Reflect: He must throw out 20% of the boat’s mass.
13.32. Set Up: FB 5 rwaterVobjg. w 5 mg 5 17.50 N and m 5 1.79 kg.
Solve: T 1 FB 2 mg 5 0. FB 5 mg 2 T 5 17.50 N 2 11.20 N 5 6.30 N.
Vobj 5
FB
6.30 N
5
5 6.43 3 1024 m3.
3
rwaterg
1 1.00 3 10 kg / m3 2 1 9.80 m / s2 2
r5
1.79 kg
m
5
5 2.78 3 103 kg / m3
V
6.43 3 1024 m3
13.33. Set Up: FB 5 rwaterViceg. mice 5 riceV. rice 5 0.92 3 103 kg / m3.
Solve: The ice is floating, so FB 5 1 mice 1 mwoman 2 g. rwaterVice 5 riceVice 1 mwoman.
Vice 5
45.0 kg
mwoman
5
5 0.562 m3.
rwater 2 rice
1000 kg / m3 2 920 kg / m3
Reflect: The buoyant force must support both the slab of ice and the woman. If the slab of ice is 20 cm thick and
square on its top and bottom surfaces, its top surface measures about 1.7 m on a side. The woman would also need to
be concerned about the slab tipping to one side, because of unbalanced torques.
13.34. Set Up: The pressure increase at a depth h below the surface of a liquid of density r is rgh. The upper face
of the block is 1.50 cm below the surface of the oil.
Solve: (a) pgauge 5 r oilg 1 1.50 3 1022 m 2 5 1 790 kg / m3 2 1 9.80 m / s2 2 1 1.50 3 1022 m 2 5 116 Pa
(b) pgauge 5 roilg 1 0.100 m 2 1 rwaterg 1 0.0150 m 2
pgauge 5 1 9.80 m / s2 2 1 3 790 kg / m3 4 3 0.100 m 4 1 3 1000 kg / m3 4 3 0.015 m 4 2 5 921 Pa
(c) Let pt be the absolute pressure at the top surface of the block. This gives a downward force pt A on the block. Let
pb be the absolute pressure at the bottom surface of the block. This gives an upward force pb A on the block.
pb A 5 pt A 1 mg.
m5
1 pgauge, b 2 pgauge, t 2 A
1 pb 2 pt 2 A
1 921 Pa 2 116 Pa 2 1 0.100 m 2 2
5
5
5 0.821 kg.
g
g
9.80 m / s2
r5
0.821 kg
m
5
5 821 kg / m3.
1 0.100 m 2 3
V
13-8
Chapter 13
13.35. Set Up: FB 5 rwaterVobjg. The net force on the sphere is zero.
Solve: (a) FB 5 1 1000 kg / m3 2 1 0.650 m3 2 1 9.80 m / s2 2 5 6.37 3 103 N
(b) FB 5 T 1 mg and
m5
FB 2 T
6.37 3 103 N 2 900 N
5
5 558 kg.
g
9.80 m / s2
(c) Now FB 5 r waterVsubg, where Vsub is the volume of the sphere that is submerged. FB 5 mg. rwaterVsub 5 mg and
Vsub 5
558 kg
m
5
5 0.558 m3.
rwater
1000 kg / m3
Vsub
0.558 m3
5
5 0.858 5 85.8%
Vobj
0.650 m3
Reflect: When the sphere is totally submerged, the buoyant force on it is greater than its weight. When it is floating,
it needs to be only partially submerged in order to produce a buoyant force equal to its weight.
13.36. Set Up: FB 5 rfluidgVsub. For a floating object, FB 5 mg, the weight of the object. Water has density
rw 5 1.00 3 103 kg / m3 and seawater has density rsw 5 1.03 3 103 kg / m3.
Solve: (a) mg 5 r fluidgVsub. Just barely floats when Vsub 5 V, the volume of the object.
V5
and V 5 1 18.0 m 2 1 15.0 m 2 h so
250,000 kg
m
5
5 243 m3
rsw
1.03 3 103 kg / m3
h5
243 m3
5 90.0 cm
1 18.0 m 2 1 15.0 m 2
(b) The volume is increased by 75% so the buoyant force is increased by 75% and can add 75% more mass, so can
carry a load of 1.88 3 105 kg. The total floating mass is 4.38 3 105 kg.
