Processing & Properties of Floor and Wall Tiles.pptx
8 - Dinâmica.pdf
1. ENAUTICA-DEM Profª Rosa Marat-Mendes 1
ESCOLA SUPERIOR NÁUTICA INFANTE D. HENRIQUE
Licenciatura em Pilotagem
Ano Letivo 2016/17
Mecânica Aplicada
Docente: Filipa Moleiro
Formulário
Formulário disponibilizado aos alunos para uso individual em Testes ou Exames de Mecânica Aplicada.
2. Estática de Partículas
Equilíbrio Estático de Partículas
- numa análise bidimensional (2D):
Σ𝐹𝐹𝑥𝑥 = 0; Σ𝐹𝐹𝑦𝑦 = 0
- numa análise tridimensional (3D):
Σ𝐹𝐹𝑥𝑥 = 0; Σ𝐹𝐹𝑦𝑦 = 0; Σ𝐹𝐹𝑧𝑧 = 0
- lei dos cossenos:
𝑅𝑅2
= 𝑃𝑃2
+ 𝑄𝑄2
− 2𝑃𝑃𝑄𝑄cos(𝐵𝐵)
- lei dos senos:
Mecânica Aplicada
Cap. 8 – Dinâmica
2ª Lei de Newton
Prof. Dra. Rosa Marat-Mendes
Escola Superior Náu0ca Infante D. Henrique
Departamento de Engenharia Marí0ma
Licenciatura em Pilotagem
2. ENAUTICA-DEM Profª Rosa Marat-Mendes 2
2ª Lei de Newton
• Se a força resultante que atua numa par+cula não for nula, a par-cula terá uma
aceleração proporcional à intensidade da resultante e na mesma direção dessa
força resultante.
• Quando a par)cula de massa m é atuada pela força 𝑭, a aceleração da par)cula
tem de sa5sfazer a equação:
• Quando uma par)cula es5ver sujeita simultaneamente a
várias forças, vem:
• Se a força que atua numa par)cula é zero, a par)cula não tem aceleração, i.e.,
ou mantém-se estacionária ou está em movimento re5líneo uniforme.
na direção e sentido da força que atua sobre ela e que as intensi-
a1, a2, a3 das acelerações são proporcionais às intensidades F1, F2,
s forças correspondentes:
alor constante obtido para a relação entre as intensidades das for-
elerações é uma característica da partícula que está sendo con-
a; ele é chamado de massa da partícula e é representado por m.
uma força F atua sobre uma partícula de massa m, a força F e a
ão a dessa partícula devem, portanto, satisfazer à relação
F ! ma (12.1)
lação fornece uma formulação completa da segunda lei de
; ela expressa não somente que as intensidades de F e a são pro-
ais, mas também (como m é um escalar positivo) que os vetores
m a mesma direção e sentido (Fig 12.2). Devemos notar que a
1) permanece válida quando F não for constante, mas varia com
em intensidade ou direção. As intensidades de F e a permane-
porcionais, e os dois vetores têm a mesma direção e sentido em
r instante dado. Entretanto, esses vetores não serão, em geral,
es à trajetória da partícula.
ndo uma partícula estiver sujeita simultaneamente a várias for-
F3
(c)
Figura 12.1
a
m
F ! ma
Figura 12.2
12.1 Newton’s Second Law and Linear Momentum 723
d (Sec. 1.3). The unit of force is a derived unit. It is called
nd is defined as the force that gives an acceleration of
of 1 kg (Fig. 12.4). From Eq. (12.1), we have
1 N 5 (1 kg)(1 m/s2
) 5 1 kg?m/s2
aid to form an absolute system of units. This means that
ts chosen are independent of the location where measure-
The meter, the kilogram, and the second may be used
earth; they may even be used on another planet. They
same meaning.
W of a body, or the force of gravity exerted on that body,
other force, be expressed in newtons. A body subjected
weight acquires an acceleration equal to the acceleration
(Be careful using the term acceleration due to gravity,
me an object accelerates with a magnitude g is during
Fig. 12.4 A force of 1 newton gives a
1-kilogram mass an acceleration of 1 m/s2
.
a = 1 m/s2
m = 1 kg
F = 1 N
12.1 Newton’s Second Law and Linear Momentum 723
rbitrarily defined (Sec. 1.3). The unit of force is a derived unit. It is called
he newton (N) and is defined as the force that gives an acceleration of
m/s2
to a mass of 1 kg (Fig. 12.4). From Eq. (12.1), we have
1 N 5 (1 kg)(1 m/s2
) 5 1 kg?m/s2
he SI units are said to form an absolute system of units. This means that
he three base units chosen are independent of the location where measure-
ments are made. The meter, the kilogram, and the second may be used
nywhere on the earth; they may even be used on another planet. They
lways have the same meaning.
The weight W of a body, or the force of gravity exerted on that body,
hould, like any other force, be expressed in newtons. A body subjected
nly to its own weight acquires an acceleration equal to the acceleration
ue to gravity g. (Be careful using the term acceleration due to gravity,
nce the only time an object accelerates with a magnitude g is during
ee-fall in the absence of drag.) It follows from Newton’s second law that
he magnitude W of the weight of a body of mass m is
Fig. 12.4 A force of 1 newton gives a
1-kilogram mass an acceleration of 1 m/s2
.
a = 1 m/s2
m = 1 kg
F = 1 N
valor constante obtido para a relação entre as intensidades das for-
acelerações é uma característica da partícula que está sendo con-
da; ele é chamado de massa da partícula e é representado por m.
do uma força F atua sobre uma partícula de massa m, a força F e a
ação a dessa partícula devem, portanto, satisfazer à relação
F ! ma (12.1)
relação fornece uma formulação completa da segunda lei de
on; ela expressa não somente que as intensidades de F e a são pro-
onais, mas também (como m é um escalar positivo) que os vetores
têm a mesma direção e sentido (Fig 12.2). Devemos notar que a
12.1) permanece válida quando F não for constante, mas varia com
po em intensidade ou direção. As intensidades de F e a permane-
proporcionais, e os dois vetores têm a mesma direção e sentido em
uer instante dado. Entretanto, esses vetores não serão, em geral,
ntes à trajetória da partícula.
uando uma partícula estiver sujeita simultaneamente a várias for-
Eq. (12.1) deve ser substituída por
"F!ma (12.2)
"F representa a soma, ou resultante, de todas as forças que atuam
a partícula.
eve-se observar que o sistema de eixos de referência em relação
al a aceleração a é determinada não é arbitrário. Esses eixos de-
er uma orientação constante em relação às estrelas e sua origem
a
m
F ! ma
Figura 12.2
1 m/s2
to a mass of 1 kg (Fig. 12.4). From Eq. (12.1), we have
1 N 5 (1 kg)(1 m/s2
) 5 1 kg?m/s2
The SI units are said to form an absolute system of units. This means that
the three base units chosen are independent of the location where measure-
ments are made. The meter, the kilogram, and the second may be used
anywhere on the earth; they may even be used on another planet. They
always have the same meaning.
The weight W of a body, or the force of gravity exerted on that body,
should, like any other force, be expressed in newtons. A body subjected
only to its own weight acquires an acceleration equal to the acceleration
due to gravity g. (Be careful using the term acceleration due to gravity,
since the only time an object accelerates with a magnitude g is during
free-fall in the absence of drag.) It follows from Newton’s second law that
the magnitude W of the weight of a body of mass m is
W 5 mg (12.6)
Recall that g 5 9.81 m/s2
, so the weight of a body of mass 1 kg
(Fig. 12.5) is
W 5 (1 kg)(9.81 m/s2
) 5 9.81 N
This value would be much less on the moon, where the acceleration due
to gravity is 1.6249 m/s2
.
Multiples and submultiples of the units of length, mass, and force
are frequently used in engineering practice. They are, respectively, the
kilometer (km) and the millimeter (mm); the megagram (Mg, which is also
called the metric ton) and the gram (g); and the kilonewton (kN). By
definition,
1 km 5 1000 m 1 mm 5 0.001 m
1 Mg 5 1000 kg 1 g 5 0.001 kg
1 kN 5 1000 N
You can convert these units to meters, kilograms, and newtons, respec-
Fig. 12.4 A force of 1 newton gives a
1-kilogram mass an acceleration of 1 m/s2
.
m = 1 kg
F = 1 N
Fig. 12.5 In the SI system, a block with m
1 kg has a weight of 9.81 N.
a = 9.81 m/s2
m = 1 kg
W = 9.81 N
3. ENAUTICA-DEM Profª Rosa Marat-Mendes 3
Equações do Movimento
12.1 Newton’s Second Law and Linear Momentum 725
Although it cannot be used as a consistent unit of mass, the mass
of the standard pound is, by definition,
1 pound-mass 5 0.4536 kg
This constant can be used to determine the mass in SI units (kilograms)
of a body that has been characterized by its weight in U.S. customary
units (pounds).
12.1D Equations of Motion
Consider a particle of mass m acted upon by several forces. Recall that
we can express Newton’s second law by the equation
oF 5 ma (12.2)
which relates the forces acting on the particle to the vector ma (Fig. 12.8).†
Two of the most important tools you will use in solving dynamics
problems, particularly those involving Newton’s second law, are the free-
body diagram and the kinetic diagram. These diagrams will help you to
model dynamic systems and apply appropriate equations of motion. The
free-body diagram shown on the left side of Fig. 12.9 is no different from
what you did in statics in Chapter 4 and consists of the following steps:
Fig. 12.8 The sum of forces applied to a
particle of mass m produces a vector ma in
the direction of the resultant force.
