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Plano de Ensino
• Parte I: Estática
• Decomposição de vetores em duas dimensões
• Centro de gravidade de figuras planas
• Momentos de Inércia de figuras planas
• Vínculos
• Reações de Apoio
• Parte II: Oscilações
• Oscilações Simples
• Oscilações Amortecidas
• Oscilações Forçadas
• Ressonância
• Parte III: Ondas
• Conceitos Fundamentais
• Ondas transversais e longitudinais
• Princípio de superposição
• Interferência
• Ondas sonoras
• Bibliografia:
• R.C. Hibbeler, Estática, 12 Ed. Pearson, São Paulo (2011)
• Tipler e Mosca, Física para Cientistas e Engenheiros, vol 1, 6a Ed, LTC, São Paulo (2009).
Avaliação e Faltas
• AV1 Professor
• AV2
• AV3 + PI
• O curso é presencial! O limite é de 20 faltas! Faltas Reprovam!
• O aluno que perder uma das AV’s tem direito a uma prova substitutiva somente mediante a entrega de atestado médico na secretaria da universidade.
• O prazo para protocolar o atestado é de 48h.
Mecânica
• É o ramo da física que estuda os movimentos e também o repouso dos sistemas físicos (corpos, partículas) que estão sujeitos à ação de forças.
• Mecânica dos corpos rígidos
• Estática (corpos em repouso ou em movimento com velocidade constante)
• Dinâmica
• Mecânica dos corpos deformáveis
• Mecânica dos Fluidos
5
Grandezas Físicas Fundamentais e o Sistema Internacional de Unidades
• Comprimento (m) com Dimensão [L]
• Tempo (s) com Dimensão [T]
• Massa (kg) com Dimensão [M]
• Força (N=kg.m/s2) com Dimensão [MLT-2]
Definições
• Uma partícula possui massa mas tem suas dimensões desprezadas
• Um corpo rígido pode ser aproximado como uma combinação de um grande número de partículas que permanece fixas, umas em relação às outras tanto antes quanto depois da aplicação de forças. Desprezamos as deformações que ocorrem em decorrência destas forças.
• Força Concentrada representa o efeito de uma carga que age em um ponto de um corpo desde que a área de contato sobre a qual ela é aplicada seja pequena comparada com o tamanho total do corpo. Exemplo: Força de contato entre uma roda e o solo
Leis de Newton
• Uma partícula originalmente em repouso ou movendo-se em linha reta com velocidade constante tende a permanecer neste estado desde que não seja submetida a uma força externa resultante desbalanceada, ou seja, diferente de zero.
• Sob a ação de uma força resultante externa não nula, uma partícula sofre uma aceleração que possui a mesma direção da força aplicada e intensidade diretamente proporcional à força e inversamente proporcional `a sua massa.
• As forças mútuas de ação e reação entre duas partículas são iguais, opostas e colineares.
~F = m~a
Propriedades Gerais dos Vetores
• Na natureza, a grosso modo, existem grandezas físicas que podem ser:
• Grandezas Escalares
• São aquelas que são caracterizadas somente por um valor numérico.
• Exemplo: Energia, Trabalho, Temperatura, Corrente Elétrica, etc.
• Grandezas Vetoriais
• São aquelas que para serem caracterizadas necessitam de, além de um valor numérico, de uma orientação (direção e sentido)!
• Exemplo: Posição, Deslocamento, Velocidade, Aceleração, Forças, etc.
Soma e subtração geométricas com vetores
Quando existem mais de dois vetores podemos agrupá-los em qualquer ordem para efetuar a soma.
~A
~B
~A+ ~B
~B + ~A
( ~A+ ~B) + ~C
Subtração (geométrica) de vetores: Subtrair um vetor de outro é a mesma coisa que somar o primeiro ao oposto do segundo.