(c) In part (b) the total mass supported by the buoyant force is 2.50 3 105 kg 1 1.88 3 105 kg 5 4.38 3 105 kg.
In fresh water the buoyant force is less by a factor of rw / r sw 5 0.971 so the total floating mass is
1 0.971 2 1 4.38 3 105 kg 2 5 4.25 3 105 kg and the extra mass that the barge can carry is 1.75 3 105 kg.
13.37. Set Up: FB 5 rfluidgVsub. For a floating object, FB 5 mg 5 robjVobjg. rice 5 0.92 3 103 kg / m3.
Solve: (a) FB 5 mg says r swgVsub 5 ricegVice. Vsub 5 0.90Vice, so 0.90rsw 5 r ice and
rsw 5
rice
5 1.02 3 103 kg / m3.
0.90
(b) rwgVsub 5 r icegVice and
0.92 3 103 kg / m3
r ice
Vsub
5
5
5 0.92
rw
Vice
1.00 3 103 kg / m3
92.0% is submerged and 8.0% is above the surface.
13.38. Set Up: 1 dyn / cm 5 1023 N / m. The excess pressure inside a liquid drop is Dp 5
Solve: (a) Dp 5
2 1 72.8 3 1023 N / m 2
5 97.1 Pa
1.50 3 1023 m
2 1 72.8 3 1023 N / m 2
(b) Dp 5
5 1.46 3 104 Pa
0.0100 3 1023 m
2g
.
R
Fluid Mechanics
13-9
13.39. Set Up: 1 dyn / cm 5 1023 N / m. The gauge pressure inside a bubble is
pgauge 5 p 2 patm 5
Solve: pgauge 5
4 1 25.0 3 1023 N / m 2
3.50 3 1022 m
4g
.
R
5 2.86 Pa.
13.40. Set Up: 1 dyn / cm 5 1023 N / m. 1 atm 5 1.013 3 105 Pa. The pressure difference for a liquid drop is
Dp 5
Solve: R 5
2g
.
R
2 1 72.8 3 1023 N / m 2
2g
5
5 7.19 3 1025 m 5 0.0719 mm.
Dp
1 0.0200 atm 2 1 1.013 3 105 Pa / atm 2
13.41. Set Up: The gauge pressure p 2 patm in a drop is related to the surface tension g of the liquid and the radius
4g
. The volume of a spherical drop is V 5 43 pR 3.
R
3V 1/3
4p 1/3
. Same gauge pressure, so gV 21/3 5 constant and gwVw21/3 5 gctVct21/3.
Solve: R 5
so p 2 patm 5 4g
4p
3V
R of the drop by p 2 patm 5
1 2
1 2
1 2 1
gw
Vw
5
gct
Vct
3
5
72.8 dynes / cm
26.8 dynes / cm
2
3
5 20.0
Reflect: Water has a larger surface tension so the water drop is larger.
13.42. Set Up: The continuity equation is A1v1 5 A2v2.
Solve: 1 18.5 m 2 1 3.75 m 2 1 2.50 cm / s 2 5 1 16.5 m 2 h 1 11.0 cm / s 2 so h 5 0.96 m, the depth of the canal at the second point.
1 2
1
13.43. Set Up: v1A1 5 v2A2
(b) v2 5 v1
1 2
A1
A2
2
A1
0.070 m2
5 1 3.50 m / s 2
5 2.33 m / s
A2
0.105 m2
0.070 m2
5 1 3.50 m / s 2
5 5.21 m / s
0.047 m2
Solve: (a) v2 5 v1
1
2
13.44. Set Up: The volume flow rate is v1A1 5 v2A2.
1.20 m3 / s
1.20 m3 / s
5
5 17.0 m / s
A1
p 1 0.150 m 2 2
2
2
(b) v1pr1 5 v2pr2 .
