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 -E±√E²-4DF 031,2 2 arctan 2D D= cos 2-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=-asinė₂ + csin04 Velocity Analysis aw, sin(8-8) 004 c sin(04-03) aw2 sin(04-0₂) 003 = b sin(03-04) Position Velocity and Acceleration Analysis: Slider-Crank Linkage AA R2 AB R3 R₁ ABA R4 ABA x AB AA A Position analysis 031 = arcsin (asin02 b (b) d = acos02-bcos03 asino2 03: =arcsin +π Velocity analysis a cos02 d= =-a02 sine₂+b03 sin03 03 -002 b cose 02 аз Acceleration Analysis aα₂ cosе₂-asin02 +bsin03 bcos03 d=-aa2 sine₂-awcos02 +bα3 sin 03 +bwcose Acceleration Analysis: Inverted Slider-Crank Linkage B AABoriali R₂ b dot AAB R1 X A X α4= aacos(0-0)+sin(03-03)]+co sin(0,-03)-2003 b+ccos(03-04) Jamboos(0-0)+cos(0-0)]+ax[sin(0-0)-csin(04 −0₂)|| - • 26cm, sin(0, −0,)−m;[b² +c² +2bccos(0, −03)] b+ccos(03-04) " k=2 Energy Equation Fk Vk + kk = mak • Vk + Σkαk ·@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 -E±√E²-4DF 031,2 2 arctan 2D D= cos 2-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=-asinė₂ + csin04 Velocity Analysis aw, sin(8-8) 004 c sin(04-03) aw2 sin(04-0₂) 003 = b sin(03-04) Position Velocity and Acceleration Analysis: Slider-Crank Linkage AA R2 AB R3 R₁ ABA R4 ABA x AB AA A Position analysis 031 = arcsin (asin02 b (b) d = acos02-bcos03 asino2 03: =arcsin +π Velocity analysis a cos02 d= =-a02 sine₂+b03 sin03 03 -002 b cose 02 аз Acceleration Analysis aα₂ cosе₂-asin02 +bsin03 bcos03 d=-aa2 sine₂-awcos02 +bα3 sin 03 +bwcose Acceleration Analysis: Inverted Slider-Crank Linkage B AABoriali R₂ b dot AAB R1 X A X α4= aacos(0-0)+sin(03-03)]+co sin(0,-03)-2003 b+ccos(03-04) Jamboos(0-0)+cos(0-0)]+ax[sin(0-0)-csin(04 −0₂)|| - • 26cm, sin(0, −0,)−m;[b² +c² +2bccos(0, −03)] b+ccos(03-04) " k=2 Energy Equation Fk Vk + kk = mak • Vk + Σkαk ·@k k=2 -,。, k=2 k=2
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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Question
I want to solve all parts of the question using rules in the picture pls very urgent
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
-E±√E²-4DF
031,2
2 arctan
2D
D= cos 2-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=-asinė₂ + csin04
Velocity Analysis
aw, sin(8-8)
004
c sin(04-03)
aw2
sin(04-0₂)
003 =
b sin(03-04)
![Position Velocity and Acceleration Analysis: Slider-Crank Linkage
AA
R2
AB
R3
R₁
ABA
R4
ABA
x
AB
AA
A
Position analysis
031
= arcsin
(asin02
b
(b)
d = acos02-bcos03
asino2
03:
=arcsin
+π
Velocity analysis
a cos02
d=
=-a02 sine₂+b03 sin03
03
-002
b cose
02
аз
Acceleration Analysis
aα₂ cosе₂-asin02 +bsin03
bcos03
d=-aa2 sine₂-awcos02 +bα3 sin 03 +bwcose
Acceleration Analysis: Inverted Slider-Crank Linkage
B
AABoriali
R₂
b dot
AAB
R1
X
A
X
α4=
aacos(0-0)+sin(03-03)]+co sin(0,-03)-2003
b+ccos(03-04)
Jamboos(0-0)+cos(0-0)]+ax[sin(0-0)-csin(04 −0₂)||
-
• 26cm, sin(0, −0,)−m;[b² +c² +2bccos(0, −03)]
b+ccos(03-04)
"
k=2
Energy Equation
Fk Vk + kk = mak • Vk + Σkαk ·@k
k=2
-,。,
k=2
k=2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F513839f2-dba8-480f-99ba-a9cdadee07a8%2Ffda06bbd-0dde-4fe0-87a7-0af8d60d2578%2Ffgfs8v5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Position Velocity and Acceleration Analysis: Slider-Crank Linkage
AA
R2
AB
R3
R₁
ABA
R4
ABA
x
AB
AA
A
Position analysis
031
= arcsin
(asin02
b
(b)
d = acos02-bcos03
asino2
03:
=arcsin
+π
Velocity analysis
a cos02
d=
=-a02 sine₂+b03 sin03
03
-002
b cose
02
аз
Acceleration Analysis
aα₂ cosе₂-asin02 +bsin03
bcos03
d=-aa2 sine₂-awcos02 +bα3 sin 03 +bwcose
Acceleration Analysis: Inverted Slider-Crank Linkage
B
AABoriali
R₂
b dot
AAB
R1
X
A
X
α4=
aacos(0-0)+sin(03-03)]+co sin(0,-03)-2003
b+ccos(03-04)
Jamboos(0-0)+cos(0-0)]+ax[sin(0-0)-csin(04 −0₂)||
-
• 26cm, sin(0, −0,)−m;[b² +c² +2bccos(0, −03)]
b+ccos(03-04)
"
k=2
Energy Equation
Fk Vk + kk = mak • Vk + Σkαk ·@k
k=2
-,。,
k=2
k=2
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