Dinámica de la partícula

Dinámica de la partícula
Javier Junquera
Bibliografía
FUENTE PRINCIPAL
Física, Volumen 1, 3° edición
Raymod A. Serway y John W. Jewett, Jr.
Ed. Thomson
ISBN: 84-9732-168-5
Capítulos 4 y 5
Física para Ciencias e Ingeniería, Volumen 1, 7° edición
Raymod A. Serway y John W. Jewett, Jr.
Cengage Learning
ISBN 978-970-686-822-0
Capítulos 5 y 6
Física, Volumen 1
R. P. Feynman, R. B. Leighton, y M. Sands
Ed. Pearson Eduación
ISBN: 968-444-350-1
Capítulo 9
Definición de dinámica y cinemática
Dinámica:
Estudio del movimiento de un objeto, y de las
relaciones de este movimiento con conceptos
físicos tales como la fuerza y la masa.
En otras palabras, estudio del movimiento
atendiendo a las causas que lo producen.
Cinemática:
Estudio del movimiento, usando los
conceptos de espacio y tiempo, sin tener
en cuenta las causas que lo producen.
Dinámica: preguntas a resolver y
conceptos básicos que vamos a introducir
¿Qué hace que un objeto se mueva o que permanezca en reposo?
¿Qué mecanismos hacen que un objeto cambie su estado de movimiento?
¿Por qué unos objetos se aceleran más que otros?
Dos conceptos básicos que vamos a introducir en este tema:
- Fuerza
- Masa
Concepto de fuerza
Puede definirse una fuerza como toda acción o influencia capaz de
modificar el estado de movimiento o de reposo de un cuerpo
(imprimiéndole una aceleración que modifica el módulo, la dirección,
o el sentido de su velocidad), o bien de deformarlo.
La fuerza es todo agente capaz de modificar el momentum de un objeto.
La fuerza es una magnitud vectorial. Por lo tanto, tiene:
- módulo (en el SI, la unidad es el Newton, N)
- dirección
- sentido
(se les aplica todas las leyes del álgebra vectorial).
Tipos de fuerza:
de contacto y de acción a distancia
Contact forces
S E C T I O N 5 . 1 • The Concept of Force
Field forces
m
(a)
M
(d)
–q
(b)
Si se examina el origen
de las fueras a una
escala atómica, la
separación entre fuerzas
de contacto y fuerzas de
campo no es tan clara
+Q
(e)
Iron
(c)
113
N
S
(f)
Figure 5.1 Some examples of applied forces. In each case a force is exerted on the object within the boxed area. Some agent in the environment external to the boxed area
exerts a force on the object.
Fuerzas de campo: no implican un
contacto físico entre dos objetos.
Another class of forces, known as field forces, do not involve physical contact beActúan
travésforce
delofespacio
vacío
tween two objects but instead act through empty
space. Theagravitational
at-
Fuerzas de contacto: implican un
contacto físico entre dos objetos
traction between two objects, illustrated in Figure 5.1d, is an example of this class of
force. This gravitational force keeps objects bound to the Earth and the planets in or-
Tipos de interacción desde un
punto de vista fundamental
Nuclear fuerte
Electromagnética
Nuclear débil
Gravitatoria
Tipos de interacción desde un
punto de vista fundamental
Nuclear fuerte
Electromagnética
Nuclear débil
Gravitatoria
Únicas relevantes en Física Clásica
Medir la intensidad de una fuerza
mediante la deformación de un muelle
C H A P T E R 5 • The Laws of Motion
0
1
2
3
4
0
2
4
3
0
1
2
3
4
0
1
2
3
4
1
114
F2
θ
F1
F1
Isaac Newton,
F2
English physicist and
mathematician
(1642–1727)
F2
(a)
(b)
(c)
(d)
Aplicamos Isaac
una
fuerza
vertical sobre el muelle.
Newton
was one of the
most brilliant scientists in history.
Como consecuencia,
el muelle se deforma.
Before the age of 30, he
formulated the basic concepts
and laws of mechanics,
discovered the law of universal
gravitation, and invented the
mathematical methods of
calculus. As a consequence of
his theories, Newton was able to
explain the motions of the
planets, the ebb and flow of the
tides, and many special features
of the motions of the Moon and
the Earth. He also interpreted
many fundamental observations
concerning the nature of light.
His contributions to physical
theories dominated scientific
thought for two centuries and
remain important today.
F
F1
Si ahora aplicamos una fuerza de magnitud
doble que la fuerza de referencia, el muelle se
$ (2.00 cm) " 2.24 cm.
deformará el doble
Figure 5.2 The vector nature of a force is tested with a spring scale. (a) A downward
force F1 elongates the spring 1.00 cm. (b) A downward force F2 elongates the spring
2.00 cm. (c) When F1 and F2 are applied simultaneously, the spring elongates by
3.00 cm. (d) When F1 is downward and F2 is horizontal, the combination of the two
forces elongates the spring √ (1.00 cm)2
2
El efecto combinado de dos fuerzas colineares
Se puede medir el valor de una
fuerza aplicada
quantity, we use the bold-faced symbol F.) If we now apply a different downward force
esF , la
suma
efectos de las fuerzas
is twice that of the reference force
as seen
in Figurede
5.2b, los
the
F whose
mirando el puntero sobre
lamagnitude
escala.
pointer moves to 2.00 cm. Figure 5.2c shows that the combined effect of the two
individuales
collinear forces is the sum of the effects of the individual forces.
2
1
Now suppose the two forces are applied simultaneously with F1 downward and F2
horizontal, as illustrated in Figure 5.2d. In this case, the pointer reads
√5.00 cm2 " 2.24 cm. The single force F that would produce this same reading is the
sum of the two vectors F1 and F2, as described in Figure 5.2d. That is,
! F ! " √F12 $ F22 " 2.24 units, and its direction is ! " tan# 1(# 0.500) " # 26.6°.
Because forces have been experimentally verified to behave as vectors, you must
use the rules of vector addition to obtain the net force on an object.
Calibramos el muelle definiendo una fuerza de
referencia
como la fuerza que produce una
elongación del muelle de una unidad
(Giraudon/Art Resource)
5.2
Como se ha verificado experimentalmente que
las fuerzas se comportan como vectores, se
deben utilizar las leyes de la adición de vectores
para conocer la fuerza neta sobre un objeto
Newton’s First Law and Inertial Frames
Primera ley del movimiento de Newton:
ley o principio de inercia
En un sistema inercial, y en ausencia de fuerzas externas, un objeto en reposo
permanece en reposo y un objeto en movimiento continúa en movimiento con una
velocidad constante (es decir, con una celeridad constante según una línea recta).
Si sobre un cuerpo no actúa ninguna fuerza, su aceleración es cero.
Un objeto tiende a mantener su estado original de movimiento en ausencia de fuerzas.
Parece contraintuitivo: en la vida ordinaria, parece que el estado natural de los cuerpos es el reposo
(sin embargo, tenemos que tener en cuenta las fuerzas de rozamiento).
Requirió una cierta imaginación darse cuenta de este principio, y el esfuerzo inicial se lo debemos a
Galileo Galilei.
La resistencia de un objeto a cambiar su velocidad se conoce con el nombre de inercia
Definición de sistema de referencia inercial
Un sistema inercial de referencia es aquel cuyo comportamiento está regulado por
la primera ley de Newton.
Cualquier sistema de referencia que se mueva con una velocidad constante respecto de
un sistema inercial será, el mismo, un sistema inercial.
Definición de masa inerte
La masa inerte (o masa inercial) es la medida de la resistencia de un objeto a que se
produzca una variación en su movimiento como respuesta a una fuerza externa.
La masa es una magnitud escalar (unidades en el SI: el kg)
Definición de masa inerte:
la masa depende de la velocidad
La masa inerte (o masa inercial) es la medida de la resistencia de un objeto a que se
produzca una variación en su movimiento como respuesta a una fuerza externa.
A velocidades pequeñas comparadas con la velocidad de la luz, la masa se puede
considerar como una propiedad inherente al objeto, independiente del entorno que rodee
al objeto y del método utilizado para medirla.
En Mecánica Relativista, la masa depende de la velocidad del objeto
¿Qué ocurre cuando la velocidad de un objeto se acerca a la de la luz?
Definición de masa inerte:
masa y peso son magnitudes diferentes
La masa inerte (o masa inercial) es la medida de la resistencia de un objeto a que se
produzca una variación en su movimiento como respuesta a una fuerza externa.
La masa y el peso son magnitudes diferentes.
El peso es el módulo de la fuerza gravitatoria.
Un objeto con la misma masa no pesa lo mismo en la Tierra que en la Luna (cambia el valor de g).
Segunda ley del movimiento de Newton:
(caso general)
La fuerza es la razón de cambio (derivada) del momento con respecto al tiempo,
entendiendo por momento el producto de la masa por la velocidad.
En sistemas en los que la masa no cambia con el tiempo ni con la velocidad
Segunda ley del movimiento de Newton:
(caso no relativista)
En un sistema de referencia inercial, la aceleración de un objeto es directamente
proporcional a la fuerza neta que actua sobre él, e inversamente proporcional a su masa.
Si sobre un cuerpo actúa más de una fuerza externa, debemos calcular primero la
resultante (suma vectorial) de todas las fuerzas externas.
Desglosando en componentes:
Unidades y magnitudes de la fuerza
En el sistema internacional, la unidad de fuerza es el Newton.
Se define como la fuerza necesaria que hay que aplicar a un
cuerpo de masa 1 kg para que adquiera una aceleración de 1 m/s2
En el sistema cgs, la unidad es la dina
Dimensiones de la fuerza: [F] = MLT-2
Fuerza gravitacional y peso
La fuerza atractiva que la Tierra ejerce sobre un objeto se denomina fuerza gravitacional
- Dirección: vertical
- Sentido: hacia el centro de la Tierra
- Módulo: peso
Un objeto en caída libre (aquel que se mueve únicamente bajo la acción de la gravedad)
experimenta un movimiento rectilíneo uniformemente acelerado con aceleración
Como sólo actúa la gravedad, la suma
de todas las fuerzas externas se
reduce a un solo término
Si el objeto tiene una masa m
Peso: módulo de la fuerza gravitacional
Fuerza gravitacional y peso:
algunas sutilezas
Peso: módulo de la fuerza gravitacional
El peso depende de la posición geográfica y altura
La masa es una propiedad inherente del sistema.
El peso no. El peso es una propiedad de un sistema de elementos (ej: el cuerpo y la Tierra)
El kg es una unidad de masa, no de peso
Laws of Motion
Tercera ley de Newton:
(principio de acción y reacción)
ple 5.2
How Much Do You Weigh in an Elevator?
Solution No, your weight is unchanged. To provide the acceleration upward, the floor or scale must exert on your feet
an upward force that is greater in magnitude than your
weight. It is this greater force that you feel, which you interpret as feeling heavier. The scale reads this upward force,
not your weight, and so its reading increases.
y had the experience of standing in an
ates upward as it moves toward a higher
ou feel heavier. In fact, if you are standscale at the time, the scale measures a
nitude that is greater than your weight.
e and measured evidence that leads you
eavier in this situation. Are you heavier?
5.6
Newton’s Third Law
If you press against a corner of this textbook with your fingertip, the book pushes back
Si dos
objetos
interactúan,
labook
fuerza
and makes
a small dent in your
skin. If you push harder, the
does the sameF
and
12 ejercida por el objeto 1 sobre el 2 es igual en módulo y
the dent in your skin is a little larger. This simple experiment illustrates a general principle of critical importance
knownopuesta
as Newton’s third law:
dirección,
pero
en sentido, a la fuerza F21 ejercida por el objeto 2 sobre el objeto 1.
If two objects interact, the force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F21 exerted by object 2 on object 1:
F12 ! "F21
(5.7)
When it is important to designate forces as interactions between two objects, we will
force producen
exerted by a on b.” The third
use this subscript
notation, where
F means “these
Las
fuerzas
siempre
por parejas. No puede existir una única fuerza aislada.
law, which is illustrated in Figure 5.5a, is equivalent to stating that forces always occur
ab
in pairs, or that a single isolated force cannot exist. The force that object 1 exerts
on object 2 may be called the action force and the force of object 2 on object 1 the reaction force. In reality, either force can be labeled the action or reaction force. The action
force is equal in magnitude to the reaction force and opposite in direction. In
all cases, the action and reaction forces act on different objects and must be of
the same type. For example, the force acting on a freely falling projectile is the gravitational force exerted by the Earth on the projectile Fg ! FEp (E ! Earth, p ! projectile), and the magnitude of this force is mg. The reaction to this force is the gravitational force exerted by the projectile on the Earth FpE ! " FEp. The reaction force FpE
must accelerate the Earth toward the projectile just as the action force FEp accelerates
the projectile toward the Earth. However, because the Earth has such a large mass, its
acceleration due to this reaction force is negligibly small.
En todos los casos, las fuerzas de acción y reacción actúan sobre objetos diferentes,
y deben ser del mismo tipo.
F12 = –F21
2
F12
F21
1
(a)
Notación
Fnh
John Gillmoure /corbisstockmarket.com
Fhn
(b)
Figure 5.5 Newton’s third law. (a) The force F12 exerted by object 1 on object 2 is
equal in magnitude and opposite in direction to the force F21 exerted by object 2 on
Fuerza ejercida por a sobre b
Ejemplo del principio de
acción y reacción
n = Ftm
S E C T I O N 5 . 6 • Newton’s Third Law
n = Ftm
Fg = FEm
Fg = FEm
Fmt
FmE
▲ PITFALL PREVENTION
Hay dos pares de fuerzas:
(a)
(b)
- De Figure
la Tierra
sobre
monitor
(elonpeso
monitor)
, ythedel monitor5.6
sobre
la Tierra
n Does
Not Alway
5.6 (a)
When ael
computer
monitor is at rest
a table,del
the forces
acting on
to n is the sobre la mesa
are the
normalel
force
n and the gravitational
Fg . The reaction
Equal mg
- Demonitor
la mesa
sobre
monitor
(la force
normal),
y del monitor
force Fmt exerted by the monitor on the table. The reaction to Fg is the force FmE
the tener
situationen
shown in F
De estas cuatro,
actúan
sobre
el monitor,
y son
lasmonitor.