Solve: (a) v1 5
r2 5 r1
17.0 m / s
v1
5 1 0.150 m 2
5 0.317 m
Å v2
Å 3.80 m / s
1 2
1
2
13.45. Set Up: The continuity equation is A1v1 5 A2v2. A 5 12 pd 2, where d is the pipe diameter.
d1 2
2.50 in. 2
1 6.00 cm / s 2 5 37.5 cm / s
v1 5
d2
1.00 in.
Reflect: To achieve the same volume flow rate the water flows faster in the smaller diameter pipe. Note that the pipe
diameters entered in a ratio so there was no need to convert units.
Solve: 12 pd12v1 5 12 pd22v2 and v2 5
13-10
Chapter 13
13.46. Set Up: The equation of continuity says A1v1 5 A2v2.
Solve: (a) d S d / 2 says A S A / 4. To keep vA constant the speed becomes 4 times as great.
(b) v S v / 5 requires A S 5A. You would have to make the cross-sectional area 5 times larger.
(c) v S 5v requires A S A / 5. A 5 12 pd 2 so to make A a factor of 5 times smaller make d a factor of "5 times
smaller, dnew 5 dold / "5 .
13.47. Set Up: Toricelli’s theorem says the speed of efflux is v 5 "2gh , where h is the distance of the small hole
below the surface of the water in the tank.
Solve: v 5 "2 1 9.80 m / s2 2 1 14.0 m 2 5 16.6 m / s
13.48. Set Up: Let point 1 be at the surface of the water in the tank. Let point 2 be in the emerging stream of water.
Since the hole is small, v1 < 0. p1 2 pair 5 3.00 atm 5 3.04 3 105 Pa. p2 5 pair. y1 2 y2 5 11.0 m. The density of
seawater is r 5 1.03 3 103 kg / m3.
Solve: p1 1 rgy1 1 12 rv12 5 p2 1 rgy2 1 12 rv22. p1 2 pair 1 rg 1 y1 2 y2 2 5 12 rv22.
v2 5
2 1 p1 2 pair 2
2 1 3.04 3 105 Pa 2
1 2g 1 y1 2 y2 2 5
1 2 1 9.80 m / s2 2 1 11.0 m 2 5 28.4 m / s
r
Å
Å 1.03 3 103 kg / m3
13.49. Set Up: Let point 1 be in the mains and point 2 be in the emerging stream at the end of the fire hose. v1 < 0.
y1 2 y2 5 0. p2 5 pair. After the water emerges from the hose, with upward velocity, it moves in free fall.
Solve: p1 1 rgy1 1 12 rv12 5 p2 1 rgy2 1 12 rv22 gives p1 2 pair 5 12 rv22. The motion of a drop of water after it
leaves the hose gives 12mv22 5 mgh.
v22 5 2gh 5 2 1 9.80 m / s2 2 1 15.0 m 2 5 294 m2 / s2.
p1 2 pair 5 12 1 1000 kg / m3 2 1 294 m2 / s2 2 5 1.47 3 105 Pa
Reflect: This is the gauge pressure required to support a column of water 15.0 m high.
13.50. Set Up: v1A1 5 v2A2 relates v2 and v1. r2 5 2r1.
Solve: p1 1 rgy1 1 12 rv12 5 p2 1 rgy2 1 12 rv22. We can subtract air pressure pair from both sides of the equation,
and the pressures become gauge pressures.
v2 5 v1
12
A1
pr12
1
5 v1 2 5 1 3.00 m / s 2
A2
2
pr2
p2 5 p1 1 rg 1 y1 2 y2 2 1 12 r 1 v12 2 v22 2
2
5 0.750 m / s. y1 2 y2 5 111.0 m.
p2 5 4.00 3 104 Pa 1 1 1000 kg / m3 2 1 9.80 m / s2 2 1 11.0 m 2 1 12 1 1000 kg / m3 2 1 3 3.00 m / s 4 2 2 3 0.750 m / s 4 2 2
p2 5 1.52 3 105 Pa
13.51. Set Up: Let point 1 be at the top surface and point 2 be at the bottom surface. Neglect the rg 1 y1 2 y2 2 term
in Bernoulli’s equation. In calculating the net force, gFy, take 1y to be upward.
