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![Page 1: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/1.jpg)
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Design of Machinery An Introduction to the Synthesis and Analysis of
Mechanisms and Machines Fourth Edition
Robert L. Norton
Chapter 4
Image Slides
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No se puede mostrar la imagen. Puede que su equipo no tenga suficiente memoria para abrir la imagen o que ésta esté dañada. Reinicie el equipo y, a continuación, abra el archivo de nuevo. Si sigue apareciendo la x roja, puede que tenga que borrar la imagen e insertarla de nuevo.
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No se puede mostrar la imagen. Puede que su equipo no tenga suficiente memoria para abrir la imagen o que ésta esté dañada. Reinicie el equipo y, a continuación, abra el archivo de nuevo. Si sigue apareciendo la x roja, puede que tenga que borrar la imagen e insertarla de nuevo.
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R1 +R4 = RSR2 +R3 = RS
R1 = diR2 = acos(θ2 )i + asin(θ2 )j
R3 = bcos(θ3)i + bsin(θ3)j R4 = cj
R2 +R3 = R1 +R4
![Page 11: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/11.jpg)
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R2 +R3 = R1 +R4
acos(θ2 )+ bcos(θ3) = d
asin(θ2 )+ bsin(θ3) = c
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R2 +R3 = R1 +R4
f1(θ3,d) = acos(θ2 )+ bcos(θ3)− d = 0
f2 (θ3,d) = asin(θ2 )+ bsin(θ3)− c = 0
![Page 13: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/13.jpg)
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R2 +R3 = R1 +R4
f1(θ3,n+1,dn+1) ! f1(θ3,n ,dn )+
∂ f1(θ3,n ,dn )∂θ3
(θ3,n+1 −θ3,n )+∂ f1(θ3,n ,dn )
∂d(dn+1 − d3,n )
f2 (θ3,n+1,dn+1) ! f2 (θ3,n ,dn )+
∂ f2 (θ3,n ,dn )∂θ3
(θ3,n+1 −θ3,n )+∂ f2 (θ3,n ,dn )
∂d(dn+1 − d3,n )
![Page 14: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/14.jpg)
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∂ f1(θ3,n ,dn )∂θ3
∂ f1(θ3,n ,dn )∂d
∂ f2 (θ3,n ,dn )∂θ3
∂ f2 (θ3,n ,dn )∂d
⎡
⎣
⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥
(θ3,n+1 −θ3,n )(dn+1 − d3,n )
⎡
⎣⎢⎢
⎤
⎦⎥⎥= −
f1(θ3,n ,dn )f2 (θ3,n ,dn )
⎡
⎣⎢⎢
⎤
⎦⎥⎥
![Page 15: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/15.jpg)
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−bsin(θ3,n ) −1bcos(θ3,n ) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
(θ3,n+1 −θ3,n )(dn+1 − dn )
⎡
⎣⎢⎢
⎤
⎦⎥⎥= −
f1(θ3,n ,dn )f2 (θ3,n ,dn )
⎡
⎣⎢⎢
⎤
⎦⎥⎥
f1(θ3,n ,d) = acos(θ2 )+ bcos(θ3,n )− dn = 0
f2 (θ3,n ,dn ) = asin(θ2 )+ bsin(θ3,n )− c = 0
![Page 16: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/16.jpg)
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−bsin(30) −1bcos(30) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
(θ3,1 −θ3,0 )(d1 − d0 )
⎡
⎣⎢⎢
⎤
⎦⎥⎥= −
f1(θ3,0 ,d0 )f2 (θ3,0 ,d0 )
⎡
⎣⎢⎢
⎤
⎦⎥⎥
θ2 = 61º
f2 (θ3,0 ,d0 ) = asin(61)+ bsin(30)− (5 + 5 3)
a = 10b = 10
c = 10sin(60)+10cos(30) = 5 + 5 3θ3,0 = 30º
d0 = 10cos(60)+10cos(30) = 5 3 + 5
f1(θ3,0 ,d0 ) = acos(61)+ bcos(30)− (5 + 5 3)
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a = 10b = 10
c = 10sin(60)+10cos(30) = 5 + 5 3θ3,0 = 30º
d0 = 10cos(60)+10cos(30) = 5 3 + 5
1 2 3 4 theta3 29.43140546692
6996 29.433015864732752
29.433015877501717
29.433015877501717
d 13.557969473862578
13.557404108920629
13.557404104385316
13.557404104385316
θ2 = 61º
![Page 18: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/18.jpg)
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a = 10b = 10
c = 10sin(60)+10cos(30) = 5 + 5 3θ3,0 = 29.433015877501717ºd0 = 13.557404104385316
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a = 10b = 10
c = 10sin(60)+10cos(30) = 5 + 5 3θ3,0 = 29.433015877501717ºd0 = 13.557404104385316
1 2 3 4 theta3 28.88515064025
