Problemas resueltos-cap-14-fisica-sears-zemansky

1
PROBLEMAS RESUELTOS
FISICA UNIVERSITARIA
CAPITULO 14 VOLUMEN 1
EDICION 11 SEARS ZEMANSKY
Sección 14.1 Densidad
Sección 14.2 Presión de un fluido
Sección 14.3 Flotación
Sección 14.4 Fuerzas de flotación y principio de Arquímedes
Sección 14.5 Ecuación de BERNOULLI
Sección 14.6 Viscosidad y turbulencia
Erving Quintero Gil
Ing. Electromecánico
Bucaramanga – Colombia
2010
Para cualquier inquietud o consulta escribir a:
quintere@hotmail.com
quintere@gmail.com
quintere2006@yahoo.com

Problema 14.1 Sears Zemansky
En un trabajo de medio tiempo, un supervisor le pide traer del almacén una varilla cilíndrica de acero de
85,8 cm de longitud y 2,85 cm de diámetro. ¿Necesitara usted un carrito? (Para contestar calcule el
peso de la varilla)
g*V*g*mw ρ==
ρ = 7,8 * 103
kg/m3
(densidad del acero)
V = volumen de la varilla de acero
g = gravedad = 9,8 m/seg2
V = área de la varilla * longitud de la varilla
d = diámetro de la varilla = 2,85 cm = 0,0285 metros
l = longitud de la varilla = 85,8 cm = 0,858 m
24-
22
m10*6,3793
4
0,002551
4
0,00081225*3,14
4
(0,0285)*3.14
4
d
varillaladearea =====
π
V = 6,3793 * 10-4
* 0,858 = 5,4735 * 10-4
m3
g*V*w ρ=
Newton41,83
seg
m
9,8*m10*5,4735*
m
kg
10*7,8w 2
34-
3
3
==
No es necesario el carrito.
El radio de la luna es de 1740 km. Su masa es de 7,35 * 1022
kg. Calcule su densidad media?
V = volumen de la luna
R = radio de la luna = 1740 km = 174 * 104
m
2

3
r**
3
4
V π=
( )34
10*174*3,14*
3
4
V =
31212
m10*222066647,310*5268024*3,14*
3
4
V ==
m = masa de la luna = 7,35 * 1022
kg.
( ) .mkg1033.3
m10*32,22066647
kg1035,7 33
312
22
×=
×
==
V
m
ρ
Imagine que compra una pieza rectangular de metal de 5 * 15 * 30 mm y masa de 0,0158 kg. El
vendedor le dice que es de oro. Para verificarlo, usted calcula la densidad media de la pieza. Que valor
obtiene? ¿Fue una estafa?
V = volumen de pieza rectangular
V = 5 * 15 * 30 mm = 2250 mm3
( )
( )
3m0,00000225
910
3m2250
3mm1000
3m1
*3mm2250V ===
m = masa de la pieza rectangular de metal = 0,0158 kg.
( ) .3mkg310022,7
3m0,00000225
kg0158.0
×===
V
mρ
La densidad del oro = 19,3 * 103
kg/m3
La densidad de la pieza rectangular de metal 7,022 * 103
kg/m3
Por lo anterior fue engañado.
Un secuestrador exige un cubo de platino de 40 kg como rescate . ¿Cuánto mide por lado?
L = longitud del cubo
V = volumen del cubo.
V = L * L * L = L3
3 VL =
m = masa del platino = 40 kg.
ρ = la densidad del platino = 21,4 * 103
kg/m3
m
V
ρ
=
3m3-10*1,869158
3m
kg310*21,4
kg40m
V ===
ρ
3

3 VL =
m0,123183 3m3-10*1,869158L ==
L = longitud del cubo = 12,31 cm
Una esfera uniforme de plomo y una de aluminio tienen la misma masa. ¿Qué relación hay entre el radio
de la esfera de aluminio y el de la esfera de plomo?
mplomo = masa esfera de plomo
maluminio = masa esfera de aluminio
mplomo = Volumen de la esfera de plomo * densidad del plomo
3
plomo
r**
3
4
plomoV π=
plomo*3
plomo
r**
3
4
plomom ρπ=
3
aluminio
r**
3
4
aluminioV π=
aluminio*3
aluminio
r**
3
4
aluminiom ρπ=
mplomo = maluminio
aluminio*3
aluminio
r**
3
4
plomo*3
plomo
r**
3
4
ρπρπ =
Cancelando términos semejantes
aluminio*3
aluminio
rplomo*3
plomo
r ρρ =
Despejando
( )
( )
3
plomor
aluminior
3
plomor
3
aluminior
aluminio
plomo
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
==
ρ
ρ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
plomor
aluminior3
1
)
aluminio
plomo
(
ρ
ρ
ρ aluminio = 2,7 * 103
kg/m3
(densidad del aluminio)
ρ plomp = 11,3 * 103
kg/m3
(densidad del plomo)
4

⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
plomor
aluminior3
1
)
310*7,2
310*3,11
(
( ) ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
plomor
aluminior
3
1
1851,4
1,61
plomor
aluminior
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
a) Calcule la densidad media del sol. B) Calcule la densidad media de una estrella de neutrones que
tiene la misma masa que el sol pero un radio de solo 20 km.
msol = masa del sol = 1,99 * 1030
kg.
Vsol = volumen del sol
rsol = radio del sol = 6,96 * 108
m
( )3
solr**
3
4
solV π=
3
810*6,96*3,14*
3
4
V ⎟
⎠
⎞⎜
⎝
⎛=
3m2410*1412,26542410*337,15*3,14*
3
4
V ==
a)
3m
kg310*1,409
3m2410*2654,1412
kg301099.1
solV
sol =
×
==
m
ρ
m estrella = masa de la estrella de neutrones = 1,99 * 1030
kg.
V estrella = volumen de la estrella de neutrones
rsol = radio de la estrella de neutrones = 20000 m
( )3
estrellar**
3
4
estrellaV π=
( )320000*3,14*
3
4
V =
3m1210*33,5103211210*8*3,14*
3
4
V ==
b) 3mkg181005938.0
3m1210*510321,33
kg301099.1
×=
×
=D
3mkg1610*93.5=
5

¿A que profundidad del mar hay una presión manométrica de 1 * 105
Pa?
h*g*0pp ρ=−
ρ = 1,03 * 103
kg/m3
(densidad del agua de mar)
m906,9
3m
Newton
10,094
2m
Newton
100000
)2sm80,9(*)3mkg310*03,1(
Pa510*1
g*
0pp
h ===
−
=
ρ
En la alimentación intravenosa, se inserta una aguja en una vena del brazo del paciente y se conecta un
tubo entre la aguja y un depósito de fluido (densidad 1050 kg/m3
) que esta a una altura h sobre el brazo.
El deposito esta abierto a la atmosfera por arriba. Si la presión manométrica dentro de la vena es de
5980 Pa, ¿Qué valor minino de h permite que entre fluido en la vena?. Suponga que el diámetro de la
aguja es lo bastante grande para despreciar la viscosidad (sección 14.6) del fluido.
La diferencia de presión entre la parte superior e inferior del tubo debe ser de al menos 5980 Pa para
forzar el líquido en la vena
p – p0 =
Pa5980h*g* =ρ
ρ = 1050 kg/m3
(densidad del fluido)
m581,0
3m
Newton
10290
2m
Newton
5980
)2sm80,9()3mkg1050(
2mN5980
g*
Pa5980
h ====
ρ
h = 58,1 cm
Un barril contiene una capa de aceite (densidad de 600 kg/m3
) de 0,12 m sobre 0,25 m de agua.
a) Que presión manométrica hay en la interfaz aceite-agua?
b) ¿Qué presión manométrica hay en el fondo del barril?
ρ aceite = 600 kg/m3
(densidad del aceite)
a) ( ) Pa.706m12,02sm80,93mkg600h*g* =⎟
⎠
⎞⎜
⎝
⎛⎟
⎠
⎞⎜
⎝
⎛=ρ
ρ agua = 1000 kg/m3
(densidad del agua)
6
h aceite = 0,12 m
h agua = 0,25 m