únicas que habría In
que
exertedsólo
by the dos
monitor
on the Earth.
(b) The
free-body diagram
for the
5.6 and in many others, w
cuenta a la hora de estudiar posibles cambios en su movimiento
that n ! mg (the normal
has the same magnitude a
Tipos de fuerzas
Fuerzas de restricción
Fuerzas elásticas
Fuerzas de fricción
Fuerzas de fricción en fluídos
Fuerzas en movimientos curvilíneos
Fuerzas ficticias
Tipos de fuerzas:
fuerzas de restricción
Limitan el movimiento
Surgen como oposición a otra fuerza
Son ilimitadas
Fuerzas normales: se definen como la fuerza de igual magnitud y dirección, pero
diferente sentido, que ejerce una superficie sobre un cuerpo apoyado sobre la misma.
Esta fuerza impide que el objeto caiga a través de la superficie.
Puede tomar cualquier valor necesario hasta el límite de ruptura de la superficie.
Tipos de fuerzas:
tensiones en cuerdas
Cuerda: cualquier dispositivo capaz de trasmitir una fuerza
Normalmente vamos a considerar despreciable las masas de las cuerdas, y
que estas son inextensibles (longitud constante)
Cuando un objeto está siendo arrastrado por una cuerda, ésta ejerce una fuerza sobre el objeto.
Al módulo de esta fuerza se le denomina tensión
Esta fuerza tiene la dirección de la propia cuerda y se ejerce en sentido saliente con
respecto al objeto.
particle experiences an acceleration, then there
n the object. Consider a crate being pulled to the
ace, as in Figure 5.8a. Suppose you are asked to
d the force the floor exerts on it. First, note that
to the crate acts through the rope. The magnihe rope. The forces acting on the crate are illusigure 5.8b. In addition to the force T, the freeSupongamos
he gravitational force Fg and the normal force n
Tipos de fuerzas:
tensiones en cuerdas
una superficie horizontal sin rozamiento
¿Cuánto vale la aceleración de la caja?
nd law in component form to the crate. The only
Applying !Fx ! max to the horizontal motion
ma x
or
ax !
T
m
ection. Applying !Fy ! may with ay ! 0 yields
!0
or
n ! Fg
Paso 1: Aislamos el objeto cuyo movimiento vamos a analizar
e magnitude as the gravitational force but acts in
Paso 2: Dibujamos el diagrama de fuerzas
(a) que actúan sobre el objeto
cceleration ax ! T/m also is constant. Hence, the
nematics from Chapter 2 can be used to obtain
as functions of time. Because ax ! T/m ! conritten as
! vxi %
" #
" mT # t
y
T
x
T
t
m
% vxit % 12
n
2
magnitude of the normal force n is equal to the
ys the case. For example, suppose a book is lying
Fg
(b)
Figure 5.8 (a) A crate being
(si tuviéramos más de un objeto,
dibujaríamos un diagrama de
fuerzas para cada uno de los
objetos por separado)
particle experiences an acceleration, then there
n the object. Consider a crate being pulled to the
ace, as in Figure 5.8a. Suppose you are asked to
d the force the floor exerts on it. First, note that
to the crate acts through the rope. The magnihe rope. The forces acting on the crate are illusigure 5.8b. In addition to the force T, the freeSupongamos
he gravitational force Fg and the normal force n
Tipos de fuerzas:
tensiones en cuerdas
una superficie horizontal sin rozamiento
¿Cuánto vale la aceleración de la caja?
nd law in component form to the crate. The only
Applying !Fx ! max to the horizontal motion
ma x
or
ax !
T
m
ection. Applying !Fy ! may with ay ! 0 yields
!0
or
n ! Fg
Paso 1: Aislamos el objeto cuyo movimiento vamos a analizar
e magnitude as the gravitational force but acts in
Paso 2: Dibujamos el diagrama de fuerzas
(a) que actúan sobre el objeto
cceleration ax ! T/m also is constant. Hence, the
nematics from Chapter 2 can be used to obtain
as functions of time. Because ax ! T/m ! conritten as
! vxi %
" #
" mT # t
y
T
x
T
t
m
% vxit % 12
n
2
magnitude of the normal force n is equal to the
ys the case. For example, suppose a book is lying
Fg
(b)
Figure 5.8 (a) A crate being
Paso 3: Elegimos unos ejes de
coordenadas convenientes para
analizar el movimiento de cada
uno de los objetos
m
ection. Applying !Fy ! may with ay ! 0 yields
!0
or
F
Tipos
den !fuerzas:
e magnitude as the gravitational force but acts in
tensiones
en cuerdas
g
cceleration ax ! T/m also is constant. Hence, the
nematics from Chapter 2 can be used to obtain
as functions of time. Because ax ! T/m ! conritten as
! vxi %
" #
" mT # t
n
y
T
x
T
t
m
% vxit % 12
(a)
2
Fg
magnitude of the normal force n is equal to the
(b)
ys the case. For example, suppose a book is lying
Paso
la segunda
ley5.8de
descompuesta en componentes
Figure
(a)Newton
A crate being
book with a force
F, as4:
in Aplicamos
Figure 5.9. Because
the
pulled to the right on a frictionless
lerating, !Fy ! 0, which gives n $ Fg $ F ! 0, or
surface. (b) The free-body diagram
mal force
is greater than
the force
of gravity.
Other sobre
Dirección
x: sólo
actúa
una fuerza
la the external
Dirección
representing
forces y: la partícula está en equilibrio, por lo
ed later.
acting on the crate. tanto su aceleración es cero y la fuerza externa
partícula
neta actuando sobre la partícula en esta
dirección tiene que anularse
Si la tensión es constante, entonces la caja seguirá
un movimiento rectilíneo uniformemente acelerado
Precaución: la normal no siempre es igual al peso
124
C H A P T E R 5 • The Laws of Motion
F
P R O B L E M - S O LV I N G H I N T S
Applying Newton’s Laws
The following procedure is recommended when dealing
Newton’s laws:
•
•
Fg
n
Figure 5.9 When one object
pushes downward on another
object with a force F, the normal
force n is greater than the
gravitational force: n ! Fg " F.
•
Draw a simple, neat diagram of the system to help con
Categorize the problem: if any acceleration componen
in equilibrium in this direction and !F ! 0. If not, t
an acceleration, the problem is one of nonequilibriu
!F ! ma.
Analyze the problem by isolating the object whose m
analyzed. Draw a free-body diagram for this object.
more than one object, draw separate free-body diagr
Dirección y: la partícula está en equilibrio,
por
lo in the free-body diagram forces exer
Do not
include
surroundings.
tanto su aceleración es cero y la fuerza externa
Establish convenient coordinate axes for each objec
• esta
neta actuando sobre la partícula en
of the forces along these axes. Apply N
dirección tiene que anularse components
!F ! m a, in component form. Check your dimensio
•
terms have units of force.
Solve the component equations for the unknowns. R
have as many independent equations as you have un
complete solution.
El módulo
de la normal es mayor
by making sure your results are consistent wi
• Finalize
de
la gravedad
Alsoque
check la
thefuerza
predictions
of your
solutions for extr
variables. By doing so, you can often detect errors in
•
•
n
Draw a simple, neat diagram of the system to help conceptualize the problem.
Categorize the problem: if any acceleration component is zero, the particle is
in equilibrium in this direction and !F ! 0. If not, the particle is undergoing
Example 5.4 AFTraffic
Light at Rest
n
g
an acceleration, the problem is one of nonequilibrium
in this direction, and
!F ! ma.
•
•
Draw a simple, neat diagram of the system to help conceptualize the problem.
Categorize the problem: if any acceleration component is zero, the particle is
in equilibrium in this direction and !F ! 0. If not, the particle is undergoing
an acceleration, the problem is one of nonequilibrium in this direction, and
!F ! ma.
Si el número
de objetos en el sistema
es mayor que uno, hay
•
•
que analizar los diagramas de fuerzas por separado
A traffic light weighing 122 N hangs from a cable tied to two
hen one object
ard on another
orce F, the normal
er than the
rce: n ! Fg " F.
5.4
•
•
•
Solution We conceptualize the problem by inspecting the
Analyze the problem by isolating the object
whose
motion
isobject
being
Figure
5.9 When
Analyze
theThe
problem drawing
by isolating
object5.10a.
whoseLet
motion
is being that the cables do
other cables
toone
a support,
as in Figure
5.10a.
in the
Figure
us assume
analyzed. Draw a free-body diagram
forpushes
thisfastened
object.
For
systems
downward on anothercontaining
analyzed. Draw a free-body diagram for this object. For systems containing
ofeach
37.0°
and 53.0° with the horinot break so that there is no acceleration of any sort in this
objectmake
with
a angles
force F,for
the
normal
more than one object, drawupper
separatecables
free-body
diagrams
object.
more than one object, draw separate free-body diagrams for each object.
force
n
is exerted
greater
than
the
zontal.
These
upper
cables
areobject
not as
as the vertical
problem in any direction. This allows us to categorize the
Do not include in the free-body
diagram
forces
by the
onstrong
its
Do not include in the free-body diagram forces exerted by the object on its
n !tension
Fg " F. in them exceeds
cable, andgravitational
will breakforce:
if the
100
N.
problem as one of equilibrium. Because the acceleration of
surroundings.
surroundings.
Willaxes
the for
traffic
remain
hanging
in this situation, or will
the system is zero, we know that the net force on the light
Establish convenient coordinate
eachlight
object
and find
the
Establish convenient coordinate axes for each object and find the
onethese
of the
cables
break?
and the net force on the knot are both zero. To analyze the
components of the forces along
axes.
Apply
Newton’s second law,
components of the forces along these axes. Apply Newton’s second law,
!F ! m a, in component form. Check your dimensions to make sure that all
!F ! m a, in component form. Check your dimensions to make sure that all
terms have units of force.
T3
terms have units of force.
y
T2
Solve the component equations for the unknowns. Remember that you must
Solve the component equations for the unknowns. Remember
that you must
37.0°
have as many independent equations as you have unknowns to
obtain a
53.0°
have as many independent equations
as
you
have
unknowns
to
obtain a
T1
complete solution.
complete solution.
•
Ejemplo: semáforo en equilibrio
•
Finalize by making sure your results are consistent with the free-body
T1 diagram.
Also check the predictions of your solutions for extreme values of the
variables. By doing so, you can often detect errors in your results.
•
T3
A Traffic Light at Rest
Example 5.4
ht weighing 122 N hangs from a cable tied to two
s fastened to a support, as in Figure 5.10a. The
es make angles of 37.0° and 53.0° with the horise upper cables are not as strong as the vertical
will break if the tension in them exceeds 100 N.
ffic light remain hanging in this situation, or will
ables break?
T2
Finalize
by making sure your results are consistent with the free-body diagram.
Also check the predictions of your solutions for extreme values of the
variables. By doing so, you can often detect
your results.
37.0° errors in53.0°
x
A Traffic Light at Rest
Solution We conceptualize
the
problem
by 122
inspecting
A traffic
light
weighing
N hangsthe
from a cable tied to two
Solution We conceptualize the problem by inspecting the
drawing in Figure 5.10a.
us assume
cables as
do in Figure 5.10a. The
otherLet
cables
fastenedthat
to athe
support,
drawing in Figure 5.10a. Let us assume that the cables do
not break so that there
is nocables
acceleration
of anyofsort
in this
upper
make angles
37.0°
and 53.0° with the horinot break so that there is no Tacceleration of any sort in this
Fg
3
problem in any direction.
to categorize
zontal. This
Theseallows
upperuscables
are not the
as strong as the vertical
problem in any direction. This allows us to categorize the
problem as one of equilibrium.
Because
the
acceleration
of
cable, and will break if the tension
in them exceeds 100 N. (b) problem as one of equilibrium.
Because the acceleration of
(c)
(a)light
the system is zero, weWill
know
thelight
net remain
force on
the
the that
traffic
hanging
in this situation, or will
the system is zero, we know that the net force on the light
and the net force onone
the of
knot
both
zero.
analyze the5.4) (a) A traffic light suspended
Figure
5.10To(Example
by force
cables.
diagram
theare
cables
break?
and the net
on(b)
theFree-body
knot are both
zero. To analyze the
Diagrama de fuerzas sobre el semáforo
Diagrama de fuerzas sobre el nudo
for the traffic light. (c) Free-body diagram for the knot where the three cables are joined.
T3
37.0°
53.0°
T1
y
T3
T2
37.0°
T1
T2
T1
53.0°
37.0°
T3
(b)
x
(c)
Figure 5.10 (Example 5.4) (a) A traffic light suspended by cables. (b) Free-body diagram
for the traffic light. (c) Free-body diagram for the knot where the three cables are joined.
T2
T1
T2
53.0°
37.0°
T3
T3
Fg
(a)
53.0°
y
T3
Fg
(a)
(b)
x
(c)
Figure 5.10 (Example 5.4) (a) A traffic light suspended by cables. (b) Free-body diagram
for the traffic light. (c) Free-body diagram for the knot where the three cables are joined.
Elegir siempre el sistema de coordenadas
más adecuado para nuestro problema
126
C H A P T E R 5 • The Laws of Motion
Ejemplo: coche en un plano inclinado
y
n
m g sin u
m g cos u
x
u
u
Fg = m g
(a)
(b)
Figure 5.11 (Example 5.6) (a) A car of mass m sliding down a frictionless incline.
(b) The free-body diagram for the car. Note that its acceleration along the incline is
g sin".