( ~B + ~C) + ~A ( ~A+ ~B) + ~C = ( ~B + ~C) + ~A
~A� ~B = ~A+ (� ~B)
~A� ~B
20 CH A P T E R 2 FO R C E VE C T O R S
2
FR ! F1 " F2
FRFR
F1 F1 F1
F2 F2
F2
(c)(b)(a)
v
Fig. 2–7
2.3 Vector Addition of Forces
Experimental evidence has shown that a force is a vector quantity sinceit has a specified magnitude, direction, and sense and it adds according tothe parallelogram law. Two common problems in statics involve eitherfinding the resultant force, knowing its components, or resolving a knownforce into two components. We will now describe how each of theseproblems is solved using the parallelogram law.
Finding a Resultant Force. The two component forces F1 and F2acting on the pin in Fig. 2–7a can be added together to form the resultantforce FR = F1 + F2, as shown in Fig. 2–7b. From this construction, or usingthe triangle rule, Fig. 2–7c, we can apply the law of cosines or the law ofsines to the triangle in order to obtain the magnitude of the resultantforce and its direction.
Finding the Components of a Force. Sometimes it isnecessary to resolve a force into two components in order to study itspulling or pushing effect in two specific directions. For example, inFig. 2–8a, F is to be resolved into two components along the twomembers, defined by the u and axes. In order to determine themagnitude of each component, a parallelogram is constructed first, bydrawing lines starting from the tip of F, one line parallel to u, and theother line parallel to . These lines then intersect with the and u axes,forming a parallelogram. The force components Fu and F are thenestablished by simply joining the tail of F to the intersection points onthe u and axes, Fig. 2–8b. This parallelogram can then be reduced to atriangle, which represents the triangle rule, Fig. 2–8c. From this, the law ofsines can then be applied to determine the unknown magnitudes of thecomponents.
v
v
vv
vFu
u
vFv
F
FR
F2F1
Using the parallelogram law force Fcaused by the vertical member can beresolved into components acting alongthe suspension cables a and b.
The parallelogram law must be used todetermine the resultant of the twoforces acting on the hook.
20 CH A P T E R 2 FO R C E VE C T O R S
2
FR ! F1 " F2
FRFR
F1 F1 F1
F2 F2
F2
(c)(b)(a)
v
Fig. 2–7
2.3 Vector Addition of Forces
Experimental evidence has shown that a force is a vector quantity sinceit has a specified magnitude, direction, and sense and it adds according tothe parallelogram law. Two common problems in statics involve eitherfinding the resultant force, knowing its components, or resolving a knownforce into two components. We will now describe how each of theseproblems is solved using the parallelogram law.
Finding a Resultant Force. The two component forces F1 and F2acting on the pin in Fig. 2–7a can be added together to form the resultantforce FR = F1 + F2, as shown in Fig. 2–7b. From this construction, or usingthe triangle rule, Fig. 2–7c, we can apply the law of cosines or the law ofsines to the triangle in order to obtain the magnitude of the resultantforce and its direction.
Finding the Components of a Force. Sometimes it isnecessary to resolve a force into two components in order to study itspulling or pushing effect in two specific directions. For example, inFig. 2–8a, F is to be resolved into two components along the twomembers, defined by the u and axes. In order to determine themagnitude of each component, a parallelogram is constructed first, bydrawing lines starting from the tip of F, one line parallel to u, and theother line parallel to . These lines then intersect with the and u axes,forming a parallelogram. The force components Fu and F are thenestablished by simply joining the tail of F to the intersection points onthe u and axes, Fig. 2–8b. This parallelogram can then be reduced to atriangle, which represents the triangle rule, Fig. 2–8c. From this, the law ofsines can then be applied to determine the unknown magnitudes of thecomponents.
v
v
vv
vFu
u
vFv
F
FR
F2F1
Using the parallelogram law force Fcaused by the vertical member can beresolved into components acting alongthe suspension cables a and b.
The parallelogram law must be used todetermine the resultant of the twoforces acting on the hook.