b) ( ) Pa.3156Pa2450Pa706m25,02sm8,93mkg1000Pa706 =+=⎟
⎠
⎞⎜
⎝
⎛⎟
⎠
⎞⎜
⎝
⎛+
Una vagoneta vacía pesa 16,5 KN. Cada neumático tiene una presión manométrica de 205 KPa (29,7
lb/pulg2
).
Calcule el área de contacto total de los neumáticos con el suelo. (Suponga que las paredes del
neumático son flexibles de modo que la presión ejercida por el neumático sobre el suelo es igual a la
presión de aire en su interior.)
Con la misma presión en los neumáticos, calcule el área después de que el auto se carga con 9,1 KN de
pasajeros y carga.
The pressure used to find the area is the gauge pressure, and so the total area is
W = Peso de la vagoneta vacia = 16,5 KN. = 16500 Newton
P= presión manométrica de cada neumático = 205 KPa = 205000 Pa.
A
W
P =
⋅=== 2m8040,0
)Pa205000(
)N16500(
P
W
A
A = 804 cm2
b)
A
W
P =
( )
( )
⋅===
+
== 2cm1248
2m1
2cm100
*2m0,1248
2m
N
205000
N25600
)Pa205000(
)N9100N16500(
P
W
A
A = 1248 cm2
Se esta diseñando una campana de buceo que resista la presión del mar a 250 m de profundidad.
a) cuanto vale la presión manométrica a esa profundidad.
b) A esta profundidad, ¿Qué fuerza neta ejercen el agua exterior y el aire interior sobre una
ventanilla circular de 30 cm de diámetro si la presión dentro de la campana es la que hay en la
superficie del agua? (desprecie la pequeña variación de presión sobre la superficie de la
ventanilla)
a) Pa.1052.2)m250)(sm80.9)(mkg1003.1( 6233
×=×=ρgh
b) The pressure difference is the gauge pressure, and the net force due to the water and the air is
N.1078.1))m15.0()(Pa1052.2( 526
×=× π
¿Qué presión barométrica (en Pa y atm. ) debe producir una bomba para subir agua del fondo del Gran
cañón (elevación 730 m) a Indian Gardens (elevación 1370 m)?
atm.61.9Pa1027.6)m640)(sm80.9)(mkg1000.1( 6233
=×=×== ρghp
7

El liquido del manómetro de tubo abierto de la fig 14.8 a es mercurio, y1 = 3 cm y y2 7 cm. La presion
atmosferica es de 980 milibares.
a) ¿Qué presion absoluta hay en la base del tubo en U?
b) Y en el tubo abierto 4 cm debajo de la superficie libre?
c) Que presion absoluta tiene el aire del tanque?
d) ¿Qué presion manométrica tiene el gas en pascales?
a) =××+×=+ −
)m1000.7)(sm80.9)(mkg106.13(Pa10980 22332
2a ρgyp
Pa.1007.1 5
× b) Repeating the calcultion with 00.412 =−= yyy cm instead of
c) The absolute pressure is that found in part (b), 1.03
Pa.101.03gives 5
2 ×y
Pa.10 5
×
d) (this is not the same as the difference between the results of parts (a) and
(b) due to roundoff error).
Pa1033.5)( 3
12 ×=− ρgyy
Hay una profundidad máxima la que un buzo puede respirar por un snorkel (Fig. 14.31) pues, al
aumentar la profundidad, aumenta la diferencia de presión que tiende a colapsar los pulmones del buzo.
Dado que el snorkel conecta los pulmones con la atmosfera, la presión en ellos es la atmosférica.
Calcule la diferencia de presión interna-externa cuando los pulmones del buzo están a 6,1 m de
profundidad. Suponga que el buzo esta en agua dulce. (un buzo que respira el aire comprimido de un
tanque puede operar a mayores profundidades que uno que usa snorkel, por que la presión del aire
dentro de los pulmones aumenta hasta equilibrar la presión externa del agua)
Pa.100.6)m1.6)(sm80.9)(mkg1000.1( 4233
×=×=ρgh
Un cilindro alto con área transversal de 12 cm2
se lleno parcialmente con mercurio hasta una altura de 5
cm. Se vierte lentamente agua sobre el mercurio (los dos líquidos no se mezclan). ¿Qué volumen de
agua deberá añadirse para aumentar al doble la presión manométrica en la base del cilindro.
With just the mercury, the gauge pressure at the bottom of the cylinder is ⋅+= mm0 ghppp With the water
to a depth , the gauge pressure at the bottom of the cylinder iswh .wwmm0 ghpghρpp ++= If this is to be
double the first value, then m.mww ghρghρ =
m680.0)1000.1106.13)(m0500.0()( 33
wmmw =××== ρρhh
The volume of water is 33424
cm816m108.16)m10m)(12.0(0.680A =×=×== −−
hV
8

Un recipiente cerrado se llena parcialmente con agua. en un principio, el aire arriba del agua esta a
presion atmosferica (1,01 * 105
Pa) y la presion manometrica en la base del recipiente es de 2500 Pa.
Despues, se bombea aire adicional al interior aumentando la presion del aire sobre el agua en 1500 Pa)
9

a) Gauge pressure is the excess pressure above atmospheric pressure. The pressure difference
between the surface of the water and the bottom is due to the weight of the water and is still 2500 Pa
after the pressure increase above the surface. But the surface pressure increase is also transmitted to
the fluid, making the total difference from atmospheric 2500 Pa+1500 Pa = 4000 Pa.
b) The pressure due to the water alone is 2500 Pa .ρgh= Thus
m255.0
)sm80.9()mkg1000(
mN2500
23
2
==h
To keep the bottom gauge pressure at 2500 Pa after the 1500 Pa increase at the surface, the pressure
due to the water’s weight must be reduced to 1000 Pa:
m102.0
)sm80.9)(mkg1000(
mN1000
23
2
==h
Thus the water must be lowered by m0.153m102.0m255.0 =−
The force is the difference between the upward force of the water and the downward forces of the air and
the weight. The difference between the pressure inside and out is the gauge pressure, so
N.1027.2N300)m75.0(m)30()sm80.9()1003.1()( 5223
×=−×=−= wAρghF
[ ] )m(2.00Pa1093m)2.14)(sm71.3)(mkg1000.1(Pa10130 232333
×−×+×
N.1079.1 5
×=
The depth of the kerosene is the difference in pressure, divided by the product ,V
mg
ρg =
m.14.4
)m250.0()smkg)(9.80205(
Pa1001.2)m(0.0700N)104.16(
32
523
=
×−×
=h
atm.64.1Pa1066.1
m)15.0(
)smkg)(9.801200(
)2(
5
2
2
2
=×====
πdπ
mg
A
F
p
10