Solve: p1 1 rgy1 1 12 rv12 5 p2 1 rgy2 1 12 rv22.
p2 2 p1 5 12 r 1 v12 2 v22 2 5 12 1 1.20 kg / m3 2 1 3 70.0 m / s 4 2 2 3 60.0 m / s 4 2 2 5 780 Pa
gFy 5 p2A 2 p1A 2 mg 5 1 780 Pa 2 1 16.2 m2 2 2 1 1340 kg 2 1 9.80 m / s2 2 5 2500 N.
The net force is 500 N, downward.
Reflect: The pressure is lower where the fluid speed is higher.
Fluid Mechanics
13-11
13.52. Set Up: A1v1 5 A2v2. A2 5 0.60A1. p1 1 rgy1 1 12 rv12 5 p2 1 rgy2 1 12 rv22. The fluid is water, so
1 2
1
2
r 5 1.00 3 103 kg / m3.
A1
A1
5 1 8.5 cm / s 2
5 14.2 cm / s
A2
0.60A1
(b) y1 5 y2 so p1 2 p2 5 12 r 1 v22 2 v12 2 5 12 1 1.00 3 103 kg / m3 2 1 3 0.142 m / s 4 2 2 3 0.085 m / s 4 2 2 5 6.47 Pa
Solve: (a) v2 5 v1
13.53. Set Up: y1 5 y2. v1A1 5 v2A2 5 465 3 1026 m3 / s.
Solve: p1 1 rgy1 1 12 rv12 5 p2 1 rgy2 1 12 rv22.
v1 5
Then v2 5
465 3 1026 m3 / s
pr12
5
465 3 1026 m3 / s
5 0.352 m / s
p 1 2.05 3 1022 m 2 2
2
2
1 p 2 p2 2 1 v12 5
1 0.40 3 105 Pa 2 1 1 0.352 m / s 2 2 5 8.95 m / s.
År 1
Å 1000 kg / m3
v1pr12 5 v2pr22 and
r2 5 r1
0.352 m / s
v1
5 1 2.05 cm 2
5 4.07 mm.
Å v2
Å 8.95 m / s
13.54. Set Up: p1 1 rgy1 1 12 rv12 5 p2 1 rgy2 1 12 rv22. The two points are close together so we can neglect
rg 1 y1 2 y2 2 . r 5 1.03 3 103 kg / m3. A1v1 5 A2v2.
Solve: p1 2 p2 1 12 rv12 5 12 rv22
v2 5
2 1 p1 2 p2 2
2 1 1.20 3 104 Pa 2 1.15 3 104 Pa 2
1 v12 5
1 1 0.300 m / s 2 2
r
Å
Å
1.05 3 103 kg / m3
v2 5 1.0 m / s 5 100 cm / s
30 cm / s
v1
A2
5 0.30.
5
5
v
A1
100 cm / s
2
A2 5 0.30A1, so 70% is blocked.
13.55. Set Up: y1 5 y2. v1A1 5 v2A2. A2 5 2A1.
Solve: p1 1 rgy1 1 12 rv12 5 p2 1 rgy2 1 12 rv22.
v2 5 v1
1 2
1 2
A1
A1
5 1 2.50 m / s 2
5 1.25 m / s.
A2
2A1
p2 5 p1 1 12 r 1 v12 2 v22 2 5 1.80 3 104 Pa 1 12 1 1000 kg / m3 2 1 3 2.50 m / s 4 2 2 3 1.25 m / s 4 2 2 5 2.03 3 104 Pa
Reflect: The gauge pressure is higher at the second point because the water speed is less there.
13.56. Set Up: At the terminal speed, a 5 0 and the forces balance. The viscous drag force is 6phrvt. The weight
of the ball bearing is mg 5 43 pr 3rg. The density of steel is r 5 7.8 3 103 kg / m3.
2 1 1.00 3 1023 m 2 2 1 9.80 m / s2 2 1 7.8 3 103 kg / m3 2
2r 2gr
5 0.113 m / s.
5
Solve: vt 5
9h
9 1 0.150 N # s / m2 2
13.57. Set Up: The viscous drag force is F 5 6phrvt. The weight of the ball bearing is mg 5 43 pr 3rg.
Solve: F 5 14 mg gives 6phrvt 5 13 pr 3rg.
vt 5
1 3.00 3 1023 m 2 2 1 9.80 m / s2 2 1 19.3 3 103 kg / m3 2
r 2gr
5
5 0.0959 m / s 5 9.59 cm / s.