2387 28.886612382393817
28.886612392681030
28.886612392681030
d 13.451011999239871
13.450489889694857
13.450489885978037
13.450489885978035
θ2 = 62º
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a = 10b = 10
c = 10sin(60)+10cos(30) = 5 + 5 3θ3,0 = 29.433015877501717ºd0 = 13.557404104385316
θ2 = 62º
![Page 21: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/21.jpg)
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f1(θ3,d) = acos(θ2 )+ bcos(θ3)− d = 0
f2 (θ3,d) = asin(θ2 )+ bsin(θ3)− c = 0
!f1 = −asin(θ2 ) !θ2 − bsin(θ3) !θ3 − !d = 0
!f2 = acos(θ2 ) !θ2 + bcos(θ3) !θ3 = 0
![Page 22: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/22.jpg)
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!f1 = −asin(θ2 ) !θ2 − bsin(θ3) !θ3 − !d = 0
!f2 = acos(θ2 ) !θ2 + bcos(θ3) !θ3 = 0
asin(θ2 ) !θ2 = −bsin(θ3) !θ3 − !d
−acos(θ2 ) !θ2 = bcos(θ3) !θ3
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−bsin(θ3) −1bcos(θ3) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
asin(θ2 )−acos(θ2 )
⎡
⎣⎢⎢
⎤
⎦⎥⎥!θ2
![Page 24: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/24.jpg)
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−bsin(θ3) −1bcos(θ3) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
asin(θ2 )−acos(θ2 )
⎡
⎣⎢⎢
⎤
⎦⎥⎥!θ2
a = 10b = 10
c = 5 + 5 3θ3 = 30º
d = 5 3 + 5θ2 = 60º
−10sin(30) −110cos(30) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
10sin(60)−10cos(60)
⎡
⎣⎢⎢
⎤
⎦⎥⎥!θ2
![Page 25: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/25.jpg)
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a = 10b = 10
c = 5 + 5 3θ3 = 30º
d = 5 3 + 5θ2 = 60º
−10sin(30) −110cos(30) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
10sin(60)−10cos(60)
⎡
⎣⎢⎢
⎤
⎦⎥⎥!θ2
!θ2 = 2π
radss
−10sin(30) −110cos(30) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
10sin(60)−10cos(60)
⎡
⎣⎢⎢
⎤
⎦⎥⎥2π
![Page 26: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/26.jpg)
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a = 10b = 10
c = 5 + 5 3θ3 = 30º
d = 5 3 + 5θ2 = 60º
!θ2 = 2π
radss
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
−10sin(30) −110cos(30) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
−110sin(60)−10cos(60)
⎡
⎣⎢⎢
⎤
⎦⎥⎥2π
![Page 27: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/27.jpg)
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−bsin(θ3) −1bcos(θ3) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
asin(θ2 )−acos(θ2 )
⎡
⎣⎢⎢
⎤
⎦⎥⎥!θ2
θ2 = 61º
−10sin(29.433015877501717º ) −110cos(29.433015877501717º ) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
10sin(61)−10cos(61)
⎡
⎣⎢⎢
⎤
⎦⎥⎥!θ2
a = 10b = 10
c = 5 + 5 3θ3 = 29.433015877501717ºd = 13.557404104385316
![Page 28: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/28.jpg)
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a = 10b = 10
c = 5 + 5 3θ3 = 29.433015877501717ºd = 13.557404104385316
θ2 = 61º
!θ2 = 2π
radss
−10sin(29.433015877501717º ) −110cos(29.433015877501717º ) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
10sin(61)−10cos(61)
⎡
⎣⎢⎢
⎤
⎦⎥⎥!θ2
−10sin(29.433015877501717º ) −110cos(29.433015877501717º ) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
10sin(61)−10cos(61)
⎡
⎣⎢⎢
⎤
⎦⎥⎥2π
![Page 29: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/29.jpg)
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a = 10b = 10c = 13.557404104385316θ3 = 29.433015877501717ºd = 13.557404104385316
θ2 = 61º
!θ2 = 2π
radss
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
−10sin(29.433015877501717º ) −110cos(29.433015877501717º ) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
−110sin(61)−10cos(61)
⎡
⎣⎢⎢
⎤
⎦⎥⎥2π
![Page 30: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/30.jpg)
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a = 10b = 10c = 13.557404104385316θ3 = 29.433015877501717ºd = 13.557404104385316