The buoyant force must be equal to the total weight; so,gg icewater mgVρVρ +=
,m563.0
mkg920mkg1000
kg0.45 3
33
water
=
−
=
−
=
iceρρ
m
V
or to two figures.3
m56.0
14.22: The buoyant force is andN,30.6N20.11N17.50 =−=B
.m1043.6
)sm80.9)(mkg1000.1(
N)30.6( 34
233
water
−
×=
×
==
gρ
B
V
The density is
.mkg1078.2
30.6
50.17
)mkg1000.1( 3333
water
water
×=⎟
⎠
⎞
⎜
⎝
⎛
×====
B
w
ρ
gρB
gw
V
m
ρ
Un objeto con densidad media ρ flota sobre un fluido de densidad ρfluido
a)?Que relacion debe haber entre las dos densidades?
b) A la luz de su respuesta a la parte(a), ¿Cómo pueden flotar barcos de acero en el agua?
c) En terminos de ρ y ρfluido ¿Qué fraccion del objeto esta sumergida y que fraccion esta sobre el fluido?
Verifique que sus respuestas den el comportamiento correcto en el limite donde ρ → ρfluido y donde ρ →
0.
d) durante un paseo en yate, su primo Tito recorta una pieza rectangular (dimensiones: 5 * 4 *3 cm) de
un salvavidas y la tira al mar, donde flota. La masa de la pieza es de 42 gr. ¿Qué porcentaje de su
volumen esta sobre la superficie ?
a) The displaced fluid must weigh more than the object, so .fluidρρ < b) If the ship does not leak, much
of the water will be displaced by air or cargo, and the average density of the floating ship is less than that
of water. c) Let the portion submerged have volume V, and the total volume be
⋅== fluid0
so,Then,. fluido0 ρ
ρ
V
V
VρVV ρ The fraction above the fluid is then 0,If.1 fluid
→− pP
P
the entire
object floats, and if fluidρρ → , none of the object is above the surface. d) Using the result of part (c),
%.3232.0
mkg1030
)m103.04.0(5.0kg)042.0(
11 3
3-6
fluid
==
×××
−=−
ρ
ρ
11

Un cable anclado al fondo de un lago de agua dulce sostiene una esfera hueca de plástico bajo la
superficie. El volumen de la esfera es de 0,650 m3
y la tensión en el cable es de 900 N.
a) Calcule la fuerza de flotación ejercida por el agua sobre la esfera.
b) ¿Qué masa tiene la esfera
b) El cable se rompe y la esfera sube a la superficie. En equilibrio, ¿Qué fracción de volumen de la
esfera estará sumergida?
a) ( )( )( ) N.6370m650.0sm80.9mkg1000.1 3233
water =×== gVρB
b) kg.5582
sm9.80
N900-N6370
==== −
g
TB
g
w
m
c) (See Exercise 14.23.) If the submerged volume is ,V ′
%.9.858590.
N6370
N5470
and
waterwater
====
′
=′
gVρ
w
V
V
gρ
w
V
Un bloque cubico de Madera de 10 cm. Por lado flota en la interfaz entre aceite y agua con su superficie
inferior 1,5 cm. Bajo la interfaz (fig 14.32). La densidad del aceite es de 790 kg/m3
a) ¿Qué presion manometrica hay en la superficie de arriba del bloque?
b) ¿y en la cara inferior
c) ¿Qué masa y densidad tiene el bloque?
a) Pa.116oiloil =ghρ
b) ( )( ) ( )( )( )( ) Pa.921sm80.9m0150.0mkg1000m100.0mkg790 233
=+
c)
( ) ( )( )
( ) kg.822.0
sm80.9
m100.0Pa805
2
2
topbottom
==
−
==
g
App
g
w
m
The density of the block is ( )
.822 33
m
kg
m10.0
kg822.0
==p Note that is the same as the average density of the fluid
displaced, ( )( ) ( ) .)mkg1000(15.0mkg79085.0 33
+
12

Un lingote de aluminio sólido pesa 89 N en el aire.
a) ¿Qué volumen tiene?
b) el lingote se cuelga de una cuerda y se sumerge por completo en agua. ¿Qué tensión hay en la
cuerda (el peso aparente del lingote en agua)?
a) Neglecting the density of the air,
( )
( )( ) ,m1036.3
mkg107.2sm80.9
N89 33
332
−
×=
×
====
gρ
w
ρ
gw
ρ
m
V
33
m104.3or −
× to two figures.
b) ( ) N.0.56
7.2
00.1
1N891
aluminum
water
water =⎟
⎠
⎞
⎜
⎝
⎛
−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=−=−=
ρ
ρ
ωVgρwBwT
a) The pressure at the top of the block is ,0 ghpp ρ+= where is the depth of the top of the block below
the surface. is greater for block Β , so the pressure is greater at the top of block Β .
h
h
b) .objfl gVB ρ= The blocks have the same volume so experience the same buoyant force.objV
c) .so0 BwTBwT −==+−
The object have the same V but is larger for brass than for aluminum so is larger for
the brass block.
.ρVgw = ρ w
B is the same for both, so T is larger for the brass block, block B.
The rock displaces a volume of water whose weight is N.10.8N28.4-N2.39 = The mass of this much
water is thus kg102.1sm9.80N8.10 2
= and its volume, equal to the rock’s volume, is
33
33
m10102.1
mkg101.00
kg102.1 −
×=
×
The weight of unknown liquid displaced is N,20.6N18.6N2.39 =− and its mass is
kg.102.2sm9.80N6.20 2
= The liquid’s density is thus 33
m101.102kg102.2 −
× ,mkg1091.1 33
×= or
roughly twice the density of water.
13

)(, 21122211 ΑΑvvΑvΑv ==
2
2
2
1 cm)10.0(20,cm)80.0( πΑπΑ ==
sm6.9
)10.0(20
(0.80)
s)m0.3( 2
2
2 ==
π
π
v
⋅===
2
3
2
2
2
1
12
sm245.0)m0700.0s)(m50.3(
ΑΑΑ
Α
vv
a) (i) s.m21.5,m047.0(ii)s.m33.2,m1050.0 2
2
22
2
2 ==== vΑvΑ
b) .m882s)3600()sm(0.245 33
2211 === tΑυtΑv
a) .98.16
)m150.0(
)sm20.1(
2
3
===
πA
dtdV
v
b) .m317.0)( 22112 === πvdtdVvvrr
a) From the equation preceding Eq. (14.10), dividing by the time interval dt gives Eq. (14.12). b) The
volume flow rate decreases by 1.50% (to two figures).
Un tanque sellado que contiene agua de mar hasta una altura de 11 metros contiene también aire sobre
el agua a una presión manométrica de 3 atmósferas. Sale agua del tanque a través de un agujero
pequeño en el fondo. Calcule la rapidez de salida del agua
The hole is given as being “small,”and this may be taken to mean that the velocity of the seawater at the
top of the tank is zero, and Eq. (14.18) gives
El agujero se da como "pequeño", y esto puede interpretarse en el sentido que la velocidad del agua de
mar en la parte superior del tanque es cero, y la ecuación. (14.18) es
))((2 ρpgyv +=
14

