34
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: [email protected] [email protected] [email protected]

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Page 1: PROBLEMAS RESUELTOS FISICA UNIVERSITARIA · PDF file1 PROBLEMAS RESUELTOS FISICA UNIVERSITARIA CAPITULO 14 VOLUMEN 1 EDICION 11 SEARS ZEMANSKY Sección 14.1 Densidad Sección 14.2

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: [email protected]@gmail.com

[email protected]

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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 * m w 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

m 10 * 6,3793 4

0,002551 4

0,00081225 * 3,14 4

(0,0285) * 3.14 4d varillala de area =====

π

V = 6,3793 * 10-4 * 0,858 = 5,4735 * 10-4 m3

g * V * w ρ=

Newton 41,83 seg

m 9,8 * m 10 * 5,4735 *mkg 10 * 7,8 w 2

34-3

3 ==

No es necesario el carrito. Problema 14.2 Sears Zemansky 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

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3r * * 34 V π=

( )3410*174 * 3,14 * 34 V =

31212 m 10* 222066647,3 10 * 5268024 * 3,14 * 34 V ==

m = masa de la luna = 7,35 * 1022 kg.

( ) .mkg 1033.3

m 10*32,22066647kg 1035,7 3 3

312

22

×=×

==Vmρ

Problema 14.3 Sears Zemansky 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

( )( )

3m 0,00000225 910

3m 2250 3mm 1000

3m 1 * 3mm 2250 V ===

m = masa de la pieza rectangular de metal = 0,0158 kg. ( ) .3mkg 310022,73m 0,00000225

kg 0158.0 ×=== Vmρ

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. Problema 14.4 Sears Zemansky 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 V L =

m = masa del platino = 40 kg. ρ = la densidad del platino = 21,4 * 103 kg/m3

m Vρ

=

3m 3-10 * 1,869158

3m

kg310 * 21,4

kg 40 m V ===ρ

3

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3 V L =

m 0,12318 3 3m 3-10 * 1,869158 L == L = longitud del cubo = 12,31 cm Problema 14.5 Sears Zemansky 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

3plomor * *

34 plomoV π=

plomo * 3plomor * *

34 plomom ρπ=

3aluminior * *

34 aluminioV π=

aluminio * 3aluminior * *

34 aluminiom ρπ=

mplomo = maluminio

aluminio * 3aluminior * *

34 plomo * 3

plomor * * 34 ρπρπ =

Cancelando términos semejantes

aluminio * 3aluminior plomo * 3

plomor ρρ =

Despejando

( )( )

3

plomoraluminior

3

plomor

3aluminior

aluminio

plomo⎟⎟

⎜⎜

⎛==

ρ

ρ

⎟⎟

⎜⎜

⎛=

plomoraluminior

31

)aluminio

plomo (ρρ

ρ aluminio = 2,7 * 103 kg/m3 (densidad del aluminio) ρ plomp = 11,3 * 103 kg/m3 (densidad del plomo)

4

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⎟⎟

⎜⎜

⎛=

plomoraluminior

31

)310 * 7,2

310 * 3,11(

( )⎟⎟

⎜⎜

⎛=

plomoraluminior

31

1851,4

1,61 plomor

aluminior =

⎟⎟

⎜⎜

Problema 14.6 Sears Zemansky 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

( )3solr * * 34 solV π=

3810*6,96 * 3,14 * 34 V ⎟

⎠⎞⎜

⎝⎛=

3m 2410* 1412,2654 2410 * 337,15 * 3,14 * 34 V ==

a) 3m

kg 310 * 1,409 3m 2410 * 2654,1412

kg 301099.1

solVsol =

×==

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

( )3estrellar * * 34 estrellaV π=

( )320000 * 3,14 * 34 V =

3m 1210 * 33,510321 1210 * 8 * 3,14 * 34 V ==

b) 3mkg181005938.03m 1210 * 510321,33

kg301099.1×=

×=D

3mkg1610* 93.5=

5

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Problema 14.7 Sears Zemansky ¿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) g = gravedad = 9,8 m/seg2

m 906,9

3m

Newton 10,094

2m

Newton 100000

)2sm80,9( * )3mkg310 * 03,1(

Pa510* 1g * 0pp

h ===−

Problema 14.8 Sears Zemansky 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 =