θ2 = 61º
!θ2 = 2π
radss
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥= -3.497234014429895
-37.773686031467101)⎡
⎣⎢⎢
⎤
⎦⎥⎥
![Page 31: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/31.jpg)
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−bsin(θ3) −1bcos(θ3) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
asin(θ2 )−acos(θ2 )
⎡
⎣⎢⎢
⎤
⎦⎥⎥!θ2
θ2 = 62º
−10sin(28.886612392681030º ) −110cos(28.886612392681030º ) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
10sin(62)−10cos(62)
⎡
⎣⎢⎢
⎤
⎦⎥⎥!θ2
a = 10b = 10
c = 5 + 5 3θ3 = 28.886612392681030d = 13.450489885978035
![Page 32: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/32.jpg)
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a = 10b = 10
c = 5 + 5 3θ3 = 28.886612392681030d = 13.450489885978035
θ2 = 62º
!θ2 = 2π
radss
−10sin(28.886612392681030º ) −110cos(28.886612392681030º ) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
10sin(62)−10cos(62)
⎡
⎣⎢⎢
⎤
⎦⎥⎥!θ2
−10sin(29.433015877501717º ) −110cos(29.433015877501717º ) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
10sin(62)−10cos(62)
⎡
⎣⎢⎢
⎤
⎦⎥⎥2π
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a = 10b = 10
c = 5 + 5 3θ3 = 28.886612392681030ºd = 13.450489885978035
θ2 = 62º
!θ2 = 2π
radss
!θ3!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
−10sin(28.886612392681030º ) −110cos(28.886612392681030º ) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
−110sin(62)−10cos(62)
⎡
⎣⎢⎢
⎤
⎦⎥⎥2π
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a = 10b = 10
c = 5 + 5 3θ3 = 28.886612392681030ºd = 13.450489885978035
θ2 = 62º
!θ2 = 2π
radss
!θ3
!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
-3.368950293230393 -39.202582096901800⎡
⎣⎢
⎤
⎦⎥
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!f1 = −asin(θ2 ) !θ2 − bsin(θ3) !θ3 − !d = 0
!f2 = acos(θ2 ) !θ2 + bcos(θ3) !θ3 = 0
!!f1 = −acos(θ2 ) !θ2
2 − asin(θ2 )!!θ2 − bcos(θ3) !θ32 − bsin(θ3)!!θ3 − !!d = 0
!f2 = −asin(θ2 ) !θ2
2 + acos(θ2 )!!θ2 − bsin(θ3) !θ32 + bcos(θ3)!!θ3 = 0
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!!f1 = −acos(θ2 ) !θ2
2 − asin(θ2 )!!θ2 − bcos(θ3) !θ32 − bsin(θ3)!!θ3 − !!d = 0
!f2 = −asin(θ2 ) !θ2
2 + acos(θ2 )!!θ2 − bsin(θ3) !θ32 + bcos(θ3)!!θ3 = 0
bsin(θ3) 1−bcos(θ3) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!!θ3!!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
−acos(θ2 ) !θ22 − asin(θ2 )!!θ2 − bcos(θ3) !θ3
2
−asin(θ2 ) !θ22 + acos(θ2 )!!θ2 − bsin(θ3) !θ3
2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
![Page 37: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/37.jpg)
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bsin(θ3) 1−bcos(θ3) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!!θ3!!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
−acos(θ2 ) !θ22 − asin(θ2 )!!θ2 − bcos(θ3) !θ3
2
−asin(θ2 ) !θ22 + acos(θ2 )!!θ2 − bsin(θ3) !θ3
2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!!θ3!!d
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
bsin(θ3) 1−bcos(θ3) 0
⎡
⎣⎢⎢
⎤
⎦⎥⎥
−1−acos(θ2 ) !θ2
2 − asin(θ2 )!!θ2 − bcos(θ3) !θ32
−asin(θ2 ) !θ22 + acos(θ2 )!!θ2 − bsin(θ3) !θ3
2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
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a = 2b = 10
c = 2sin(60)+10sin(30) = 3 + 5
d = 2cos(60)+10cos(30) = 3 + 5θ2 = 60ºθ3 = 30º
!θ2 = 2πrads
!!θ2 = 0rads2
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REACCIONES Y MOMENTOS DE ACTUACIÓN
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FOx + FAx = m1acm1xFOy + FAy = m1acm1y +m1g−FAx = m2acm2x−FAy + FBy = m2acm2y +m2g
Suma de fuerzas para los cuerpos a y b
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L12sin(θ )FOx −
L12sin(θ )FAx −
L12cos(θ )FOy +
L12cos(θ )FAy = Icm1!!θ
− L22sin(β )FAx +