20 CH A P T E R 2 FO R C E VE C T O R S
2
FR ! F1 " F2
FRFR
F1 F1 F1
F2 F2
F2
(c)(b)(a)
v
Fig. 2–7
2.3 Vector Addition of Forces
Experimental evidence has shown that a force is a vector quantity sinceit has a specified magnitude, direction, and sense and it adds according tothe parallelogram law. Two common problems in statics involve eitherfinding the resultant force, knowing its components, or resolving a knownforce into two components. We will now describe how each of theseproblems is solved using the parallelogram law.
Finding a Resultant Force. The two component forces F1 and F2acting on the pin in Fig. 2–7a can be added together to form the resultantforce FR = F1 + F2, as shown in Fig. 2–7b. From this construction, or usingthe triangle rule, Fig. 2–7c, we can apply the law of cosines or the law ofsines to the triangle in order to obtain the magnitude of the resultantforce and its direction.
Finding the Components of a Force. Sometimes it isnecessary to resolve a force into two components in order to study itspulling or pushing effect in two specific directions. For example, inFig. 2–8a, F is to be resolved into two components along the twomembers, defined by the u and axes. In order to determine themagnitude of each component, a parallelogram is constructed first, bydrawing lines starting from the tip of F, one line parallel to u, and theother line parallel to . These lines then intersect with the and u axes,forming a parallelogram. The force components Fu and F are thenestablished by simply joining the tail of F to the intersection points onthe u and axes, Fig. 2–8b. This parallelogram can then be reduced to atriangle, which represents the triangle rule, Fig. 2–8c. From this, the law ofsines can then be applied to determine the unknown magnitudes of thecomponents.
v
v
vv
vFu
u
vFv
F
FR
F2F1
Using the parallelogram law force Fcaused by the vertical member can beresolved into components acting alongthe suspension cables a and b.
The parallelogram law must be used todetermine the resultant of the twoforces acting on the hook.
20 CH A P T E R 2 FO R C E VE C T O R S
2
FR ! F1 " F2
FRFR
F1 F1 F1
F2 F2
F2
(c)(b)(a)
v
Fig. 2–7
2.3 Vector Addition of Forces
Experimental evidence has shown that a force is a vector quantity sinceit has a specified magnitude, direction, and sense and it adds according tothe parallelogram law. Two common problems in statics involve eitherfinding the resultant force, knowing its components, or resolving a knownforce into two components. We will now describe how each of theseproblems is solved using the parallelogram law.
Finding a Resultant Force. The two component forces F1 and F2acting on the pin in Fig. 2–7a can be added together to form the resultantforce FR = F1 + F2, as shown in Fig. 2–7b. From this construction, or usingthe triangle rule, Fig. 2–7c, we can apply the law of cosines or the law ofsines to the triangle in order to obtain the magnitude of the resultantforce and its direction.
Finding the Components of a Force. Sometimes it isnecessary to resolve a force into two components in order to study itspulling or pushing effect in two specific directions. For example, inFig. 2–8a, F is to be resolved into two components along the twomembers, defined by the u and axes. In order to determine themagnitude of each component, a parallelogram is constructed first, bydrawing lines starting from the tip of F, one line parallel to u, and theother line parallel to . These lines then intersect with the and u axes,forming a parallelogram. The force components Fu and F are thenestablished by simply joining the tail of F to the intersection points onthe u and axes, Fig. 2–8b. This parallelogram can then be reduced to atriangle, which represents the triangle rule, Fig. 2–8c. From this, the law ofsines can then be applied to determine the unknown magnitudes of thecomponents.
v
v
vv
vFu
u
vFv
F
FR
F2F1
Using the parallelogram law force Fcaused by the vertical member can beresolved into components acting alongthe suspension cables a and b.
The parallelogram law must be used todetermine the resultant of the twoforces acting on the hook.