Pa 5980h * g * =ρ ρ = 1050 kg/m3 (densidad del fluido) g = gravedad = 9,8 m/seg2

m581,0

3m

Newton 10290

2m

Newton 5980

)2sm80,9( )3mkg1050(

2mN5980g * Pa 5980h ====

ρ

h = 58,1 cm Problema 14.9 Sears Zemansky 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) g = gravedad = 9,8 m/seg2

a) ( ) Pa. 706m12,0 2sm80,9 3mkg600h * g * =⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛=ρ

ρ agua = 1000 kg/m3 (densidad del agua) g = gravedad = 9,8 m/seg2

6

h aceite = 0,12 m

h agua = 0,25 m

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b) ( ) Pa. 3156 Pa 2450 Pa 706 m 25,0 2sm8,9 3mkg1000Pa706 =+=⎟⎠⎞⎜

⎝⎛⎟

⎠⎞⎜

⎝⎛+

Problema 14.10 Sears Zemansky 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.

AW P =

⋅=== 2m 8040,0 )Pa 205000(

)N 16500( PW A

A = 804 cm2

b) AW P =

( )( )

⋅===+

== 2cm 1248 2m 1

2cm 100 * 2m 0,1248

2m

N 205000

N 25600 )Pa 205000(

)N 9100 N 16500( PW A

A = 1248 cm2

Problema 14.11 Sears Zemansky 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)(mkg 1003.1( 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))m 15.0()(Pa 1052.2( 526 ×=× π Problema 14.12 Sears Zemansky ¿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. 61.9Pa 1027.6)m 640)(sm 80.9)(mkg 1000.1( 6233 =×=×== ρghp

7

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Problema 14.13 Sears Zemansky 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) =××+×=+ − )m 1000.7)(sm 80.9)(mkg 106.13(Pa 10980 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. 101.03 gives 52 ×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)( 312 ×=− ρgyy

Problema 14.14 Sears Zemansky 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)m 1.6)(sm 80.9)(mkg1000.1( 4233 ×=×=ρgh Problema 14.15 Sears Zemansky 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()( 33wmmw =××== ρρhh

The volume of water is 33 424 cm 816m 108.16)m 10m)(12.0 (0.680A =×=×== −−hV Problema 14.16 Sears Zemansky

8

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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

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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(

mN 250023

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:

m 102.0)sm 80.9)(mkg 1000(

mN 100023

2

==h

Thus the water must be lowered by m0.153m102.0m255.0 =− Problema 14.17 Sears Zemansky 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 Problema 14.18 Sears Zemansky

[ ] )m (2.00 Pa 1093m) 2.14)(sm 71.3)(mkg1000.1(Pa 10130 2323 33 ×−×+× N.1079.1 5×=

Problema 14.19 Sears Zemansky The depth of the kerosene is the difference in pressure, divided by the product ,V

mgρg =

m. 14.4)m250.0( )smkg)(9.80205(

Pa1001.2)m(0.0700N)104.16(32

523

=×−×

=h

Problema 14.20 Sears Zemansky

atm.64.1Pa 1066.1m)15.0(

)smkg)(9.801200()2(

52

2

2 =×====πdπ

mgAFp

10

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Problema 14.21 Sears Zemansky The buoyant force must be equal to the total weight; so ,gg icewater mgVρVρ +=

,m 563.0mkg 920mkg 1000

kg 0.45 333

water

=−

=−

=iceρρ

mV

or to two figures. 3m 56.0 Problema 14.22 Sears Zemansky 14.22: The buoyant force is and N, 30.6N 20.11N 17.50 =−=B