= ))mkg10(1.03Pa)1013(3.00)(1.0m)0.11)(sm80.9((2
3352
××+
s.m4.28=
Note that y = 0 and were used at the bottom of the tank, so that p was the given gauge pressure
at the top of the tank.
app =
Tenga en cuenta que y = 0 y p = se utilizaron en la parte inferior del tanque, de modo que p es lap0
presión manométrica dada en la parte superior del tanque
Se corta un agujero circular de 6 mm de diámetro en el costado de un tanque de agua grande, 14 m
debajo del nivel del agua en el tanque. El tanque esta abierto al aire por arriba. Calcule
a) la rapidez de salida
b) el volumen descargado por unidad de tiempo.
v = velocidad en m/seg.
h = altura en metros.
h*g*22v =
seg
m
16,56
2seg
2m
274,4m14*
2seg
m
9,8*2h*g*2v ====
b) el volumen descargado por unidad de tiempo.
V = volumen en m3
/ seg
V = v * A
A = area = π * r2
r = 3 mm = 0,003 m
A = π * r2
= 3,14 * (0,003)2
= 2,8274 * 10-5
m2
V = v * A = 16,56 m/seg * 2,8274 * 10-5
m2
V = 4,68 * 10-4
m3
/seg
¿Que presión manométrica se requiere en una toma municipal de agua para que el chorro de una
manguera de bomberos conectada a ella alcance una altura vertical de 15 m?. (Suponga que la toma
tiene un diámetro mucho mayor que la manguera).
The assumption may be taken to mean that 01 =v in Eq. (14.17). At the maximum height, ,02 =v and
using gauge pressure for (the water is open to the atmosphere),0,and 221 =ppp
Pa510*1,47
2m
Newton
147000m15*
2seg
m
9,8*
3m
kg
10002y*g*1p ==== ρ
p1 = 1,47 * 105
Pa
Using 14
1
2 vv = in Eq. (14.17),
⎥
⎦
⎤
⎢
⎣
⎡
−+⎟
⎠
⎞
⎜
⎝
⎛
+=−+−+= )(
32
15
)()(
2
1
21
2
1121
2
2
2
112 yygυρpyyρgvvρpp
15

⎟
⎠
⎞
⎜
⎝
⎛
+×+×= m)0.11)(sm80.9()sm00.3(
32
15
)mkg1000.1(Pa1000.5 22334
Pa.1062.1 5
×=
Neglecting the thickness of the wing (so that 21 yy = in Eq. (14.17)), the pressure difference is
Pa.078)(2)1( 2
1
2
2 =−=Δ vvρp The net upward force is then
N.496)smkg)(9.801340()m(16.2Pa)780( 22
−=−×
a)
( )( )
s.kg30.1s0.60
kg355.0220
= b) The density of the liquid is ,mkg1000 3
m100.355
kg355.0
33 =−
×
and so the volume flow
rate is s.L1.30sm1030.1 33
mkg1000
skg30.1
3 =×= −
This result may also be obtained
from
( )( )
s.L30.1s0.60
L355.0220
= c) 24
33
m1000.2
sm1030.1
1 −
−
×
×
=v
s.m63.14s,m50.6 12 === vv
d) ( ) ( )12
2
1
2
221
2
1
yyρgvvρpp −+−+=
( )( ) ( ) ( )( )
( )( )( )m35.1sm80.9mkg1000
sm50.6sm63.1mkg100021kPa152
23
223
−+
−+=
kPa119=
The water is discharged at a rate of s.m352.023
34
m1032.1
sm1065.4
1 == −
−
×
×
v The pipe is given as horizonatal, so the
speed at the constriction is s,m95.822
12 =Δ+= ρpvv keeping an extra figure, so the cross-section
are at the constriction is ,m1019.5 25
sm95.8
sm1065.4 34
−×
×=
−
and the radius is cm.41.0== πAr
From Eq. (14.17), with ,21 yy =
( ) 2
11
2
12
11
2
2
2
112
8
3
42
1
2
1
ρvp
v
vρpvvρpp +=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+=−+=
( )( ) Pa,1003.2sm50.2mkg1000.1
8
3
Pa1080.1 42334
×=×+×=
where the continutity relation
2
1
2
v
v = has been used.
16

Let point 1 be where and point 2 be wherecm00.41 =r cm.00.22 =r The volume flow rate has the value
scm7200 3
at all points in the pipe.
sm43.1so,cm7200 1
32
1111 === vπrvAv
sm73.5so,cm7200 2
32
2222 === vπrvAv
2
222
2
111
2
1
2
1
ρvρgypρvρgyp ++=++
( ) Pa1025.2
2
1
soPa,1040.2and 52
2
2
112
5
221 ×=−+=×== vvρpppyy
a) The cross-sectional area presented by a sphere is ,4
2
D
π therefore ( ) .40
2
D
πppF −= b) The force on
each hemisphere due to the atmosphere is ( )22
m1000.5 −
×π ( )( ) .776975.0Pa10013.1 5
Ν=×
a) ( )( )( ) Pa.1010.1m1092.10sm80.9mkg1003.1 83233
×=×××=ρgh
b) The fractional change in volume is the negative of the fractional change in density.
The density at that depth is then
( ) ( ) ( )( )( )111833
0 Pa108.45Pa1016.11mkg1003.11 −−
××+×=Δ+= pkρρ
,mkg1008.1 33
×=
A fractional increase of Note that to three figures, the gauge pressure and absolute pressure are
the same.
%.0.5
a) The weight of the water is
( )( ) ( )( )( )( ) N,1088.5m0.3m0.4m00.5sm80.9mkg1000.1 5233
×=×=ρgV
or to two figures. b) Integration gives the expected result the force is what it would be if the
pressure were uniform and equal to the pressure at the midpoint;
N109.5 5
×
2
d
gAF ρ=
( )( ) ( )( )( )( ) N,1076.1m50.1m0.3m0.4sm80.9mkg1000.1 5233
×=×=
or to two figures.N108.1 5
×
17

18
)
Let the width be w and the depth at the bottom of the gate be The force on a strip of vertical thickness
at a depth is then and the torque about the hinge is
.H
dh h (wdhρghdF = ( ) ;2 dhHhρgwhdτ −=
integrating from givesHhh == to0 m.N1061.212 43
⋅×== Hgωρτ
a) See problem 14.45; the net force is fromdF∫ ,22,to0 2
gAHHgFHhh ρωρ ==== where
.HA ω= b) The torque on a strip of vertical thickness about the bottom is
and integrating from
dh
( ) ( ) ,dhhHgwhhHdFdτ −ρ=−= Hhh == to0 gives .66 23
ρgAHρgwHτ ==
c) The force depends on the width and the square of the depth, and the torque about the bottom
depends on the width and the cube of the depth; the surface area of the lake does not affect either result
(for a given width).
The acceleration due to gravity on the planet is
d
p
ρd
p
g
V
m
Δ
=
Δ
=
and so the planet’s mass is
mGd
pVR
G
gR
M
22
Δ
==
The cylindrical rod has mass radius and length,M ,R L with a density that is proportional to the
square of the distance from one end, .2
Cx=ρ
a) The volume element Then the integral becomes
Integrating gives
.2
dVCxdVM ∫=∫= ρ .2
dxπRdV =
.22
0 dxRCxM L
π∫= .
3
3
22
0
2 L
RCdxxRCM L
ππ =∫= Solving for .3, 32
LRMCC π=
b) The density at the end isLx = ( )( ) ( ).232
3232
LπR
M
LπR
M
LCxρ === The denominator is just the total
volume so,V ,3 VM=ρ or three times the average density, .VM So the average density is one-third
the density at the end of the rod.Lx =
a) At the model predicts,0=r 3
mkg700,12== Aρ and at ,Rr = the model predicts