Cuando se trabaja con planos inclinados es conveniente escoger un eje de
coordenadas con el eje
x paralelo al plano inclinado y el eje y perpendicular al mismo
Solving (1) for a , we see that the acceleration along the inx
cline is caused by the component of Fg directed down the
incline:
(3)
a x ! g sin "
(4)
t!
√
2d
!
ax
√
2d
g sin "
Using Equation 2.13, with vxi ! 0, we find that
v xf 2 ! 2a x d
Elegir siempre el sistema de coordenadas
más adecuado para nuestro problema
126
C H A P T E R 5 • The Laws of Motion
Ejemplo: coche en un plano inclinado
y
n
mg sin u
m g cos u
x
u
u
Fg = mg
(a)
(b)
Figure
5.11 componente
(Example 5.6) (a) A car of
sliding down
El peso va a tener ahora
una
a mass
lo mlargo
dela frictionless
eje x yincline.
una componente a lo largo del eje y
(b) The free-body diagram for the car. Note that its acceleration along the incline is
g sin".
Solving (1) for a x , we see that the acceleration along the incline is caused by the component of Fg directed down the
incline:
Aceleración independiente de la masa
(3)
a x ! g sin "
(4)
t!
√
2d
!
ax
√
2d
g sin "
La normal no es igual al peso
Using Equation 2.13, with vxi ! 0, we find that
v xf 2 ! 2a x d
to m 2 gives
force acting on the system, we have
!Fx(system) ! F ! (m 1 & m 2)ax
(2)
También es importante definir el Fsistema
(1)
a !
m &m
objeto de nuestro problema
x
1
!Fx ! P12 ! m 2a
Substituting the value of the acceleration
into (2) gives
2
"
m2
To finalize this part, note that this would be the same acceler(3)
P12 ! m 2 a x !
m1 &
that of aque
single
object ofamass
equal
to thesuperficie
comEjemplo:ation
un as
bloque
empuja
otro
sobre
sin fricción
bined masses of the two blocks in Figure 5.12a and subject
to the same force.
To finalize the problem, we see from
the contact force P12 is less than the appl
is consistent with the fact that the fo
F
Asumimos que la fuerza
m1 m
2
accelerate block 2 alone must be less th
es constante
quired to produce the same acceleration f
system.
(a)
To finalize further, it is instructive to c
n1
sion for P12 by considering the forces acti
¿Cuánto vale la aceleración del
n2 sistema?
in Figure 5.12b. The horizontal forces acti
y
applied force F to the right and the contac
P21
F
P12
left (the force exerted by m 2 on m 1). From
x
m1
m 2 la misma aceleración:
law, P21 is the reaction to P12, so P21 ! P1
Los dos bloques deben experimentar
ton’s second law to m 1 gives
m g
- están en contacto
m1g
2
- permanecen en contacto
a lo largo
(4)
(b)
(c) de todo el movimiento
!Fx ! F ' P21 ! F ' P1
Active Figure 5.12 (Example 5.7) A force is applied to a block
of mass m 1, which pushes on a second block of mass m 2. (b) The
free-body diagram for m 1. (c) The free-body diagram for m 2.
At the Active Figures link at http://www.pse6.com,
you can study the forces involved in this two-block
system.
Substituting into (4) the value of ax from (1
P12 ! F ' m 1a x ! F ' m 1
"m
F
1 & m2
#!
This agrees with (3), as it must.
Es la misma aceleración que experimentaría un objeto de masa igual
a la suma de las masas y
que estuviera sometido a la misma fuerza
(B) Dete
tomm1 and
Two blocks of masses
m 2, with m 1 % m 2, are placed in
2 gives
the two b
contact with each other on a frictionless, horizontal surface,
force F is applied
to P12 ! m 2a
Fx !
Fx(system) ! F ! (m 1 & m 2)ax as in Figure 5.12a. A constant horizontal(2)
Solution
m 1 as shown. (A) Find the magnitude of the acceleration of
ternal to
the system.
Substituting the value of the acceleration
force by
intothe
(2)situation
gives using Figure 5.12a and
F
single pa
Solution Conceptualize
(1)
ax !
dividuall
realizing that both blocks must experience the same accelerm1 & m2
force. To
ation because they are in contact with each other and rediagram
main in contact throughout the motion. We categorize this as
m2
To finalize this part, note that this would be the same
acceler(3)
P
!
m
a
!
5.12c,
wh
12
2
x
a Newton’s second law problem because we have a force apm
&
1
ation as that of a single object of mass equal toplied
thetocom5.12c we
a system and we are looking for an acceleration. To
is the co
analyze
the
problem,
we
first
address
the
combination
of
two
bined masses of the two blocks in Figure 5.12a and subject
blocks as a system. Because F is the only external horizontal
to the same force.
To finalize the problem, we see which
fromis
to m 2 giv
force acting on the system, we have
force acting on the system, we have
!
!
También es importante definir el sistema
objeto de nuestro problema
"
Ejemplo: un bloque que empuja a otro sobre superficie sin fricción
the contact force P12 is less than the appl
fact that the fo
F
Asumimos que la fuerza
m1 m
2
accelerate block 2 alone must be Substitu
less th
es constante
into (2)f
F
quired
to produce
the same acceleration
(1)
ax !
m1 & m2
system.
(a)
To finalize further, it is instructive to c
To finalize this part, note that this would be the same accelern1
sion for
P12of by
considering
ation as that of a single
object
mass
equal to the the
com-forces acti
¿Cuál es la fuerza que el objeto de 1 ejerce
sobre
el
objeto
2?
n2
Figure
The
horizontal
bined masses of theintwo
blocks5.12b.
in Figure
5.12a
and subjectforces acti
y
to the same force. applied force F to the right and the contac
To fina
P21
F
P12
theFrom
con
left (the force exerted by m 2 on m 1).
x
is
consi
F
m1
m
law, P21 is them1reaction
to P12, so P21accelera
! P1
m2
Es una fuerza
interna 2al sistema.
ton’s second law to m 1 gives
quired t
m g
! F ! (m 1 with
& m 2)athe
x
!Fx(system)
is consistent
2
m1g
No podemos calcular esta fuerza considerando
el sistema
completo como una
sola partícula
system.
(a)
(b)
Active Figure 5.12 (Example 5.7) A force is applied to a block
y
pushesde
on a second block of mass m 2. (b) The
of mass de
m 1, which
diagrama
fuerzas
free-body diagram for m 1. (c) The free-body diagram for m 2.
x
Dibujamos el
cuerpo aislado para cada bloque
At the Active Figures link at http://www.pse6.com,
you can study the forces involved in this two-block
system.
La única fuerza horizontal que actúa sobre
el bloque 2 es la fuerza de contacto
P1fi
!Fx ! F ' P21 ! Fsion'Tofor
(4)
(c)
n1
n2
in Figur
Substituting into (4) the value of ax from
(1
applied
P21
F
m1
P12
m2
P12 ! F ' m 1a x ! F ' m 1
m1g
(b)
m 2g
"
left (the
#
F law, P21
!
ton’s sec
m1 & m
2
(c)
This agrees with (3), as it must.
Active Figure 5.12 (Example 5.7) A force is applied to a block
of mass m 1, which pushes on a second block of mass m 2. (b) The
free-body diagram for m 1. (c) The free-body diagram for m 2.
At the Active Figures link at http://www.pse6.com,
you can study the forces involved in this two-block
Substitu
P12 ! F
eiling of an elevator, as illustrated in Figure 5.13.
t if the elevator accelerates either upward or downe spring scale gives a reading that is different from
ht of the fish.
tor is either at rest or moving at constant velocity, the fish
does not accelerate, and so #Fy " T $ Fg " 0 or
T " Fg " mg. (Remember that the scalar mg is the weight
of the fish.)
If the elevator moves with an acceleration a relative to
an observer standing outside the elevator in an inertial
frame (see Fig. 5.13), Newton’s second law applied to the
fish gives the net force on the fish:
También es importante definir el sistema
objeto de nuestro problema
!
n Conceptualize by noting that the reading on the
elated to the extension of the spring in the scale,
related to the force on the end of the spring as in
2. Imagine that a string is hanging from the end
pring, so that the magnitude of the force exerted
pring is equal to the tension T in the string. Thus,
oking for T. The force T pulls down on the string
s up on the fish. Thus, we can categorize this probne of analyzing the forces and acceleration associh the fish by means of Newton’s second law. To anproblem, we inspect the free-body diagrams for
in Figure 5.13 and note that the external forces
n the fish are the downward gravitational force
(1)
Fy " T $ mg " ma y
where we have chosen upward as the positive y direction.
Thus, we conclude from (1) that the scale reading T is
greater than the fish’s weight mg if a is upward, so that ay is
positive, and that the reading is less than mg if a is downward, so that ay is negative.
For example, if the weight of the fish is 40.0 N and a is upward, so that ay " % 2.00 m/s2, the scale reading from (1) is
Ejemplo: se pesa un objeto con la ayuda de una báscula suspendida
del techo de un ascensor
a
Demostrar que si el ascensor acelera la
báscula indica un peso diferente del peso
real del pescado
a
T
T
mg
mg
(a)
(b)
Observer in
inertial frame
Figure 5.13 (Example 5.8) Apparent weight versus true weight. (a) When the elevator
accelerates upward, the spring scale reads a value greater than the weight of the fish.
(b) When the elevator accelerates downward, the spring scale reads a value less than
the weight of the fish.
Un observador dentro del ascensor no se
encuentra en un sistema inercial.
Analizaremos la situación en un sistema inercial,
desde un punto fijo en el suelo
eiling of an elevator, as illustrated in Figure 5.13.
t if the elevator accelerates either upward or downe spring scale gives a reading that is different from
ht of the fish.
tor is either at rest or moving at constant velocity, the fish
does not accelerate, and so #Fy " T $ Fg " 0 or
T " Fg " mg. (Remember that the scalar mg is the weight
of the fish.)
If the elevator moves with an acceleration a relative to
an observer standing outside the elevator in an inertial
frame (see Fig. 5.13), Newton’s second law applied to the
fish gives the net force on the fish:
También es importante definir el sistema
objeto de nuestro problema
!
n Conceptualize by noting that the reading on the
elated to the extension of the spring in the scale,
related to the force on the end of the spring as in
2. Imagine that a string is hanging from the end
pring, so that the magnitude of the force exerted
pring is equal to the tension T in the string. Thus,
oking for T. The force T pulls down on the string
s up on the fish. Thus, we can categorize this probne of analyzing the forces and acceleration associh the fish by means of Newton’s second law. To anproblem, we inspect the free-body diagrams for
in Figure 5.13 and note that the external forces
n the fish are the downward gravitational force
(1)
Fy " T $ mg " ma y
where we have chosen upward as the positive y direction.
Thus, we conclude from (1) that the scale reading T is
greater than the fish’s weight mg if a is upward, so that ay is
positive, and that the reading is less than mg if a is downward, so that ay is negative.
For example, if the weight of the fish is 40.0 N and a is upward, so that ay " % 2.00 m/s2, the scale reading from (1) is
Ejemplo: se pesa un objeto con la ayuda de una báscula suspendida
del techo de un ascensor
a
Demostrar que si el ascensor acelera la
báscula indica un peso diferente del peso
real del pescado
a
T
T
mg
mg
(a)
El peso medido está relacionado con la extensión
del muelle que, a su vez, está relacionado con la
fuerza que se ejerce sobre el extremo del muelle
(b)
Observer in
inertial frame
Figure 5.13 (Example 5.8) Apparent weight versus true weight. (a) When the elevator
accelerates upward, the spring scale reads a value greater than the weight of the fish.
(b) When the elevator accelerates downward, the spring scale reads a value less than
the weight of the fish.
Esta fuerza es igual a la tensión T en el muelle.
La fuerza
empuja hacia abajo el muelle y
empuja hacia arriba al pescado.
eiling of an elevator, as illustrated in Figure 5.13.
t if the elevator accelerates either upward or downe spring scale gives a reading that is different from
ht of the fish.
tor is either at rest or moving at constant velocity, the fish
does not accelerate, and so #Fy " T $ Fg " 0 or
T " Fg " mg. (Remember that the scalar mg is the weight
of the fish.)
If the elevator moves with an acceleration a relative to
an observer standing outside the elevator in an inertial
frame (see Fig. 5.13), Newton’s second law applied to the
fish gives the net force on the fish:
También es importante definir el sistema
objeto de nuestro problema
!
n Conceptualize by noting that the reading on the
elated to the extension of the spring in the scale,
related to the force on the end of the spring as in
2. Imagine that a string is hanging from the end
pring, so that the magnitude of the force exerted
pring is equal to the tension T in the string. Thus,
oking for T. The force T pulls down on the string
s up on the fish. Thus, we can categorize this probne of analyzing the forces and acceleration associh the fish by means of Newton’s second law. To anproblem, we inspect the free-body diagrams for
in Figure 5.13 and note that the external forces
n the fish are the downward gravitational force
(1)
Fy " T $ mg " ma y
where we have chosen upward as the positive y direction.
Thus, we conclude from (1) that the scale reading T is
greater than the fish’s weight mg if a is upward, so that ay is
positive, and that the reading is less than mg if a is downward, so that ay is negative.
For example, if the weight of the fish is 40.0 N and a is upward, so that ay " % 2.00 m/s2, the scale reading from (1) is
Ejemplo: se pesa un objeto con la ayuda de una báscula suspendida
del techo de un ascensor
a
Demostrar que si el ascensor acelera la
báscula indica un peso diferente del peso
real del pescado
a
T
T
mg
mg
(a)
Sobre el pescado actúan dos fuerzas:
- su peso
- la fuerza ejercida por el muelle
(b)
Observer in
inertial frame
Figure 5.13 (Example 5.8) Apparent weight versus true weight. (a) When the elevator
accelerates upward, the spring scale reads a value greater than the weight of the fish.
(b) When the elevator accelerates downward, the spring scale reads a value less than
the weight of the fish.