Regra do Paralelogramo
Decomposição de vetores em 2D
sen(✓) =ay|~a|
cos(✓) =a
x
|~a|a
x
= |~a|cos✓
ay = |~a|sen✓
Exercício: Determine o ângulo theta e o módulo do vetor a a partir das equações anteriores.
tan(✓) =ay
ax
✓ = arctan(ay
ax
)
Módulo de um vetor: a2 = a2x
+ a2y
! a =q
a2x
+ a2y
18 | C H A P T E R 1 Measurement and Vectors
vector is called resolving the vector into its components. The components of a vec-tor along the x, y, and z directions, illustrated in Figure 1-13 for a vector in the xyplane, are called the rectangular (or Cartesian) components. Note that the compo-nents of a vector do depend on the coordinate system used, although the vector it-self does not.
We can use right-triangle geometry to find the rectangular components of a vec-tor. If is the angle measured counterclockwise* from the direction to the di-rection of (see Figure 1-13), then
1-2x COMPONENT OF A VECTOR
and
1-3y COMPONENT OF A VECTOR
where A is the magnitude of If we know and we can find the angle from
1-4
and the magnitude A from the Pythagorean theorem:
1-5a
In three dimensions,
1-5b
Components can be positive or negative. The x component of a vector is posi-tive if the x coordinate of an ant as it walks from the tail to the head of the vectorincreases. Thus, if points in the positive x direction, then is positive, and if points in the negative x direction, then is negative.
It is important to note that in Equation 1-4, the inverse tangent function is mul-tiple valued. This issue is clarified in Example 1-9.
PRACTICE PROBLEM 1-6A car travels 20.0 km in a direction north of west. Let east be the direction andnorth be the direction, as in Figure 1-14. Find the x and y components of the dis-placement vector of the car.
Once we have resolved a vector into its components, we can manipulate the in-dividual components. Consider two vectors and that lie in the plane. The rectangular components of each vector and those of the sum areshown in Figure 1-15. We see that the rectangular components of each vector andthose of the sum are equivalent to the two component equations
1-6aand
1-6b
In other words, the sum of the x components equals the x component of the re-sultant, and the sum of the y components equals the y component of the resultant.The angle and magnitude of the resultant vector can be found using Equations 1-4and 1-5a, respectively.
Cy ! Ay " By
Cx ! Ax " Bx
CS
! AS
" BS
CS
! AS
" BS
xyBS
AS
"y"x30.0°
Ax
AS
AxAS
A ! 3Ax2 " Ay2 " Az
2
A ! 3Ax2 " Ay2
tan u !AyAx
u ! tan#1AyAx
uAyAx
AS
.
Ay ! A sinu
Ax ! A cosu
AS
"xu
N
S
EW30.0
2.0 km
°
A
F I G U R E 1 - 1 4
y
x
Ax
Ay
Bx
By
Cx
CyA
B
C
F I G U R E 1 - 1 5
* This assumes the direction is counterclockwise from the direction."x90°"y
y
xAx
θ
AyAy = A sin θ
Ax = A cos θ
A
F I G U R E 1 - 1 3 The rectangularcomponents of a vector. is the angle betweenthe direction of the vector and the direction. The angle is positive if it ismeasured counterclockwise from the direction, as shown.
"x
"xu
~A = Ax
ı̂+Ay
|̂
A =qA2
x
+A2y
Lembrando ainda queA
x
= Acos(✓)
A
y
= Asen(✓)
Adição e Subtração Algébrica de Vetores no plano
~A = Ax
ı̂+Ay
|̂
~B = Bx
ı̂+By
|̂
Adição
Subtração
~A+ ~B = (Ax
+Bx
)̂ı+ (Ay
+By
)|̂
~A� ~B = (Ax
�Bx
)̂ı+ (Ay
�By
)|̂
Decomposição de Forças
2.3 VECTOR ADDITION OF FORCES 21
2
F
u
(b)
F
FuFu
(c)
F
u
(a)
v v
Fv
Fv
Fig. 2–8
Addition of Several Forces. If more than two forces are to beadded, successive applications of the parallelogram law can be carriedout in order to obtain the resultant force. For example, if three forces F1,F2, F3 act at a point O, Fig. 2–9, the resultant of any two of the forces isfound, say, F1 + F2—and then this resultant is added to the third force,yielding the resultant of all three forces; i.e., FR = (F1 + F2)+F3. Using the parallelogram law to add more than two forces, as shown here, oftenrequires extensive geometric and trigonometric calculation to determinethe numerical values for the magnitude and direction of the resultant.Instead, problems of this type are easily solved by using the “rectangular-component method,” which is explained in Sec. 2.4.
