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.