=
m m
ma
F1
F2
y
Axes
Applied force
Free-body diagram Kinetic diagram
Body force
Considere uma par)cula de massa m sob a ação de diversas forças, pela 2ª Lei de
Newton:
of a body that has been characterized by its weight in U.S. customary
units (pounds).
12.1D Equations of Motion
Consider a particle of mass m acted upon by several forces. Recall that
we can express Newton’s second law by the equation
oF 5 ma (12.2)
which relates the forces acting on the particle to the vector ma (Fig. 12.8).†
Two of the most important tools you will use in solving dynamics
problems, particularly those involving Newton’s second law, are the free-
body diagram and the kinetic diagram. These diagrams will help you to
model dynamic systems and apply appropriate equations of motion. The
free-body diagram shown on the left side of Fig. 12.9 is no different from
what you did in statics in Chapter 4 and consists of the following steps:
Fig. 12.8
particle of
the directi
m
F2
y
Axes
Applied force
Free-body diagram Kinetic diagram
Dimensions
Body Body
Body force
Inertial term
W = 785 N
30°
P
ma
Componentes cartesianas:
iagram and kinetic diagram to write oF 5 ma directly in component
orm. Examples of using these diagrams to help you write your equations
f motion are shown in the Sample Problems, and you can get extra
ractice by solving the Free-Body Problems 12.F1 through 12.F12.
As mentioned, it is usually more convenient to replace Eq. (12.2)
ith equivalent equations involving scalar quantities. As we saw in
hapter 11, we can resolve these vectors into components using several
ifferent coordinate systems (e.g., Cartesian, tangential and normal, or
adial and transverse), depending on the type of problem we are solving.
ectangular Components. Resolving each force F and the accele-
ation a into rectangular components, we have
o(Fxi 1 Fyj 1 Fzk) 5 m(a
xi 1 a
yj 1 a
zk)
follows from this equation that
oFx 5 ma
x oFy 5 ma
y oFz 5 ma
z (12.8)
ecall from Sec. 11.4C that the components of the acceleration are equal
o the second derivatives of the coordinates of the particle. This gives us
oFx 5 mẍ oFy 5 mÿ oFz 5 mz̈ (12.89)
Consider, as an example, the motion of a projectile. If we neglect
r resistance, the only force acting on the projectile after it has been fired
its weight W 5 2Wj. The equations defining the motion of the projectile
re therefore
mẍ 5 0 mÿ 5 2W mz̈ 5 0
diagram and kinetic diagram to write oF 5 ma directly in component
form. Examples of using these diagrams to help you write your equations
of motion are shown in the Sample Problems, and you can get extra
practice by solving the Free-Body Problems 12.F1 through 12.F12.
As mentioned, it is usually more convenient to replace Eq. (12.2)
with equivalent equations involving scalar quantities. As we saw in
Chapter 11, we can resolve these vectors into components using several
different coordinate systems (e.g., Cartesian, tangential and normal, or
radial and transverse), depending on the type of problem we are solving.
Rectangular Components. Resolving each force F and the accele-
ration a into rectangular components, we have
o(Fxi 1 Fyj 1 Fzk) 5 m(a
xi 1 a
yj 1 a
zk)
It follows from this equation that
oFx 5 ma
x oFy 5 ma
y oFz 5 ma
z (12.8)
Recall from Sec. 11.4C that the components of the acceleration are equal
to the second derivatives of the coordinates of the particle. This gives us
oFx 5 mẍ oFy 5 mÿ oFz 5 mz̈ (12.89)
Consider, as an example, the motion of a projectile. If we neglect
air resistance, the only force acting on the projectile after it has been fired
is its weight W 5 2Wj. The equations defining the motion of the projectile
are therefore
mẍ 5 0 mÿ 5 2W mz̈ 5 0
and the components of the acceleration of the projectile are
12.1
When a problem involves two or more bodies, you should write
equations of motion for each of the bodies (see Sample Probs. 12.3
through 12.5). Recall from Sec. 12.1A that all accelerations should be
measured with respect to a newtonian frame of reference. In most
engineering applications, you can determine accelerations with respect to
axes attached to the earth, but relative accelerations measured with respect
to moving axes, such as axes attached to an accelerated body, cannot be
substituted for a in the equations of motion.
Tangential and Normal Components. We can also resolve the
forces and the acceleration of the particle into components along the
tangent to the path (in the direction of motion) and the normal (toward
the inside of the path) (Fig. 12.10). Substituting into Eq. (12.2), we obtain
the two scalar equations of
oFt 5 mat oFn 5 man (12.9)
Fig. 12.10 The net force acting on a particle moving in
a curvilinear path can be resolved into components
tangent to the path and normal to the path, producing
=
man
mat
n
m
t
n
m
t
ΣFn
ΣFt
Componentes tangencial e normal:
12.1 Newton’s Second Law and Linear Momentum 727
hen a problem involves two or more bodies, you should write
ns of motion for each of the bodies (see Sample Probs. 12.3
12.5). Recall from Sec. 12.1A that all accelerations should be
d with respect to a newtonian frame of reference. In most
ring applications, you can determine accelerations with respect to
ached to the earth, but relative accelerations measured with respect
ng axes, such as axes attached to an accelerated body, cannot be
ed for a in the equations of motion.
ntial and Normal Components. We can also resolve the
nd the acceleration of the particle into components along the
to the path (in the direction of motion) and the normal (toward
de of the path) (Fig. 12.10). Substituting into Eq. (12.2), we obtain
scalar equations of
oFt 5 mat oFn 5 man (12.9)
=
man
mat
n
m
t
n
m
t
ΣFn
ΣFt
tached to the earth, but relative accelerations measured with respect
ing axes, such as axes attached to an accelerated body, cannot be
uted for a in the equations of motion.
ential and Normal Components. We can also resolve the
and the acceleration of the particle into components along the
t to the path (in the direction of motion) and the normal (toward
ide of the path) (Fig. 12.10). Substituting into Eq. (12.2), we obtain
o scalar equations of
oFt 5 mat oFn 5 man (12.9)
Photo 12.3 A fighter jet making a sharp
turn has a large normal component of
acceleration, often equal to several g. As a
result, the pilot experiences a large normal
force, which in extreme cases, can cause
blackouts.
Fig. 12.10 The net force acting on a particle moving in
a curvilinear path can be resolved into components
tangent to the path and normal to the path, producing
tangential and normal components of acceleration.
=
man
mat
n
m
t
n
m
t
ΣFn
ΣFt
ubstituting for at and an from Eqs. (11.39), we have
oFt 5 m
dv
dt
oFn 5 m
v2
ρ
(12.99)
n solve these equations for two unknowns.
l and Transverse Components. Consider a particle P,
olar coordinates r and θ, that moves in a plane under the action of
forces. Resolving the forces and the acceleration of the particle
4. ENAUTICA-DEM Profª Rosa Marat-Mendes 4
Passos para a resolução de exercícios
body diagram and the kinetic diagram. These diagrams will help you to
model dynamic systems and apply appropriate equations of motion. The
free-body diagram shown on the left side of Fig. 12.9 is no different from
what you did in statics in Chapter 4 and consists of the following steps:
Fig. 12.9 Steps in drawing a free-body diagram and a kinetic
diagram for solving dynamics problems.
=
x
y
m = 80 kg
Axes
Applied force
Free-body diagram Kinetic diagram
Dimensions
Support forces
Body Body
Body force
Inertial term
W = 785 N
F
N
30°
P
ma
Body: Define your system by isolating the body (or bodies) of
interest. If a problem has multiple bodies (such as in Sample
Problems 12.3 through 12.5), you may have to draw multiple free-
body diagrams and kinetic diagrams.
Axes: Draw an appropriate coordinate system (e.g., Cartesian,
normal and tangential, or radial and transverse).