.mkg103.15m)1037.6)(mkg1050.1(mkg700,12 336433
×=××−=−= −
BRAρ
b), c)
∫ ∫ ⎥⎦
⎤
⎢⎣
⎡
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=⎥
⎦
⎤
⎢
⎣
⎡
−=−==
R
BR
A
πRBRAR
πdrrBrAπdmM
0
343
2
4
3
3
4
43
4][4
⎥
⎦
⎤
⎢
⎣
⎡ ××
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
=
−
4
m)1037.6)(mkg1050.1(3
mkg700,12
3
m)1037.6(4 643
3
36
π
kg,1099.5 24
×=
which is within 0.36% of the earth’s mass. d) If is used to denote the mass contained in a sphere
of radius
m )(r
,r then .)( 2
rrGmg = Using the same integration as that in part (b), with an upper limit of
r instead of R gives the result.
e) )kgmN10673.6()(,atand,0at0 22112
⋅×===== −
RRGmgRrgrg
.sm85.9m)10(6.37kg)1099.5( 22624
=××
f) ;
2
3
3
4
4
3
3
4 2
⎥⎦
⎤
⎢⎣
⎡
−⎟
⎠
⎞
⎜
⎝
⎛
=⎥
⎦
⎤
⎢
⎣
⎡
−⎟
⎠
⎞
⎜
⎝
⎛
=
Br
A
πGBr
Ar
dr
dπG
dr
dg
setting ths equal to zero gives m1064.532 6
×== BAr , and at this radius
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
B
A
BA
B
AπG
g
3
2
4
3
3
2
3
4
B
πGA
9
4 2
=
.sm02.10
)mkg1050.1(9
)mkg700,12()kgmN10673.6(4 2
43
232211
=
×
⋅×
= −
−
π
a) Equation (14.4), with the radius r instead of height becomes,y .)(s drRrρgdrρgdp −=−= This
form shows that the pressure decreases with increasing radius. Integrating, with ,at0 Rrp ==
∫ ∫ −==−=
r
R r
rR
R
ρg
drr
R
ρg
drr
R
ρg
p
R
22sss
).(
2
b) Using the above expression with ,and0 3
4
3
R
M
V
M
r π
ρ ===
Pa.1071.1
m)1038.6(8
)smkg)(9.801097.5(3
)0( 11
26
224
×=
×
×
=
π
p
19

c) While the same order of magnitude, this is not in very good agreement with the estimated value. In
more realistic density models (see Problem 14.49 or Problem 9.99), the concentration of mass at lower
radii leads to a higher pressure.
a) Pa.1047.1)m100.15)(sm80.9)(mkg1000.1( 32233
waterwater ×=××= −
ghρ
b) The gauge pressure at a depth of 15.0 h−cm below the top of the mercury column must be that found
in part (a); which is solved for),cm0.15()cm(15.0 waterHg gρhgρ =− cm.9.13=h
Following the hint,
∫ ==
h
o
2
))(2( ρgπRhdyπRρgyF
where R and h are the radius and height of the tank (the fact that hR =2 is more or less coincidental).
Using the given numerical values gives N.1007.5 8
×=F
14.53: For the barge to be completely submerged, the mass of water displaced would need to be
kg.101.056)m1240)(22mkg10(1.00 7333
water ×=×××=Vρ The mass of the barge itself is
kg,1039.7)m100.4)402212)4022(2(()mkg108.7( 53233
×=×××+×+× −
so the barge can hold of coal. This mass of coal occupies a solid volume of
which is less than the volume of the interior of the barge
kg1082.9 6
×
,m1055.6 33
× ),m1006.1( 34
×
but the coal must not be too loosely packed.
The difference between the densities must provide the “lift” of 5800 N (see Problem 14.59). The average
density of the gases in the balloon is then
.mkg96.0
)m2200)(sm80.9(
)N5800(
mkg23.1 3
32
3
ave =−=ρ
a) The submerged volume so,is water gρ
w
V′
20

⋅==
×
===
′
%3030.0
)m(3.0)mkg1000.1(
)kg900(
333
water
water
Vρ
m
V
gρw
V
V
b) As the car is about to sink, the weight of the water displaced is equal to the weight of the car plus the
weight of the water inside the car. If the volume of water inside the car is V ′′ ,
⋅==−=−=
′′
′′+= %7070.030.011or,
gVp
w
V
V
gpVwgVρ
water
waterwater
a) The volume displaced must be that which has the same weight and mass as the
ice,
3
cmgm00.1
gm70.9
cm70.93 = (note that the choice of the form for the density of water avoids conversion of
units). b) No; when melted, it is as if the volume displaced by the of melted ice displaces the
same volume, and the water level does not change. c)
gm70.9
⋅= 3
cmgm05.1
gm70.9
cm24.93 d) The melted water takes
up more volume than the salt water displaced, and so flows over. A way of considering this
situation (as a thought experiment only) is that the less dense water “floats” on the salt water, and as
there is insufficient volume to contain the melted ice, some spills over.
3
cm46.0
The total mass of the lead and wood must be the mass of the water displaced, or
;)( waterwoodPbwoodwoodPbPb ρVVρVρV +=+
solving for the volume ,PbV
waterPb
woodwater
woodPb
ρρ
ρρ
−
−
= VV
3333
333
32
mkg1000.1mkg103.11
mkg600mkg1000.1
)m102.1(
×−×
−×
×= −
,m1066.4 34−
×=
which has a mass of 5.27 kg.
The fraction f of the volume that floats above the fluid is ,1
fluidρ
ρ
−=f where is the average density of
the hydrometer (see Problem 14.23 or Problem 14.55), which can be expressed as
ρ
.
1
1
fluid
f
ρρ
−
=
21

Thus, if two fluids are observed to have floating fraction .
1
1
,and
2
1
1221
f
f
ρρff
−
−
= In this form, it’s clear
that a larger corresponds to a larger density; more of the stem is above the fluid. Using2f
097.0,242.0 )cm2.13(
)cmcm)(0.40020.3(
2)cm2.13(
)cmcm)(0.40000.8(
1 3
2
3
2
==== ff gives .mkg839)839.0( 3
wateralcohol == ρρ
a) The “lift” is from which,)( 2Hair gρρV −
.m100.11
)sm80.9)(mkg0899.0mkg(1.20
N000,120 33
233
×=
−
=V
b) For the same volume, the “lift” would be different by the ratio of the density differences,
N.102.11N)000,120( 4
Hair
Heair
2
×=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
ρρ
ρρ
This increase in lift is not worth the hazards associated with use of hydrogen.
a) Archimedes’ principle states .so,
A
M
LMggLA
ρ
ρ ==
b) The buoyant force is ,)( FMgxLgA +=+ρ and using the result of part (a) and solving for x
gives .ρgA
F
x =
c) The “spring constant,” that is, the proportionality between the displacement x and the applied
force F, is and the period of oscillation is,ρgAk =
.22
ρgA
M
π
k
M
πT ==
1: a)
( )
( ) ( )
m.107.0
m450.0mkg1003.1
kg0.70
233
=
×
====
πρA
m
ρgA
mg
ρgA
w
x
b) Note that in part (c) of Problem is the mass of the buoy, not the mass of the man, andM,60.14
A is the cross-section area of the buoy, not the amplitude. The period is then
22