L22cos(β )FAy +
L22cos(β )FOy = Icm2 !!β
Suma de momentos en los cuerpos a y b
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1 1 0 0 0 00 0 1 1 0 00 −1 0 0 0 00 0 0 −1 1 0
L12sin(θ ) − L1
2sin(θ ) − L1
2cos(θ ) L1
2cos(θ ) 0 1
0 − L22sin(β ) 0 L2
2cos(β ) L2
2cos(β ) 0
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥
FOxFAxFOyFAyFByM
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥
=
m1acm1xm1acm1y +m1g
m2acm2xm2acm2y +m2g
Icm1!!θ
Icm2 !!β
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥
Suma de momentos en los cuerpos a y b
![Page 53: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/53.jpg)
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FOxFAxFOyFAyFByM
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥
=
1 1 0 0 0 00 0 1 1 0 00 −1 0 0 0 00 0 0 −1 1 0
L12sin(θ ) − L1
2sin(θ ) − L1
2cos(θ ) L1
2cos(θ ) 0 1
0 − L22sin(β ) 0 L2
2cos(β ) L2
2cos(β ) 0
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥
−1
m1acm1xm1acm1y +m1g
m2acm2xm2acm2y +m2g
Icm1!!θ
Icm2 !!β
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥
Suma de momentos en los cuerpos a y b
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FOx + FAx = m1acm1xFOy + FAy = m1acm1y +m1g−FAx + FBx = m2acm2x−FAy + FBy = m2acm2y +m2g−FBx + FCx = m3acm3x−FBy + FCy = m3acm2y +m3g
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L12sin(θ )FOx −
L12sin(θ )FAx −
L12cos(θ )FOy +
L12cos(θ )FAy = Icm1!!θ
− L22sin(β )FAx +
L22cos(β )FAy −
L22sin(β )FBx +
L22cos(β )FBy = Icm2 !!β
L32sin(α )FBx −
L32cos(α )FCy +
L32sin(α )FDx −
L32cos(α )FDy = Icm3 !!α
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1 1 0 0 0 0 0 0 00 0 1 1 0 0 0 0 00 −1 0 0 1 0 0 0 00 0 0 −1 0 0 1 0 00 0 0 0 −1 1 0 0 00 0 0 0 0 0 −1 1 0
L12sin(θ ) − L1
2sin(θ ) − L1
2cos(θ ) L1
2cos(θ ) 0 0 0 0 1
0 0 0 0 L32sin(α ) L3
2sin(α ) − L3
2cos(α ) − L3
2cos(α ) 0
0 − L22sin(β ) 0 L2
2cos(β ) − L2
2sin(β ) 0 L2
2cos(β ) 0 0
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
FOxFAxFOyFAyFBxFCxFByFCyM
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
=
m1acm1xm1acm1y +m1g
m2acm2xm2acm2y +m2g
m3acm3xm3acm3y +m3g
Icm1!!θIcm3 !!α
Icm2 !!β
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
![Page 58: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/58.jpg)
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FOxFAxFOyFAyFBxFCxFByFCyM
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
=
1 1 0 0 0 0 0 0 00 0 1 1 0 0 0 0 00 −1 0 0 1 0 0 0 00 0 0 −1 0 0 1 0 00 0 0 0 −1 1 0 0 00 0 0 0 0 0 −1 1 0
L12sin(θ ) − L1
2sin(θ ) − L1
2cos(θ ) L1
2cos(θ ) 0 0 0 0 1
0 0 0 0 L32sin(α ) L3
2sin(α ) − L3
2cos(α ) − L3
2cos(α ) 0
0 − L22sin(β ) 0 L2
2cos(β ) − L2
2sin(β ) 0 L2
2cos(β ) 0 0
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
−1
m1acm1xm1acm1y +m1g
m2acm2xm2acm2y +m2g
m3acm3xm3acm3y +m3g
Icm1!!θIcm3 !!α
Icm2 !!β
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
![Page 59: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/59.jpg)
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f1(θ2,r2 ) = r1 cos(θ1)+ r2 cos(θ2 )− Lf2 (θ2,r2 ) = r1 sin(θ1)− r2 sin(θ2 )
cos(θ2,n ) −r2,n sin(θ2,n )−sin(θ2,n ) −r2,n cos(θ2,n )
⎡
⎣⎢⎢
⎤
⎦⎥⎥
r2,n+1 − r2,nθ2,n+1 −θ2,n
⎡
⎣⎢⎢
⎤
⎦⎥⎥= −
r1 cos(θ1)+ r2,n cos(θ2,n )− Lr1 sin(θ1)− r2,n sin(θ2,n )
⎡
⎣⎢⎢
⎤
⎦⎥⎥
Análisis de posición
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f1(θ2,r2 ) = r1 cos(θ1)+ r2 cos(θ2 )− Lf2 (θ2,r2 ) = r1 sin(θ1)− r2 sin(θ2 )
!f1(θ2,r2 ) = −r1 sin(θ1) !θ1 + !r2 cos(θ2 )− r2 sin(θ2 ) !θ2 = 0!f2 (θ2,r2 ) = r1 cos(θ1) !θ1 − !r2 sin(θ2 )− r2 cos(θ2 ) !θ2 = 0
!r2 cos(θ2 )− r2 sin(θ2 ) !θ2 = r1 sin(θ1) !θ1− !r2 sin(θ2 )− r2 cos(θ2 ) !θ2 = r1 cos(θ1) !θ1
cos(θ2 ) −r2 sin(θ2 )−sin(θ2 ) −r2 cos(θ2 )
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!r2!θ2
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