Support Forces: Replace supports or constraints with appropriate
forces (e.g., two perpendicular forces for a pin, normal forces,
Diagrama de corpo livre Diagrama ciné1co
1. Desenhar o D.C.L. com as forças aplicadas, dimensões, eixos;
2. Desenhar o Diagrama ciné5co do corpo com os termos inerciais (𝑚𝒂);
3. Escrever as equações de equilíbrio dinâmico, ex:
4. Se necessário usar as equações da cinemá5ca para relacionar 𝑥, 𝑣, 𝑎 e 𝑡:
Ì A S 7zŠIÌ 6 ^ Z q GŠ Y B ÌvMÌ7Ì {7Š Y B c I 0 Ì LvBIÌ7 q D Ì8BBGbG7ŠY v p Ì
h۳
7#Ȭ Ȭ
M CH-G:A@FMA3M AG9A@M2/G-@6H=-DM
AAD19@-G2FM
S ɻ(ɻ( V ɻɻ (ɻ ɻ(ɻ( ɻ$Ɵҳ֏ (ɻɻ V8ɻ
ɻVɻ(Vɻɻ ɻ( V 8ɻ(ɻ ɻ(ɻ ɻ(VV ( 1ɻV(ɻɻ ɻ
ɻ ɻɻ ɻO;ҳK;ҳPҳV 1ɻ¨ɻ#ҳ ɻ ɻb( ɻɻ
8ɻɻ (ɻ
i^ҳј Mġҳ iJ=OҳŸ iJVKҳŸ iJĖPҳј =OҳŸVKҳŸĖPҳ
¨ ɻ ɻb(ɻ ɻɻ( 8ɻ ɻVɻ„FKFҳPҳVɻɻ ɻ
ɻ ɻɻb(ɻ ɻV ɻVɻɻ ɻ ɻ ɻ
Wb 1ɻɻ(ɻ ɻ ɻ ɻ ɻV( (ɻb(®ɻ
ÄJ=ҳ, =ҳ
iJVҳј Vҳ
iJjҳ, jҳ
b?ҳ
Pɻ( V (8ɻɻ ɻ( V ɻɻV ( ɻɻɻɻɻ ɻ$ (8ɻ
ɻ ɻ ɻɻɻ ɻb(ɻ( ɻ ɻ ɻVɻ ɻɻ
Ŵ
Ì A S 7zŠIÌ 6 ^ Z q GŠ Y B ÌvMÌ7Ì {7Š Y B c I 0 Ì LvBIÌ7 q D Ì8BBGbG7ŠY v p Ì
h۳
È۳
7#Ȭ Ȭ
M CH-G:A@FMA3M AG9A@M2/G-@6H=-
AAD19@-G2FM
S ɻ(ɻ( V ɻɻ (ɻ ɻ(ɻ( ɻ$Ɵҳ֏ (ɻɻ
ɻVɻ(Vɻɻ ɻ( V 8ɻ(ɻ ɻ(ɻ ɻ(VV ( 1ɻV(ɻɻ
ɻ ɻɻ ɻO;ҳK;ҳPҳV 1ɻ¨ɻ#ҳ ɻ ɻb(
8ɻɻ (ɻ
i^ҳј Mġҳ iJ=OҳŸ iJVKҳŸ iJĖPҳј =OҳŸVKҳŸĖPҳ
¨ ɻ ɻb(ɻ ɻɻ( 8ɻ ɻVɻ„FKFҳPҳVɻ
ɻ ɻɻb(ɻ ɻV ɻVɻɻ ɻ
Wb 1ɻɻ(ɻ ɻ ɻ ɻ ɻV( (ɻb(®ɻ
ÄJ=ҳ, =ҳ
iJVҳј Vҳ
iJjҳ, jҳ
Pɻ( V (8ɻɻ ɻ( V ɻɻV ( ɻɻɻɻɻ ɻ$
ɻ ɻ ɻɻɻ ɻb(ɻ( ɻ ɻ ɻVɻ ɻ
Ŵ
Ì A S 7zŠIÌ 6 ^ Z q GŠ Y B ÌvMÌ7Ì {7Š Y B c I 0 Ì LvBIÌ7 q D Ì8BBGbG7ŠY v p Ì
h۳
È۳
7#Ȭ Ȭ
M CH-G:A@FMA3M AG9A@M2/
AAD19@-G2FM
S ɻ(ɻ( V ɻɻ (ɻ ɻ(ɻ( ɻ$Ɵҳ֏
ɻVɻ(Vɻɻ ɻ( V 8ɻ(ɻ ɻ(ɻ ɻ(VV (
ɻ ɻɻ ɻO;ҳK;ҳPҳV 1ɻ¨ɻ#ҳ
8ɻɻ (ɻ
i^ҳј Mġҳ iJ=OҳŸ iJVKҳŸ iJĖPҳј =OҳŸVKҳ
¨ ɻ ɻb(ɻ ɻɻ( 8ɻ ɻVɻ„FKFҳPҳV
ɻ ɻɻb(ɻ ɻV ɻVɻ
Wb 1ɻɻ(ɻ ɻ ɻ ɻ ɻV( (
ÄJ=ҳ, =ҳ
iJVҳј Vҳ
iJjҳ, jҳ
Pɻ( V (8ɻɻ ɻ( V ɻɻV ( ɻɻɻ
ɻ ɻ ɻɻɻ ɻb(ɻ( ɻ ɻ ɻV
6. ENAUTICA-DEM Profª Rosa Marat-Mendes 6
Exemplo 2 – Dinâmica com movimentos dependentes
100 kg
300 kg
A
B
D
C
PROBLEMA RESOLVIDO 12.2
Os dois blocos mostrados na figura partem do repouso. Não há atrito no
plano horizontal nem na roldana, e a roldana é assumida como tendo massa
desprezível. Determine a aceleração de cada bloco e a tração em cada corda.
!
A
WA
T1
T2
N
mAaA
mA ! 100 kg
mB ! 300 kg
SOLUÇÃO
Cinemática. Notamos que se o bloco A se move de xA para a direita, o
bloco B se move para baixo por meio de
Diferenciando duas vezes em relação a t, temos
(1)
Cinética. Aplicamos a segunda lei de Newton sucessivamente ao bloco A,
ao bloco B e à roldana C.
Bloco A. Representando por T1 a tração na corda ACD, escrevemos
(2)
730 Kinetics of Particles: Newton’s Second Law
Sample Problem 12.3
The two blocks shown start from rest. The horizontal plane and the pulley
are frictionless, and the pulley is assumed to be of negligible mass. Deter-
mine the acceleration of each block and the tension in each cord.
STRATEGY: You are interested in finding the tension in the rope and
the acceleration of the two blocks, so use Newton’s second law. The two
blocks are connected by a cable, indicating that you need to relate their
accelerations using the techniques discussed in Chapter 11 for objects with
dependent motion.
MODELING: Treat both blocks as particles and assume that the pulley
is massless and frictionless. Since there are two masses, you need two
systems: block A by itself and block B by itself. The free-body and kinetic
diagrams for these objects are shown in Figs. 1 and 2. To help determine
the forces acting on block B, you can also isolate the massless pulley C
as a system (Fig. 3).
ANALYSIS: You can start with either kinetics or kinematics. The key
is to make sure you keep track of your equations and unknowns.
Kinetics. Apply Newton’s second law successively to block A, block B,
and pulley C.
Block A. Denote the tension in cord ACD by T1 (Fig. 1). Then you have
y
1
oFx 5 mAa
A: T1 5 100a
A (1)
Block B. Observe that the weight of block B is
WB 5 mBg5 (300 kg)(9.81 m/s2
) 5 2940 N
Denote the tension in cord BC by T2 (Fig. 2). Then
1
woFy 5 mBa
B: 2940 2 T2 5 300a
B (2)
Pulley C. Assuming mC is zero, you have (Fig. 3)
1
woFy 5 mCa
C 5 0: T2 2 2T1 5 0 (3)
At this point, you have three equations, (1), (2), and (3), and four unknowns,
T1, T2, a
B, and a
A. Therefore, you need one more equation, which you can
get from kinematics.
Kinematics. It is important to make sure that the directions you
100 kg
300 kg
A
B
D
C
=
A
WA
T1
N
mAaA
mA = 100 kg
y
x
Fig. 1 Free-body diagram and kinetic
diagram for A.
=
B
WB = 2940 N
T2
mBaB
mB = 300 kg
y
x
Fig. 2 Free-body diagram and kinetic
diagram for B.
Fig. 3 Free-body diagram and kinetic
diagram for the pulley.
=
T1 T1
T2
0
C
730 Kinetics of Particles: Newton’s Second Law
Sample Problem 12.3
The two blocks shown start from rest. The horizontal plane and the pulley
are frictionless, and the pulley is assumed to be of negligible mass. Deter-
mine the acceleration of each block and the tension in each cord.
STRATEGY: You are interested in finding the tension in the rope and
the acceleration of the two blocks, so use Newton’s second law. The two
blocks are connected by a cable, indicating that you need to relate their
accelerations using the techniques discussed in Chapter 11 for objects with
dependent motion.
MODELING: Treat both blocks as particles and assume that the pulley
is massless and frictionless. Since there are two masses, you need two
systems: block A by itself and block B by itself. The free-body and kinetic
diagrams for these objects are shown in Figs. 1 and 2. To help determine
the forces acting on block B, you can also isolate the massless pulley C
as a system (Fig. 3).
ANALYSIS: You can start with either kinetics or kinematics. The key
is to make sure you keep track of your equations and unknowns.
Kinetics. Apply Newton’s second law successively to block A, block B,
and pulley C.
Block A. Denote the tension in cord ACD by T1 (Fig. 1). Then you have
y
1
oFx 5 mAa
A: T1 5 100a
A (1)
Block B. Observe that the weight of block B is
WB 5 mBg5 (300 kg)(9.81 m/s2
) 5 2940 N
Denote the tension in cord BC by T2 (Fig. 2). Then
1
woFy 5 mBa
B: 2940 2 T2 5 300a
B (2)
Pulley C. Assuming mC is zero, you have (Fig. 3)
1
woFy 5 mCa
C 5 0: T2 2 2T1 5 0 (3)
At this point, you have three equations, (1), (2), and (3), and four unknowns,
T1, T2, a
B, and a
A. Therefore, you need one more equation, which you can
100 kg
300 kg
A
B
D
C
=
A
WA
T1
N
mAaA
mA = 100 kg
y
x
Fig. 1 Free-body diagram and kinetic
diagram for A.