( )
( )( ) ( )
s.42.2
m450.0sm80.9mkg101.03
kg950
2 2233
=
×
=
π
πT
To save some intermediate calculation, let the density, mass and volume of the life preserver
be ,and,0 vmρ and the same quantities for the person be Then, equating the buoyant force
and the weight, and dividing out the common factor of
.and,1 VMρ
,g
( )( ) ,80.0 10water VρvρvVρ +=+
Eliminating V in favor of and and eliminating m in favor of and1ρ ,M 0ρ ,v
( ) .80.0
1
water0 ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+=+ v
ρ
M
ρMvρ
Solving for ,0ρ
( ) ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+= Mv
ρ
M
ρ
v
ρ
1
water0 80.0
1
( ) ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−=
1
water
water 80.01
ρ
ρ
v
M
ρ
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
−−×= 3
33
3
33
mkg980
mkg101.03
(0.80)1
m0.400
kg0.75
mkg1003.1
⋅= 3
mkg732
To the given precision, the density of air is negligible compared to that of brass, but not compared to that
of the wood. The fact that the density of brass may not be known the three-figure precision does not
matter; the mass of the brass is given to three figures. The weight of the brass is the difference between
the weight of the wood and the buoyant force of the air on the wood, and canceling a common factor of
and)(, brass,airwoodwood MρρVg =−
1
wood
air
brass
airwood
wood
brasswoodwoodwood 1
−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=
−
==
ρ
ρ
M
ρρ
ρ
MVρM
kg.0958.0
mkg150
mkg20.1
1)kg0950.0(
1
3
3
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=
−
23

The buoyant force on the mass A, divided by kg70.4kg80.1kg1.00kg7.50bemust, =−−g (see
Example 14.6), so the mass block is kg.8.20kg3.50kg70.4 =+ a) The mass of the liquid displaced by
the block is so the density of the liquid iskg,70.4 .mkg1024.1 33
m103.80
kg70.4
33- ×=×
b) Scale D will read the
mass of the block, as found above. Scale E will read the sum of the masses of the beaker and
liquid,
kg,20.8
kg.80.2
Neglecting the buoyancy of the air, the weight in air is
N.0.45)( A1A1AuAu =+ VρVρg
and the buoyant force when suspended in water is
N.6.0N39.0N45.0)( A1Auwater =−=+ gVVρ
These are two equations in the two unknowns Multiplying the second by.and A1Au VV A1ρ and the first by
waterρ and subtracting to eliminate the term givesA1V
N)0.6(N)0.45()( A1waterA1AuAuwater ρρρρgVρ −=−
))0.6(N)0.45((
)(
Auwater
A1Auwater
Au
AuAuAu ρρ
ρρρ
ρ
gVρw −
−
==
N))0.6)(7.2(N)0.45)(00.1((
)7.23.19)(00.1(
)3.19(
−
−
=
N.5.33=
Note that in the numerical determination of specific gravities were used instead of densities.,Auw
The ball’s volume is
333
cm7238cm)0.12(
3
4
3
4
=== ππrV
As it floats, it displaces a weight of water equal to its weight. a) By pushing the ball under water, you
displace an additional amount of water equal to 84% of the ball’s volume or
This much water has a mass of.cm6080)cm7238)(84.0( 33
= kg080.66080 =g and weighs
N,6.59)smkg)(9.80080.6( 2
= which is how hard you’ll have to push to submerge the ball.
b) The upward force on the ball in excess of its own weight was found in part (a): The ball’s
mass is equal to the mass of water displaced when the ball is floating:
N.6.59
kg,158.1g1158)cmg00.1)(cm7238)(16.0( 33
==
24

and its acceleration upon release is thus
2net
sm5.51
kg1.158
N6.59
===
m
F
a
a) The weight of the crown of its volume V is gVρw crown= , and when suspended the apparent weight is
the difference between the weight and the buoyant force,
.)( watercrowncrown gVρρgVfρfw −==
Dividing by the common factors leads to
.
1
1
or
water
crown
crowncrownwater
fρ
ρ
fρρρ
−
==+−
As the apparent weight approaches zero, which means the crown tends to float; from the above
result, the specific gravity of the crown tends to 1. As the apparent weight is the same as the
weight, which means that the buoyant force is negligble compared to the weight, and the specific gravity
of the crown is very large, as reflected in the above expression. b) Solving the above equations for f in
terms of the specific gravity,
,0→f
,1→f
,1 crown
water
ρ
ρ
f −= and so the weight of the crown would be
( )( )( ) N.2.12N9.123.1911 =− c) Approximating the average density by that of lead for a “thin” gold
plate, the apparent weight would be ( )( )( ) N.8.11N9.123.1111 =−
a) See problem 14.67. Replacing f with, respectively, wwwater and wwfluid gives
,
- fluidfluid
steel
ww
w
ρ
ρ
= ,
- waterfluid
steel
ww
w
ρ
ρ
=
and dividing the second of these by the first gives
.
-
-
water
fluid
water
fluid
ww
ww
ρ
ρ
=
b) When is greater than the term on the right in the above expression is less than one,
indicating that the fluids is less dense than water, and this is consistent with the buoyant force when
suspended in liquid being less than that when suspended in water. If the density of the fluid is the same
as that of water , as expected. Similarly, if is less than , the term on the right in
the above expression is greater than one, indicating the the fluid is denser than water. c) Writing the
result of part (a) as
fluidw water,w
=fluidw waterw fluidw waterw
25

.
1
1
water
fluid
water
fluid
f
f
ρ
ρ
−
−
=
and solving for ,fluidf
( ) ( )( ) %.4.84844.0128.0220.1111 water
water
fluid
fluid ==−=−−= f
ρ
ρ
f
a) Let the total volume be V; neglecting the density of the air, the buoyant force in terms of the weight is
,
)(
0
m
waterwater ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+== V
ρ
gw
gρgVρB
or
⋅
ρ
−=
g
w
gρ
B
V
wwater
0
b) .m1052.2 34
Cuwater
−
×=− gρ
w
gρ
B
Since the total volume of the casting is ,water
B
gρ the cavities are
12.4% of the total volume.
a) Let d be the depth of the oil layer, h the depth that the cube is submerged in the water, and L be the
length of a side of the cube. Then, setting the buoyant force equal to the weight, canceling the common
factors of g and the cross-section area and supressing units,
−==++=+ LhLLhdLhdLdh (0.65)so,(0.35)byrelatedareand,.)550()750()1000(
d. Substitution into the first relation gives m.040.05.00
2
(750)(1000)
(550)0)(0.65)(100
=== −
− L
Ld b) The gauge pressure at
the lower face must be sufficient to support the block (the oil exerts only sideways forces directly on the
block), and Pa.539m)100.0)(sm80.9)(mkg(550 23
wood ==ρ= gLp As a check, the gauge pressure,
found from the depths and densities of the fluids, is
Pa.539)sm80.9))(mkgm)(1000025.0()mkgm)(750040.0(( 233
=+
The ship will rise; the total mass of water displaced by the barge-anchor combination must be the same,
and when the anchor is dropped overboard, it displaces some water and so the barge itself displaces
less water, and so rises.
To find the amount the barge rises, let the original depth of the barge in the water be
( ) ( ) abwaterab0 andwhere, mmAmmh ρ+= are the masses of the barge and the anchor, and A is the area
26