sin(θ1)cos(θ1)
⎡
⎣⎢⎢
⎤
⎦⎥⎥r1 !θ1
Análisis de velocidad
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Análisis de aceleración
cos(θ2 ) −r2 sin(θ2 )sin(θ2 ) r2 cos(θ2 )
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!r2!θ2
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
cos(θ2 )sin(θ2 )
⎡
⎣⎢⎢
⎤
⎦⎥⎥r2 !θ2
2 +cos(θ1)−sin(θ1)
⎡
⎣⎢⎢
⎤
⎦⎥⎥r1 !θ1
2 +2sin(θ2 )−2cos(θ2 )
⎡
⎣⎢⎢
⎤
⎦⎥⎥!r2 !θ2 +
sin(θ1)cos(θ1)
⎡
⎣⎢⎢
⎤
⎦⎥⎥r1!!θ1
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Análisis de fuerzas Fx∑ = macmx
Fy∑ = macmy
Rueda 1 Fx∑ = FAx + FB sin(θ2 ) = 0
Rueda 2 Fx∑ = FCx − FB sin(θ2 ) = 0
Fy∑ = FAy −m1g + FB cos(θ2 ) = 0
Fy∑ = FCy −m2g − FB cos(θ2 ) = 0
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Análisis de momentos
M∑ = I !!θ
Rueda 1
M∑ = (r1 cos(θ1)i + r1 sin(θ1) j)× (FB sin(θ2 )i + FB cos(θ2 ) j)+M = I1!!θ1
Rueda 2
M∑ = (−r2 cos(θ2 )i + r2 sin(θ2 ) j)× (−FB sin(θ2 )i − FB cos(θ2 ) j) = I2 !!θ2
M∑ = r1FB cos(θ1)cos(θ2 )− r1FB sin(θ1)sin(θ2 )+M = I1!!θ1
M∑ = r2 cos(θ2 )FB = I2 !!θ2
![Page 65: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/65.jpg)
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Ecuaciones de fuerzas y momentos
1 0 0 0 sin(θ2 ) 00 1 0 0 cos(θ2 ) 00 0 1 0 −sin(θ2 ) 00 0 0 1 −cos(θ2 ) 00 0 0 0 r1 cos(θ1)cos(θ2 )− r1 sin(θ1)sin(θ2 ) 10 0 0 0 r2 0
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥
FAxFAyFCxFCyFBM
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥
=
0000I1!!θ1I2 !!θ2
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥
![Page 66: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/66.jpg)
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Ecuaciones de fuerzas y momentos
FAxFAyFCxFCyFBM
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥
=
1 0 0 0 sin(θ2 ) 00 1 0 0 cos(θ2 ) 00 0 1 0 −sin(θ2 ) 00 0 0 1 −cos(θ2 ) 00 0 0 0 r1 cos(θ1)cos(θ2 )− r1 sin(θ1)sin(θ2 ) 10 0 0 0 r2 0
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥
−1
0000I1!!θ1I2 !!θ2
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥
![Page 67: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/67.jpg)
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Ecuaciones de fuerzas y momentos
¿Cómo se implementa en Matlab?
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Péndulo simple
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Péndulo simple
Ft = −mgsin(θ ) = mat∑
at = l !!θ
−mgsin(θ ) = ml !!θ
!!θ + g
lsin(θ ) = 0
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Péndulo simple
!!θ + g
lsin(θ ) = 0
Suponemos que el ángulo siempre será
pequeño
θ ≈ sin(θ )
La ecuación diferencial la
podemos escribir como sigue
!!θ + g
lθ = 0
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Péndulo simple
!!θ + g
lθ = 0
La solución de la ecuación diferencial
esta dada por:
θ(t) = Asin(ωt +φ)
A,φ
Dependen de
θ(0), !θ(0)
ω = gl
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Péndulo simple
!!θ + g
lθ = 0
θ(t) = Asin(ωt +φ)
θ(0) = π6
!θ(0) = 0π6= Asin(φ)
Aω cos(φ) = 0
A = π6
φ = π2
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Péndulo simple
!!θ + g
lθ = 0
θ(t) = π6sin(ωt + π
2)
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Péndulo simple
!!θ + g
lθ = 0
θ(t) = π6sin(ωt + π
2)
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Péndulo simple
!!θ + g
lsin(θ ) = 0
¿Qué pasa cuando el
ángulo no es pequeño?
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Péndulo simple
!!θ + g
lsin(θ ) = 0
¿Qué pasa cuando el
ángulo no es pequeño?
x = x0 + v0t +12at 2
Recuerdan la siguiente ecuación?
![Page 77: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/77.jpg)
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Péndulo simple
!!θ + g
lsin(θ ) = 0
¿Qué pasa cuando el
ángulo no es pequeño?
Recuerdan la siguiente ecuación?