.m 1043.6)sm 80.9)(mkg 1000.1(

N) 30.6( 34233

water

−×=×

==gρ

BV

The density is

.mkg 1078.230.650.17)mkg1000.1( 333 3

waterwater

×=⎟⎠⎞

⎜⎝⎛×====

Bwρ

gρBgw

Vmρ

Problema 14.23 Sears Zemansky 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 ρρ

VVVρV V 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),

%.3232.0mkg1030

)m103.04.0(5.0kg)042.0(11 3

3-6

fluid

==×××

−=−ρρ

Problema 14.24 Sears Zemansky 11

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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.6370m650.0 sm80.9 mkg1000.1 3233

water =×== gVρB b) kg. 5582sm 9.80

N 900-N 6370 ==== −g

TBgwm

c) (See Exercise 14.23.) If the submerged volume is ,V ′

%.9.858590.N 6370N5470 and

waterwater

====′

=′gVρ

wVV

gρwV

Problema 14.25 Sears Zemansky

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. 116oil oil =ghρ b) ( )( ) ( )( )( )( ) Pa. 921sm 80.9 m 0150.0 mkg 1000m 100.0 mkg 790 233 =+

c) ( ) ( ) ( )

( ) kg. 822.0sm 80.9

m 100.0 Pa 8052

2topbottom ==

−==

gApp

gwm

The density of the block is ( ) .822 33 mkg

m 10.0kg 822.0 ==p Note that is the same as the average density of the fluid

displaced, ( ) ( ) ( ) . )mkg 1000( 15.0mkg 790 85.0 33 +

12

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Problema 14.26 Sears Zemansky 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,

( )( )( ) ,m 1036.3

mkg107.2 sm80.9N89 33

332−×=

×====

gρw

ρgw

ρmV

33 m104.3or −× to two figures.

b) ( ) N. 0.567.2

00.11N 891aluminum

waterwater =⎟

⎠⎞

⎜⎝⎛ −=⎟⎟

⎞⎜⎜⎝

⎛−=−=−=ρρωVgρwBwT

Problema 14.27 Sears Zemansky 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 .

hh

b) .objfl gVB ρ= The blocks have the same volume so experience the same buoyant force. objV c) . so 0 BwTBwT −==+− The object have the same V but is larger for brass than for aluminum so is larger for the brass block.

.ρVgw = ρ wB is the same for both, so T is larger for the brass block, block B.

Problema 14.28 Sears Zemansky 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

3333 m10102.1

mkg 101.00kg 102.1 −×=

×

The weight of unknown liquid displaced is N, 20.6N 18.6N 2.39 =− and its mass is

kg. 102.2sm 9.80N 6.20 2 = The liquid’s density is thus 33 m101.102kg102.2 −× ,mkg1091.1 33×= or roughly twice the density of water.

13Problema 14.29 Sears Zemansky

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)(, 21122211 ΑΑvvΑvΑv ==

22

21 cm) 10.0(20 ,cm)80.0( πΑπΑ ==

sm6.9)10.0(20

(0.80)s)m0.3( 2

2

2 ==ππv

Problema 14.30 Sears Zemansky

⋅===2

3

2

2

2

112

sm 245.0)m 0700.0s)(m 50.3(ΑΑΑ

Αvv

a) (i) s.m 21.5 ,m047.0 (ii) s.m 33.2 ,m 1050.0 22

222

2 ==== vΑvΑ b) .m882s)3600( )sm(0.245 33

2211 === tΑυtΑv Problema 14.31 Sears Zemansky

a) .98.16)m150.0()sm20.1(2

3

===πA

dtdVv

b) .m 317.0)( 22112 === πvdtdVvvrr Problema 14.32 Sears Zemansky 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). Problema 14.33 Sears Zemansky 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

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= ))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 la p0 presión manométrica dada en la parte superior del tanque Problema 14.34 Sears Zemansky 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. g = gravedad = 9,8 m/seg2

h = altura en metros.

h * g * 2 2v =

segm 16,56

2seg

2m 274,4 m 14 * 2seg

m 9,8 * 2 h * g * 2 v ====

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 Problema 14.35 Sears Zemansky ¿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