Si el acelerador está en reposo o se mueve con
velocidad constante, el pescado no se acelera
eiling of an elevator, as illustrated in Figure 5.13.
t if the elevator accelerates either upward or downe spring scale gives a reading that is different from
ht of the fish.
tor is either at rest or moving at constant velocity, the fish
does not accelerate, and so #Fy " T $ Fg " 0 or
T " Fg " mg. (Remember that the scalar mg is the weight
of the fish.)
If the elevator moves with an acceleration a relative to
an observer standing outside the elevator in an inertial
frame (see Fig. 5.13), Newton’s second law applied to the
fish gives the net force on the fish:
También es importante definir el sistema
objeto de nuestro problema
!
n Conceptualize by noting that the reading on the
elated to the extension of the spring in the scale,
related to the force on the end of the spring as in
2. Imagine that a string is hanging from the end
pring, so that the magnitude of the force exerted
pring is equal to the tension T in the string. Thus,
oking for T. The force T pulls down on the string
s up on the fish. Thus, we can categorize this probne of analyzing the forces and acceleration associh the fish by means of Newton’s second law. To anproblem, we inspect the free-body diagrams for
in Figure 5.13 and note that the external forces
n the fish are the downward gravitational force
(1)
Fy " T $ mg " ma y
where we have chosen upward as the positive y direction.
Thus, we conclude from (1) that the scale reading T is
greater than the fish’s weight mg if a is upward, so that ay is
positive, and that the reading is less than mg if a is downward, so that ay is negative.
For example, if the weight of the fish is 40.0 N and a is upward, so that ay " % 2.00 m/s2, the scale reading from (1) is
Ejemplo: se pesa un objeto con la ayuda de una báscula suspendida
del techo de un ascensor
a
Demostrar que si el ascensor acelera la
báscula indica un peso diferente del peso
real del pescado
a
T
T
mg
mg
(a)
Sobre el pescado actúan dos fuerzas:
- su peso
- la fuerza ejercida por el muelle
(b)
Observer in
inertial frame
Figure 5.13 (Example 5.8) Apparent weight versus true weight. (a) When the elevator
accelerates upward, the spring scale reads a value greater than the weight of the fish.
(b) When the elevator accelerates downward, the spring scale reads a value less than
the weight of the fish.
Si el acelerador acelera con respecto a un
observador inercial
eiling of an elevator, as illustrated in Figure 5.13.
t if the elevator accelerates either upward or downe spring scale gives a reading that is different from
ht of the fish.
tor is either at rest or moving at constant velocity, the fish
does not accelerate, and so #Fy " T $ Fg " 0 or
T " Fg " mg. (Remember that the scalar mg is the weight
of the fish.)
If the elevator moves with an acceleration a relative to
an observer standing outside the elevator in an inertial
frame (see Fig. 5.13), Newton’s second law applied to the
fish gives the net force on the fish:
También es importante definir el sistema
objeto de nuestro problema
!
n Conceptualize by noting that the reading on the
elated to the extension of the spring in the scale,
related to the force on the end of the spring as in
2. Imagine that a string is hanging from the end
pring, so that the magnitude of the force exerted
pring is equal to the tension T in the string. Thus,
oking for T. The force T pulls down on the string
s up on the fish. Thus, we can categorize this probne of analyzing the forces and acceleration associh the fish by means of Newton’s second law. To anproblem, we inspect the free-body diagrams for
in Figure 5.13 and note that the external forces
n the fish are the downward gravitational force
(1)
Fy " T $ mg " ma y
where we have chosen upward as the positive y direction.
Thus, we conclude from (1) that the scale reading T is
greater than the fish’s weight mg if a is upward, so that ay is
positive, and that the reading is less than mg if a is downward, so that ay is negative.
For example, if the weight of the fish is 40.0 N and a is upward, so that ay " % 2.00 m/s2, the scale reading from (1) is
Ejemplo: se pesa un objeto con la ayuda de una báscula suspendida
del techo de un ascensor
a
Demostrar que si el ascensor acelera la
báscula indica un peso diferente del peso
real del pescado
a
T
T
mg
mg
(a)
(b)
Observer in
inertial frame
Figure 5.13 (Example 5.8) Apparent weight versus true weight. (a) When the elevator
accelerates upward, the spring scale reads a value greater than the weight of the fish.
(b) When the elevator accelerates downward, the spring scale reads a value less than
the weight of the fish.
Si acelera hacia arriba, la tensión es mayor
y la báscula marcará un peso mayor
Si acelera hacia abajo, la tensión es menor
y la báscula marcará un peso menor
¿Qué pasa si se rompe la sujeción del
ascensor y este cae en caída libre?
Answer If the elevator falls freely, its acceleration is
When two objects of unequal mass
hungWe
vertically
over a (2) downward.
Because
the objects
connected
by an inexay are
! "g.
see from
that the scale
reading
T isare
zero
in
frictionless pulley of negligible mass,
as
in
Figure
5.14a,
the
tensible
string,
their
accelerations
must
be
of
equal
magnithis case; that is, the fish appears to be weightless.
31.8 N
arrangement is called an Atwood machine. The device is
sometimes used in the laboratory to measure the free-fall acMachineceleration. Determine the magnitude of the acceleration of
the two objects and the tension in the lightweight cord.
tude. The objects in the Atwood machine are subject to the
gravitational force as well as to the forces exerted by the
Interactive
.9 The Atwood
strings connected to them—thus,
we can categorize this as a
Newton’s second law problem. To analyze the situation, the
free-body diagrams for the two objects are shown in Figure
bjects of unequal mass are Solution
hung vertically
over a
downward. Because the objects
are connected by an inexConceptualize the situation pictured in Figure
5.14b. Two forces act on each object: the upward force T expulley of negligible mass, as5.14a—as
in Figure
5.14a,moves
the upward,
tensible
string,
magnione object
the other
objecttheir
movesaccelerations
erted by themust
string be
andof
theequal
downward
gravitational force. In
t is called an Atwood machine. The device is
tude. The objects in the Atwood
areinsubject
to pulley
the is modeled as
problemsmachine
such as this
which the
massless
and
frictionless,
the
tension
in
the string on both
gravitational force as well as to the forces exerted by the
used in the laboratory to measure the free-fall acsides of the pulley is the same. If the pulley has mass and/or
Determine the magnitude of the acceleration of
strings connected to them—thus, we can categorize this as a
is subject to friction, the tensions on either side are not the
cts and the tension in the lightweight cord.
Newton’s second law problem.
Tothe
analyze
therequires
situation,
the we will learn in
same and
situation
techniques
free-body diagrams for the
two objects
are shown in Figure
Chapter
10.
We
must
be
very
careful with
signs
in problems such as
onceptualize the situation pictured in Figure
5.14b. Two forces act on each object: the upward
force
T exIn Figure 5.14a, notice that if object 1 accelerates upne object moves upward, the other object moves
erted by the string and thethis.
downward
gravitational force. In
ward, then object 2 accelerates downward. Thus, for consisproblems such as this intency
which
is modeled
asdirection as posiwiththe
signs,pulley
if we define
the upward
massless and frictionless, tive
thefor
tension
object 1,inwethe
muststring
define on
the both
downward direction as
m
a
1
positive
for
object
2.
With
this
sign
convention, both obsides of the pulley is the same. If the pulley has mass and/or
jects accelerate in the same direction as defined by the
is subject to friction, the tensions on either side are not the
choice of sign. Furthermore, according to this sign convena
m2
same
and
the situation requires
wethe
will
in
tion, the techniques
y component of
netlearn
force exerted
on object 1
Chapter
10.
is
T
"
m
g,
and
the
y
component
of
the
net
force
exerted
1
(a)
on object
2 is min
T. Notice such
that we
2g "
We must be very careful
with signs
problems
ashave chosen the
signs
of
the
forces
to
be
consistent
with
T
this. In Figure 5.14a, notice that if object 1 accelerates up- the choices of
signs for up and down for each object. If we assume that
T
ward, then object 2 accelerates
Thus, for consism 2 $ mdownward.
1, then m 1 must accelerate upward, while m 2 must
tency with signs, if we define
the upward
accelerate
downward.direction as posim1
Whenthe
Newton’s
seconddirection
law is applied
tive for object 1, we must define
downward
as to object 1, we
m
obtain
2
m1
a
positive
for object 2. With this sign convention, both ob-
La máquina de Atwood
Dos objetos con masas diferentes se cuelgan verticalmente de una
polea sin rozamiento de masa despreciable
Cuando uno se mueve hacia arriba el otro
se mueve hacia abajo
Como la cuerda es inextensible, las dos
aceleraciones tienen que tener el mismo módulo
Dibujamos los diagramas de cuerpo aislado
m2
(a)
T
T
Con nuestras aproximaciones, la tensión de la
thecuerda
same direction
(1) as defined
mde
!m
y ! T "by
1gthe
#Flados
a ambos
la1aypolea es la misma
jects accelerate in
choice of sign. Furthermore,
according
this
sign convena
Similarly,
for objectto2 we
find
tion,
the y component of the net force exerted on object 1
m2g
g " T ! m 2a y
y ! m 2exerted
is T " m 1g, and the y component of(2)the net#Fforce
(b)
(2) that
is added
(1), T cancels
we have
on object 2 is m g " T. When
Notice
weto have
chosenandthe
Active Figure 5.14 (Example 5.9) The Atwood machine. (a)2 Two
signs inextensible
of the forces
with
the choices of
cord overto be consistent "m
objects (m 2 $ m 1) connected by a massless
1g # m 2g ! m 1a y # m 2a y
a frictionless pulley. (b) Free-body diagrams
twoand
objects.
signs for
fortheup
down for each object. If we assume that
mhttp://www.pse6.com,
must
m 22 "
m1
2 $ m 1, then m 1 must accelerate upward, while m
At the Active Figures link at
(3)
ay !
g
you can adjust the masses of the
objects ondownward.
the Atwood
accelerate
m1 # m2
m1g
!
"
Answer If the elevator falls freely, its acceleration is
When two objects of unequal mass
hungWe
vertically
over a (2) downward.
Because
the objects
connected
by an inexay are
! "g.
see from
that the scale
reading
T isare
zero
in
frictionless pulley of negligible mass,
as
in
Figure
5.14a,
the
tensible
string,
their
accelerations
must
be
of
equal
magnithis case; that is, the fish appears to be weightless.
31.8 N
arrangement is called an Atwood machine. The device is
sometimes used in the laboratory to measure the free-fall acMachineceleration. Determine the magnitude of the acceleration of
the two objects and the tension in the lightweight cord.
tude. The objects in the Atwood machine are subject to the
gravitational force as well as to the forces exerted by the
Interactive
.9 The Atwood
strings connected to them—thus,
we can categorize this as a
Newton’s second law problem. To analyze the situation, the
free-body diagrams for the two objects are shown in Figure
bjects of unequal mass are Solution
hung vertically
over a
downward. Because the objects
are connected by an inexConceptualize the situation pictured in Figure
5.14b. Two forces act on each object: the upward force T expulley of negligible mass, as5.14a—as
in Figure
5.14a,moves
the upward,
tensible
string,
magnione object
the other
objecttheir
movesaccelerations
erted by themust
string be
andof
theequal
downward
gravitational force. In
t is called an Atwood machine. The device is
tude. The objects in the Atwood
areinsubject
to pulley
the is modeled as
problemsmachine
such as this
which the
massless
and
frictionless,
the
tension
in
the string on both
gravitational force as well as to the forces exerted by the
used in the laboratory to measure the free-fall acsides of the pulley is the same. If the pulley has mass and/or
Determine the magnitude of the acceleration of
strings connected to them—thus, we can categorize this as a
is subject to friction, the tensions on either side are not the
cts and the tension in the lightweight cord.
Newton’s second law problem.
Tothe
analyze
therequires
situation,
the we will learn in
same and
situation
techniques
free-body diagrams for the
two objects
are shown in Figure
Chapter
10.
We
must
be
very
careful with
signs
in problems such as
onceptualize the situation pictured in Figure
5.14b. Two forces act on each object: the upward
force
T exIn Figure 5.14a, notice that if object 1 accelerates upne object moves upward, the other object moves
erted by the string and thethis.
downward
gravitational force. In
ward, then object 2 accelerates downward. Thus, for consisproblems such as this intency
which
is modeled
asdirection as posiwiththe
signs,pulley
if we define
the upward
massless and frictionless, tive
thefor
tension
object 1,inwethe
muststring
define on
the both
downward direction as
m
a
1
positive
for
object
2.
With
this
sign
convention, both obsides of the pulley is the same. If the pulley has mass and/or
jects accelerate in the same direction as defined by the
is subject to friction, the tensions on either side are not the
choice of sign. Furthermore, according to this sign convena
m2
same
and
the situation requires
wethe
will
in
tion, the techniques
y component of
netlearn
force exerted
on object 1
Chapter
10.
is
T
"
m
g,
and
the
y
component
of
the
net
force
exerted
1
(a)
on object
2 is min
T. Notice such
that we
2g "
We must be very careful
with signs
problems
ashave chosen the
signs
of
the
forces
to
be
consistent
with
T
this. In Figure 5.14a, notice that if object 1 accelerates up- the choices of
signs for up and down for each object. If we assume that
T
ward, then object 2 accelerates
Thus, for consism 2 $ mdownward.