=
B
WB = 2940 N
T2
mBaB
mB = 300 kg
y
x
Fig. 2 Free-body diagram and kinetic
diagram for B.
Fig. 3 Free-body diagram and kinetic
=
T1 T1
T2
0
C
mine the acceleration of each block and the tension in each cord.
STRATEGY: You are interested in finding the tension in the rope and
the acceleration of the two blocks, so use Newton’s second law. The two
blocks are connected by a cable, indicating that you need to relate their
accelerations using the techniques discussed in Chapter 11 for objects with
dependent motion.
MODELING: Treat both blocks as particles and assume that the pulley
is massless and frictionless. Since there are two masses, you need two
systems: block A by itself and block B by itself. The free-body and kinetic
diagrams for these objects are shown in Figs. 1 and 2. To help determine
the forces acting on block B, you can also isolate the massless pulley C
as a system (Fig. 3).
ANALYSIS: You can start with either kinetics or kinematics. The key
is to make sure you keep track of your equations and unknowns.
Kinetics. Apply Newton’s second law successively to block A, block B,
and pulley C.
Block A. Denote the tension in cord ACD by T1 (Fig. 1). Then you have
y
1
oFx 5 mAa
A: T1 5 100a
A (1)
Block B. Observe that the weight of block B is
WB 5 mBg5 (300 kg)(9.81 m/s2
) 5 2940 N
Denote the tension in cord BC by T2 (Fig. 2). Then
1
woFy 5 mBa
B: 2940 2 T2 5 300a
B (2)
Pulley C. Assuming mC is zero, you have (Fig. 3)
1
woFy 5 mCa
C 5 0: T2 2 2T1 5 0 (3)
At this point, you have three equations, (1), (2), and (3), and four unknowns,
T1, T2, a
B, and a
A. Therefore, you need one more equation, which you can
get from kinematics.
Kinematics. It is important to make sure that the directions you
assumed for the kinetic diagrams are consistent with the kinematic
analysis. Note that if block A moves through a distance xA to the right,
block B moves down through a distance
xB 5
1
2
xA
Differentiating twice with respect to t, you have
a
B 5
1
a
A (4)
100 kg
300 kg B
C
=
A
WA
T1
N
mAaA
mA = 100 kg
y
x
Fig. 1 Free-body diagram and kinetic
diagram for A.
=
B
WB = 2940 N
T2
mBaB
mB = 300 kg
y
x
Fig. 2 Free-body diagram and kinetic
diagram for B.
Fig. 3 Free-body diagram and kinetic
diagram for the pulley.
=
T1 T1
T2
0
C
730 Kinetics of Particles: Newton’s Second Law
Sample Problem 12.3
The two blocks shown start from rest. The horizontal plane and the pulley
are frictionless, and the pulley is assumed to be of negligible mass. Deter-
mine the acceleration of each block and the tension in each cord.
STRATEGY: You are interested in finding the tension in the rope and
the acceleration of the two blocks, so use Newton’s second law. The two
blocks are connected by a cable, indicating that you need to relate their
accelerations using the techniques discussed in Chapter 11 for objects with
dependent motion.
MODELING: Treat both blocks as particles and assume that the pulley
is massless and frictionless. Since there are two masses, you need two
systems: block A by itself and block B by itself. The free-body and kinetic
diagrams for these objects are shown in Figs. 1 and 2. To help determine
the forces acting on block B, you can also isolate the massless pulley C
as a system (Fig. 3).
ANALYSIS: You can start with either kinetics or kinematics. The key
is to make sure you keep track of your equations and unknowns.
Kinetics. Apply Newton’s second law successively to block A, block B,
and pulley C.
Block A. Denote the tension in cord ACD by T1 (Fig. 1). Then you have
y
1
oFx 5 mAa
A: T1 5 100a
A (1)
Block B. Observe that the weight of block B is
WB 5 mBg5 (300 kg)(9.81 m/s2
) 5 2940 N
Denote the tension in cord BC by T2 (Fig. 2). Then
1
woFy 5 mBa
B: 2940 2 T2 5 300a
B (2)
Pulley C. Assuming mC is zero, you have (Fig. 3)
1
woFy 5 mCa
C 5 0: T2 2 2T1 5 0 (3)
At this point, you have three equations, (1), (2), and (3), and four unknowns,
T1, T2, a
B, and a
A. Therefore, you need one more equation, which you can
100 kg
300 kg
A
B
D
C
=
A
WA
T1
N
mAaA
mA = 100 kg
y
x
Fig. 1 Free-body diagram and kinetic
diagram for A.
=
B
WB = 2940 N
T2
mBaB
mB = 300 kg
y
x
Fig. 2 Free-body diagram and kinetic
diagram for B.
Fig. 3 Free-body diagram and kinetic
=
T1 T1
T2
0
C
The two blocks shown start from rest. The horizontal plane and the pulley
are frictionless, and the pulley is assumed to be of negligible mass. Deter-
mine the acceleration of each block and the tension in each cord.
STRATEGY: You are interested in finding the tension in the rope and
the acceleration of the two blocks, so use Newton’s second law. The two
blocks are connected by a cable, indicating that you need to relate their
accelerations using the techniques discussed in Chapter 11 for objects with
dependent motion.
MODELING: Treat both blocks as particles and assume that the pulley
is massless and frictionless. Since there are two masses, you need two
systems: block A by itself and block B by itself. The free-body and kinetic
diagrams for these objects are shown in Figs. 1 and 2. To help determine
the forces acting on block B, you can also isolate the massless pulley C
as a system (Fig. 3).
ANALYSIS: You can start with either kinetics or kinematics. The key
is to make sure you keep track of your equations and unknowns.
Kinetics. Apply Newton’s second law successively to block A, block B,
and pulley C.
Block A. Denote the tension in cord ACD by T1 (Fig. 1). Then you have
y
1
oFx 5 mAa
A: T1 5 100a
A (1)
Block B. Observe that the weight of block B is
WB 5 mBg5 (300 kg)(9.81 m/s2
) 5 2940 N
Denote the tension in cord BC by T2 (Fig. 2). Then
1
woFy 5 mBa
B: 2940 2 T2 5 300a
B (2)
Pulley C. Assuming mC is zero, you have (Fig. 3)
1
woFy 5 mCa
C 5 0: T2 2 2T1 5 0 (3)
At this point, you have three equations, (1), (2), and (3), and four unknowns,
T1, T2, a
B, and a
A. Therefore, you need one more equation, which you can
get from kinematics.
Kinematics. It is important to make sure that the directions you
assumed for the kinetic diagrams are consistent with the kinematic
analysis. Note that if block A moves through a distance xA to the right,
block B moves down through a distance
100 kg
300 kg
A
B
D
C
=
A
WA
T1
N
mAaA
mA = 100 kg
y
x
Fig. 1 Free-body diagram and kinetic
diagram for A.
=
B
WB = 2940 N
T2
mBaB
mB = 300 kg
y
x
Fig. 2 Free-body diagram and kinetic
diagram for B.
Fig. 3 Free-body diagram and kinetic
diagram for the pulley.
=
T1 T1
T2
0
C
mine the acceleration of each block and the tension in each cord.
STRATEGY: You are interested in finding the tension in the rope and
the acceleration of the two blocks, so use Newton’s second law. The two
blocks are connected by a cable, indicating that you need to relate their
accelerations using the techniques discussed in Chapter 11 for objects with
dependent motion.
MODELING: Treat both blocks as particles and assume that the pulley
is massless and frictionless. Since there are two masses, you need two
systems: block A by itself and block B by itself. The free-body and kinetic
diagrams for these objects are shown in Figs. 1 and 2. To help determine
the forces acting on block B, you can also isolate the massless pulley C
as a system (Fig. 3).
ANALYSIS: You can start with either kinetics or kinematics. The key
is to make sure you keep track of your equations and unknowns.
Kinetics. Apply Newton’s second law successively to block A, block B,
and pulley C.
Block A. Denote the tension in cord ACD by T1 (Fig. 1). Then you have
y
1
oFx 5 mAa
A: T1 5 100a
A (1)
Block B. Observe that the weight of block B is
WB 5 mBg5 (300 kg)(9.81 m/s2
) 5 2940 N
Denote the tension in cord BC by T2 (Fig. 2). Then
1
woFy 5 mBa
B: 2940 2 T2 5 300a
B (2)
Pulley C. Assuming mC is zero, you have (Fig. 3)
1
woFy 5 mCa
C 5 0: T2 2 2T1 5 0 (3)
At this point, you have three equations, (1), (2), and (3), and four unknowns,
T1, T2, a
B, and a
A. Therefore, you need one more equation, which you can
get from kinematics.
100 kg
300 kg B
C
=
A
WA
T1
N
mAaA
mA = 100 kg
y
x
Fig. 1 Free-body diagram and kinetic
diagram for A.
=
B
WB = 2940 N
T2
mBaB
mB = 300 kg
y
x
Fig. 2 Free-body diagram and kinetic
diagram for B.
Fig. 3 Free-body diagram and kinetic
diagram for the pulley.
=
T1 T1
T2
0
C
Sample Problem 12.3
The two blocks shown start from rest. The horizontal plane and the pulley
are frictionless, and the pulley is assumed to be of negligible mass. Deter-
mine the acceleration of each block and the tension in each cord.