Pa 510 * 1,47 2m

Newton 147000 m 15 * 2seg

m 9,8 * 3m

kg 1000 2y * g * 1p ==== ρ

p1 = 1,47 * 105 Pa Problema 14.36 Sears Zemansky Using 14

12 vv = in Eq. (14.17),

⎥⎦

⎤⎢⎣

⎡−+⎟

⎠⎞

⎜⎝⎛+=−+−+= )(

3215)()(

21

2121121

22

2112 yygυρpyyρgvvρpp

15

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⎟⎠⎞

⎜⎝⎛ +×+×= m)0.11)(sm80.9()sm00.3(

3215 )mkg1000.1(Pa1000.5 22334

Pa.1062.1 5×= Problema 14.37 Sears Zemansky Neglecting the thickness of the wing (so that 21 yy = in Eq. (14.17)), the pressure difference is

Pa.078)(2)1( 21

22 =−=Δ vvρp The net upward force is then

N.496)smkg)(9.801340()m(16.2Pa)780( 22 −=−× Problema 14.38 Sears Zemansky a) ( )( ) s.kg30.1s 0.60

kg355.0 220 = b) The density of the liquid is ,mkg 1000 3m 100.355

kg 355.033 =−×

and so the volume flow

rate is s.L1.30sm1030.1 33mkg1000skg30.1

3 =×= − This result may also be obtained

from ( )( ) s.L30.1s 0.60L 355.0220 = c) 24

33

m1000.2sm 1030.1

1 −

×

×=v

s.m63.14s,m50.6 12 === vv

d) ( ) ( )1221

2221 2

1 yyρgvvρpp −+−+=

( ) ( ) ( ) ( )( )

( )( )( )m35.1 sm80.9 mkg1000 sm50.6sm63.1 mkg1000 21 kPa 152

23

223

−+

−+=

kPa119= Problema 14.39 Sears Zemansky The water is discharged at a rate of s.m352.023

34

m1032.1sm1065.4

1 == −

×

×v The pipe is given as horizonatal, so the

speed at the constriction is s,m95.82212 =Δ+= ρpvv keeping an extra figure, so the cross-section

are at the constriction is ,m1019.5 25sm95.8

sm1065.4 34 −× ×=−

and the radius is cm.41.0== πAr Problema 14.40 Sears Zemansky From Eq. (14.17), with ,21 yy =

( ) 211

212

1122

2112 8

342

121 ρvpvvρpvvρpp +=⎟⎟

⎞⎜⎜⎝

⎛−+=−+=

( )( ) Pa,1003.2sm50.2mkg1000.183Pa1080.1 42334 ×=×+×=

where the continutity relation 21

2vv = has been used.

Problema 14.41 Sears Zemansky

16

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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.

sm43.1 so ,cm7200 132

1111 === vπrvAv sm73.5 so ,cm7200 2

322222 === vπrvAv

2222

2111 2

121 ρvρgypρvρgyp ++=++

( ) Pa 1025.221 so Pa,1040.2 and 52

22112

5221 ×=−+=×== vvρpppyy

Problema 14.42 Sears Zemansky a) The cross-sectional area presented by a sphere is ,4

2Dπ therefore ( ) .402DπppF −= b) The force on

each hemisphere due to the atmosphere is ( )22 m1000.5 −×π ( )( ) .776975.0 Pa10013.1 5 Ν=× Problema 14.43 Sears Zemansky a) ( )( )( ) Pa.1010.1m1092.10 sm80.9mkg1003.1 83233 ×=×××=ρgh b) The fractional change in volume is the negative of the fractional change in density. The density at that depth is then

( ) ( ) ( )( )( )1118330 Pa108.45 Pa1016.11 mkg1003.11 −−××+×=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

Problema 14.44 Sears Zemansky a) The weight of the water is

( )( ) ( ) ( ) ( )( ) N,1088.5m0.3 m0.4 m00.5 sm 80.9 mkg1000.1 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×