of the bottom of the barge. When the anchor is dropped, the buoyant force on the barge is less than what
it was by an amount equal to the buoyant force on the anchor; symbolically,
( ) ,watersteelawater0water gmAghAgh ρρρρ −=′
which is solved for
( )
( )( ) m,1057.5
m00.8mkg7860
kg0.35 4
23
steel
a
0
−
×===′−=Δ
A
m
hhh
ρ
or about 0.56 mm.
a) The average density of a filled barrel is ,mkg875mkg750 3
m0.120
kg0.153
oil 3 =+=+ V
m
ρ which is less than
the density of seawater, so the barrel floats.
b) The fraction that floats (see Problem 14.23) is
%.0.15150.0
mkg1030
mkg875
11 3
3
water
ave
==−=−
ρ
ρ
c) The average density is 333
m
kg
m0.120
kg0.32
m
kg
1172910 =+ which means the barrel sinks. In order to lift it,
a tension N173)80.9)(m120.0)(1030()80.9)(m120.0)(1177( 232
s
m3
m
kg
s
m3
m
kg
=−= 3T is required.
a) See Exercise 14.23; the fraction of the volume that remains unsubmerged is .1 ρL
ρB
− b) Let the depth of
the liquid be x and the depth of the water be y. Then .and LyxgLwgyLgx B =+=+ ρρρ Therefore
andyLx −= .)(
ωL
BL
ρρ
Lρρ
y −
−
= c) m.046.0m)10.0(0.16.13
8.76.13
== −
−
y
a) The change is height is related to the displaced volumeyΔ ,by
A
V
yV
Δ
=ΔΔ where A is the surface
area of the water in the lock. is the volume of water that has the same weight as the metal, soVΔ
27

gAρ
w
A
gρw
A
V
y
water
water
==
Δ
=Δ
m.213.0
m))m)(20.00.60)((sm80.9)(mkg(1.00x10
N)1050.2(
233
6
=
×
=
b) In this case, is the volume of the metal; in the above expression,VΔ waterρ is replaced by
,00.9 watermetal ρρ = which gives m;189.0and, 9
8
9
=Δ=′Δ−Δ=′Δ Δ
yyyy y
the water sinks by this amount.
a) Consider the fluid in the horizontal part of the tube. This fluid, with mass ,Alρ is subject to a net force
due to the pressure difference between the ends of the tube, which is the difference between the gauge
pressures at the bottoms of the ends of the tubes. This difference is ),( RL yyρg − and the net force on
the horizontal part of the fluid is
,)( RL AlaAyyg ρρ =−
or
.)( RL l
g
a
yy =−
b) Again consider the fluid in the horizontal part of the tube. As in part (a), the fluid is accelerating; the
center of mass has a radial acceleration of magnitude ,22
rad la ω= and so the difference in heights
between the columns is .2))(2( 222
glgll ωω =
Anticipating Problem, 14.77, an equivalent way to do part (b) is to break the fluid in the horizontal part of
the tube into elements of thickness dr; the pressure difference between the sides of this piece is
(see Problem 14.78), and integrating fromdrrdp )( 2
ωρ= ,2givesto0 22
lplrr ρω=Δ== giving the
same result.
c) At any point, Newton’s second law gives pAdladpA = from which the area A cancels out. Therefore
the cross-sectional area does not affect the result, even if it varies. Integrating the above result from 0 to l
gives between the ends. This is related to the height of the columns throughpalp =Δ ypgp Δ=Δ
from which p cancels out.
a) The change in pressure with respect to the vertical distance supplies the force necessary to keep a
fluid element in vertical equilibrium (opposing the weight). For the rotating fluid, the change in pressure
with respect to radius supplies the force necessary to keep a fluid element accelerating toward the axis;
28

specifically, ,ρa drdrdp p
p
== ∂
∂
and using givesrωa 2
= .2
rρωp
p
=∂
∂
b) Let the pressure at
be (atmospheric pressure); integrating the expression for0,0 == ry ap p
p
∂
∂
from part (a) gives
( ) .2
2
a
2
0, r
ρω
pyrp +==
c) In Eq. as found in part (b),)0,(,),5.14( 1a2 === yrpppp ),(yand0 21 rhy == the height of the liquid
above the plane. Using the result of part (b) gives0=y .2)( 22
grωrh =
a) The net inward force is ( ) ,AdppAAdpp =−+ and the mass of the fluid element is .rρAd ′ Using
Newton’s second law, with the inward radial acceleration of gives b) Integrating
the above expression,
,'2
rω .2
rdrdp ′′ρω=
∫ ∫ ′′=
p
p
r
r
rdrdp
0 0
2
ρω
( ),
2
0
22
2
0 rr
ρω
pp −⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=−
which is the desired result. c) Using the same reasoning as in Section 14.3 (and Problem 14.78), the net
force on the object must be the same as that on a fluid element of the same shape. Such a fluid element
is accelerating inward with an acceleration of magnitude and so the force on the object is
d) If the inward force is greater than that needed to keep the object moving
in a circle with radius at angular frequency , and the object moves inward. If , the
net force is insufficient to keep the object in the circular motion at that radius, and the object moves
outward. e) Objects with lower densities will tend to move toward the center, and objects with higher
densities will tend to move away from the center.
,cm
2
Rω
.cm
2
RρVω ,obcmR cmobRρρ >
cmobR ω ,obcm cmobRρρR <
(Note that increasing x corresponds to moving toward the back of the car.)
a) The mass of air in the volume element is ρAdxρdV = , and the net force on the element in the
forward direction is From Newton’s second law,( ) .AdppAAdpp =−+ ,)( adxρAAdp = from which
b) With given to be constant, and with.adxρdp = ρ .,0 00 ρaxppxatpp +=== c) Using
29

3
kg/m1.2=ρ in the result of part (b) gives ( )( )( ) atmp-523
1015~Pa0.15m5.2sm0.5mkg2.1 ×= , so the
fractional pressure difference is negligble. d) Following the argument in Section 14-4, the force on the
balloon must be the same as the force on the same volume of air; this force is the product of the mass
and the acceleration, or e) The acceleration of the balloon is the force found in part (d) divided
by the mass
ρV .ρVa
( ., balbal aρρorVρ ) The acceleration relative to the car is the difference between this
acceleration and the car’s acceleration, ( )[ ] .1balrel aρρa −= f) For a balloon filled with air, ( ) 1bal <ρρ (air
balloons tend to sink in still air), and so the quantity in square brackets in the result of part (e) is negative;
the balloon moves to the back of the car. For a helium balloon, the quantity in square brackets is positive,
and the balloon moves to the front of the car.
If the block were uniform, the buoyant force would be along a line directed through its geometric center,
and the fact that the center of gravity is not at the geometric center does not affect the buoyant force.
This means that the torque about the geometric center is due to the offset of the center of gravity, and is
equal to the product of the block’s weight and the horizontal displacement of the center of gravity from
the geometric center, .2m)075.0( The block’s mass is half of its volume times the density of water, so
the net torque is
m,N02.7
2
m075.0
)sm80.9(
2
)mkg1000()m30.0( 2
33
⋅=
or to two figures. Note that the buoyant force and the block’s weight form a couple, and the
torque is the same about any axis.
mN0.7 ⋅
a) As in Example 14.8, the speed of efflux is .2gh After leaving the tank, the water is in free fall, and
the time it takes any portion of the water to reach the ground is ,)(2
g
hH
t −
= in which time the water
travels a horizontal distance ).(2 hHhvtR −==
b) Note that if ,)()(, hhHhHhhHh −=′−′−=′ and so hHh −=′ gives the same range. A hole
below the water surface is a distance h above the bottom of the tank.hH −
The water will rise until the rate at which the water flows out of the hole is the rate at which water is
added;
,2
dt
dV
ghA =
30