STRATEGY: You are interested in finding the tension in the rope and
the acceleration of the two blocks, so use Newton’s second law. The two
blocks are connected by a cable, indicating that you need to relate their
accelerations using the techniques discussed in Chapter 11 for objects with
dependent motion.
MODELING: Treat both blocks as particles and assume that the pulley
is massless and frictionless. Since there are two masses, you need two
systems: block A by itself and block B by itself. The free-body and kinetic
diagrams for these objects are shown in Figs. 1 and 2. To help determine
the forces acting on block B, you can also isolate the massless pulley C
as a system (Fig. 3).
ANALYSIS: You can start with either kinetics or kinematics. The key
is to make sure you keep track of your equations and unknowns.
Kinetics. Apply Newton’s second law successively to block A, block B,
and pulley C.
Block A. Denote the tension in cord ACD by T1 (Fig. 1). Then you have
y
1
oFx 5 mAa
A: T1 5 100a
A (1)
Block B. Observe that the weight of block B is
WB 5 mBg5 (300 kg)(9.81 m/s2
) 5 2940 N
Denote the tension in cord BC by T2 (Fig. 2). Then
1
woFy 5 mBa
B: 2940 2 T2 5 300a
B (2)
Pulley C. Assuming mC is zero, you have (Fig. 3)
1
woFy 5 mCa
C 5 0: T2 2 2T1 5 0 (3)
At this point, you have three equations, (1), (2), and (3), and four unknowns,
T1, T2, a
B, and a
A. Therefore, you need one more equation, which you can
get from kinematics.
Kinematics. It is important to make sure that the directions you
assumed for the kinetic diagrams are consistent with the kinematic
100 kg
300 kg
A
B
D
C
=
A
WA
T1
N
mAaA
mA = 100 kg
y
x
Fig. 1 Free-body diagram and kinetic
diagram for A.
=
B
WB = 2940 N
T2
mBaB
mB = 300 kg
y
x
Fig. 2 Free-body diagram and kinetic
diagram for B.
Fig. 3 Free-body diagram and kinetic
diagram for the pulley.
=
T1 T1
T2
0
C
Estudo cinemá1co, movimentos dependentes:
!
!
!
B
A
WA
WB ! 2.940 N
T1
T1 T1
T2
T2
N
0
mAaA
mBaB
mA ! 100 kg
mB ! 300 kg
C
SOLUÇÃO
Cinemática. Notamos que se o bloco A se move de xA para a
bloco B se move para baixo por meio de
Diferenciando duas vezes em relação a t, temos
Cinética. Aplicamos a segunda lei de Newton sucessivamente a
ao bloco B e à roldana C.
Bloco A. Representando por T1 a tração na corda ACD, escreve
Bloco B. Observando que o peso do bloco B é
WB ! mBg ! (300 kg)(9,81 m/s
2
) ! 2.940 N
e representando por T2 a tração na corda BC, escrevemos
ou, substituindo para aB de (1),
!
!
!
B
A
WA
WB ! 2.940 N
T1
T1 T1
T2
T2
N
0
mAaA
mBaB
mA ! 100 kg
mB ! 300 kg
C
SOLUÇÃO
Cinemática. Notamos que se o bloco A se move de xA para a d
bloco B se move para baixo por meio de
Diferenciando duas vezes em relação a t, temos
Cinética. Aplicamos a segunda lei de Newton sucessivamente ao
ao bloco B e à roldana C.
Bloco A. Representando por T1 a tração na corda ACD, escrevem
Bloco B. Observando que o peso do bloco B é
WB ! mBg ! (300 kg)(9,81 m/s2
) ! 2.940 N
e representando por T2 a tração na corda BC, escrevemos
ou, substituindo para aB de (1),
Roldana C. Já que mC é assumida como sendo zero, temos
12 - 16
• Escrever as equações do movimento para os blocos
e para a roldana:
:
A
A
x a
m
F =
∑
( ) A
a
T kg
100
1 =
:
B
B
y a
m
F =
∑
( )( ) ( )
( ) B
B
B
B
B
a
T
a
T
a
m
T
g
m
kg
300
-
N
2940
kg
300
s
m
81
.
9
kg
300
2
2
2
2
=
=
−
=
−
:
0
=
=
∑ C
C
y a
m
F
0
2 1
2 =
− T
T
• Escrever as relações cinemáticas para os
movimentos dependentes e acelerações dos
blocos:
A
B
A
B a
a
x
y 2
1
2
1 =
=
x
y
O
12 - 16
• Escrever as equações do movimento para os blocos
e para a roldana:
:
A
A
x a
m
F =
∑
( ) A
a
T kg
100
1 =
:
B
B
y a
m
F =
∑
( )( ) ( )
( ) B
B
B
B
B
a
T
a
T
a
m
T
g
m
kg
300
-
N
2940
kg
300
s
m
81
.
9
kg
300
2
2
2
2
=
=
−
=
−
:
0
=
=
∑ C
C
y a
m
F
0
2 1
2 =
− T
T
• Escrever as relações cinemáticas para os
movimentos dependentes e acelerações dos
blocos:
A
B
A
B a
a
x
y 2
1
2
1 =
=
x
y
O
12 - 16
• Escrever as equações do movimento para os blocos
e para a roldana:
:
A
A
x a
m
F =
∑
( ) A
a
T kg
100
1 =
:
B
B
y a
m
F =
∑
( )( ) ( )
( ) B
B
B
B
B
a
T
a
T
a
m
T
g
m
kg
300
-
N
2940
kg
300
s
m
81
.
9
kg
300
2
2
2
2
=
=
−
=
−
:
0
=
=
∑ C
C
y a
m
F
0
2 1
2 =
− T
T
• Escrever as relações cinemáticas para os
movimentos dependentes e acelerações dos
blocos:
A
B
A
B a
a
x
y 2
1
2
1 =
=
x
y
O
s
m
20
.
4
s
m
40
.
8
2
2
1
2
=
=
=
a
a
a
A
B
A
• Combinar as relações cinemáticas com as
equações do movimento para resolver em ordem
às acelerações e força de tracção no cabo:
A
B
A
B a
a
x
y 2
1
2
1 =
=
( ) A
a
T kg
100
1 =
( )
( )( )
A
B
a
a
T
2
1
2
kg
300
-
N
2940
kg
300
-
N
2940
=
=
( ) ( ) 0
kg
100
2
kg
150
N
2940
0
2 1
2
=
−
−
=
−
A
A a
a
T
T
x
y
O
s
m
40
.
8 2
=
aA
• Combinar as relações cinemáticas com as
equações do movimento para resolver em ordem
às acelerações e força de tracção no cabo:
A
B
A
B a
a
x
y 2
1
2
1 =
=
( ) A
a
T kg
100
1 =
( )
( )( )
A
B
a
a
T
2
1
2
kg
300
-
N
2940
kg
300
-
N
2940
=
=
( ) ( ) 0
kg
100
2
kg
150
N
2940
0
2 1
2
=
−
−
=
−
A
A a
a
T
T
x
y
O
12.1 Newton’s Second Law and Linear Momentum 731
You now have four equations and four unknowns, so you can solve this
problem. You can do this using a computer, a calculator, or by hand. To
solve these equations by hand, you can substitute for aB from Eq. (4) into
Eq. (2) for
2940 2 T2 5 300(1
2aA)
T2 5 2940 2 150aA (5)
Now substitute for T1 and T2 from Eqs. (1) and (5), respectively, into
Eq. (3).
2940 2 150aA 2 2(100aA) 5 0
2940 2 350aA 5 0 aA 5 8.40 m/s2
b
Then substitute the value obtained for aA into Eqs. (4) and (1).
aB 5 1
2 aA 5 1
2(8.40 m/s2
) aB 5 4.20 m/s2
b
12.1 Newton’s Second Law and Linear Momentum 731
You now have four equations and four unknowns, so you can solve this
problem. You can do this using a computer, a calculator, or by hand. To
solve these equations by hand, you can substitute for aB from Eq. (4) into
Eq. (2) for
2940 2 T2 5 300(1
2aA)
T2 5 2940 2 150aA (5)
Now substitute for T1 and T2 from Eqs. (1) and (5), respectively, into
Eq. (3).
2940 2 150aA 2 2(100aA) 5 0
2940 2 350aA 5 0 aA 5 8.40 m/s2
b
Then substitute the value obtained for aA into Eqs. (4) and (1).
aB 5 1
2 aA 5 1
2(8.40 m/s2
) aB 5 4.20 m/s2
b
T1 5 100aA 5 (100 kg)(8.40 m/s2
) T1 5 840 N b
Recalling Eq. (3), you have
T2 5 2T1 T2 5 2(840 N) T2 5 1680 N b
REFLECT and THINK: Note that the value obtained for T2 is not equal
12.1 Newton’s Second Law and Linear Momentum 731
You now have four equations and four unknowns, so you can solve this
problem. You can do this using a computer, a calculator, or by hand. To
solve these equations by hand, you can substitute for aB from Eq. (4) into
Eq. (2) for
2940 2 T2 5 300(1
2aA)
T2 5 2940 2 150aA (5)
Now substitute for T1 and T2 from Eqs. (1) and (5), respectively, into
Eq. (3).