2dgAF ρ=

( )( ) ( ) ( )( ) ( ) N,1076.1m50.1 m0.3 m0.4 sm 80.9 mkg1000.1 5233 ×=×= or to two figures. N108.1 5× Problema 14.45 Sears Zemansky

17

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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 gives Hhh == to0 m.N1061.212 43 ⋅×== Hgωρτ Problema 14.46 Sears Zemansky a) See problem 14.45; the net force is from dF∫ ,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). Problema 14.47 Sears Zemansky The acceleration due to gravity on the planet is

dp

ρdpg

Vm

Δ=

Δ=

and so the planet’s mass is

mGdpVR

GgRM

22 Δ==

Problema 14.48 Sears Zemansky 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, . 2Cx=ρ a) The volume element Then the integral becomes

Integrating gives

.2dVCxdVM ∫=∫= ρ .2dxπRdV =

.220 dxRCxM L π∫= .

3

322

02 LRCdxxRCM L ππ =∫= Solving for .3, 32 LRMCC π=

b) The density at the end is Lx = ( )( ) ( ).2323232

LπRM

LπRM 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 = Problema 14.49 Sears Zemansky a) At the model predicts ,0=r 3mkg700,12== Aρ and at ,Rr = the model predicts

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.mkg103.15m)1037.6)(mkg1050.1(mkg700,12 336433 ×=××−=−= −BRAρ b), c)

∫ ∫ ⎥⎦⎤

⎢⎣⎡ −⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥

⎤⎢⎣

⎡−=−==

R BRAπRBRARπdrrBrAπdmM0

3432

43

34

434][4

⎥⎦

⎤⎢⎣

⎡ ××−⎟⎟

⎞⎜⎜⎝

⎛ ×=

4m)1037.6)(mkg1050.1(3mkg700,12

3m)1037.6(4 643

336π

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 .)( 2rrGmg = Using the same integration as that in part (b), with an upper limit of

r instead of R gives the result. e) )kgmN10673.6()( ,at and ,0at 0 22112 ⋅×===== −RRGmgRrgrg

.sm85.9m)10(6.37kg)1099.5( 22624 =××

f) ; 2

33

44

33

4 2

⎥⎦⎤

⎢⎣⎡ −⎟⎠⎞

⎜⎝⎛=⎥

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛=

BrAπGBrArdrdπG

drdg

setting ths equal to zero gives m1064.532 6×== BAr , and at this radius

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

BABA

BAπGg

32

43

32

34

B

πGA9

4 2

=

.sm02.10)mkg1050.1(9

)mkg700,12()kgmN10673.6(4 243

232211

=×⋅×

= −

−π

Problema 14.50 Sears Zemansky 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 ,at 0 Rrp ==

∫ ∫ −==−=r

R rrR

Rρgdrr

Rρgdrr

Rρgp

R 22sss ).(2

b) Using the above expression with , and 0 34

3RM

VMr

πρ ===

Pa.1071.1m)1038.6(8

)smkg)(9.801097.5(3)0( 1126

224

×=×

×=

πp

19

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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. Problema 14.51 Sears Zemansky a) Pa. 1047.1)m100.15)(sm 80.9)(mkg 1000.1( 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 Problema 14.52 Sears Zemansky 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 7333water ×=×××=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. Problema 14.54 Sears Zemansky 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(mkg23.1 332

3ave =−=ρ

Problema 14.55 Sears Zemansky a) The submerged volume so , is

water gρwV ′

20

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⋅==×

===′

%3030.0)m(3.0 )mkg1000.1(

)kg900(333

water

water

Vρm

Vgρw

VV

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

wVVgpVwgVρ

waterwaterwater

Problema 14.56 Sears Zemansky a) The volume displaced must be that which has the same weight and mass as the ice, 3

cmgm00.1gm70.9 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

cmgm05.1gm70.9 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.

