Position Velocity and Acceleration Analysis: Slider-Crank Linkage AB R3 AA R2 R₁ ABA R4 ABA x AB (b) AA A Position analysis 031 = arcsin (asin02 b d = acos02-bcos03 asino2 03: =arcsin +π Velocity analysis a cose₂ d= =-a02 sine₂+b03 sin03 03 -002 b cosey 02 аз Acceleration Analysis aα₂ cosе₂-asin02 +bsin03 bcos03 d=-a2 sine₂-aw cos02 +bα3 sin03 +bwcos03 Acceleration Analysis: Inverted Slider-Crank Linkage R₂ R₁ b dot AAB AAB Corioli X α4= aacos(0-0)+sin(0-0)]+co sin(0,-03)-2003 b+ccos(03-04) amboos(0-0)+ccos(0-0)]+ax[sin(@2-03)-csin(04 −0₂)|| - • 26cm, sin(0, −0,)−m;[b² +c² +2bccos(0,−03)] b+ccos(03-04) " k=2 Energy Equation -,。, FV + 1 = a + k k=2 k=2 k=2 For the mechanism shown below, the input link (AB) rotates CCW with a constant angular velocity of = 1.5 rad/s. At the position shown 02 = 55": a) Draw clearly the vector loop (CBD). b) Find the angular velocity of link (CD) and the slip velocity of slider B. c) Find the linear velocity of point D. d) Using Energy Method, determine the required moment M needed by the input crank (AB) to withstand the exerted force F = 545 N if ẞ = 20". Neglect the mass of all members. The dimensions are: a = 0.17 m, c = 0.24 m, CD = 0.5 m F Y B 02 b 03 C Position, Velocity and Acceleration Analysis: Pin-jointed Fourbar linkage VBA VBA 0412 R3 03 R4 R1 -B±√B²-4AC Position Analysis = 2arctan 2A A cose₂-K₁ - K₂cos02 + K3 B = -2sin02 C K₁ (K₂+1)cos02 + K3 K₁ = • K₂ K3 2ac 004 031,2 2 arctan -E±√√E²-4DF 2D ⚫D=cos02-K₁ - K₁cos02 + K5 • E = -2sin02 FK₁+(K4-1)cos02 + K5 K₁ = . c2-d2-a²-b2 Ks 2ab PS: if is calculated before you can use one of the equations below to solve for 03 bcos03=-acos₂+ccos04+d bsin03=-asino₂+ csin04 Velocity Analysis aw, sin(0-8) 004 c sin(04-03) aw2 sin(04-0₂) 003 = b sin(03-04)
Position Velocity and Acceleration Analysis: Slider-Crank Linkage AB R3 AA R2 R₁ ABA R4 ABA x AB (b) AA A Position analysis 031 = arcsin (asin02 b d = acos02-bcos03 asino2 03: =arcsin +π Velocity analysis a cose₂ d= =-a02 sine₂+b03 sin03 03 -002 b cosey 02 аз Acceleration Analysis aα₂ cosе₂-asin02 +bsin03 bcos03 d=-a2 sine₂-aw cos02 +bα3 sin03 +bwcos03 Acceleration Analysis: Inverted Slider-Crank Linkage R₂ R₁ b dot AAB AAB Corioli X α4= aacos(0-0)+sin(0-0)]+co sin(0,-03)-2003 b+ccos(03-04) amboos(0-0)+ccos(0-0)]+ax[sin(@2-03)-csin(04 −0₂)|| - • 26cm, sin(0, −0,)−m;[b² +c² +2bccos(0,−03)] b+ccos(03-04) " k=2 Energy Equation -,。, FV + 1 = a + k k=2 k=2 k=2 For the mechanism shown below, the input link (AB) rotates CCW with a constant angular velocity of = 1.5 rad/s. At the position shown 02 = 55": a) Draw clearly the vector loop (CBD). b) Find the angular velocity of link (CD) and the slip velocity of slider B. c) Find the linear velocity of point D. d) Using Energy Method, determine the required moment M needed by the input crank (AB) to withstand the exerted force F = 545 N if ẞ = 20". Neglect the mass of all members. The dimensions are: a = 0.17 m, c = 0.24 m, CD = 0.5 m F Y B 02 b 03 C Position, Velocity and Acceleration Analysis: Pin-jointed Fourbar linkage VBA VBA 0412 R3 03 R4 R1 -B±√B²-4AC Position Analysis = 2arctan 2A A cose₂-K₁ - K₂cos02 + K3 B = -2sin02 C K₁ (K₂+1)cos02 + K3 K₁ = • K₂ K3 2ac 004 031,2 2 arctan -E±√√E²-4DF 2D ⚫D=cos02-K₁ - K₁cos02 + K5 • E = -2sin02 FK₁+(K4-1)cos02 + K5 K₁ = . c2-d2-a²-b2 Ks 2ab PS: if is calculated before you can use one of the equations below to solve for 03 bcos03=-acos₂+ccos04+d bsin03=-asino₂+ csin04 Velocity Analysis aw, sin(0-8) 004 c sin(04-03) aw2 sin(04-0₂) 003 = b sin(03-04)