1, then m 1 must accelerate upward, while m 2 must
tency with signs, if we define
the upward
accelerate
downward.direction as posim1
Whenthe
Newton’s
seconddirection
law is applied
tive for object 1, we must define
downward
as to object 1, we
m
obtain
2
m1
a
positive
for object 2. With this sign convention, both ob-
La máquina de Atwood
Dos objetos con masas diferentes se cuelgan verticalmente de una
polea sin rozamiento de masa despreciable
m2
(a)
T
T
reemplazando
ecuaciones
jects accelerate inYthe
same direction
defined
(1) asen
m 1gthe
! m 1a y
y ! T "by
#Flas
choice of sign. Furthermore,
according
this
sign convena
Similarly,
for objectto2 we
find
tion,
the y component of the net force exerted on object 1
m2g
g " T ! m 2a y
y ! m 2exerted
is T " m 1g, and the y component of(2)the net#Fforce
(b)
(2) that
is added
(1), T cancels
we have
on object 2 is m g " T. When
Notice
weto have
chosenandthe
Active Figure 5.14 (Example 5.9) The Atwood machine. (a)2 Two
signs inextensible
of the forces
with
the choices of
cord overto be consistent "m
objects (m 2 $ m 1) connected by a massless
1g # m 2g ! m 1a y # m 2a y
a frictionless pulley. (b) Free-body diagrams
twoand
objects.
signs for
fortheup
down for each object. If we assume that
mhttp://www.pse6.com,
must
m 22 "
m1
2 $ m 1, then m 1 must accelerate upward, while m
At the Active Figures link at
(3)
ay !
g
you can adjust the masses of the
objects ondownward.
the Atwood
accelerate
m1 # m2
m1g
!
"
de movimiento
Solution Conceptualize the motion in Figure 5.15. If m 2
moves down the incline, m 1 moves upward. Because the obFxconnected
# 0 by a cord (which we assume does not
ects are
stretch), their accelerations have the same magnitude. We
can identify
of the a
two#
objects
and we are
F #forces
T !onmeach
mthis
1g #som
y
1a as a Newooking yfor an acceleration,
we1 categorize
ton’s second-law problem. To analyze the problem, coner
the balldiagrams
to accelerate
upward,
it is
siderfor
the free-body
shown in Figures
5.15b and
5.15c. Applying Newton’s second law in component form
$ m g. In (2), we replaced ay with
a beto the1 ball, choosing the upward direction
as positive,
ation
yields has only a y component.
#
#
Dos
#Fy& # n ! m 2 g cos % # 0
(4)
In (3) we replaced ax& with a because the two objects have
Whenofthis
for a is(1)substituted
into (2),
accelerations
equalexpression
magnitude a. Equations
and (4)
provide no information regarding the acceleration. However, if we solve (2) for T and then substitute this value for T
into (3) and solve for a, we obtain
m m g (sin % " 1)
cuerpos unidos por una cuerda
(6)
(5)
a#
1
T#
we find
2
m1 " m2
m 2 g sin % ! m 1 g
m1 " m2
To finalize
the problem,están
note that
the block
accelerates
it is convenient
theobjetos
positive x&con
Dos
masas
diferentes
unidos
por
una cuerda,
(1)
0 choose
#Fx #to
When this expression for a is substituted into (2), we find
down
the reposa
incline only
if m 2 un
cline, (2)
as in Figure
consistency
sin %plano
$ m 1. Ifinclinado
m 1 $ m 2 sin %,
m 1g # m For
uno de
ellos
sobre
1a y # m 1a
#Fy # T !5.15c.
Note that in order for the ball to accelerate upward, it is
necessary that T $ m1g. In (2), we replaced ay with a because the acceleration has only a y component.
For the block it is convenient to choose the positive x&
axis along the incline, as in Figure 5.15c. For consistency
a
(6)
T#
y
y′
m2
a
m1
T
T
θ
(a)
m1
θ
Cuando uno se mueve hacia abajo por el
plano inclinado, el otro se mueve hacia arriba
T
n
a
m2
y′
To finalize the problem, note that the block accelerates
n
down the incline only if m 2 sin % $ m 1. If m 1 $ m 2 sin %,
y
m1
m 1m 2 g (sin % " 1)
m1 " m2
y
m1
T
x
x
m 2g cos θ
m 1g
(b)
m2g sin θ
Como la cuerda es inextensible, las dos
m g sin θ
θ
aceleraciones
tienen que tener el mismo módulo
x′
m 1g
2
mθ2g cos θ
m 2g
(c)
(b)
Figure 5.15 (a)
(Example 5.10) (a) Two objects connected by
a lightweight cord strung
over a frictionless pulley. (b) Free-body diagram for the ball. (c) Free-body diagram for
5.15 the
(Example
5.10)is (a)
Two objects connected by a lightweight
block. (The incline
frictionless.)
x′
m 2g
Dibujamos
los diagramas de cuerpo aislado
(c)
Figure
cord strung
over a frictionless pulley. (b) Free-body diagram for the ball. (c) Free-body
for
Para eldiagram
cuerpo
he block. (The incline is frictionless.)
1
Para el cuerpo 2
Solution Conceptualize the motion in Figure 5.15. If m 2
moves down the incline, m 1 moves upward. Because the obFxconnected
# 0 by a cord (which we assume does not
ects are
stretch), their accelerations have the same magnitude. We
can identify
of the a
two#
objects
and we are
F #forces
T !onmeach
mthis
1g #som
y
1a as a Newooking yfor an acceleration,
we1 categorize
ton’s second-law problem. To analyze the problem, coner
the balldiagrams
to accelerate
upward,
it is
siderfor
the free-body
shown in Figures
5.15b and
5.15c. Applying Newton’s second law in component form
$ m g. In (2), we replaced ay with
a beto the1 ball, choosing the upward direction
as positive,
ation
yields has only a y component.
#
#
Dos
#Fy& # n ! m 2 g cos % # 0
(4)
In (3) we replaced ax& with a because the two objects have
Whenofthis
for a is(1)substituted
into (2),
accelerations
equalexpression
magnitude a. Equations
and (4)
provide no information regarding the acceleration. However, if we solve (2) for T and then substitute this value for T
into (3) and solve for a, we obtain
m m g (sin % " 1)
cuerpos unidos por una cuerda
(6)
(5)
a#
1
T#
we find
2
m1 " m2
m 2 g sin % ! m 1 g
m1 " m2
To finalize
the problem,están
note that
the block
accelerates
it is convenient
theobjetos
positive x&con
Dos
masas
diferentes
unidos
por
una cuerda,
(1)
0 choose
#Fx #to
When this expression for a is substituted into (2), we find
down
the reposa
incline only
if m 2 un
cline, (2)
as in Figure
consistency
sin %plano
$ m 1. Ifinclinado
m 1 $ m 2 sin %,
m 1g # m For
uno de
ellos
sobre
1a y # m 1a
#Fy # T !5.15c.
Note that in order for the ball to accelerate upward, it is
necessary that T $ m1g. In (2), we replaced ay with a because the acceleration has only a y component.
For the block it is convenient to choose the positive x&
axis along the incline, as in Figure 5.15c. For consistency
a
(6)
T#
y
y′
m2
a
m1
T
T
θ
(a)
m1
θ
Cuando uno se mueve hacia abajo por el
plano inclinado, el otro se mueve hacia arriba
T
n
a
m2
y′
To finalize the problem, note that the block accelerates
n
down the incline only if m 2 sin % $ m 1. If m 1 $ m 2 sin %,
y
m1
m 1m 2 g (sin % " 1)
m1 " m2
y
m1
T
x
x
m 2g cos θ
m 1g
(b)
m2g sin θ
Como la cuerda es inextensible, las dos
m g sin θ
θ
aceleraciones
tienen que tener el mismo módulo
x′
m 1g
2
mθ2g cos θ
m 2g
(c)
(b)
Figure 5.15 (a)
(Example 5.10) (a) Two objects connected by
a lightweight cord strung
over a frictionless pulley. (b) Free-body diagram for the ball. (c) Free-body diagram for
5.15 the
(Example
5.10)is (a)
Two objects connected by a lightweight
block. (The incline
frictionless.)
x′
m 2g
Dibujamos
los diagramas de cuerpo aislado
(c)
Figure
cord strung
over a frictionless pulley.
(b) Free-body diagram
for the ball. (c)y
Free-body
diagramde
for las
Despejando
la aceleración
la tensión
he block. (The incline is frictionless.)
anteriores ecuaciones
El bloque 2 se acelerará hacia abajo de la rampa si y sólo si
El bloque 1 se acelerará verticalmente hacia abajo si
Tipos de fuerzas:
fuerzas elásticas
La fuerza elástica es la ejercida por objetos tales como resortes, que tienen una
posición normal, fuera de la cual almacenan energía potencial y ejercen fuerzas.
Tipos de fuerzas:
fuerzas de fricción
Cuando un objeto se mueve sobre una superficie, o a través de un medio viscoso,
existe una resistencia al movimiento debida a que el objeto interactúa con su entorno.
Éstas son las fuerzas de rozamiento.
Se debe a la naturaleza de las dos superficies (rugosidad, composición) y de la
superficie de contacto
water, there is resistance to the motion because the object interacts with its surroundings. We call such resistance a force of friction. Forces of friction are very important
in our everyday lives. They allow us to walk or run and are necessary for the motion of
wheeled vehicles.
Imagine that you are working in your garden and have filled a trash can with yard clippings. You then try to drag the trash can across the surface of your concrete patio, as in
Figure 5.16a. This is a real surface, not an idealized, frictionless surface. If we apply an external horizontal force F to the trash can, acting to the right, the trash can remains staCuando un objeto
se mueve
sobre
unathat
superficie,
a través
detrash
un can
medio
tionary
if F is small.
The force
counteracts o
F and
keeps the
from viscoso,
moving
As long as the trash
not
acts toal
themovimiento
left and is calleddebida
the force aofque
staticelfriction
fs . interactúa
Force of sta
existe una resistencia
objeto
concan
suisentorno.
moving, fsÉstas
! F. Thus,
F is increased,
increases. Likewise, if F decreases, fs also
s also
sonif las
fuerzas fde
rozamiento.
Tipos de fuerzas:
fuerzas de fricción
La fuerza que
n
n
F
fs
Motion
F
fk
mg
mg
(a)
(b)
Si aplicamos una fuerza externa horizontal al cubo que actúe
Active Figure 5.16 The direction of the
hacia la derecha,
el
cubo
permanecerá
inmóvil
si
es
pequeña
tween
a trash can and a rough surface is
|f|
of the applied force F. Because both surf
contrarresta a
e impide que el cubo se mueva es la fuerza
de only
rozamiento
is made
at a few points,estático
as illustrated
fs,max
view. (a) For small applied forces, the ma
static friction equals the magnitude of th
Mientras el cubo esté quieto, si aumenta
también aumentará
(b) When the magnitude of the applied f
magnitude of the maximum force of stati
F
can breaks free. The applied force is now
f s=
of kinetic friction and the trash can accel
fk = µkn
water, there is resistance to the motion because the object interacts with its surroundings. We call such resistance a force of friction. Forces of friction are very important
in our everyday lives. They allow us to walk or run and are necessary for the motion of
wheeled vehicles.
Imagine that you are working in your garden and have filled a trash can with yard clippings. You then try to drag the trash can across the surface of your concrete patio, as in
Figure 5.16a. This is a real surface, not an idealized, frictionless surface. If we apply an external horizontal force F to the trash can, acting to the right, the trash can remains staCuando un objeto
se mueve
sobre
unathat
superficie,
a través
detrash
un can
medio
tionary
if F is small.
The force
counteracts o
F and
keeps the
from viscoso,
moving
As long as the trash
not
acts toal
themovimiento
left and is calleddebida
the force aofque
staticelfriction
fs . interactúa
Force of sta
existe una resistencia
objeto
concan
suisentorno.
moving, fsÉstas
! F. Thus,
F is increased,
increases. Likewise, if F decreases, fs also
s also
sonif las
fuerzas fde
rozamiento.
Tipos de fuerzas:
fuerzas de fricción
n
n
F
fs
F
fk
mg
mg
(a)
(b)
Si aumentamos el módulo de
Motion
el cubo de basura puede llegar
moverse
ActiveaFigure
5.16 The direction of the
tween a trash can and a rough surface is
of the applied force F. Because both surf
Cuando el cubo de basura está a punto de comenzar a deslizarse,
el módulo de
is made only at a few points, as illustrated
fs,max
toma su valor máximo
view. (a) For small applied forces, the ma
static friction equals the magnitude of th
Cuando el módulo de
es mayor que
el cubo de basura
se empieza
a of the applied f
(b) When
the magnitude
magnitude of the maximum force of stati
mover y adquiere
una aceleración hacia la derecha.
F
can breaks free. The applied force is now
f s=
of kinetic friction and the trash can accel
fk = µkn
|f|
water, there is resistance to the motion because the object interacts with its surroundings. We call such resistance a force of friction. Forces of friction are very important
in our everyday lives. They allow us to walk or run and are necessary for the motion of
wheeled vehicles.
Imagine that you are working in your garden and have filled a trash can with yard clippings. You then try to drag the trash can across the surface of your concrete patio, as in
Figure 5.16a. This is a real surface, not an idealized, frictionless surface. If we apply an external horizontal force F to the trash can, acting to the right, the trash can remains staCuando un objeto
se mueve
sobre
unathat
superficie,
a través
detrash
un can
medio
tionary
if F is small.
The force
counteracts o
F and
keeps the
from viscoso,
moving
As long as the trash
not
acts toal
themovimiento
left and is calleddebida
the force aofque
staticelfriction
fs . interactúa
Force of sta
existe una resistencia
objeto
concan
suisentorno.
moving, fsÉstas
! F. Thus,
F is increased,
increases. Likewise, if F decreases, fs also
s also
sonif las
fuerzas fde
rozamiento.
Tipos de fuerzas:
fuerzas de fricción
La
n
n
F
fs
Motion
F
fk
mg
mg
(a)
(b)
Mientras el cubo de basura está en movimiento, la fuerza de rozamiento
menor
que
Activees
Figure
5.16 The
direction of the
tween a trash can and a rough surface is
the applied force F. Because both surf
fuerza de rozamiento de un objeto en movimiento se denomina fuerzaisofde
rozamiento
dinámico
made
only at a few points,
as illustrated
fs,max
view. (a) For small applied forces, the ma
static friction equals the magnitude of th
La fuerza neta en la dirección x,
, produce una aceleración
lamagnitude
derecha
(b)hacia
When the
of the applied f
magnitude of the maximum force of stati
F
can breaks free. The applied force is now
f s=
of kinetic friction and the trash can accel
fk = µkn
|f|
water, there is resistance to the motion because the object interacts with its surroundings. We call such resistance a force of friction. Forces of friction are very important
in our everyday lives. They allow us to walk or run and are necessary for the motion of
wheeled vehicles.