θn+1 = θn + !θnt +
12!!θnt
2
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Péndulo simple
!!θ + g
lsin(θ ) = 0
!θn+1 =θn+1 −θn
Δtθn+1 = θn + !θn+1Δt
!!θn =!θn+1 − !θn
Δt!θn+1 = !θn + !!θnΔtθn+1 = θn + ( !θn + !!θnΔt)Δtθn+1 = θn + !θnΔt + !!θnΔt
2
![Page 79: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/79.jpg)
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Péndulo simple
!!θ + g
lsin(θ ) = 0
θn+1 = θn + !θnΔt + !!θnΔt2
!!θn = − glsin(θn )
De donde
θn+1 = θn + !θnΔt −
glsin(θn )Δt
2
Obtuvimos una ecuación que nos permite obtener una aproximación
de la solución de la ecuación
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Péndulo simple
θn+1 = θn + !θnΔt −
glsin(θn )Δt
2
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Péndulo simple
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Péndulo físico
Considere el péndulo como un cuerpo rígido, que tiene masa e inercia, plantee las ecuaciones y mire la diferencia en comparación al péndulo
simple
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Péndulo físico
Fr∑ = Rr −mgcos(θ ) = 0
Ft∑ = Rt −mgsin(θ )+ F = mat = m!!θL2
M∑ = −RtL2+ F L
2= I !!θ
Rr = mgcos(θ )
Rt = m!!θL2− F +mgsin(θ )
−RtL2+ F L
2= I !!θ
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Péndulo físico
Rr = mgcos(θ )
Rt = m !θ2 L2− F +mgsin(θ )
−RtL2+ F L
2= I !!θ
I !!θ +m!!θ L
2
4+mg L
2sin(θ ) = 3F L
2
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Péndulo físico
Rr = mgcos(θ )
Rt = m !θ2 L2− F +mgsin(θ )
−RtL2+ F L
2= I !!θ
I +m L2
4⎛⎝⎜
⎞⎠⎟!!θ +mg L
2sin(θ ) = 3F L
2
!!θ +mg L
2
I +m L2
4⎛⎝⎜
⎞⎠⎟
sin(θ ) =3F L2
I +m L2
4⎛⎝⎜
⎞⎠⎟
![Page 86: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/86.jpg)
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Péndulo físico
!!θ +mg L
2
m L2
12+m L2
4⎛⎝⎜
⎞⎠⎟
sin(θ ) =3F L2
m L2
12+m L2
4⎛⎝⎜
⎞⎠⎟
!!θ +mg L
2
m L2
3
sin(θ ) =3F L2
m L2
3
!!θ + 32gLsin(θ ) = 9
2FmL
![Page 87: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/87.jpg)
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Péndulo físico
!!θ + 32gLsin(θ ) = 9
2F cos(ωt)
mL
Función coseno
ω n =32gL
Frecuencia natural
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Péndulo físico
!!θ + 32gLsin(θ ) = 9
2F cos(ωt)
mL
ω <ω n =32gL
![Page 89: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/89.jpg)
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Péndulo físico
!!θ + 32gLsin(θ ) = 9
2F cos(ωt)
mL ω <ω n =32gL
![Page 90: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/90.jpg)
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Péndulo físico
!!θ + 32gLsin(θ ) = 9
2F cos(ωt)
mL ω =ω n =32gL
![Page 91: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/91.jpg)
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Péndulo físico
!!θ + 32gLsin(θ ) = 9
2F cos(ωt)
mL ω >ω n =32gL
![Page 92: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/92.jpg)
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Péndulo físico
!!θ + 32gLsin(θ ) = 9
2Fe−kt
mL
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Péndulo físico
!!θ + 32gLsin(θ ) = 9
2Fe−kt
mL
![Page 94: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/94.jpg)
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Péndulo doble
x1 = l1 sin(θ1)y1 = −l1 cos(θ1)
x2 = l1 sin(θ1)+ l2 sin(θ2 )y2 = −l1 cos(θ1)− l2 cos(θ2 )
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
Energía potencial
![Page 95: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/95.jpg)
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Péndulo doble
x1 = l1 sin(θ1)y1 = −l1 cos(θ1)
x2 = l1 sin(θ1)+ l2 sin(θ2 )y2 = −l1 cos(θ1)− l2 cos(θ2 )
Energía cinética
!x1 = l1 cos(θ1) !θ1!y1 = l1 sin(θ1) !θ1
!x2 = l1 cos(θ1) !θ1 + l2 cos(θ2 ) !θ2!y2 = l1 sin(θ1) !θ1 + l2 sin(θ2 ) !θ2