F1
FRF1 ! F2
F33
F1F3
F2
F1
F2
F1 ! F2 FR
F3O
Fig. 2–9
The resultant force on the hookrequires the addition of , then thisresultant is added to .F3
F1 + F2
FR
Como desenhar as componentes de F? 2.3 VECTOR ADDITION OF FORCES 21
2
F
u
(b)
F
FuFu
(c)
F
u
(a)
v v
Fv
Fv
Fig. 2–8
Addition of Several Forces. If more than two forces are to beadded, successive applications of the parallelogram law can be carriedout in order to obtain the resultant force. For example, if three forces F1,F2, F3 act at a point O, Fig. 2–9, the resultant of any two of the forces isfound, say, F1 + F2—and then this resultant is added to the third force,yielding the resultant of all three forces; i.e., FR = (F1 + F2)+F3. Using the parallelogram law to add more than two forces, as shown here, oftenrequires extensive geometric and trigonometric calculation to determinethe numerical values for the magnitude and direction of the resultant.Instead, problems of this type are easily solved by using the “rectangular-component method,” which is explained in Sec. 2.4.
F1
FRF1 ! F2
F33
F1F3
F2
F1
F2
F1 ! F2 FR
F3O
Fig. 2–9
The resultant force on the hookrequires the addition of , then thisresultant is added to .F3
F1 + F2
FR
O paralelogramo é reduzido então a um triângulo no qual a lei dos senos pode ser aplicada para se determinar as intensidades desconhecidas das componentes.
2.3 VECTOR ADDITION OF FORCES 21
2
F
u
(b)
F
FuFu
(c)
F
u
(a)
v v
Fv
Fv
Fig. 2–8
Addition of Several Forces. If more than two forces are to beadded, successive applications of the parallelogram law can be carriedout in order to obtain the resultant force. For example, if three forces F1,F2, F3 act at a point O, Fig. 2–9, the resultant of any two of the forces isfound, say, F1 + F2—and then this resultant is added to the third force,yielding the resultant of all three forces; i.e., FR = (F1 + F2)+F3. Using the parallelogram law to add more than two forces, as shown here, oftenrequires extensive geometric and trigonometric calculation to determinethe numerical values for the magnitude and direction of the resultant.Instead, problems of this type are easily solved by using the “rectangular-component method,” which is explained in Sec. 2.4.
F1
FRF1 ! F2
F33
F1F3
F2
F1
F2
F1 ! F2 FR
F3O
Fig. 2–9
The resultant force on the hookrequires the addition of , then thisresultant is added to .F3
F1 + F2
FR
Alguma relações importantes
22 CH A P T E R 2 FO R C E VE C T O R S
2
A
C
B
b
(c)
c
a
Sine law:
sin a sin b sin cA B! ! C
Cosine law:C ! A2 " B2 # 2AB cos c
FR
F1
F2
F
Fu
u
(b)
(a)
v
Fv
Fig. 2–10
Procedure for Analysis
Problems that involve the addition of two forces can be solved asfollows:
Parallelogram Law.
• Two “component” forces F1 and F2 in Fig. 2–10a add according tothe parallelogram law, yielding a resultant force FR that forms thediagonal of the parallelogram.
• If a force F is to be resolved into components along two axes uand , Fig. 2–10b, then start at the head of force F and constructlines parallel to the axes, thereby forming the parallelogram. Thesides of the parallelogram represent the components, F and F .
• Label all the known and unknown force magnitudes and theangles on the sketch and identify the two unknowns as themagnitude and direction of FR, or the magnitudes of itscomponents.
Trigonometry.
• Redraw a half portion of the parallelogram to illustrate thetriangular head-to-tail addition of the components.