2940 2 150aA 2 2(100aA) 5 0
2940 2 350aA 5 0 aA 5 8.40 m/s2
b
Then substitute the value obtained for aA into Eqs. (4) and (1).
aB 5 1
2 aA 5 1
2(8.40 m/s2
) aB 5 4.20 m/s2
b
T1 5 100aA 5 (100 kg)(8.40 m/s2
) T1 5 840 N b
Recalling Eq. (3), you have
T2 5 2T1 T2 5 2(840 N) T2 5 1680 N b
REFLECT and THINK: Note that the value obtained for T2 is not equal
to the weight of block B. Rather than choosing B and the pulley as sepa-
rate systems, you could have chosen the system to be B and the pulley.
In this case, T2 would have been an internal force.
You now have four equations and four unknowns, so you can solve this
problem. You can do this using a computer, a calculator, or by hand. To
solve these equations by hand, you can substitute for aB from Eq. (4) into
Eq. (2) for
2940 2 T2 5 300(1
2aA)
T2 5 2940 2 150aA (5)
Now substitute for T1 and T2 from Eqs. (1) and (5), respectively, into
Eq. (3).
2940 2 150aA 2 2(100aA) 5 0
2940 2 350aA 5 0 aA 5 8.40 m/s2
b
Then substitute the value obtained for aA into Eqs. (4) and (1).
aB 5 1
2 aA 5 1
2(8.40 m/s2
) aB 5 4.20 m/s2
b
T1 5 100aA 5 (100 kg)(8.40 m/s2
) T1 5 840 N b
Recalling Eq. (3), you have
T2 5 2T1 T2 5 2(840 N) T2 5 1680 N b
REFLECT and THINK: Note that the value obtained for T2 is not equal
to the weight of block B. Rather than choosing B and the pulley as sepa-
rate systems, you could have chosen the system to be B and the pulley.
In this case, T2 would have been an internal force.
Sample Problem 12.4
Collar A has a ramp that is welded to it and a force P 5 5 lb applied as
7. ENAUTICA-DEM Profª Rosa Marat-Mendes 7
Exemplo 3 – Dinâmica com movimento curvilíneo
30°
2 m
O
m
=
T = 2.5 mg
W = mg
man
n
t
mat
30°
Fig. 1 Free-body diagram and
kinetic diagram for the bob.
man
PROBLEMA RESOLVIDO 12.4
A extremidade de um pêndulo de 2 m de comprimento descreve um arco
de circunferência em um plano vertical. Se a tração na corda é 2,5 vezes o
peso do pêndulo para a posição mostrada na figura, encontre a velocidade e
a aceleração do pêndulo nessa posição.
SOLUÇÃO
O peso do pêndulo é W ! mg; a tração na corda é, portanto, 2,5 mg. Recor-
dando que an é dirigido para O e assumindo at como mostrado na figura,
aplicamos a segunda lei de Newton e obtemos
mg sen 30° ! mat
at ! g sen 30° ! 4,90 m/s
2
at ! 4,90 m/s
2
o !
2,5 mg # mg cos 30° ! man
an ! 1,634 g ! 16,03 m/s
2
an ! 16,03 m/s
2
r !
Como an ! v
2
/!, temos v
2
! !an ! (2 m)(16,03 m/s
2
).
v ! $5,66 m/s v ! 5,66 m/s (para cima ou para baixo) !
PROBLEMA RESOLVIDO 12.5
12.1 Newton’s Second Law and Linear Momentum 735
Sample Problem 12.6
The bob of a 2-m pendulum describes an arc of a circle in a vertical plane.
If the tension in the cord is 2.5 times the weight of the bob for the position
shown, find the velocity and the acceleration of the bob in that position.
STRATEGY: The most direct approach is to use Newton’s law with
tangential and normal components.
MODELING: Choose the bob as your system; if its radius is small, you
can model it as a particle. Draw the free-body and kinetic diagrams for
the bob knowing that the weight of the bob is W 5 mg
; the tension in the
cord is 2.5mg
. The normal acceleration a n is directed toward O, and you
can assume that a t is in the direction shown in Fig. 1.
ANALYSIS: You can obtain scalar equations by applying Newton’s
second law in the normal and tangential directions. Hence,
1b oFt 5 mat: mgsin 30° 5 mat
at 5 gsin 30° 5 14.90 m/s2
a t 5 4.90 m/s2
b b
1a oFn 5 man: 2.5mg2 mgcos 30° 5 man
an 5 1.634g5 116.03 m/s2
a n 5 16.03 m/s2
a b
Since an 5 v2
/ρ, you have v2
5 ρan 5 (2 m)(16.03 m/s2
). Thus,
v 5 65.66 m/s v 5 5.66 m/s
G
(up or down) b
REFLECT and THINK: If you look at these equations for an angle of
zero instead of 30°, you will see that when the bob is straight below
point O, the tangential acceleration is zero, and the velocity is a maximum.
The normal acceleration is not zero because the bob has a velocity at this
point.
30°
2 m
O
m
=
T = 2.5 mg
W = mg
man
n
t
mat
30°
Fig. 1 Free-body diagram and
kinetic diagram for the bob.
30°
2 m
O
m
!
T ! 2,5 mg
W ! mg
man
n
t
mat
30°
PROBLEMA RESOLVIDO 12.4
A extremidade de um pêndulo de 2 m de comprimento descreve um arco
de circunferência em um plano vertical. Se a tração na corda é 2,5 vezes o
peso do pêndulo para a posição mostrada na figura, encontre a velocidade e
a aceleração do pêndulo nessa posição.
SOLUÇÃO
O peso do pêndulo é W ! mg; a tração na corda é, portanto, 2,5 mg. Recor-
dando que an é dirigido para O e assumindo at como mostrado na figura,
aplicamos a segunda lei de Newton e obtemos
mg sen 30° ! mat
at ! g sen 30° ! 4,90 m/s
2
at ! 4,90 m/s
2
o !
2,5 mg # mg cos 30° ! man
an ! 1,634 g ! 16,03 m/s2
an ! 16,03 m/s2
r !
Como an ! v
2
/!, temos v
2
! !an ! (2 m)(16,03 m/s
2
).
v ! $5,66 m/s v ! 5,66 m/s (para cima ou para baixo) !
PROBLEMA RESOLVIDO 12.5
Determine a velocidade de segurança calculada para uma curva de rodovia
de raio ! ! 120 m, inclinada a um ângulo ! 18°. A velocidade de segu-
rança calculada de uma curva de uma rodovia com declive é a velocidade
escalar na qual um carro deve trafegar sem que nenhuma força de atrito
lateral seja exercida em suas rodas.
30°
2 m
O
m
!
T ! 2,5 mg
W ! mg
man
n
t
mat
30°
PROBLEMA RESOLVIDO 12.4
A extremidade de um pêndulo de 2 m de comprimento descreve um arco
de circunferência em um plano vertical. Se a tração na corda é 2,5 vezes o
peso do pêndulo para a posição mostrada na figura, encontre a velocidade e
a aceleração do pêndulo nessa posição.
SOLUÇÃO
O peso do pêndulo é W ! mg; a tração na corda é, portanto, 2,5 mg. Recor-
dando que an é dirigido para O e assumindo at como mostrado na figura,
aplicamos a segunda lei de Newton e obtemos
mg sen 30° ! mat
at ! g sen 30° ! 4,90 m/s
2
at ! 4,90 m/s
2
o !
2,5 mg # mg cos 30° ! man
an ! 1,634 g ! 16,03 m/s
2
an ! 16,03 m/s
2
r !
Como an ! v
2
/!, temos v
2
! !an ! (2 m)(16,03 m/s
2
).
v ! $5,66 m/s v ! 5,66 m/s (para cima ou para baixo) !
PROBLEMA RESOLVIDO 12.5
Determine a velocidade de segurança calculada para uma curva de rodovia
de raio ! ! 120 m, inclinada a um ângulo ! 18°. A velocidade de segu-
Sample Problem 12.6
The bob of a 2-m pendulum describes an arc of a circle in a vertical plane.
If the tension in the cord is 2.5 times the weight of the bob for the position
shown, find the velocity and the acceleration of the bob in that position.
STRATEGY: The most direct approach is to use Newton’s law with
tangential and normal components.
MODELING: Choose the bob as your system; if its radius is small, you
can model it as a particle. Draw the free-body and kinetic diagrams for
the bob knowing that the weight of the bob is W 5 mg
; the tension in the
cord is 2.5mg
. The normal acceleration a n is directed toward O, and you
can assume that a t is in the direction shown in Fig. 1.
ANALYSIS: You can obtain scalar equations by applying Newton’s
second law in the normal and tangential directions. Hence,
1b oFt 5 mat: mgsin 30° 5 mat
at 5 gsin 30° 5 14.90 m/s2
a t 5 4.90 m/s2
b b
1a oFn 5 man: 2.5mg2 mgcos 30° 5 man
an 5 1.634g5 116.03 m/s2
a n 5 16.03 m/s2
a b
Since an 5 v2
/ρ, you have v2
5 ρan 5 (2 m)(16.03 m/s2
). Thus,
v 5 65.66 m/s v 5 5.66 m/s
G
(up or down) b
REFLECT and THINK: If you look at these equations for an angle of
zero instead of 30°, you will see that when the bob is straight below
point O, the tangential acceleration is zero, and the velocity is a maximum.
The normal acceleration is not zero because the bob has a velocity at this
point.
2 m
m
=
2.5 mg
man
mat
30°
e-body diagram and
ram for the bob.