3cm46.0

Problema 14.57 Sears Zemansky 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

woodwaterwoodPb ρρ

ρρ−−

=VV

3333

33332

mkg1000.1mkg103.11mkg600mkg1000.1)m102.1(

×−×−×

×= −

,m1066.4 34−×= which has a mass of 5.27 kg. Problema 14.58 Sears Zemansky

The fraction f of the volume that floats above the fluid is ,

1fluidρρ

−=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 fρρ

−=

21

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Thus, if two fluids are observed to have floating fraction .11, and

2

11221 f

fρρff−−

= In this form, it’s clear

that a larger corresponds to a larger density; more of the stem is above the fluid. Using 2f

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( 3wateralcohol == ρρ

Problema 14.59 Sears Zemansky a) The “lift” is from which ,)(

2Hair gρρV −

.m100.11)sm80.9)(mkg0899.0mkg(1.20

N000,120 33233 ×=

−=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

×=⎟⎟⎠

⎞⎜⎜⎝

−−ρρρρ

This increase in lift is not worth the hazards associated with use of hydrogen. Problema 14.60 Sears Zemansky

a) Archimedes’ principle states . so ,A

MLMggLAρ

ρ ==

b) The buoyant force is ,)( FMgxLgA +=+ρ and using the result of part (a) and solving for x gives .ρgA

Fx = 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 =

.22ρgAMπ

kMπT ==

Problema 14.61 Sears Zemansky

1: a)( )

( ) ( )m.107.0

m450.0mkg1003.1kg0.70

233 =×

====πρA

mρgAmg

ρgAwx

b) Note that in part (c) of Problem is the mass of the buoy, not the mass of the man, and M ,60.14A is the cross-section area of the buoy, not the amplitude. The period is then

22

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( )( )( ) ( )

s.42.2m450.0sm80.9 mkg101.03

kg9502 2233 =×

πT

Problema 14.62 Sears Zemansky 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 and 1ρ ,M 0ρ ,v

( ) .80.01

water0 ⎟⎟⎠

⎞⎜⎜⎝

⎛+=+ v

ρMρMvρ

Solving for ,0ρ

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎞⎜⎜⎝

⎛+= Mv

ρMρ

1water0 80.01

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

1

waterwater 80.01

ρρ

vMρ

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×−−×= 3

33

333

mkg980mkg101.03(0.80)1

m0.400kg0.75mkg1003.1

⋅= 3mkg732 Problema 14.63 Sears Zemansky 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

airbrass

airwood

woodbrasswoodwoodwood 1

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−==

ρρM

ρρρMVρM

kg.0958.0mkg150mkg20.11)kg0950.0(

1

3

3

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

Problema 14.64 Sears Zemansky

23

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The buoyant force on the mass A, divided by kg70.4kg80.1kg1.00kg7.50 bemust , =−−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 is kg,70.4 .mkg1024.1 33m103.80

kg70.433- ×=

×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,20.8kg.80.2

Problema 14.65 Sears Zemansky 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 gives A1V N)0.6(N)0.45()( A1waterA1AuAuwater ρρρρgVρ −=−

))0.6(N)0.45(()( Auwater

A1Auwater

AuAuAuAu ρρ

ρρρρgVρw −

−==

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 Problema 14.66 Sears Zemansky The ball’s volume is

333 cm7238cm)0.12(34

34

=== 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 = kg080.6 6080 =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,158.1g 1158)cmg00.1)(cm7238)(16.0( 33 ==

24

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and its acceleration upon release is thus

2net sm5.51kg1.158N6.59

===m

Fa

Problema 14.67 Sears Zemansky 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

1or water

crowncrowncrown water fρ

ρfρρρ−

==+−

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

,1crown

waterρρf −= and so the weight of the crown would be

( )( ) ( ) N. 2.12N 9.12 3.1911 =− c) Approximating the average density by that of lead for a “thin” gold plate, the apparent weight would be ( )( ) ( ) N. 8.11N 9.12 3.1111 =− Problema 14.68 Sears Zemansky a) See problem 14.67. Replacing f with, respectively, wwwater and wwfluid gives

