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 Consider the inverted slider-crank linkage as shown in the following figure. Given are: Link lengths: Link O₂A = 10 cm, Link O4B = 15 cm, Link 0204 = 20 cm. Positions: y=90, 02 = 140°, 03198.8", 04 = 108.8, b = AB = 24.113 cm. Angular velocities: (02-15 rad/s (CW), 003 004 = -3.22 rad/s (CW) Velocity of slip: b = -80 cm/s = ⚫ Link O₂A rotates with a constant angular velocity. a) Calculate the angular acceleration of links 4 and 3 and the acceleration of slip at point B. b) Calculate the magnitude and direction of the acceleration of point B. YA 02 Ꮎ, 03 B X 04 04 Position, Velocity and Acceleration Analysis: Pin-jointed Fourbar linkage Y VBA R3 R4 R2 d R₁ -B±√B²-4AC 0412 =2arctan 2A Position Analysis x 04 VBA -E±√E²-4DF 03,2 = 2 arctan 2D D A cose₂-K₁ - K₂cos02 + K3 B = -2sin02 ⚫C=K₁- (K₂+1)cos0₂+K3 cose₂-K₁ - K₁cose₂+ Ks E = -2sin02 F • K₁ = K₁+(K-1)cos02 + K5 . K₁₁ = • K₂ a²-b²+c²+d² • K3 2ac • K₁₁ = c2-d²-a²-b2 ⚫ K5 2ab PS: if 0 is calculated before you can use one of the equations below to solve for 03 bcos03-acos02+ccos04+d bsin03=-asin02 + csin04 Velocity Analysis aw2 sin(02-03) 004 c sin(04-03) a 003 sin(04-0₂) 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 Consider the inverted slider-crank linkage as shown in the following figure. Given are: Link lengths: Link O₂A = 10 cm, Link O4B = 15 cm, Link 0204 = 20 cm. Positions: y=90, 02 = 140°, 03198.8", 04 = 108.8, b = AB = 24.113 cm. Angular velocities: (02-15 rad/s (CW), 003 004 = -3.22 rad/s (CW) Velocity of slip: b = -80 cm/s = ⚫ Link O₂A rotates with a constant angular velocity. a) Calculate the angular acceleration of links 4 and 3 and the acceleration of slip at point B. b) Calculate the magnitude and direction of the acceleration of point B. YA 02 Ꮎ, 03 B X 04 04 Position, Velocity and Acceleration Analysis: Pin-jointed Fourbar linkage Y VBA R3 R4 R2 d R₁ -B±√B²-4AC 0412 =2arctan 2A Position Analysis x 04 VBA -E±√E²-4DF 03,2 = 2 arctan 2D D A cose₂-K₁ - K₂cos02 + K3 B = -2sin02 ⚫C=K₁- (K₂+1)cos0₂+K3 cose₂-K₁ - K₁cose₂+ Ks E = -2sin02 F • K₁ = K₁+(K-1)cos02 + K5 . K₁₁ = • K₂ a²-b²+c²+d² • K3 2ac • K₁₁ = c2-d²-a²-b2 ⚫ K5 2ab PS: if 0 is calculated before you can use one of the equations below to solve for 03 bcos03-acos02+ccos04+d bsin03=-asin02 + csin04 Velocity Analysis aw2 sin(02-03) 004 c sin(04-03) a 003 sin(04-0₂) 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 just using the rules attached in picture i nees her solving 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%2Fe359f13c-963c-481b-92b5-0e1770e78ef7%2Fnul9dbw_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:Consider the inverted slider-crank linkage as shown in the following figure. Given are:
Link lengths: Link O₂A = 10 cm, Link O4B = 15 cm, Link 0204 = 20 cm.
Positions: y=90, 02 = 140°, 03198.8", 04 = 108.8, b = AB = 24.113 cm.
Angular velocities: (02-15 rad/s (CW), 003 004 = -3.22 rad/s (CW)
Velocity of slip: b = -80 cm/s
=
⚫ Link O₂A rotates with a constant angular velocity.
a) Calculate the angular acceleration of links 4 and 3 and the acceleration of slip at point B.
b) Calculate the magnitude and direction of the acceleration of point B.
YA
02
Ꮎ,
03
B
X
04
04
Position, Velocity and Acceleration Analysis: Pin-jointed Fourbar linkage
Y
VBA
R3
R4
R2
d
R₁
-B±√B²-4AC
0412
=2arctan
2A
Position Analysis
x
04
VBA
-E±√E²-4DF
03,2 = 2 arctan
2D
D
A
cose₂-K₁ - K₂cos02 + K3
B = -2sin02
⚫C=K₁- (K₂+1)cos0₂+K3
cose₂-K₁ - K₁cose₂+ Ks
E = -2sin02
F
• K₁ =
K₁+(K-1)cos02 + K5
.
K₁₁ =
• K₂
a²-b²+c²+d²
• K3
2ac
• K₁₁ =
c2-d²-a²-b2
⚫ K5
2ab
PS: if 0 is calculated before you can use one of the equations below to solve for 03
bcos03-acos02+ccos04+d
bsin03=-asin02 + csin04
Velocity Analysis
aw2 sin(02-03)
004
c sin(04-03)
a
003
sin(04-0₂)
b sin(03-04)
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