18h
18 1 0.986 N # s / m2 2
13-12
Chapter 13
13.58. Set Up: At the terminal speed a 5 0 and the forces balance. The viscous drag force is 6phrvt. The weight
of the sphere is mg 5 43 pr 3rg.
3 3 0.20 3 1023 kg 4 1/3
3m 1/3
5
5 1.75 3 1023 m. 6phrvt 5 mg and
Solve: The radius of the sphere is r 5
4pr
4p 3 8900 kg / m3 4
h5
1 2 1
2
1 0.20 3 1023 kg 2 1 9.80 m / s2 2
mg
5
5 0.990 N # s / m2.
6prvt
6p 1 1.75 3 1023 m 2 1 6.0 3 1022 m / s 2
1
2
1
2
13.59. Set Up: The flow rate, DV / Dt, is related to the radius R or diameter D of the artery by Poiseuille’s law:
pR 4 p1 2 p2
pD 4 p1 2 p2
DV
5
5
.
Dt
8h
L
128h
L
Assume the pressure gradient 1 p1 2 p2 2 / L in the artery remains the same.
Solve: 1 DV / Dt 2 / D 4 5
2 1 DV / Dt 2 old and Dold
1
2
p p1 2 p2
5 constant, so
128h
L
5 D. This gives
Dnew 5 Dold S
1 DV / Dt 2 old / Dold4 5 1 DV / Dt 2 new / Dnew4.
1 DV / Dt 2 new 5
1 DV / Dt 2 new 1/4
T 5 21/4D 5 1.19D.
1 DV / Dt 2 old
Reflect: Since the flow rate is proportional to D 4, a 19% increase in D doubles the flow rate.
13.60. Set Up: Seawater has density r 5 1.03 3 103 kg / m3. The bulk modulus of water is B 5 2.2 3 109 Pa
(Example 11-2). pair 5 1.01 3 105 Pa.
Solve: (a)
p0 5 pair 1 rgh 5 1.01 3 105 Pa 1 1 1.03 3 103 kg / m3 2 1 9.80 m / s2 2 1 10.92 3 103 m 2 5 1.10 3 108 Pa
(b) At the surface 1.00 m3 of seawater has mass 1.03 3 103 kg. At a depth of 10.92 km the change in volume is
DV 5 2
1 Dp 2 V0
1 1.10 3 108 Pa 2 1 1.00 m3 2
52
5 20.050 m3.
B
2.2 3 109 Pa
The volume of this mass of water at this depth therefore is V 5 V0 1 DV 5 0.950 m3.
r5
1.03 3 103 kg
m
5 1.08 3 103 kg / m3.
5
V
0.950 m3
13.61. Set Up: rHg 5 13.6 3 103 kg / m3. Let x be the height of the mercury surface in the right arm above the level
of the mercury-water interface in the left arm.
Solve: (a) p 2 pair 5 rwatergh 5 1 1.00 3 103 kg / m3 2 1 9.80 m / s2 2 1 0.150 m 2 5 1.47 3 103 Pa
(b) The gauge pressure a distance x below the mercury surface in the right arm equals the gauge pressure at the
mercury-water interface, since these two points are at the same height in the mercury. r Hggx 5 1.47 3 103 Pa and
x5
1.47 3 103 Pa
5 1.1 cm.
1 13.6 3 103 kg / m3 2 1 9.80 m / s2 2
h 1 x 5 15.0 cm, so h 5 13.9 cm.
Reflect: The weight of the water pushes the mercury down in the left-hand arm. A 1.1 cm column of mercury produces the same pressure as a 15.0 cm column of water.
Fluid Mechanics
13-13
13.62. Set Up: Steel has density 7.8 3 103 kg / m3. The maximum buoyant force is FB 5 rwaterVbargeg.
Solve: The volume of steel in the barge is
1 2 3 22 m 3 12 m 1 2 3 40 m 3 12 m 1 22 m 3 40 m 2 3 1 0.040 m 2 5 94.7 m3.