,- fluidfluid

steel

www

ρρ

= ,- waterfluid

steel

www

ρρ

=

and dividing the second of these by the first gives

.--

water

fluid

water

fluid

wwww

ρρ

=

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

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.11

water

fluid

water

fluid

ff

ρρ

−−

=

and solving for ,fluidf

( ) ( ) ( ) %.4.84844.0128.0 220.1111 waterwater

fluidfluid ==−=−−= f

ρρf

Problema 14.69 Sears Zemansky a) Let the total volume be V; neglecting the density of the air, the buoyant force in terms of the weight is

,)(0

mwaterwater ⎟⎟

⎞⎜⎜⎝

⎛+== V

ρgwgρgVρB

or

⋅ρ

−=g

wgρ

BVwwater

0

b) .m1052.2 34Cuwater

−×=− gρw

gρB Since the total volume of the casting is ,

water

Bgρ the cavities are

12.4% of the total volume. Problema 14.70 Sears Zemansky 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)by related are and ,.)550()750()1000(

d. Substitution into the first relation gives m.040.05.002

(750)(1000)(550)0)(0.65)(100 === −

− LLd 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 =+ Problema 14.71 Sears Zemansky 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 and where, mmAmmh ρ+= are the masses of the barge and the anchor, and A is the area

26

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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

m00.8 mkg7860kg0.35 4

23steel

a0

−×===′−=ΔA

mhhhρ

or about 0.56 mm. Problema 14.72 Sears Zemansky a) The average density of a filled barrel is ,mkg875mkg750 3

m0.120kg0.153

oil 3 =+=+ Vmρ which is less than

the density of seawater, so the barrel floats. b) The fraction that floats (see Problem 14.23) is

%.0.15150.0mkg1030mkg87511 3

3

water

ave ==−=−ρρ

c) The average density is 333 m

kgm0.120kg0.32

mkg 1172 910 =+ which means the barrel sinks. In order to lift it,

a tension N173) 80.9)(m120.0)( 1030() 80.9)(m120.0)( 1177( 232 sm3

mkg

sm3

mkg =−= 3T is required.

Problema 14.73 Sears Zemansky 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

and yLx −= .)(ωL

BLρρ

Lρρy −−= c) m.046.0m)10.0(0.16.13

8.76.13 == −−y

Problema 14.74 Sears Zemansky

a) The change is height is related to the displaced volume yΔ ,by AVyV 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, so VΔ

27

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gAρ

wA

gρwAVy

water

water ==Δ

m.213.0m))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;189.0 and , 98

9 =Δ=′Δ−Δ=′Δ Δ yyyy y the water sinks by this amount. Problema 14.75 Sears Zemansky 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 lgayy =−

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 from drrdp )( 2ωρ= ,2 gives to0 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 through palp =Δ ypgp Δ=Δ from which p cancels out. Problema 14.76 Sears Zemansky 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

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specifically, ,ρa drdrdpp

p == ∂

∂ and using gives rωa 2= .2rρωp

p =∂

∂ b) Let the pressure at

be (atmospheric pressure); integrating the expression for 0,0 == ry app

p

∂ from part (a) gives

( ) .22

a 20, rρωpyrp +==

c) In Eq. as found in part (b), )0,(,),5.14( 1a2 === yrpppp ),(y and 0 21 rhy == the height of the liquid

above the plane. Using the result of part (b) gives 0=y .2)( 22 grωrh = Problema 14.77 Sears Zemansky 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

rrdrdp

0 0

2ρω

( ),2

022

2

0 rrρωpp −⎟⎟⎠

⎞⎜⎜⎝

⎛=−

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.