Sample Problem 12.6
The bob of a 2-m pendulum describes an arc of a circle in a vertical plane.
If the tension in the cord is 2.5 times the weight of the bob for the position
shown, find the velocity and the acceleration of the bob in that position.
STRATEGY: The most direct approach is to use Newton’s law with
tangential and normal components.
MODELING: Choose the bob as your system; if its radius is small, you
can model it as a particle. Draw the free-body and kinetic diagrams for
the bob knowing that the weight of the bob is W 5 mg
; the tension in the
cord is 2.5mg
. The normal acceleration a n is directed toward O, and you
can assume that a t is in the direction shown in Fig. 1.
ANALYSIS: You can obtain scalar equations by applying Newton’s
second law in the normal and tangential directions. Hence,
1b oFt 5 mat: mgsin 30° 5 mat
at 5 gsin 30° 5 14.90 m/s2
a t 5 4.90 m/s2
b b
1a oFn 5 man: 2.5mg2 mgcos 30° 5 man
an 5 1.634g5 116.03 m/s2
a n 5 16.03 m/s2
a b
Since an 5 v2
/ρ, you have v2
5 ρan 5 (2 m)(16.03 m/s2
). Thus,
v 5 65.66 m/s v 5 5.66 m/s
G
(up or down) b
REFLECT and THINK: If you look at these equations for an angle of
zero instead of 30°, you will see that when the bob is straight below
point O, the tangential acceleration is zero, and the velocity is a maximum.
The normal acceleration is not zero because the bob has a velocity at this
point.
2 m
m
=
5 mg
man
mat
30°
ody diagram and
m for the bob.
30°
2 m
O
m
!
T ! 2,5 mg
W ! mg
man
n
t
mat
30°
PROBLEMA RESOLVIDO 12.4
A extremidade de um pêndulo de 2 m de comprimento descreve um arco
de circunferência em um plano vertical. Se a tração na corda é 2,5 vezes o
peso do pêndulo para a posição mostrada na figura, encontre a velocidade e
a aceleração do pêndulo nessa posição.
SOLUÇÃO
O peso do pêndulo é W ! mg; a tração na corda é, portanto, 2,5 mg. Recor-
dando que an é dirigido para O e assumindo at como mostrado na figura,
aplicamos a segunda lei de Newton e obtemos
mg sen 30° ! mat
at ! g sen 30° ! 4,90 m/s2
at ! 4,90 m/s2
o !
2,5 mg # mg cos 30° ! man
an ! 1,634 g ! 16,03 m/s2
an ! 16,03 m/s2
r !
Como an ! v
2
/!, temos v
2
! !an ! (2 m)(16,03 m/s
2
).
v ! $5,66 m/s v ! 5,66 m/s (para cima ou para baixo) !
30°
2 m
O
m
!
T ! 2,5 mg
mg
man
mat
30°
PROBLEMA RESOLVIDO 12.4
A extremidade de um pêndulo de 2 m de comprimento descreve um arco
de circunferência em um plano vertical. Se a tração na corda é 2,5 vezes o
peso do pêndulo para a posição mostrada na figura, encontre a velocidade e
a aceleração do pêndulo nessa posição.
SOLUÇÃO
O peso do pêndulo é W ! mg; a tração na corda é, portanto, 2,5 mg. Recor-
dando que an é dirigido para O e assumindo at como mostrado na figura,
aplicamos a segunda lei de Newton e obtemos
mg sen 30° ! mat
at ! g sen 30° ! 4,90 m/s2
at ! 4,90 m/s2
o !
2,5 mg # mg cos 30° ! man
an ! 1,634 g ! 16,03 m/s2
an ! 16,03 m/s2
r !
Como an ! v
2
/!, temos v
2
! !an ! (2 m)(16,03 m/s
2
).
v ! $5,66 m/s v ! 5,66 m/s (para cima ou para baixo) !
8. ENAUTICA-DEM Profª Rosa Marat-Mendes 8
Exemplo 4 – Dinâmica com atrito
P
30°
80 kg
PROBLEMA RESOLVIDO 12.1
Um bloco de 80-kg está em repouso sobre um plano horizontal. Encontre
a intensidade da força P necessária para dar ao bloco uma aceleração de
2,5 m/s
2
para a direita. O coeficiente de atrito cinético entre o bloco e o
plano é !k ! 0,25.
SOLUÇÃO
O peso do bloco é
W ! mg ! (80 kg)(9,81 m/s
2
) ! 785 N
!
P
30°
N
F
W ! 785 N
ma
m ! 80 kg
Notamos que F ! !kN ! 0,25N e que a ! 2,5 m/s
2
. Expressando que as
forças que atuam no bloco são equivalentes ao vetor ma, escrevemos
(1)
(2)
Resolvendo (2) para N e substituindo o resultado em (1), obtemos
N ! P sen 30 785 N
P
30°
80 kg
PROBLEMA RESOLVIDO 12.1
Um bloco de 80-kg está em repouso sobre um plano horizontal. Encontre
a intensidade da força P necessária para dar ao bloco uma aceleração de
2,5 m/s
2
para a direita. O coeficiente de atrito cinético entre o bloco e o
plano é !k ! 0,25.
SOLUÇÃO
O peso do bloco é
W ! mg ! (80 kg)(9,81 m/s
2
) ! 785 N
!
P
30°
N
F
W ! 785 N
ma
m ! 80 kg
Notamos que F ! !kN ! 0,25N e que a ! 2,5 m/s
2
. Expressando que as
forças que atuam no bloco são equivalentes ao vetor ma, escrevemos
(1)
(2)
Resolvendo (2) para N e substituindo o resultado em (1), obtemos
PROBLEMA RESOLVIDO 12.1
Um bloco de 80-kg está em repouso sobre um plano horizontal. Encontre
a intensidade da força P necessária para dar ao bloco uma aceleração de
2,5 m/s
2
para a direita. O coeficiente de atrito cinético entre o bloco e o
plano é !k ! 0,25.
SOLUÇÃO
O peso do bloco é
W ! mg ! (80 kg)(9,81 m/s
2
) ! 785 N
!
P
30°
N
F
W ! 785 N
ma
m ! 80 kg
Notamos que F ! !kN ! 0,25N e que a ! 2,5 m/s
2
. Expressando que as
forças que atuam no bloco são equivalentes ao vetor ma, escrevemos
(1)
(2)
Resolvendo (2) para N e substituindo o resultado em (1), obtemos
N ! P sen 30 785 N
P cos 30° # 0,25(P sen 30° 785 N) ! 200 N P ! 535 N !
30°
80 kg
Um bloco de 80-kg está em repouso sobre um plano horizontal. Encontr
a intensidade da força P necessária para dar ao bloco uma aceleração d
2,5 m/s
2
para a direita. O coeficiente de atrito cinético entre o bloco e
plano é !k ! 0,25.
SOLUÇÃO
O peso do bloco é
W ! mg ! (80 kg)(9,81 m/s
2
) ! 785 N
!
P
30°
N
F
W ! 785 N
ma
m ! 80 kg
Notamos que F ! !kN ! 0,25N e que a ! 2,5 m/s
2
. Expressando que a
forças que atuam no bloco são equivalentes ao vetor ma, escrevemos
(1
(2
Resolvendo (2) para N e substituindo o resultado em (1), obtemos
N ! P sen 30 785 N
P cos 30° # 0,25(P sen 30° 785 N) ! 200 N P ! 535 N !
plano é !k ! 0,25.
SOLUÇÃO
O peso do bloco é
W ! mg ! (80 kg)(9,81 m/s
2
) ! 785 N
!
P
30°
N
F
W ! 785 N
ma
m ! 80 kg
Notamos que F ! !kN ! 0,25N e que a ! 2,5 m/s
2
. Expressando que as
forças que atuam no bloco são equivalentes ao vetor ma, escrevemos
(1)
(2)
Resolvendo (2) para N e substituindo o resultado em (1), obtemos
N ! P sen 30 785 N
P cos 30° # 0,25(P sen 30° 785 N) ! 200 N P ! 535 N !
plano é !k ! 0,25.
SOLUÇÃO
O peso do bloco é
W ! mg ! (80 kg)(9,81 m/s
2
) ! 785 N
!
P
30°
N
F
W ! 785 N
ma
m ! 80 kg
Notamos que F ! !kN ! 0,25N e que a ! 2,5 m/s
2
. Expressando que as
forças que atuam no bloco são equivalentes ao vetor ma, escrevemos
(1)
(2)
Resolvendo (2) para N e substituindo o resultado em (1), obtemos
N ! P sen 30 785 N
P cos 30° # 0,25(P sen 30° 785 N) ! 200 N P ! 535 N !
OLUÇÃO
peso do bloco é
W ! mg ! (80 kg)(9,81 m/s
2
) ! 785 N
!
P
30°
N
F
W ! 785 N
ma
m ! 80 kg
otamos que F ! !kN ! 0,25N e que a ! 2,5 m/s
2
. Expressando que as
rças que atuam no bloco são equivalentes ao vetor ma, escrevemos
(1)
(2)
esolvendo (2) para N e substituindo o resultado em (1), obtemos
N ! P sen 30 785 N
P cos 30° # 0,25(P sen 30° 785 N) ! 200 N P ! 535 N !