• From this triangle, the magnitude of the resultant force can bedetermined using the law of cosines, and its direction isdetermined from the law of sines. The magnitudes of two forcecomponents are determined from the law of sines. The formulasare given in Fig. 2–10c.
vu
v
Important Points
• A scalar is a positive or negative number.
• A vector is a quantity that has a magnitude, direction, and sense.
• Multiplication or division of a vector by a scalar will change themagnitude of the vector.The sense of the vector will change if thescalar is negative.
• As a special case, if the vectors are collinear, the resultant isformed by an algebraic or scalar addition.
A
sen(a)=
B
sen(b)=
C
sen(c)
C =p
A
2 +B
2 � 2ABcos(b)
Lei dos Senos
Lei dos Cosenos
Exemplo 2.1 (Hibbeler, 12a Ed.)
O gancho na figura abaixo está sujeito a duas forças F1 e F2. Determine a intensidade e a direção da força resultante.
The screw eye in Fig. 2–11a is subjected to two forces, F1 and F2.Determine the magnitude and direction of the resultant force.
SOLUTIONParallelogram Law. The parallelogram is formed by drawing a linefrom the head of F1 that is parallel to F2, and another line from thehead of F2 that is parallel to F1.The resultant force FR extends to wherethese lines intersect at point A, Fig. 2–11b. The two unknowns are themagnitude of FR and the angle (theta).
Trigonometry. From the parallelogram, the vector triangle isconstructed, Fig. 2–11c. Using the law of cosines
Ans.
Applying the law of sines to determine ,
Thus, the direction (phi) of FR, measured from the horizontal, is
Ans.
NOTE: The results seem reasonable, since Fig. 2–11b shows FR to havea magnitude larger than its components and a direction that isbetween them.
f = 39.8° + 15.0° = 54.8°
f
u = 39.8°
sin u = 150 N212.6 N
(sin 115º)150 Nsin u
= 212.6 Nsin 115°
u
= 213 N
= 210 000 + 22 500 - 30 000(-0.4226) = 212.6 N
FR = 2(100 N)2 + (150 N)2 - 2(100 N)(150 N) cos 115°
u
EXAMPLE 2.1
2.3 VECTOR ADDITION OF FORCES 23
F1 ! 100 N
F2 ! 150 N10"
15"
(a)
Fig. 2–11
(c)
FR 150 N
100 N15"
115"u
f
2
FR
90" # 25" ! 65"
10"
15"
100 N
A
65"115"
150 N
(b)
! 115"360 # 2(65")
2
u
Solução: Desenhando o paralelogramo e as forças que o geram, determinamos os respectivos ângulos (geometria)The screw eye in Fig. 2–11a is subjected to two forces, F1 and F2.
Determine the magnitude and direction of the resultant force.
SOLUTIONParallelogram Law. The parallelogram is formed by drawing a linefrom the head of F1 that is parallel to F2, and another line from thehead of F2 that is parallel to F1.The resultant force FR extends to wherethese lines intersect at point A, Fig. 2–11b. The two unknowns are themagnitude of FR and the angle (theta).
Trigonometry. From the parallelogram, the vector triangle isconstructed, Fig. 2–11c. Using the law of cosines
Ans.
Applying the law of sines to determine ,
Thus, the direction (phi) of FR, measured from the horizontal, is
Ans.
NOTE: The results seem reasonable, since Fig. 2–11b shows FR to havea magnitude larger than its components and a direction that isbetween them.
f = 39.8° + 15.0° = 54.8°
f
u = 39.8°
sin u = 150 N212.6 N
(sin 115º)150 Nsin u
= 212.6 Nsin 115°
u
= 213 N
= 210 000 + 22 500 - 30 000(-0.4226) = 212.6 N
FR = 2(100 N)2 + (150 N)2 - 2(100 N)(150 N) cos 115°
u
EXAMPLE 2.1
2.3 VECTOR ADDITION OF FORCES 23
F1 ! 100 N
F2 ! 150 N10"
15"
(a)
Fig. 2–11
(c)
FR 150 N
100 N15"
115"u
f
2
FR
90" # 25" ! 65"
10"
15"
100 N
A
65"115"
150 N
(b)
! 115"360 # 2(65")
2
u
The screw eye in Fig. 2–11a is subjected to two forces, F1 and F2.Determine the magnitude and direction of the resultant force.