which is solved for
cm.1.13
)sm80.9(2
1
m10501
sm1040.2
2
1
2
2
24
342
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
×
=⎟
⎠
⎞
⎜
⎝
⎛
= −
−
.gA
dtdV
h
Note that the result is independent of the diameter of the bucket.
14.82: a) .sm200.0)m0160.0()m00.8)(sm80.9(2)(2 322
33133 ==−= AyygAv
b) Since is atmospheric, the gauge pressure at point 2 is3p
),(
9
8
1
2
1
)(
2
1
31
2
2
32
3
2
2
2
32 yyρg
A
A
vvvp −=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=−= ρρ
using the expression for found above. Subsititution of numerical values gives Pa.3υ 4
2 1097.6 ×=p
The pressure difference, neglecting the thickness of the wing, is ),()21( 2
bottom
2
top vvρp −=Δ and solving
for the speed on the top of the wing gives
.sm133)mkg20.1(Pa)2000(2s)m120( 32
top =+=v
The pressure difference is comparable to that due to an altitude change of about so ignoring the
thickness of the wing is valid.
m,200
a) Using the constancy of angular momentum, the product of the radius and speed is constant, so the
speed at the rim is about h.km17
350
30
h)km200( =⎟
⎠
⎞
⎜
⎝
⎛
b) The pressure is lower at the eye, by an
amount
Pa.108.1
hkm6.3
sm1
)h)km17()hkm200(()mkg2.1(
2
1 3
2
223
×=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=Δp
c) m1602
2
=g
v
to two figures. d) The pressure at higher altitudes is even lower.
31

The speed of efflux at point D is ,2 1gh and so is 18gh at C. The gauge pressure at C is then
and this is the gauge pressure at E. The height of the fluid in the column is1,11 34 ρghρghρgh −=− .3 1h
a) ,A
dtdV
v = so the speeds are
.sm50.1
m100.40
sm1000.6
andsm00.6
m100.10
sm1000.6
24
33
24
33
=
×
×
=
×
×
−
−
−
−
b) Pa1069.1orPa,10688.1)( 442
2
2
12
1
××=−=Δ vvp ρ to three figures.
c) cm.7.12)sm80.9)(mkg106.13(
Pa)10688.1(
H 233
4
g
===Δ ×
×Δ
g
p
h ρ
a) The speed of the liquid as a function of the distance y that it has fallen is ,22
0 gyvv += and the
cross-section area of the flow is inversely proportional to this speed. The radius is then inversely
proportional to the square root of the speed, and if the radius of the pipe is the radius r of the stream a
distance y below the pipe is
,0r
.
2
1
)2(
41
2
0
0412
0
00
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+=
+
=
v
gy
r
gyv
vr
r
b) From the result of part (a), the height is found from or,2)21( 412
0 =+ vgy
m.10.1
)sm80.9(2
s)m2.1(15
2
15
2
22
0
===
g
v
y
a) The volume V of the rock is
32

.m1057.8
)sm80.9)(mkg1000.1(
N)0.21)smkg)(9.8000.3(( 34
233
2
waterwater
−
×=
×
−
=
−
==
g
Tw
g
B
V
ρρ
In the accelerated frames, all of the quantities that depend on g (weights, buoyant forces, gauge
pressures and hence tensions) may be replaced by ,agg +=′ with the positive direction taken upward.
Thus, the tension is ,)( 0 g
g
TgVmBgmT
′
=′−=′−′= ρ where N.0.210 =T
N.4.26N)0.21(,sm50.2for;b) 9.80
2.509.802
===+=′ +
Taagg
N.6.15N)0.21(,sm50.2Forc) 80.9
50.280.92
==−= −
Ta
.0and0,Ifd) ==′−= Tgga
a) The tension in the cord plus the weight must be equal to the buoyant force, so
)mkg180mkg1000)(sm80.9)(m50.0(m)20.0)(21(
)(
3322
foamwater
−=
−= ρρVgT
b) The depth of the bottom of the styrofoam is not given; let this depth be Denote the length of the
piece of foam by L and the length of the two sides by l. The pressure force on the bottom of the foam is
then
.0h
( )lLghp 2)( 00 ρ+ and is directed up. The pressure on each side is not constant; the force can be
found by integrating, or using the result of Problem 14.44 or Problem 14.46. Although these problems
found forces on vertical surfaces, the result that the force is the product of the average pressure and the
area is valid. The average pressure is ))),22((( 00 lhρgp −+ and the force on one side has magnitude
Lllhρgp )))22((( 00 −+
and is directed perpendicular to the side, at an angle of from the vertical. The force on the other
side has the same magnitude, but has a horizontal component that is opposite that of the other side. The
horizontal component of the net buoyant force is zero, and the vertical component is
°0.45
,
2
)))22(()(0.45cos(22)(
2
0000
Ll
gLllhgpLlghpB ρρρ =−+°−+=
the weight of the water displaced.
When the level of the water is a height y above the opening, the efflux speed is ,2gy and
.2)2( 2
gydπdt
dV
= As the tank drains, the height decreases, and
33
N.4.80=

.2
)2(
2)2(
2
2
2
gy
D
d
D
gyd
A
dtdV
dt
dy
⎟
⎠
⎞
⎜
⎝
⎛
−=−=−=
π
π
This is a separable differential equation, and the time T to drain the tank is found from
,2
2
dtg
D
d
y
dy
⎟
⎠
⎞
⎜
⎝
⎛
−=
which integrates to
[ ] ,22
2
0
Tg
D
d
y H ⎟
⎠
⎞
⎜
⎝
⎛
−=
or
.
2
2
2
22
g
H
d
D
g
H
d
D
T ⎟
⎠
⎞
⎜
⎝
⎛
=⎟
⎠
⎞
⎜
⎝
⎛
=
a) The fact that the water first moves upwards before leaving the siphon does not change the efflux
speed, .2gh b) Water will not flow if the absolute (not gauge) pressure would be negative. The hose is
open to the atmosphere at the bottom, so the pressure at the top of the siphon is ),(a hHρgp +− where
the assumption that the cross-section area is constant has been used to equate the speed of the liquid at
the top and bottom. Setting and solving for H gives0=p ( ) .hρgpH a −=
Any bubbles will cause inaccuracies. At the bubble, the pressure at the surfaces of the water will be the
same, but the levels need not be the same. The use of a hose as a level assumes that pressure is the
same at all point that are at the same level, an assumption that is invalidated by the bubble.
34

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