17
Estática dos Sólidos Prof. Ettore Baldini-Neto [email protected]

Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

Embed Size (px)

Citation preview

Page 1: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

Estática dos Sólidos

Prof. Ettore [email protected]

Page 2: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

Plano de Ensino

• Equilíbrio de uma partícula

• Diagrama de Corpo Livre, Sistemas de Forças 2D e 3D

• Resultante de um sistema de forças (2D e 3D)

• Momento: formulações escalar e vetorial

• Princípio dos momentos

• Momento em relação a um eixo, momento binário

• Propriedades Geométricas de Figuras Planas

• Carregamentos

• Equilíbrio de um corpo rígido

• Diagrama de Corpo Livre, equações de equilíbrio

• Análise Estrutural

• (Treliças simples, método dos nós...)

Page 3: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

Bibliografia

• Hibbeler, R. C. Estática 10aed/12aed, Pearson SP (2005/2011)

• Schmidt, R. J.; Boresi, A. P., Estática: Thomson, (2003)

• Bedford & Fowler; Engineering mechanics-Statics 3, NJ, Prentice Hall (2002).

• Beer, F. P., Johnston Jr, E. R., Mecânica vetorial para engenheiros: estátcia 5ed. SP Makron Books, (1991).

Page 4: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

Avaliação e Faltas

• AV1 Professor

• AV2 integrada

• AV3 projeto integrador

• 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.

Page 5: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

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

Page 6: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

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]

Page 7: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

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

Page 8: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

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

Page 9: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

Popriedades 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.

Page 10: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

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

Page 11: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

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

Page 12: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

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

Page 13: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

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

Page 14: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

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)

Lei dos Senos

Lei dos Cosenos

C =p

A

2 +B

2 � 2ABcos(c)

Page 15: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

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

Page 16: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

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

Page 17: Estática dos Sólidos - estatica2015.weebly.comestatica2015.weebly.com/uploads/3/7/5/7/37571293/aula01-estática.pdf · • Momento: formulações escalar e vetorial • Princípio

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