,cm2Rω

.cm2RρVω ,obcmR cmobRρρ >

cmobR ω ,obcm cmobRρρR <

Problema 14.78 Sears Zemansky (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

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3kg/m 1.2=ρ in the result of part (b) gives ( )( )( ) atmp-523 1015 ~ Pa 0.15m 5.2 sm 0.5 mkg 2.1 ×= , 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. Problema 14.79 Sears Zemansky 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,N02.72

m075.0)sm80.9(2

)mkg1000()m30.0( 233

⋅=

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.

mN 0.7 ⋅

Problema 14.80 Sears Zemansky 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 ,)(2g

hHt −= 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 − Problema 14.81 Sears Zemansky The water will rise until the rate at which the water flows out of the hole is the rate at which water is added;

,2dtdVghA =

30

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which is solved for

cm.1.13)sm80.9(2

1m10501

sm1040.221

2

2

24

342

=⎟⎟⎠

⎞⎜⎜⎝

⎛××

=⎟⎠⎞

⎜⎝⎛= −

.gAdtdVh

Note that the result is independent of the diameter of the bucket. Problema 14.82 Sears Zemansky 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 is 3p

),(981

21) (

21

31

2

2

323

22

232 yyρg

AAvvvp −=

⎟⎟

⎜⎜

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−=−= ρρ

using the expression for found above. Subsititution of numerical values gives Pa. 3υ4

2 1097.6 ×=p Problema 14.83 Sears Zemansky The pressure difference, neglecting the thickness of the wing, is ),( )21( 2

bottom2top vvρp −=Δ and solving

for the speed on the top of the wing gives

.sm133)mkg20.1(Pa)2000(2s)m120( 32top =+=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

Problema 14.84 Sears Zemansky 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.km1735030h)km200( =⎟

⎠⎞

⎜⎝⎛ b) The pressure is lower at the eye, by an

amount

Pa. 108.1hkm6.3

sm1)h)km17()hkm200(()mkg2.1(21 3

2223 ×=⎟⎟

⎞⎜⎜⎝

⎛−=Δp

c) m16022=g

v to two figures. d) The pressure at higher altitudes is even lower.

31

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Problema 14.85 Sears Zemansky 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 is 1,11 34 ρghρghρgh −=− .3 1h Problema 14.86 Sears Zemansky a) ,A

dtdVv = so the speeds are

.sm50.1m100.40

sm1000.6 and sm00.6m100.10

sm1000.624

33

24

33

=××

=××

b) Pa1069.1or Pa,10688.1)( 442

2212

1 ××=−=Δ vvp ρ to three figures.

c) cm.7.12)sm80.9)(mkg106.13(

Pa)10688.1(H 233

4

g===Δ

×

×Δg

ph ρ

Problema 14.87 Sears Zemansky 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

.21)2(

41

20

04120

00−

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

vgyr

gyvvr

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(152

152

220 ===

gvy

Problema 14.88 Sears Zemansky a) The volume V of the rock is

32

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.m1057.8)sm80.9)(mkg1000.1(

N)0.21)smkg)(9.8000.3(( 34233

2

waterwater

−×=×

−=

−==

gTw

gBV

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 ggTgVmBgmT ′=′−=′−′= ρ where N.0.210 =T

N.4.26N)0.21(,sm50.2for ; b) 9.802.509.802 ===+=′ +Taagg

N.6.15N)0.21(,sm50.2For c) 80.950.280.92 ==−= −Ta

.0 and 0, If d) ==′−= Tgga Problema 14.89 Sears Zemansky 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

0000LlgLllhgpLlghpB ρρρ =−+°−+=

the weight of the water displaced. Problema 14.90 Sears Zemansky When the level of the water is a height y above the opening, the efflux speed is ,2gy and

.2)2( 2 gydπdtdV = As the tank drains, the height decreases, and

33

N.4.80=

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.2)2(2)2( 2

2

2

gyDd

Dgyd

AdtdV

dtdy

⎟⎠⎞

⎜⎝⎛−=−=−=

ππ

This is a separable differential equation, and the time T to drain the tank is found from

,2 2

dtgDd

ydy

⎟⎠⎞

⎜⎝⎛−=

which integrates to

[ ] ,2 22

0Tg

Ddy H ⎟⎠⎞

⎜⎝⎛−=

or

.22

2 22

gH

dD

gH

dDT ⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

Problema 14.91 Sears Zemansky 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 gives 0=p ( ) .hρgpH a −= Problema 14.92 Sears Zemansky 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