SOLUÇÃO
O peso do bloco é
W ! mg ! (80 kg)(9,81 m/s
2
) ! 785 N
!
P
30°
N
F
W ! 785 N
ma
m ! 80 kg
Notamos que F ! !kN ! 0,25N e que a ! 2,5 m/s
2
. Expressando que as
forças que atuam no bloco são equivalentes ao vetor ma, escrevemos
(1)
(2)
Resolvendo (2) para N e substituindo o resultado em (1), obtemos
N ! P sen 30 785 N
P cos 30° # 0,25(P sen 30° 785 N) ! 200 N P ! 535 N !
plano é !k ! 0,25.
SOLUÇÃO
O peso do bloco é
W ! mg ! (80 kg)(9,81 m/s
2
) ! 785 N
!
P
30°
N
F
W ! 785 N
ma
m ! 80 kg
Notamos que F ! !kN ! 0,25N e que a ! 2,5 m/s
2
. Expressand
forças que atuam no bloco são equivalentes ao vetor ma, escrevemo
Resolvendo (2) para N e substituindo o resultado em (1), obtemos
N ! P sen 30 785 N
P cos 30° # 0,25(P sen 30° 785 N) ! 200 N P ! 53
⟺
2,5 m/s
2
para a direita. O coeficiente de atrito cinético entre o bloco e o
plano é !k ! 0,25.
SOLUÇÃO
O peso do bloco é
W ! mg ! (80 kg)(9,81 m/s
2
) ! 785 N
!
P
30°
N
F
W ! 785 N
ma
m ! 80 kg
Notamos que F ! !kN ! 0,25N e que a ! 2,5 m/s
2
. Expressando que as
forças que atuam no bloco são equivalentes ao vetor ma, escrevemos
(1)
(2)
Resolvendo (2) para N e substituindo o resultado em (1), obtemos
N ! P sen 30 785 N
P cos 30° # 0,25(P sen 30° 785 N) ! 200 N P ! 535 N !
10. ENAUTICA-DEM Profª Rosa Marat-Mendes 10
Problemas
12.12 Os dois blocos mostrados na figura estão originalmente em repouso.
Desprezando as massas das roldanas e o efeito do atrito nessas rolda-
nas e considerando que os coeficientes de atrito entre ambos o bloco
A e a superfície horizontal são !s ! 0,25 e !k ! 0,20, determine (a) a
aceleração de cada bloco e (b) a tração no cabo.
12.13 Os coeficientes de atrito entre a carga e o reboque de piso plano mos-
trado na figura são !s ! 0,40 e !k ! 0,30. Sabendo que velocidade
escalar do equipamento é 72 km/h, determine (a) a menor distância
na qual o equipamento pode ser parado se a carga não pode se movi-
mentar.
3 m
Figura P12.13
12.14 Um caminhão-baú está viajando a 96 km/h quando o motorista acio-
na os freios. Sabendo que as forças de frenagem do caminhão e do
baú são de 18 kN e 68 kN, respectivamente, determine (a) a distância
percorrida pelo caminhão-baú antes de ele chegar ao repouso e (b) o
componente horizontal da força no acoplamento entre o caminhão e
o baú enquanto eles estão diminuindo a velocidade.
8700 kg
7500 kg
A
25 kg
B
Figura P12.11 e P12.12
Fig. P12.10
12.11 The c
show
rig is
be br
12.12 A lig
brake
25 M
30 kN
befor
cars
12.13 The t
of th
block
(b) th
12.14 Solve
4 m
Fig. P12.11
A
200 lb
350 lb
B
30°
para engenheiros: dinâmica
12.15 O bloco A tem a massa de 40 kg e o bloco B tem a massa de 8 kg.
Os coeficientes de atrito entre todas as superfícies de contato são
!s ! 0,20 e !k ! 0,15. Se P! 0, determine (a) a aceleração do bloco
B e (b) a tração na corda.
12.16 O bloco A tem a massa de 40 kg e o bloco B tem a massa de 8 kg.
Os coeficientes de atrito entre todas as superfícies de contato são
!s ! 0,20 e !k ! 0,15. Se P! 40 N n, determine (a) a aceleração do
bloco B e (b) a tração na corda.
12.17 As caixas A e B estão em repouso sobre uma esteira transportadora
que está inicialmente em repouso. A esteira é ligada de repente num
sentido de movimento para cima, de modo que ocorre escorregamen-
to entre a esteira e as caixas. Sabendo que os coeficientes de atrito
cinético entre a esteira e as caixas são de (!k)A ! 0,30 e (!k)B ! 0,32,
B
A
P
749
A
A A 2100 lb
2200 lb
200 lb
200 lb
(1) (2) (3)
Fig. P12.15
12.16 Boxes A and B are at rest on a conveyor belt that is initially at
rest. The belt is suddenly started in an upward direction so that
slipping occurs between the belt and the boxes. Knowing that the
coefficients of kinetic friction between the belt and the boxes are
(µk)A 5 0.30 and (µk)B 5 0.32, determine the initial acceleration
of each box.
12.17 A 5000-lb truck is being used to lift a 1000-lb boulder B that is on
a 200-lb pallet A. Knowing the acceleration of the truck is 1 ft/s2
,
determine (a) the horizontal force between the tires and the ground,
(b) the force between the boulder and the pallet.
a
A
B
Fig. P12.17
12.18 Block A has a mass of 40 kg, and block B has a mass of 8 kg. The
coefficients of friction between all surfaces of contact are µs 5 0.20
and µk 5 0.15. If P 5 0, determine (a) the acceleration of block B,
(b) the tension in the cord.
12.19 Block A has a mass of 40 kg, and block B has a mass of 8 kg. The
coefficients of friction between all surfaces of contact are µs 5 0.20
and µk 5 0.15. If P 5 40 N, determine (a) the acceleration of block
B, (b) the tension in the cord.
A
B
100 lb
80 lb
15 °
Fig. P12.16
B
P
25 °
A
Fig. P12.18 and P12.19
bee87342_ch12_718-794.indd 749
bee87342_ch12_718-794.indd 749 11/26/14 11:44 AM
11/26/14 11:44 AM
11. ENAUTICA-DEM Profª Rosa Marat-Mendes 11
Problemas
ara engenheiros: dinâmica
12.32 O bloco B de 15 kg é sustentado pelo bloco A de 25 kg e está preso a
uma corda à qual é aplicada uma força horizontal de 225 N, tal como
mostra a figura. Desprezando o atrito, determine (a) a aceleração do
bloco A, (b) a aceleração do bloco B em relação a A.
A
B
15 kg
25 kg
25º
225 N
Figura P12.32
12.33 O bloco B de massa 10 kg repousa na superfície superior de uma
cunha de 22 kg como mostra na figura. Sabendo que o sistema é libe-
rado do repouso e desprezando o atrito, determine (a) a aceleração
de B, (b) a velocidade de B em relação a A em t ! 0,5 s.
716 Mecânica vetorial para engenheiros: dinâmica
12.32 O bloco B de 15 kg é sustentado pelo bloco A de 25 kg e está preso a
uma corda à qual é aplicada uma força horizontal de 225 N, tal como
mostra a figura. Desprezando o atrito, determine (a) a aceleração do
bloco A, (b) a aceleração do bloco B em relação a A.
A
B
15 kg
25 kg
25º
225 N
Figura P12.32
12.33 O bloco B de massa 10 kg repousa na superfície superior de uma
cunha de 22 kg como mostra na figura. Sabendo que o sistema é libe-
rado do repouso e desprezando o atrito, determine (a) a aceleração
de B, (b) a velocidade de B em relação a A em t ! 0,5 s.
12.34 Um painel deslizante de 40 kg é suportado pelos roletes em B e
C. Um contrapeso A de 25 kg é fixado por cabo como mostrado na
figura e, nos casos a e c, estão em contato com a borda vertical do
painel. Desprezando o atrito, determine em cada caso mostrado a
aceleração do painel e a tração na corda imediatamente depois do
sistema sair do repouso.
C
B
C
B
A A
C
B
A
B
30º
A
20º
Figura P12.33
Ŵ
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'
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12. ENAUTICA-DEM Profª Rosa Marat-Mendes 12
Problemas
Figura P12.46
12.47 Um trecho de uma pista de tobogã mostrada na figura está contido
em um plano vertical. As seções AB e CD têm raios de curvatura com
indicado e a seção BC é uma linha reta e forma um ângulo de 20° com
a horizontal. Sabendo que o coeficiente de atrito cinético entre o trenó
e a pista é 0,10 e que a velocidade escalar do trenó é 7 m/s em B, deter-
mine a componente tangencial da aceleração do trenó (a) exatamente
antes dele alcançar B, (b) exatamente depois dele passar por C.
A
D
18 m
40 m
B
C
12 m
!
Figura P12.46
12.47 Um trecho de uma pista de tobogã mostrada na figura está co
em um plano vertical. As seções AB e CD têm raios de curvatura
indicado e a seção BC é uma linha reta e forma um ângulo de 20°
a horizontal. Sabendo que o coeficiente de atrito cinético entre o
e a pista é 0,10 e que a velocidade escalar do trenó é 7 m/s em B, d
mine a componente tangencial da aceleração do trenó (a) exatam
antes dele alcançar B, (b) exatamente depois dele passar por C.
A
D
18 m
40 m
B
C
12 m
Figura P12.47