K = 1
2m1( !x1
2 + !y12 )+ 1
2m2 ( !x2
2 + !y22 )
![Page 96: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/96.jpg)
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Péndulo doble Energía cinética
K = 12m1l1
2 !θ12 (sin2(θ1)+ cos
2(θ1))+12m2 ( !x2
2 + !y22 )
!x22 = (l1 cos(θ1) !θ1 + l2 cos(θ2 ) !θ2 )
2
= l12 !θ1
2 cos2(θ1)+ l22 !θ2
2 cos2(θ2 )+ 2l1l2 cos(θ1)cos(θ2 ) !θ1 !θ2!y22 = (l1 sin(θ1) !θ1 + l2 sin(θ2 ) !θ2 )
2
= l12 !θ1
2 sin2(θ1)+ l22 !θ2
2 sin2(θ2 )+ 2l1l2 sin(θ1)sin(θ2 ) !θ1 !θ2!x22 + !y2
2 = l12 !θ1
2 + l22 !θ2
2 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 )
!x1 = l1 cos(θ1) !θ1!y1 = l1 sin(θ1) !θ1
!x2 = l1 cos(θ1) !θ1 + l2 cos(θ2 ) !θ2!y2 = l1 sin(θ1) !θ1 + l2 sin(θ2 ) !θ2
![Page 97: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/97.jpg)
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Péndulo doble Energía cinética
K = 12m1l1
2 !θ12 (sin2(θ1)+ cos
2(θ1))
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
![Page 98: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/98.jpg)
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Péndulo doble Energía cinética Y
Potencial
K = 12m1l1
2 !θ12 (sin2(θ1)+ cos
2(θ1))
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
ℓ = K −V
![Page 99: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/99.jpg)
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Péndulo doble Energía cinética Y
Potencial
K = 12m1l1
2 !θ12
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
∂K∂ !θ1
= m1l12 !θ1 +m2l1
2 !θ1 +m2l1l2 !θ2 cos(θ1 −θ2 )
∂K∂ !θ2
= m2l22 !θ2 +m2l1l2 !θ1 cos(θ1 −θ2 )
![Page 100: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/100.jpg)
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Péndulo doble Energía cinética Y
Potencial
K = 12m1l1
2 !θ12
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
∂∂t
∂K∂ !θ1
⎛⎝⎜
⎞⎠⎟= m1l1
2 !!θ1 +m2l12 !!θ1 +m2l1l2 !!θ2 cos(θ1 −θ2 )
∂∂t
∂K∂ !θ2
⎛⎝⎜
⎞⎠⎟= m2l2
2 !!θ2 +m2l1l2 !!θ1 cos(θ1 −θ2 )
![Page 101: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/101.jpg)
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Péndulo doble Energía cinética Y
Potencial
K = 12m1l1
2 !θ12
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
∂K∂θ1
= −m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )
∂K∂θ2
= m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )
![Page 102: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/102.jpg)
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Péndulo doble Energía cinética Y
Potencial
K = 12m1l1
2 !θ12
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
∂V∂θ1
= m1gl1 sin(θ1)+m2gl1 sin(θ1)
∂V∂θ2
= m2gl2 sin(θ2 )
![Page 103: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/103.jpg)
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Péndulo doble Energía cinética Y
Potencial
K = 12m1l1
2 !θ12
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
∂∂t
∂K∂ !θ1
⎛⎝⎜
⎞⎠⎟− ∂∂t
∂V∂ !θ1
⎛⎝⎜
⎞⎠⎟− ∂K∂θ1
+ ∂V∂θ1
= 0
m1l12 !!θ1 +m2l1
2 !!θ1 +m2l1l2 !!θ2 cos(θ1 −θ2 )+m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )+m1gl1 sin(θ1)+m2gl1 sin(θ1) = 0
![Page 104: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/104.jpg)
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Péndulo doble Energía cinética Y
Potencial
K = 12m1l1
2 !θ12
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
∂∂t
∂K∂ !θ2
⎛⎝⎜
⎞⎠⎟− ∂∂t
∂V∂ !θ2
⎛⎝⎜
⎞⎠⎟− ∂K∂θ2
+ ∂V∂θ2
= 0
m2l22 !!θ2 +m2l1l2 !!θ1 cos(θ1 −θ2 )−m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )+m2gl2 sin(θ2 ) = 0
![Page 105: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/105.jpg)
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Péndulo doble Energía cinética Y
Potencial
K = 12m1l1
2 !θ12
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
m1l12 !!θ1 +m2l1