Imagine that you are working in your garden and have filled a trash can with yard clippings. You then try to drag the trash can across the surface of your concrete patio, as in
Figure 5.16a. This is a real surface, not an idealized, frictionless surface. If we apply an external horizontal force F to the trash can, acting to the right, the trash can remains staCuando un objeto
se mueve
sobre
unathat
superficie,
a través
detrash
un can
medio
tionary
if F is small.
The force
counteracts o
F and
keeps the
from viscoso,
moving
As long as the trash
not
acts toal
themovimiento
left and is calleddebida
the force aofque
staticelfriction
fs . interactúa
Force of sta
existe una resistencia
objeto
concan
suisentorno.
moving, fsÉstas
! F. Thus,
F is increased,
increases. Likewise, if F decreases, fs also
s also
sonif las
fuerzas fde
rozamiento.
Tipos de fuerzas:
fuerzas de fricción
n
n
F
fs
F
fk
mg
mg
(a)
(b)
La fuerza neta en la dirección x,
Motion
, produce una aceleración
hacia
la 5.16
derecha
Active
Figure
The direction of the
tween a trash can and a rough surface is
of the applied force F. Because both surf
Si
el objeto se moverá hacia la derecha con celeridad
constante
is made
only at a few points, as illustrated
fs,max
view. (a) For small applied forces, the ma
Si se elimina la fuerza alicada, la fuerza de rozamiento que actúastatic
hacia
la equals
izquierda
friction
the magnitude of th
(b)
When
the
magnitude
of the applied f
proporciona al cubo una aceleración en la dirección –x y hace que el cubo se detenga
magnitude of the maximum force of stati
F
can breaks free. The applied force is now
f s=
of kinetic friction and the trash can accel
fk = µkn
|f|
5.8
Tipos de fuerzas:
fuerzas de fricción
Forces of Friction
When an object is in motion either on a surface or in a viscous medium such as air or
water, there is resistance to the motion because the object interacts with its surroundings. We call such resistance a force of friction. Forces of friction are very important
in our everyday lives. They allow us to walk or run and are necessary for the motion of
wheeled vehicles.
Imagine that you are working in your garden and have filled a trash can with yard clippings. You then try to drag the trash can across the surface of your concrete patio, as in
Figure 5.16a. This is a real surface, not an idealized, frictionless surface. If we apply an external horizontal force F to the trash can, acting to the right, the trash can remains stationary if F is small. The force that counteracts F and keeps the trash can from moving
acts to the left and is called the force of static friction fs . As long as the trash can is not
moving, fs ! F. Thus, if F is increased, fs also increases. Likewise, if F decreases, fs also
Cuando un objeto se mueve sobre una superficie, o a través de un medio viscoso,
existe una resistencia al movimiento debida a que el objeto interactúa con su entorno.
Éstas son las fuerzas de rozamiento.
Force of static friction
Se debe a la naturaleza de las dos superficies (rugosidad, composición) y de la
superficie de contacto
n
n
Motion
Se pueden clasificar en:
fs
- fuerzas de rozamiento estático
F
F
fk
(cuando el objeto está parado)
mg
- fuerzas de rozamiento dinámico
mg
(cuando
el objeto está en movimiento)
(a)
(b)
Active Figure 5.16 The direction of the force of friction f between a trash can and a rough surface is opposite the direction
of the applied force F. Because both surfaces are rough, contact
is made only at a few points, as illustrated in the “magnified”
view. (a) For small applied forces, the magnitude of the force of
static friction equals the magnitude of the applied force.
(b) When the magnitude of the applied force exceeds the
magnitude of the maximum force of static friction, the trash
can breaks free. The applied force is now larger than the force
of kinetic friction and the trash can accelerates to the right.
(c) A graph of friction force versus applied force. Note that
fs,max # fk .
|f|
fs,max
fs
=F
fk = µkn
0
F
Kinetic region
Static region
(c)
At the Active Figures link at http://www.pse6.com
you can vary the applied force on the trash can and
practice sliding it on surfaces of varying roughness.
Note the effect on the trash can’s motion and the corresponding behavior of the graph in (c).
Tipos de fuerzas:
fuerzas de fricción, dirección, sentido y módulo
La dirección de la fuerza de rozamiento sobre un objeto es opuesta al movimiento del objeto,
respecto de la superficie con la que se encuentra en contacto, o
La dirección de la fuerza de rozamiento se opone al deslizamiento de una superficie sobre otra
El módulo de la fuerza de rozamiento
Igualdad en el umbral de deslizamiento:
- estático:
Situación de movimiento inminente
(o equilibrio estricto)
- dinámico:
dónde µs y µk son unas constantes adimensionales denominadas, respectivamente
los coeficientes de rozamiento estático y dinámico,
n es el módulo de la fuerza normal.
Tipos de fuerzas:
fuerzas de fricción, coeficientes de rozamiento
Generalmente µk es menor que µs.
Supondremos que µk es independiente de la velocidad relativa de las superficies.
Tipos de fuerzas:
fuerzas de fricción en un plano inclinado
Descomposición del peso en una componente normal y otra tangencial al plano
Módulo de la componente normal que el plano ejerce sobre el objeto
Fuerzas de rozamiento:
force fhorse exerted by the Earth and the backward tenforce T exerted by the sled (Fig. 5.18c). The resultant
ese two forces causes the horse to accelerate.
he force that accelerates the system (horse plus sled) is
et force fhorse " fsled. When fhorse balances fsled, the sysmoves with constant velocity.
Determinación experimental de los
coeficientes de rozamiento
Un bloque se coloca sobre una superficie rugosa inclinada con respecto a la horizontal
T
T
El ángulo de inclinación
b)
xample 5.11)
aumenta hasta que el objeto comienza a moverse
fhorse
¿Cómo se relaciona
el coeficiente de rozamiento estático con el ángulo crítico
(c)
para que el bloque comience a moverse?
Seleccionamos un sistema de coordenadas con un
eje x positivo paralelo al plano inclinado
y
n
Mientras que el bloque no se mueve, las fuerzas se
compensan y el bloque se encuentra en equilibrio
x
f
mg sin θ
mg cos θ
θ
mg
θ
De la 2 Ecuación
e 5.19 (Example 5.12) The external forces exerted on a
lying on a rough incline are the gravitational force mg, the
al force n, and the force of friction f. For convenience, the
ational force is resolved into a component along the incline
n ! and a component perpendicular to the incline mg cos !.
Sustituyendo en la 1 Ecuación
force fhorse exerted by the Earth and the backward tenforce T exerted by the sled (Fig. 5.18c). The resultant
ese two forces causes the horse to accelerate.
he force that accelerates the system (horse plus sled) is
et force fhorse " fsled. When fhorse balances fsled, the sysmoves with constant velocity.
Determinación experimental de los
coeficientes de rozamiento
Un bloque se coloca sobre una superficie rugosa inclinada con respecto a la horizontal
T
T
El ángulo de inclinación
b)
xample 5.11)
aumenta hasta que el objeto comienza a moverse
fhorse
¿Cómo se relaciona
el coeficiente de rozamiento estático con el ángulo crítico
(c)
para que el bloque comience a moverse?
En el ángulo crítico, el bloque se encuentra en el
umbral de deslizamiento, la fuerza de rozamiento
tiene su módulo máximo
y
n
x
f
mg sin θ
mg cos θ
θ
mg
θ
e 5.19 (Example 5.12) The external forces exerted on a
lying on a rough incline are the gravitational force mg, the
al force n, and the force of friction f. For convenience, the
ational force is resolved into a component along the incline
n ! and a component perpendicular to the incline mg cos !.
force fhorse exerted by the Earth and the backward tenforce T exerted by the sled (Fig. 5.18c). The resultant
ese two forces causes the horse to accelerate.
he force that accelerates the system (horse plus sled) is
et force fhorse " fsled. When fhorse balances fsled, the sysmoves with constant velocity.
Determinación experimental de los
coeficientes de rozamiento
Un bloque se coloca sobre una superficie rugosa inclinada con respecto a la horizontal
T
T
El ángulo de inclinación
b)
xample 5.11)
aumenta hasta que el objeto comienza a moverse
fhorse
¿Cómo se relaciona
el coeficiente de rozamiento estático con el ángulo crítico
(c)
para que el bloque comience a moverse?
Si el ángulo es mayor que el ángulo crítico,
el bloque comienza a moverse, con un movimiento
acelerado por el plano inclinado
y
n
x
f
mg sin θ
mg cos θ
Hay que sustituir el coeficiente de rozamiento
estático por el coeficiente de rozamiento dinámico
(que es más pequeño)
θ
mg
θ
e 5.19 (Example 5.12) The external forces exerted on a
lying on a rough incline are the gravitational force mg, the
al force n, and the force of friction f. For convenience, the
ational force is resolved into a component along the incline
n ! and a component perpendicular to the incline mg cos !.
Si una vez que el bloque ha comenzado a moverse
volvemos al ángulo crítico, el objeto seguirá acelerando
por el plano inclinado (la fuerza de rozamiento es menor
cuando se mueve que cuando está parado)
force fhorse exerted by the Earth and the backward tenforce T exerted by the sled (Fig. 5.18c). The resultant
ese two forces causes the horse to accelerate.
he force that accelerates the system (horse plus sled) is
et force fhorse " fsled. When fhorse balances fsled, the sysmoves with constant velocity.
Determinación experimental de los
coeficientes de rozamiento
Un bloque se coloca sobre una superficie rugosa inclinada con respecto a la horizontal
T
T
El ángulo de inclinación
b)
xample 5.11)
aumenta hasta que el objeto comienza a moverse
fhorse
¿Cómo se relaciona
el coeficiente de rozamiento estático con el ángulo crítico
(c)
para que el bloque comience a moverse?
y
n
x
f
mg sin θ
mg cos θ
θ
mg
θ
e 5.19 (Example 5.12) The external forces exerted on a
lying on a rough incline are the gravitational force mg, the
al force n, and the force of friction f. For convenience, the
ational force is resolved into a component along the incline
n ! and a component perpendicular to the incline mg cos !.
Para volver a la situación de equilibrio habrá que
replantear las ecuaciones de movimiento
sustituyendo
por
y reducir el ángulo a un
valor tal que el bloque se deslice hacia abajo con
velocidad constante
Aceleración de dos objetos unidos por una
cuerda en el caso de que exista fricción
Determinar la aceleración del sistema asumiendo cuerda inextensible de masa
despreciable, polea sin rozamiento y sin masa, y coeficiente de rozamiento dinámico
Asumimos que el módulo de la fuerza no es lo suficientemente grande como
para levantar al objeto de la superficie
A P T E R 5 • The Laws of Motion
Cuerpo 1
y
a
m1
x
F
θ
n
T
Cuerpo 2
F sin θ
F
θ
T
F cos θ
fk
m2
a
m 2g
m2
(a)
(b)
m 1g
(c)
Figure 5.21 (Example 5.14) (a) The external force F applied as shown can cause the
block to accelerate to the right. (b) and (c) The free-body diagrams assuming that the
block accelerates to the right and the ball accelerates upward. The magnitude of the
force of kinetic friction in this case is given by f % # n % # (m g $ F sin !).
of radius r experiences an acceleration that has a magnitude
ac !
v2
r
Tipos de fuerzas:
The acceleration is called centripetal acceleration because a is directed toward the center
fuerzas
en
movimientos
to v. (If there were a component
of the circle.
Furthermore,
a is always perpendicularcurvilíneos
c
c
m
Fr
r
Fr
constante)
Mike Powell / Allsport / Getty Images
of acceleration parallel to v, the particle’s speed would be changing.)
Consider a ball of mass m that is tied to a string of length r and is being whirled at
Caso
de
un movimiento
constant speed in a horizontal circular
path, as
illustrated
in Figure 6.1. Itscircular
weight is uniforme
supported by a frictionless table. Why does the ball move in a circle? According to
(partícula
en
trayectoria
circular
Newton’s first law,
the ball tendsmoviéndose
to move in a straight
line;
however, the string
prevents con celeridad
Partícula que se mueve en una trayectoria circular de radio r
con velocidad uniforme v experimenta una aceleración
centrípeta dirigida hacia el centro del círculo de módulo
An athlete in the process of
throwing the hammer at the 1996
Olympic Games in Atlanta, Georgia.
The force exerted by the chain
causes the centripetal acceleration
of the hammer. Only when the
Figure 6.1 Overhead view of a ball moving in a
aceleración
es releases
perpendicular
athlete
the hammer willal
it vector
circularEl
pathvector
in a horizontal
plane. A forcesiempre
Fr
move along a straight-line path
directed toward the center of the circle keeps
tangent to the circle.
the ball moving in its circular path.
velocidad
151
¿Qué hace que la partícula se mueva con trayectoria
circular?
Si hay una aceleración, hay una fuerza neta (segunda ley de Newton)
Si la aceleración hacia el centro del círculo, la fuerza hacia el centro del círculo
of radius r experiences an acceleration that has a magnitude
ac !
v2
r
Tipos de fuerzas:
The acceleration is called centripetal acceleration because a is directed toward the center
fuerzas
en
movimientos
to v. (If there were a component
of the circle.
Furthermore,
a is always perpendicularcurvilíneos
c
c
of acceleration parallel to v, the particle’s speed would be changing.)