SOLUTIONParallelogram Law. The parallelogram is formed by drawing a linefrom the head of F1 that is parallel to F2, and another line from thehead of F2 that is parallel to F1.The resultant force FR extends to wherethese lines intersect at point A, Fig. 2–11b. The two unknowns are themagnitude of FR and the angle (theta).
Trigonometry. From the parallelogram, the vector triangle isconstructed, Fig. 2–11c. Using the law of cosines
Ans.
Applying the law of sines to determine ,
Thus, the direction (phi) of FR, measured from the horizontal, is
Ans.
NOTE: The results seem reasonable, since Fig. 2–11b shows FR to havea magnitude larger than its components and a direction that isbetween them.
f = 39.8° + 15.0° = 54.8°
f
u = 39.8°
sin u = 150 N212.6 N
(sin 115º)150 Nsin u
= 212.6 Nsin 115°
u
= 213 N
= 210 000 + 22 500 - 30 000(-0.4226) = 212.6 N
FR = 2(100 N)2 + (150 N)2 - 2(100 N)(150 N) cos 115°
u
EXAMPLE 2.1
2.3 VECTOR ADDITION OF FORCES 23
F1 ! 100 N
F2 ! 150 N10"
15"
(a)
Fig. 2–11
(c)
FR 150 N
100 N15"
115"u
f
2
FR
90" # 25" ! 65"
10"
15"
100 N
A
65"115"
150 N
(b)
! 115"360 # 2(65")
2
u
FR =qF
21 + F
22 � 2F1.F2.cos(115)
=p(100)2 + (150)2 � 2.100.150.cos(115)
= 213N
Falta determinar a direção desta força resultante, ou seja, o ângulo ø da figura. Para isto, vamos utilizar a lei dos senos para o triângulo formado pela força resultante e suas componentes.
The screw eye in Fig. 2–11a is subjected to two forces, F1 and F2.Determine the magnitude and direction of the resultant force.
SOLUTIONParallelogram Law. The parallelogram is formed by drawing a linefrom the head of F1 that is parallel to F2, and another line from thehead of F2 that is parallel to F1.The resultant force FR extends to wherethese lines intersect at point A, Fig. 2–11b. The two unknowns are themagnitude of FR and the angle (theta).
Trigonometry. From the parallelogram, the vector triangle isconstructed, Fig. 2–11c. Using the law of cosines
Ans.
Applying the law of sines to determine ,
Thus, the direction (phi) of FR, measured from the horizontal, is
Ans.
NOTE: The results seem reasonable, since Fig. 2–11b shows FR to havea magnitude larger than its components and a direction that isbetween them.
f = 39.8° + 15.0° = 54.8°
f
u = 39.8°
sin u = 150 N212.6 N
(sin 115º)150 Nsin u
= 212.6 Nsin 115°
u
= 213 N
= 210 000 + 22 500 - 30 000(-0.4226) = 212.6 N
FR = 2(100 N)2 + (150 N)2 - 2(100 N)(150 N) cos 115°
u
EXAMPLE 2.1
2.3 VECTOR ADDITION OF FORCES 23
F1 ! 100 N
F2 ! 150 N10"
15"
(a)
Fig. 2–11
(c)
FR 150 N
100 N15"
115"u
f
2
FR
90" # 25" ! 65"
10"
15"
100 N
A
65"115"
150 N
(b)
! 115"360 # 2(65")
2
u
Finalmente, da figura temos
Exemplos Resolvidos em sala (Hibbeler): 2.2, 2.3 e 2.4
150
sen(✓)=
FR
sen(115)✓ = 39, 8o
� = 39, 8o + 15o = 54, 8o