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
pls using just the rules attached in picture i need her all. solving pls veryyy urgent
![Position Velocity and Acceleration Analysis: Slider-Crank Linkage
AB
R3
AA
R2
R₁
ABA
R4
ABA
x
AB
(b)
AA
A
Position analysis
031
= arcsin
(asin02
b
d = acos02-bcos03
asino2
03:
=arcsin
+π
Velocity analysis
a cose₂
d=
=-a02 sine₂+b03 sin03
03
-002
b cosey
02
аз
Acceleration Analysis
aα₂ cosе₂-asin02 +bsin03
bcos03
d=-a2 sine₂-aw cos02 +bα3 sin03 +bwcos03
Acceleration Analysis: Inverted Slider-Crank Linkage
R₂
R₁
b dot
AAB
AAB Corioli
X
α4=
aacos(0-0)+sin(0-0)]+co sin(0,-03)-2003
b+ccos(03-04)
amboos(0-0)+ccos(0-0)]+ax[sin(@2-03)-csin(04 −0₂)||
-
• 26cm, sin(0, −0,)−m;[b² +c² +2bccos(0,−03)]
b+ccos(03-04)
"
k=2
Energy Equation
-,。,
FV + 1 = a + k
k=2
k=2
k=2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc499f01a-345f-47dc-93f0-5d7b5f68a0c4%2F482341cd-9c3e-45af-84ba-552fc84c9297%2Fz6aik5a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Position Velocity and Acceleration Analysis: Slider-Crank Linkage
AB
R3
AA
R2
R₁
ABA
R4
ABA
x
AB
(b)
AA
A
Position analysis
031
= arcsin
(asin02
b
d = acos02-bcos03
asino2
03:
=arcsin
+π
Velocity analysis
a cose₂
d=
=-a02 sine₂+b03 sin03
03
-002
b cosey
02
аз
Acceleration Analysis
aα₂ cosе₂-asin02 +bsin03
bcos03
d=-a2 sine₂-aw cos02 +bα3 sin03 +bwcos03
Acceleration Analysis: Inverted Slider-Crank Linkage
R₂
R₁
b dot
AAB
AAB Corioli
X
α4=
aacos(0-0)+sin(0-0)]+co sin(0,-03)-2003
b+ccos(03-04)
amboos(0-0)+ccos(0-0)]+ax[sin(@2-03)-csin(04 −0₂)||
-
• 26cm, sin(0, −0,)−m;[b² +c² +2bccos(0,−03)]
b+ccos(03-04)
"
k=2
Energy Equation
-,。,
FV + 1 = a + k
k=2
k=2
k=2
Transcribed Image Text:For the mechanism shown below, the input link (AB) rotates CCW with a constant angular velocity of
= 1.5 rad/s.
At the position shown 02 = 55":
a) Draw clearly the vector loop (CBD).
b) Find the angular velocity of link (CD) and the slip velocity of slider B.
c) Find the linear velocity of point D.
d) Using Energy Method, determine the required moment M needed by the input crank (AB) to
withstand the exerted force F = 545 N if ẞ = 20".
Neglect the mass of all members.
The dimensions are: a = 0.17 m, c = 0.24 m, CD = 0.5 m
F
Y
B
02
b
03
C
Position, Velocity and Acceleration Analysis: Pin-jointed Fourbar linkage
VBA
VBA
0412
R3
03
R4
R1
-B±√B²-4AC
Position Analysis
= 2arctan
2A
A
cose₂-K₁ - K₂cos02 + K3
B = -2sin02
C K₁ (K₂+1)cos02 + K3
K₁ =
• K₂
K3
2ac
004
031,2
2 arctan
-E±√√E²-4DF
2D
⚫D=cos02-K₁ - K₁cos02 + K5
•
E = -2sin02
FK₁+(K4-1)cos02 + K5
K₁ =
.
c2-d2-a²-b2
Ks
2ab
PS: if is calculated before you can use one of the equations below to solve for 03
bcos03=-acos₂+ccos04+d
bsin03=-asino₂+ csin04
Velocity Analysis
aw, sin(0-8)
004
c sin(04-03)
aw2
sin(04-0₂)
003 =
b sin(03-04)
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