The mass of the barge is mbarge 5 rV 5 1 7.8 3 103 kg / m3 2 1 94.7 m3 2 5 7.39 3 105 kg.
FB 5 1 1000 kg / m3 2 1 22 m 3 40 m 3 12 m 2 1 9.80 m / s2 2 5 1.035 3 108 N.
FB 5 1 m barge 1 m coal 2 g and
mcoal 5
FB
2 mbarge 5 9.8 3 106 kg.
g
The volume of this mass of coal is
V5
9.8 3 106 kg
m
5
5 6.5 3 103 m3.
r
1500 kg / m3
The volume of the barge is 22 m 3 40 m 3 12 m 5 1.06 3 104 m3; this mass of coal easily fits into the barge.
13.63. Set Up: The density of lead is 11.3 3 103 kg / m3. The buoyant force when the wood sinks is FB 5 rwaterVtotg,
where Vtot is the volume of the wood plus the volume of the lead. r 5 m / V.
Solve: Vwood 5 1 0.600 m 2 1 0.250 m 2 1 0.080 m 2 5 0.0120 m3.
m wood 5 r woodVwood 5 1 600 kg / m3 2 1 0.0120 m3 2 5 7.20 kg.
FB 5 1 m wood 1 m lead 2 g. Using FB 5 rwaterVtotg and Vtot 5 Vwood 1 Vlead gives
rwater 1 Vwood 1 Vlead 2 g 5 1 mwood 1 mlead 2 g.
mlead 5 r leadVlead then gives rwaterVwood 1 r waterVlead 5 mwood 1 r leadVlead.
Vlead 5
1 1000 kg / m3 2 1 0.0120 m3 2 2 7.20 kg
r waterVwood 2 mwood
5
5 4.66 3 1024 m3.
r lead 2 rwater
11.3 3 103 kg / m3 2 1000 kg / m3
mlead 5 r leadVlead 5 5.27 kg.
Reflect: 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 FB is neglected, the calculation is simplified: r waterVwoodg 5 1 mwood 1 mlead 2 g and m lead 5 4.8 kg. The result
of this calculation is in error by about 9%.
13.64. Set Up: FB 5 rairVg, where V is the volume of the airship. The lift is FL 5 FB 2 mgasg.
Solve: (a) mgas 5 rgasV so FL 5 rairVg 2 rgasVg.
V5
FL
120 3 103 N
5
5 1.10 3 104 m3.
g 1 rair 2 r gas 2
1 9.80 m / s2 2 1 1.20 kg / m3 2 0.0838 kg / m3 2
(b) FL 5 1 rair 2 rgas 2 Vg 5 1 1.20 kg / m3 2 0.166 kg / m3 2 1 1.10 3 104 m3 2 1 9.80 m / s2 2 5 1.11 3 105 N 5 111 kN
Hydrogen is explosive, helium is not.
13.65. Set Up: Seawater has density rsw 5 1.03 3 103 kg / m3. FB 5 rswVsubg. Vsub is the submerged volume:
Vsub 5 Vlp 1 0.80Vp, where Vlp 5 0.0400 m3 is the volume of the life preserver and Vp is the volume of the person.
Solve: FB 5 mtotg. m tot 5 mlp 1 m p 5 rlpVlp 1 mp.
mp
75.0 kg
Vp 5
5
5 0.0765 m3.
rp
980 kg / m3
FB 5 mtotg gives rsw 1 Vlp 1 0.80Vp 2 g 5 1 rlpVlp 1 m p 2 g.
rlp 5
rsw 1 Vlp 1 0.80Vp 2 2 mp
Vlp
1 1.03 3 103 kg / m3 2 1 0.0400 m3 1 3 0.80 4 3 0.0765 m3 4 2 2 75.0 kg
5 731 kg / m3
rlp 5
0.0400 m3
13-14
Chapter 13
13.66. Set Up: FB 5 rVAg. Apply Newton’s second law to the beaker, liquid and block as a combined object and
also to the block as a single object. Take 1y upward.
