Position and velocity analysis: Inverted Slider-crank linkage 02 041,2 = 2arctan R3 R4 R1 d -T±√√T²-4SU 25 b dot X 02 AAB AAB Corioli AB b 03 = 0, + Y asinė₂ – csino sin03 P = asino₂siny + (acos₂ - d)cosy Q=-asino₂cos y + (acos₂-d)siny R=-csin y; SR-Q; T = 2P; U=Q+R b=- Velocity analysis 004- aw2 cos(02-03) b+ccos(04-03) -a02sine2+4(bsin03 +csin04) cose 04 Acceleration Analysis azcos(03-02)+sin(03-02)]+co sin(04-03)-2603 b+ccos(03-04) awbcos(03-02)+ccos(04-02)]+ad2[bsin(02-03)-csin(04-02)] +2bc04 sin(04-03)-[b²+c²+2bccos(04-03)] +cos(0-0) A₁ = ax2(-sin02+ jcos02)-aw(cos02+jsin02) ABA bo3 (sin03-jcos03)+bw (cos03+jsin03) +2(sine,-jcose,)-(cose,+jsine,) Ag-ca (sin0,-jcos04)-ce (cose,+jsin0) 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 O2O4 = 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. Y 02 B 02 04 ᎾᏊ X
Position and velocity analysis: Inverted Slider-crank linkage 02 041,2 = 2arctan R3 R4 R1 d -T±√√T²-4SU 25 b dot X 02 AAB AAB Corioli AB b 03 = 0, + Y asinė₂ – csino sin03 P = asino₂siny + (acos₂ - d)cosy Q=-asino₂cos y + (acos₂-d)siny R=-csin y; SR-Q; T = 2P; U=Q+R b=- Velocity analysis 004- aw2 cos(02-03) b+ccos(04-03) -a02sine2+4(bsin03 +csin04) cose 04 Acceleration Analysis azcos(03-02)+sin(03-02)]+co sin(04-03)-2603 b+ccos(03-04) awbcos(03-02)+ccos(04-02)]+ad2[bsin(02-03)-csin(04-02)] +2bc04 sin(04-03)-[b²+c²+2bccos(04-03)] +cos(0-0) A₁ = ax2(-sin02+ jcos02)-aw(cos02+jsin02) ABA bo3 (sin03-jcos03)+bw (cos03+jsin03) +2(sine,-jcose,)-(cose,+jsine,) Ag-ca (sin0,-jcos04)-ce (cose,+jsin0) 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 O2O4 = 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. Y 02 B 02 04 ᎾᏊ X
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
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I want to solve it based on the inverted slider-crank linkage rule, which is given in an image below . Please, I do not want to solve it at all using the geometric method.
![Position and velocity analysis: Inverted Slider-crank linkage
02
041,2
= 2arctan
R3
R4
R1
d
-T±√√T²-4SU
25
b dot
X
02
AAB
AAB Corioli
AB
b
03 = 0, + Y
asinė₂ – csino
sin03
P = asino₂siny + (acos₂ - d)cosy
Q=-asino₂cos y + (acos₂-d)siny
R=-csin y; SR-Q;
T = 2P;
U=Q+R
b=-
Velocity analysis
004-
aw2 cos(02-03)
b+ccos(04-03)
-a02sine2+4(bsin03 +csin04)
cose
04
Acceleration Analysis
azcos(03-02)+sin(03-02)]+co sin(04-03)-2603
b+ccos(03-04)
awbcos(03-02)+ccos(04-02)]+ad2[bsin(02-03)-csin(04-02)]
+2bc04 sin(04-03)-[b²+c²+2bccos(04-03)]
+cos(0-0)
A₁ = ax2(-sin02+ jcos02)-aw(cos02+jsin02)
ABA bo3 (sin03-jcos03)+bw (cos03+jsin03)
+2(sine,-jcose,)-(cose,+jsine,)
Ag-ca (sin0,-jcos04)-ce (cose,+jsin0)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F620cfc7b-53ee-42d9-b25a-c30e308dc613%2F547a0c7c-2879-4fb7-a7c7-ef77ba866071%2F8q3kjlg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Position and velocity analysis: Inverted Slider-crank linkage
02
041,2
= 2arctan
R3
R4
R1
d
-T±√√T²-4SU
25
b dot
X
02
AAB
AAB Corioli
AB
b
03 = 0, + Y
asinė₂ – csino
sin03
P = asino₂siny + (acos₂ - d)cosy
Q=-asino₂cos y + (acos₂-d)siny
R=-csin y; SR-Q;
T = 2P;
U=Q+R
b=-
Velocity analysis
004-
aw2 cos(02-03)
b+ccos(04-03)
-a02sine2+4(bsin03 +csin04)
cose
04
Acceleration Analysis
azcos(03-02)+sin(03-02)]+co sin(04-03)-2603
b+ccos(03-04)
awbcos(03-02)+ccos(04-02)]+ad2[bsin(02-03)-csin(04-02)]
+2bc04 sin(04-03)-[b²+c²+2bccos(04-03)]
+cos(0-0)
A₁ = ax2(-sin02+ jcos02)-aw(cos02+jsin02)
ABA bo3 (sin03-jcos03)+bw (cos03+jsin03)
+2(sine,-jcose,)-(cose,+jsine,)
Ag-ca (sin0,-jcos04)-ce (cose,+jsin0)
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 O2O4 = 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.
Y
02
B
02
04
ᎾᏊ
X
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