2 !!θ1 +m2l1l2 !!θ2 cos(θ1 −θ2 )+m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )+m1gl1 sin(θ1)+m2gl1 sin(θ1) = 0m2l2
2 !!θ2 +m2l1l2 !!θ1 cos(θ1 −θ2 )−m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )+m2gl2 sin(θ2 ) = 0
![Page 106: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/106.jpg)
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Péndulo doble Energía cinética Y
Potencial
K = 12m1l1
2 !θ12
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
(m1 +m2 )l12 !!θ1 +m2l1l2 !!θ2 cos(θ1 −θ2 )+m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )+ gl1(m1 +m2 )sin(θ1) = 0
m2l22 !!θ2 +m2l1l2 !!θ1 cos(θ1 −θ2 )−m2l1l2 !θ1 !θ2 sin(θ1 −θ2 )+m2gl2 sin(θ2 ) = 0
![Page 107: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/107.jpg)
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Péndulo doble Energía cinética Y
Potencial
K = 12m1l1
2 !θ12
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
(m1 +m2 )l1!!θ1 +m2l2 !!θ2 cos(θ1 −θ2 )+m2l2 !θ1 !θ2 sin(θ1 −θ2 )+ g(m1 +m2 )sin(θ1) = 0m2l1!!θ1 cos(θ1 −θ2 )+m2l2 !!θ2 −m2l1 !θ1 !θ2 sin(θ1 −θ2 )+m2gsin(θ2 ) = 0
![Page 108: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/108.jpg)
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Péndulo doble Energía cinética Y
Potencial
K = 12m1l1
2 !θ12
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
(m1 +m2 )l1 m2l2 cos(θ1 −θ2 )m2l1 cos(θ1 −θ2 ) m2l2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
!!θ1!!θ2
⎡
⎣⎢⎢
⎤
⎦⎥⎥+
m2l2 !θ1 !θ2 sin(θ1 −θ2 )+ g(m1 +m2 )sin(θ1)
−m2l1 !θ1 !θ2 sin(θ1 −θ2 )+m2gsin(θ2 )
⎡
⎣⎢⎢
⎤
⎦⎥⎥= 0
0⎡
⎣⎢
⎤
⎦⎥
![Page 109: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/109.jpg)
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Péndulo doble Energía cinética Y
Potencial
K = 12m1l1
2 !θ12
+ 12m2 (l1
2 !θ12 + l2
2 !θ22 + 2l1l2 !θ1 !θ2 cos(θ1 −θ2 ))
V = m1gy1 +m2gy2= −m1gl1 cos(θ1)−m2gl1 cos(θ1)−m2gl2 cos(θ2 )
!!θ1!!θ2
⎡
⎣⎢⎢
⎤
⎦⎥⎥= −
(m1 +m2 )l1 m2l2 cos(θ1 −θ2 )m2l1 cos(θ1 −θ2 ) m2l2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
−1m2l2 !θ1 !θ2 sin(θ1 −θ2 )+ g(m1 +m2 )sin(θ1)
−m2l1 !θ1 !θ2 sin(θ1 −θ2 )+m2gsin(θ2 )
⎡
⎣⎢⎢
⎤
⎦⎥⎥
![Page 110: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/110.jpg)
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Péndulo doble Discretización
diferencias finitas
!!θ1,n!!θ2,n
⎡
⎣
⎢⎢
⎤
⎦
⎥⎥= −
(m1 +m2 )l1 m2l2 cos(θ1,n −θ2,n )m2l1 cos(θ1,n −θ2,n ) m2l2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
−1m2l2 !θ1,n !θ2,n sin(θ1,n −θ2,n )+ g(m1 +m2 )sin(θ1,n )
−m2l1 !θ1,n !θ2,n sin(θ1,n −θ2,n )+m2gsin(θ2,n )
⎡
⎣
⎢⎢
⎤
⎦
⎥⎥
θ1,n+1 = θ1,n +!θ1,nΔt + !!θ1,nΔt
2
θ2,n+1 = θ2,n +!θ2,nΔt + !!θ2,nΔt
2
!θ1,n+1 = !θ1,n + !!θ1,nΔt
!θ2,n+1 = !θ2,n + !!θ2,nΔt
![Page 111: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/111.jpg)
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Péndulo doble Discretización
diferencias finitas
!!θ1,n!!θ2,n
⎡
⎣
⎢⎢
⎤
⎦
⎥⎥= −
(m1 +m2 )l1 m2l2 cos(θ1,n −θ2,n )m2l1 cos(θ1,n −θ2,n ) m2l2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
−1m2l2 !θ1,n !θ2,n sin(θ1,n −θ2,n )+ g(m1 +m2 )sin(θ1,n )
−m2l1 !θ1,n !θ2,n sin(θ1,n −θ2,n )+m2gsin(θ2,n )
⎡
⎣
⎢⎢
⎤
⎦
⎥⎥
θ1,n+1 = θ1,n +!θ1,nΔt + !!θ1,nΔt
2 θ2,n+1 = θ2,n +
!θ2,nΔt + !!θ2,nΔt2
!θ1,n+1 = !θ1,n + !!θ1,nΔt
!θ2,n+1 = !θ2,n + !!θ2,nΔt
![Page 112: Chapter 4 Image Slides - WordPress.com · Design of Machinery An Introduction to the Synthesis and Analysis of Mechanisms and Machines Fourth Edition Robert L. Norton Chapter 4 Image](https://reader030.vdocument.in/reader030/viewer/2022040605/5eae634779aaeb148c293532/html5/thumbnails/112.jpg)
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Péndulo doble
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Péndulo doble
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