Solve: Forces on the combined object: FD 1 FE 2 1 wA 1 wB 1 wC 2 5 0. wA 5 FD 1 FE 2 wB 2 wC. D and E read
mass rather than weight, so write the equation as mA 5 mD 1 m E 2 mB 2 m C. mD 5 FD / g is the reading in kg of
scale D; a similar statement applies to m E.
mA 5 3.50 kg 1 7.50 kg 2 1.00 kg 2 1.80 kg 5 8.20 kg.
Forces on A: FB 1 FD 2 wA 5 0. rVAg 1 FD 2 mAg 5 0. rVA 1 mD 5 m A.
r5
8.20 kg 2 3.50 kg
mA 2 mD
5
5 1.24 3 103 kg / m3
VA
3.80 3 1023 m3
(b) D reads the mass of A: 8.20 kg. E reads the total mass of B and C: 2.80 kg.
13.67. Set Up: The density of gold is rg 5 19.3 3 103 kg / m3 and the density of aluminum is ral 5 2.7 3
103 kg / m3. FB 5 r wVtotg.
Solve: Find the volume Vtot of the ingot: FB 1 T 2 w 5 0. r wVtotg 5 W 2 T.
Vtot 5
45.0 N 2 39.0 N
W2T
5
5 6.12 3 1024 m3.
rwg
1 1000 kg / m3 2 1 9.80 m / s2 2
The mass of the ingot is mtot 5 W / g 5 4.59 kg. The total mass is the mass of gold, mg, plus the mass of aluminum,
mal. The total volume is the volume of gold, Vg, plus the volume of aluminum, Val.
mtot 5 mg 1 m al 5 r gVg 1 ralVal 5 rgVg 1 ral 1 Vtot 2 Vg 2 .
Vg 5
4.59 kg 2 1 2700 kg / m3 2 1 6.12 3 1024 m3 2
mtot 2 r alVtot
5
5 1.77 3 1024 m3.
rg 2 r al
19.3 3 103 kg / m3 2 2.7 3 103 kg / m3
wg 5 mgg 5 rgVgg 5 1 19.3 3 103 kg / m3 2 1 1.77 3 1024 m3 2 1 9.80 m / s2 2 5 33.5 N.
Reflect: If the ingot had the same weight in air (45.0 N) and was pure gold, its volume would be 2.38 3 1024 m3
and the balance would read 42.7 N when the ingot was submerged.
13.68. Set Up: The pressure at the surface of the earth is p 5 1.013 3 105 Pa. The height of fluid in the barometer
is related to air pressure p by p 5 rgh.
p
1.013 3 105 Pa
5
5 4.73 3 103 kg / m3
Solve: (a) r 5
gh
1 9.80 m / s2 2 1 2.185 m 2
(b) p 5 rgh 5 1 4.73 3 103 kg / m3 2 A 14 B 1 9.80 m / s2 2 1 0.725 m 2 5 8.40 3 103 Pa
13.69. Set Up: Apply p1 1 rgy1 1 12 rv12 5 p2 1 rgy2 1 12 rv22, with point 1 at the surface of the acid in the tank
and point 2 in the stream as it emerges from the hole. p1 5 p2 5 pair. Since the hole is small the level in the tank
drops slowly and v1 < 0. After a drop of acid exits the hole the only force on it is gravity and it moves in projectile
motion. For the projectile motion take 1y downward, so ax 5 0 and ay 5 19.80 m / s2.
Solve: Bernoulli’s equation with p1 5 p2 and v1 5 0 gives
v2 5 "2g 1 y1 2 y2 2 5 "2 1 9.80 m / s2 2 1 0.75 m 2 5 3.83 m / s.
projectile motion: Use the vertical motion to find the time in the air. v0y 5 0, ay 5 19.80 m / s2, y 2 y0 5 1.4 m.
y 2 y0 5 v0yt 1 12 ayt 2 gives
t5
Å
2 1 1.4 m 2
2 1 y 2 y0 2
5
5 0.535 s
ay
Å 9.80 m / s2
The horizontal distance a drop travels in this time is x 2 x0 5 v0xt 1 12 axt 2 5 1 3.83 m / s 2 1 0.535 s 2 5 2.05 m.
Reflect: If the depth of acid in the tank is increased, then the velocity of the stream as it emerges from the hole
increases and the horizontal range of the stream increases.