Consider a ball of mass m that is tied to a string of length r and is being whirled at
Caso
de
un movimiento
constant speed in a horizontal circular
path, as
illustrated
in Figure 6.1. Itscircular
weight is uniforme
supported by a frictionless table. Why does the ball move in a circle? According to
Newton’s first law,
ball tends
move in a straight line;
the string prevents
Sithehay
unatoaceleración,
hayhowever,
una fuerza
neta (segunda ley
de Newton)
m
Fr
r
Fr
Mike Powell / Allsport / Getty Images
Si la aceleración hacia el centro del círculo, la fuerza hacia el centro del círculo
Tendencia natural: moverse en una línea recta con velocidad constante
La cuerda impide este movimiento, ejerciendo una fuerza radial sobre el
objeto que hace que siga una trayectoria circular
An athlete in the process of
throwing the hammer at the 1996
Olympic Games in Atlanta, Georgia.
The force exerted by the chain
causes the centripetal acceleration
of the hammer. Only when the
athlete releases the hammer will it
move along a straight-line path
tangent to the circle.
Esta fuerza es la tensión de la cuerda: orientada según la
longitud de la cuerda y se dirige hacia el centro del círculo
Figure 6.1 Overhead view of a ball moving in a
circular path in a horizontal plane. A force Fr
directed toward the center of the circle keeps
the ball moving in its circular path.
151
Independientemente de la naturaleza de la fuerza que actúe sobre el objeto con
movimiento circular, podemos aplicar la segunda ley de Newton según la dirección radial.
upward orientation—it does not invert. What is the directio
© Tom Carroll/Index
In Section 4.4 we found that a particle moving with uniform speed v in a circular path
celeration when you are at the top of the wheel? (a) upwar
of radius r experiences an acceleration that has a magnitude
possible to determine. What is the direction of your c
when you are at the bottom of the wheel? (d) upward (e) dow
determine.
v
Tipos de fuerzas:
a !
r
fuerzas The
enacceleration
movimientos
curvilíneos
center Quiz 6.2 You are riding on the Ferris wheel o
is called centripetal acceleration
because a is directed toward theQuick
2
c
Figure 6.3 (Quick Quiz 6.1 and
6.2) A Ferris wheel located on the
Navy Pier in Chicago,
Illinois.
c
the direction of the normal force exerted by the seat on you w
of the circle. Furthermore, ac is always perpendicular to v. (If there were a component
the wheel? (a) upward (b) downward (c) impossible to deter
of acceleration parallel to v, the particle’s speed would be changing.)
Caso
de
un
movimiento
circular
uniforme
tion ofatthe normal force exerted by the seat on you when you
Consider a ball of mass m that is tied to a string of length r and is being whirled
wheel?is(d) upward (e) downward (f) impossible to determine.
constant speed in a horizontal circular path, as illustrated in Figure 6.1. Its weight
Tendencia
natural:
moverse
en
una
línea
recta
con
velocidad
constante
supported by a frictionless table. Why does the ball move in a circle? According
to
Newton’s first law, the ball tends to move in a straight
however, the string prevents
PITFALLline;
PREVENTION
When the String is Cut
Fr
r
Fr
Study Figure 6.2 very carefully.
Many students (wrongly) think
m that the ball will move radially
away from the center of the circle
when the string is cut. The velocity of the ball is tangent to the circle. By Newton’s first law, the ball
continues to move in the direction that it is moving just as the
force from the string disappears.
Figure 6.1 Overhead view of a ball moving in a
circular path in a horizontal plane. A force Fr
At thetoward
Activethe
Figures
directed
centerlink
of the circle keeps
at http://www.pse6.com,
you path.
the ball moving in its circular
can “break” the string yourself
and observe the effect on the
ball’s motion.
Mike Powell / Allsport / Getty Images
▲
La cuerda impide este movimiento,
ejerciendo una fuerza radial sobre el
of Travel
objeto que 6.1
haceDirection
que siga
una trayectoria circular
r
An athlete in the process of
throwing the hammer at the 1996
Olympic Games in Atlanta, Georgia.
The force exerted by the chain
causes the centripetal acceleration
of the hammer. Only when the
athlete releases the hammer will it
move along a straight-line path
Active Figure 6.2
tangent to the circle.
moving in a circula
151When the string br
direction tangent t
Si la fuerza que actúa sobre el objeto desaparece, este se
desplazará a lo largo de una línea recta tangente al círculo.
and
that that
te T find
and find
v2 v2
tan "tan
!"!
rg rg
El péndulo cónico
v!√
g tan
v r!
√r g"tan "
Un pequeño objeto de masa m suspendido de una cuerda de longitud L.
he geometry
in Figure
6.4, 6.4,
we see
r ! Lr !
sinL"sin
; ";
om
the geometry
in Figure
we that
see that
re,
erefore,
El objeto gira con una celeridad v en un círculo de radio r.
v ! v √!Lg sin
tan""tan "
√Lg"sin
¿Cuánto vale v?
at the
is independent
of the
of the
te
thatspeed
the speed
is independent
ofmass
the mass
ofobject.
the object.
La bola está en equilibrio en la dirección vertical
La bola sigue un movimiento circular en la dirección horizonta
L
T
L
θ
θ
T cosTθcos θ
θ
T
r
Dibujamos el diagrama de cuerpo aislado
θ
Como el objeto no se acelera en la dirección vertical
r
T sin Tθ sin θ
mg mg
mg mg
La componente horizontal de la tensión es la
6.4
6.2) The
pendulum
and its
freeure (Example
6.4 (Example
6.2) conical
The conical
pendulum
and
its freeresponsable de la aceleración centrípeta
agram.
dy
diagram.
on
6.1 yields
uation
6.1 yields
(1)
T!m
v2 v2
and
that that
te T find
and find
v2 v2
tan "tan
!"!
rg rg
El péndulo cónico
v!√
g tan
v r!
√r g"tan "
Un pequeño objeto de masa m suspendido de una cuerda de longitud L.
he geometry
in Figure
6.4, 6.4,
we see
r ! Lr !
sinL"sin
; ";
om
the geometry
in Figure
we that
see that
re,
erefore,
El objeto gira con una celeridad v en un círculo de radio r.
v ! v √!Lg sin
tan""tan "
√Lg"sin
¿Cuánto vale v?
at the
is independent
of the
of the
te
thatspeed
the speed
is independent
ofmass
the mass
ofobject.
the object.
Como el objeto no se acelera en la dirección vertical
L
T
L
θ
θ
T cosTθcos θ
θ
T
r
r
La componente horizontal de la tensión es la
responsable de la aceleración centrípeta
θ
T sin Tθ sin θ
mg mg
mg mg
Dividiendo la segunda ecuación entre la primera
6.4
6.2) The
pendulum
and its
freeure (Example
6.4 (Example
6.2) conical
The conical
pendulum
and
its freeagram.
dy diagram.
Como
Independiente de la
masa del objeto
on
6.1 yields
uation
6.1 yields
(1)
T!m
v2 v2
Fuerzas sobre un piloto en un
movimiento circular
Un piloto de masa m ejecuta un loop .
Determinar la fuerza ejercida por el asiento sobre el piloto en en el fondo y en el tope del loop
R 6 • Circular Motion and Other Applications of Newton’s Laws
C HAPTE R 6 • Circular Motion and Other Applications of Newton’s Laws
n bot
Top
Analicemos el diagrama del cuerpo aislado del
piloto en la parte de debajo del loop
n bot
Top
A
A
ntop
mg
(b)
Bottom
Bottom
(a)
mg
ntop
mg
mg
(c)
(c) de la fuerza normal ejercida por el asiento
La magnitud
sobre el piloto es mayor que el peso del piloto.
(b)
(a) 6.6) (a) An aircraft executes a loop-the-loop maneuver as it
Figure 6.7 (Example
moves in a vertical circle at constant speed. (b) Free-body diagram for the pilot at the
Figurebottom
6.7 (Example
(a)
An aircraft
executes
a loop-the-loop
maneuver
of the loop.6.6)
In this
position
the pilot
experiences
an apparent weight
greater as it
moves than
in a his
vertical
circle(c)
at Free-body
constantdiagram
speed. for
(b)the
Free-body
the pilot at the
true weight.
pilot at thediagram
top of thefor
loop.
El piloto experimenta un peso aparente que es mayor que
su peso real.
bottom of the loop. In this position the pilot experiences an apparent weight greater
than his true weight. (c) Free-body diagram for the pilot at the top of the loop.
Fuerzas sobre un piloto en un
movimiento circular
Un piloto de masa m ejecuta un loop .
Determinar la fuerza ejercida por el asiento sobre el piloto en en el fondo y en el tope del loop
R 6 • Circular Motion and Other Applications of Newton’s Laws
ular Motion and Other Applications of Newton’s Laws
n bot
Top
Analicemos el diagrama del cuerpo aislado del
piloto en la parte de arriba del loop
n bot
Top
A
ntop
mg
(b)
Bottom
(a)
Bottom
ntop
mg
(c)
mg
mg
(c) de la fuerza normal ejercida por el asiento
La magnitud
sobre el piloto es menor que el peso del piloto.
(b)
Figure 6.7 (Example 6.6) (a) An aircraft executes a loop-the-loop maneuver as it
(a)
moves in a vertical circle at constant speed. (b) Free-body diagram for the pilot at the
bottom of theFigure
loop. In6.7
this position
the
an apparent
weight
greater
(Example pilot
6.6) experiences
(a) An aircraft
executes
a loop-the-loop
maneuver as it
than his truemoves
weight. in
(c)aFree-body
diagram
for
the
pilot
at
the
top
of
the
loop.
vertical circle at constant speed. (b) Free-body diagram for the pilot at the
El piloto experimenta un peso aparente que es menor que
su peso real.
bottom of the loop. In this position the pilot experiences an apparent weight greater
than his true weight. (c) Free-body diagram for the pilot at the top of the loop.
e same magnitude, the normal force at
ent weight that is greater than his true weight by a factor
Ejemplo de fuerzas de fricción:
desplazamiento de un coche en una carretera horizontal
Cuando un coche acelera en una carretera horizontal, la fuerza no equilibrada que
causa la aceleración es debida al rozamiento entre los neumáticos y la carretera
En reposo: el peso del coche está equilibrado por la fuerza
normal que el suelo ejerce sobre los neumáticos
Para que comience el movimiento: el motor del coche ejerce un
par sobre el eje de dirección
Si no hubiera rozamiento con la carretera:
las ruedas simplemente girarían sobre sí
mismas, con la superficie de los neumáticos
moviéndose hacia atrás.
Si hay rozamiento, pero el par no es lo
suficientemente grande: los neumáticos no
se deslizarán debido a la fricción estática.
Ejemplo de fuerzas de fricción:
desplazamiento de un coche en una carretera curva
La fuerza de fricción ejercida por la carretera sobre el coche tiene la dirección
hacia delante y suministra la aceleración necesaria para que el coche acelere
Si cada neumático rueda sin deslizamiento, su superficie de contacto
con la carretera se encuentra en reposo relativo con ésta.
Superficie de contacto con el suelo se mueve hacia atrás
con respecto al eje con velocidad v
El eje se desplaza hacia adelante con velocidad v con
respecto a la carretera.
El rozamiento entre las ruedas y el suelo es fricción estática
rve, as shown in Figure 6.5. If the radius of the curve is
.0
m and
coefficient
of static
frictionroad
between
the tires
500-kg
carthe
moving
on a flat,
horizontal
negotiates
a
de,dry
pavement
is
0.500,
find
the
maximum
speed
the
car
as shown in Figure 6.5. If the radius of the curve is
nm
have
make the
andand
thestill
coefficient
of turn
staticsuccessfully.
friction between the tires
speed at which it is on the verge of skidding outward. At this
point,
the friction
hascan
its have
maximum
value
fs, max is
!the
"sn.
The
maximum
speedforce
the car
around
the curve
Because
the car
shown
in verge
Figureof6.5b
is in equilibrium
the
speed
at which
it is
on the
skidding
outward. Atinthis
vertical
the magnitude
of the normal
equals
point,
thedirection,
friction force
has its maximum
value fs, force
max ! "sn.
the weight
mg) and
thus fs,6.5b
"smg.
Substituting
this
max !
Because
the (n
car!shown
in Figure
is in
equilibrium
in the
value for
fs into (1),
find that the
maximum
vertical
direction,
thewe
magnitude
of the
normalspeed
forceisequals
Fuerzas
sobre
un coche
que toma una curva en una
dry
pavement
is
0.500,
find
the
maximum
speed
the
car
olution In this case, the force that enables the car to rehave
still make
the
successfully.
ain
in and
its circular
path
isturn
the force
of static friction.
(Static
carretera
horizontal
plana
the weight (n ! mg) and thus f
! " mg. Substituting this
s, max
s
cause no
slipping
at the
point
of contact
ution
In this
case,occurs
the force
that
enables
the carbetween
to refs , max
r maximum
"sm gspeed
r
value
for
f
into
(1),
we
find
that
the
s
(2)
v
!
!
! √is"s g r
ad
and
tires.
If
this
force
of
static
friction
were
zero—for
max
n in its circular
path
is
the
force
of
static
friction.
(Static
m
m
Uncar
coche
de masa
m describe
una curva de radio r sobre una carretera horizontal plana.
ample,
the
were
icypoint
road—the
car would
conause
noifslipping
occursonatan
the
of contact
between
"sm2g)(35.0
r
! √f(0.500)(9.80
m/s
m)
s , max r
ue
in
a
straight
line
and
slide
off
the
road.)
Hence,
from
(2)
v
!
!
! carretera
"s g r
d and tires. If this
force
of
static
friction
were
zero—for
√
max
Si
el
coeficiente
de
rozamiento
estático
entre
los
neumáticos
y
la
es µ,
m
m
uation
6.1
we
have
mple, if the car were on an icy road—the car would con! 13.1 m/s
2)(35.0 m)
2road.) Hence, from
!
m/s
√(0.500)(9.80
e in a straight
line
and
slide
off
the
v
¿Cuál es
la
máxima
celeridad
que
puede
alcanzar
el
coche
para
tomar
la curva sin salirse?
(1)
fs ! m
ation 6.1 we have
r
Note that the maximum
speed
does not depend on the mass
m/s
! 13.1
2
En este
caso,
la fuerza
responsable
deneed
quemultiel coche siga
v
fs
of the
car. That
is why curved
highways do not
(1)
fs ! m
ple speed
limit
signs tospeed
cover
thefuerza
various
masses
of
r
una trayectoria
circular
es does
la
de on
rozamiento
estática
Note
that the
maximum
not depend
thevehicles
mass
using
the
road.
f
of the car. That is why curved highways do not need multi-
√
√
√
√
entre los neumáticos y la carretera
s
ple speed limit signs to cover the various masses of vehicles
What
Suppose that a car travels this curve on a wet day
using
theIf?
road.
andDibujamos
begins to skid on
curve when de
its speed
reaches
only
elthe
diagrama
cuerpo
aislado
8.00 m/s.
What can that
we say
about
thethis
coefficient
fricWhat
If? Suppose
a car
travels
curve onofastatic
wet day
tionbegins
in thisto
case?
and
skid on the curve when its speed reaches only
(a)
(a)
n
n
fs
fs
mg
(b)
gure 6.5 (Example 6.4) (a)mg
The force of static friction di-
8.00 m/s. What can we say about the coefficient of static fricAnswer The coefficient of friction between tires and a wet
tion in this case?
road should be smaller than that between tires and a dry road.
This expectation
is consistent
with experience
withand
driving,
beAnswer
The coefficient
of friction
between tires
a wet
cause
a skidbe
is smaller
more likely
a wet
road than
dryaroad.
road
should
thanon
that
between
tires aand
dry road.
To
check
our
suspicion,
we
can
solve
(2)
for
the coeffiThis expectation is consistent with experience with driving,
becient
of
friction:
Como
el
coche
está
en
equilibrio
en
la
dirección
cause a skid is more likely on a wet road than a dry road.
vertical
2
To check our suspicion, we vcan
max solve (2) for the coeffi"s !
cient of friction:
gr
2
vmax
Substituting the numerical
values,
"s !
gr
2
v max
(8.00 m/s)2 No dependen
"s !
!
! 0.187
Substituting
the gnumerical
values,
r
(9.80
m/s2)(35.0 m)
de la masa
he inside of the curve. Suppose the designated speed
ramp is to be 13.4 m/s (30.0 mi/h) and the radius
curve is 50.0 m. At what angle should the curve be
d?
The car is in equilibrium in the vertical direction. Thus,
from #Fy " 0 we have
(2)
n cos ! " mg
Fuerzas sobre un coche que
toma
curva en una
Dividing
(1) by (2) una
gives
on On a level (unbanked)
the force that causes
carretera
conroad,peralte
v
(3)
tan ! "
2
ntripetal acceleration is the force of static friction berg
car and road, as we saw in the previous example.
2
Siisla
curva
está
peraltada
con un ángulo $1la fuerza
normal
tendrá una componente
(13.4 m/s)
er, if the road
banked
at an
angle
!, as in Figure
! " tan
2) " 20.1#
hacia el centro
de la m/s
curva
(50.0 m)(9.80
e normal force n has a horizontal componentapuntando
n sin !
ng toward the center of the curve. Because the ramp
If a car rounds the curve at a speed less than 13.4 m/s, frice designed so that the force of static friction is zero,
tion is needed to keep it from sliding down the bank (to the
he component nx " n sin ! causes the centripetal
left in Fig. 6.6). A driver who attempts to negotiate the curve
at a speed greater than 13.4 m/s has to depend on friction
to keep from sliding up the bank (to the right in Fig. 6.6).
nx
The banking angle is independent of the mass of the vehicle
negotiating the curve.
!
ny
n
"
Imaginemos que se quiera diseñar la rampa de manera que un
What
If? What
if this samela
roadway
builtceleridad
on Mars in dada aún en
coche
pudiera
negociar
curvawere
a un
the future to connect different colony centers; could it be
de rozamiento
traveled at the sameausencia
speed?
Answer The reduced gravitational force on Mars would
Segunda
ley de Newton
Segunda ley de Newton
mean that the car is not pressed so tightly to the roadway.
enThe
la dirección
radial
en la dirección y
reduced normal
force results in a smaller component
θ
Fg
6.6 (Example 6.5) A car rounding a curve on a road
at an angle ! to the horizontal. When friction is ne, the force that causes the centripetal acceleration and
he car moving in its circular path is the horizontal com-
of the normal force toward the center of the circle. This
smaller component will not be sufficient to provide the centripetal acceleration associated with the original speed. The
centripetal acceleration must be reduced, which can be
done by reducing the speed v.
Equation (3) shows that the speed v is proportional to
the square root of g for a roadway of fixed radius r banked at
a fixed angle !. Thus, if g is smaller, as it is on Mars, the
speed v with which the roadway can be safely traveled is also
Tipos de fuerzas:
fuerzas de fricción en fluidos
Interacción entre el objeto y el medio a través del cual se mueve.
El medio ejerce una fuerza de resistencia
cuando este se mueve a su través.
Módulo depende de la celeridad relativa entre el objeto y el medio
Dirección y sentido de
sobre el objeto es siempre opuesta a la dirección del movimiento
Generalmente, el módulo de la fuerza aumenta a medida que aumenta el módulo de la velocidad
Tipos de fuerzas:
fuerzas de fricción en fluidos
Fuerzas de resistencia proporcional a la velocidad del objeto
Modelo válido a velocidades bajas
b es una constante, depende de las propiedades del medio y de la forma y dimensiones del objeto.
El signo menos nos dice que la fuerza de resistencia es opuesta a la velocidad.
S ECTI O N 6.4 • Motion in the Presence of Resistive Forces
163
Tipos de fuerzas: v = 0
=g
fuerzas de fricción aen
fluidos
v
Una esfera de masa m que se deja caer desde la la posición de reposo
vT
Únicas fuerzas: peso y fuerza de resistencia
R
0.632vT
(ignoramos empuje de Arquímedes. Podría incluirse
variando el peso aparente de la esfera).
v
mg
v = vT
a=0
t
τ
(a)
(b)
(c)
Active Figure 6.15 (a) A small sphere falling through a liquid. (b) Motion diagram of
the sphere as it falls. (c) Speed–time graph for the sphere. The sphere reaches a
maximum (or terminal) speed vT, and the time constant # is the time interval during
which it reaches a speed of 0.632vT.
the gravitational force Fg , let us describe its motion.1 Applying Newton’s second law
to the vertical motion, choosing the downward direction to be positive, and noting that
!Fy ! mg " bv, we obtain
mg " bv ! ma ! m
dv
dt
At the Active Figures link
at http://www.pse6.com, you
can vary the size and mass of
the sphere and the viscosity
(resistance to flow) of the
surrounding medium, then
observe the effects on the
sphere’s motion and its
speed–time graph.
Condiciones iniciales: en t = 0
(6.3)
where the acceleration dv/dt is downward. Solving this expression for the acceleration
gives
S ECTI O N 6.4 • Motion in the Presence of Resistive Forces
163
Tipos de fuerzas: v = 0
=g
fuerzas de fricción aen
fluidos
v
vT
R
0.632vT
v
Condiciones iniciales: en t = 0
v = vT
a=0
At the Active Figures link
S ECTI O N 6.4 • Motion in the Presence of Resistive Forces
163
at http://www.pse6.com, you
t
can vary the size and mass of
τ
the sphere and the viscosity
v = 0 (a)
(b)
(c)
(resistance to flow) of the
a=g
surrounding
medium,
Active Figure 6.15 (a) A small sphere fallingCuando
through a liquid.
(b) Motion la
diagram
of
t aumenta,
velocidad
aumenta,
la fuerza
dethen
observe the effects on the
the sphere as it falls. (c) Speed–time graph for the sphere. The sphere reaches a
resistencia
aumenta
y
la
aceleración
disminuye.
sphere’s
motion and its
maximum (or terminal) speed vTv, and the time constant # is the time interval during
speed–time graph.
which it reaches a speed of 0.632vT.
mg
vT
R
mg
La aceleración se hace cero cuando la fuerza de resistencia
se equilibra
con law
el peso.
describe its motion.1 Applying
Newton’s second
the gravitational force Fg , let us
to the vertical motion, choosing the downward direction to be positive, and noting that
En ese momento, el objeto alcanza la
0.632vT
!Fy ! mg " bv, we obtain
velocidad límite vT, y a
partir de ese momento se mueve con velocidad constante
v = vT
a=0
mg " bv ! ma ! m
where the acceleration dv/dt is downward.
Solving
τ
(b)
(c)
gives
6.15 (a) A small sphere falling through a liquid. (b) Motion diagram of
dv
dt
(6.3)
At the Active Figures link
at http://www.pse6.com, you
t
can
vary the
mass of
this expression
forsize
theand
acceleration
the sphere and the viscosity
(resistance to flow) of the
surrounding medium, then
S ECTI O N 6.4 • Motion in the Presence of Resistive Forces
163
Tipos de fuerzas: v = 0
=g
fuerzas de fricción aen
fluidos
v
vT
R
v
0.632vT
S E CTI O N 6.4 • Motion in the Presence of Resistive Forces
mg
163
Condiciones iniciales: en t = 0
v = vT
a=0
t
τ
v
(a)
(b)
(c)
Active Figure 6.15 (a) A small sphere falling through a liquid. (b) Motion diagram of
Solución
the sphere as it falls. (c) Speed–time graph for the sphere. The sphere reaches
a
vTmaximum (or terminal) speed vT, and the time constant # is the time interval during
which it reaches a speed of 0.632vT.
general
gravitational force Fg , let us describe its motion.1 Applying Newton’s second law
0.632vthe
T
to the vertical motion, choosing the downward direction to be positive, and noting that
!Fy ! mg " bv, we obtain
At the Active Figures link
dv
at
http://www.pse6.com,
you
mg " bv ! ma ! m
(6.3)
t
dt mass of
can vary the size and
τ
the sphere and the viscosity
where the acceleration
dv/dt is downward.
Solving this expression for the acceleration
(c)
(resistance to flow) of the
surrounding medium, then
h a liquid.gives
(b) Motion diagram of
observe the effects on the
here. The sphere reaches a
At the Active Figures link
at http://www.pse6.com, you
can vary the size and mass of
the sphere and the viscosity
(resistance to flow) of the
surrounding medium, then
observe the effects on the
sphere’s motion and its
speed–time graph.
Tipos de fuerzas:
fuerzas ficticias
Las leyes de Newton sólo son validas en sistemas de referencia inerciales
Cuando la aceleración de un objeto se mide con respecto a un sistema de referencia que a su vez
se acelera con respecto a un sistema de referencia inercial, la fuerza resultante no es igual al
producto de la masa por la aceleración
Incluso en este sistema de referencia acelerado, podemos utilizar la ley de Newton
si introducimos fuerzas ficticias o pseudofuerzas que dependan de la aceleración del sistema de
referencia
En el sistema de referencia acelerado:
Tipos de fuerzas:
fuerzas ficticias. Ejemplo 1
Se deja caer un objeto en el interior de un vagón de ferrocarril con velocidad inicial nula y
aceleración constante ac
Un observador situado en la vía ve caer
el objeto verticalmente (no hay
velocidad inicial a lo largo de x), y con
aceleración constante a lo largo de y, g
Con respecto al vagón, posee una
aceleración vertical g, y una
aceleración horizontal –ac. La bola cae
hacia la parte de atrás del vagón
En el sistema de referencia del vagón se puede utilizar la segunda ley de Newton si
introducimos una fuerza ficticia
que actúa sobre cualquier objeto de masa m
Tipos de fuerzas:
fuerzas ficticias. Ejemplo 2
Una lámpara que cuelga de una cuerda del techo de un vagón.
Para cada observador, la componente vertical de la tensión es igual al peso de la lámpara.
Con respecto al vagón, la lámpara está en
equilibrio, y no tiene aceleración. La
componente horizontal de la tensión equilibra
una fuerza ficticia que actúa sobre todos los
objetos del vagón para un observador situado
en el vagón.
Física, P. A. Tipler, Ed. Reverté, Tercera Edición, Capítulo 5
Un observador situado en la vía ve que la
lámpara se acelera hacia la derecha debido a la
acción de de la fuerza no equilibrada, la
componente horizontal de la tensión.
Tipos de fuerzas:
fuerzas ficticias. Ejemplo 3
Una plataforma giratoria.
Cada punto de la trayectoria se mueve en círculo y tiene una aceleración centrípeta.
Para un observador inercial, el bloque se mueve
en círculo con velocidad v, y está acelerado
hacia el centro del círculo, v2/r, por la fuerza no
equilibrada de la tensión de la fuerza.
Para un observador en la plataforma, el bloque
está en reposo y no acelera. Para usar la
segunda ley de Newton se debe utilizar una
fuerza fictica de magnitud v2/r y que apunte
hacia fuera del círculo, la fuerza centrífuga.
Física, P. A. Tipler, Ed. Reverté, Tercera Edición, Capítulo 5
Tipos de fuerzas:
fuerzas ficticias
Supongamos que un observador se encuentra en un sistema de referencia acelerado
(piénsese en el ascensor, un tiovivo, o la Tierra que al estar en rotación no es un sistema
inercial). Este observador realiza experimentos físicos sencillos (dejar caer un objeto,
medir la tensión de una cuerda..). Como el sistema de referencia en el que está sufre una
aceleración, sus resultados, medidos por él, no coincidirán en general con los que
obtendría en esos mismos experimentos si estuviera en reposo.
Si este observador cree firmemente en las ecuaciones de Newton, las escribirá tal y como
conocemos. Sin embargo, las aceleraciones su sistema está sufriendo, y que el desconoce
que existen, las interpretará,(para que le cuadren las ecuaciones) como una cierta fuerza.
Esta fuerza no existe como tal (no hay ninguna interacción de la naturaleza que las
genere), pero necesita creer en su existencia para que sigan siendo válidas las ecuaciones
de Newton.
Estas fuerzas, que aparecen sólo en los sistemas de referencia no inerciales se denominan
FUERZAS DE INERCIA, o fuerzas ficticias.