Fluid Mechanics
13-15
13.70. Set Up: The lift is p2 2 p1, where point 2 is at the lower wing surface and point 1 is at the upper wing
surface. Neglect rg 1 y1 2 y2 2 in Bernoulli’s equation.
Solve: p1 1 rgy1 1 12 rv12 5 p2 1 rgy2 1 12 rv22. p2 2 p1 5 12 r 1 v12 2 v22 2 .
v1 5
Å
v22 1
1
2
2
2
1 p 2 p1 2 5 1 120 m / s 2 2 1
1 2000 N / m2 2 5 133 m / s.
r 2
Å
1.20 kg / m3
13.71. Set Up: For a spherical astronomical object, g 5 G
m
. p 5 pair 1 rgh.
R2
Solve: (a) Find g on Europa:
g 5 1 6.673 3 10211 N # m2 / kg2 2
4.78 3 1022 kg
5 1.30 m / s2
1 1.565 3 106 m 2 2
The gauge pressure at a depth of 100 m would be
p 2 pair 5 rgh 5 1 1.00 3 103 kg / m3 2 1 1.30 m / s2 2 1 100 m 2 5 1.30 3 105 Pa
(b) Solve for h on earth that gives p 2 pair 5 1.30 3 105 Pa:
p 2 pair
1.30 3 105 Pa
5
5 13.3 m
rg
1 1.00 3 103 kg / m3 2 1 9.80 m / s2 2
h5
Reflect: g on Europa is less than on earth so the pressure at a water depth of 100 m is much less on Europa than it
would be on earth.
13.72. Set Up: The discharge rate is v1A1 5 v2A2. The density of mercury is 13.6 3 103 kg / m3. Let point 1 be
where A1 5 40.0 3 1024 m2 and point 2 is where A2 5 10.0 3 1024 m2. y1 5 y2.
6.00 3 1023 kg / m3
6.00 3 1023 kg / m3
Solve: (a) v1 5
5 1.50 m / s. v2 5
5 6.00 m / s
24
2
40.0 3 10 m
10.0 3 1024 m2
(b) p1 1 rgy1 1 12 rv12 5 p2 1 rgy2 1 12 rv22.
p1 2 p2 5 12 r 1 v22 2 v12 2 5 12 1 1000 kg / m3 2 1 3 6.00 m / s 4 2 2 3 1.50 m / s 4 2 2 5 1.69 3 104 Pa
(c) p1 2 p2 5 rgh and
h5
p1 2 p2
1.69 3 104 Pa
5
5 0.127 m 5 12.7 cm.
rg
1 13.6 3 103 kg / m3 2 1 9.80 m / s2 2
13.73. Set Up: p 2 pair 5 rgh, A1v1 5 A2v2 and p1 1 rgy1 1 12 rv12 5 p2 1 rgy2 1 12 rv22. y1 5 y2
Solve: (a) p1 5 pair 1 rgh1 and p2 5 pair 1 rgh2, where h1 and h2 are the heights of the liquid in the two cylinders.
p1 2 p2 5 rg 1 h1 2 h2 2 . p1 2 p2 5 Dp and h1 2 h2 5 h so Dp 5 rgh.
(b) p1 2 p2 5 12 g 1 v22 2 v12 2 .
v2 5
Using Dp 5 rgh gives
1 2
1 2
A1
A1
v1 so Dp 5 12 r S
A2
A2
(c) Once v1 is measured, v2 5
1 2
v1 5
2
2 1 T v12 and v1 5
2gh
Å 1 A1 / A2 2 2 2 1
2Dp
Å r 3 1 A1 / A2 2 2 2 1 4
.
.
A1
v.
A2 1
13.74. Set Up: Let r1 and r2 be the density of the fluid at points 1 and 2.
Solve: (a) The derivation of Eq. (13.10) leads to Dm 1 5 Dm2. Dm 1 5 r1A1v1 Dt and Dm 2 5 r 2A2v2 Dt so
r1A1v1 Dt 5 r2A2v2 Dt. Dt divides out and r1A1v1 5 r 2A2v2.
(b) If r1 5 r 2 then the density divides out and the usual continuity equation A1v1 5 A2v2 is obtained.