Consider the washing mechanism within an agricultural production line as shown in the following figure. Link lengths are shown in the figure in mm. The input link (OA) has a constant angular velocity of 10 rad/s CCW. At the instant shown, if OD = 551.77 mm: a) Draw the vector loop of the mechanism and establish the local coordinate system (x-y). b) Determine the local angular positions of output link (DB) and coupler link (AB). c) Determine the angular velocities of output link (DB) and coupler link (AB). d) Deduce, depending on part (b), the transmission angle. 100 mm 1. 80 mm -240 mm- B. 360 mm D Position, Velocity and Accleration Analysis: Pin-jointed Fourbar linkage 041,2 A R3 03 VB VBA VBA VB VA (b) 002 a VA R2 02 R1 02 b R4 x C 04 004 - X 04 Position Analysis = 2 arctan -B± √√B² - 4AC 2A -E±√E²-4DF 031,2 = 2 arctan 2D cos02 K₁ - K₂cos02 + K3 Dcos02 K₁ + K4cos02 + K5 E = -2sin02 FK₁+(K4-1)cos02 + K5 • A • B = -2sin02 • C K₁ (K2 + 1)cos02 + K3 d • K₁ = • K₂ = C a²-b²+c²+d2 • K3 = 2ac K₁ = K₁ = c²-d²-a²-b² K5 = 2ab PS: if is calculated before you can use one of the equations below to solve for 03 bcos03=-acos02 + ccos04 +d Velocity Analysis bsin03=-asin02 + csin04 aw2 sin(04-02) aw2 sin(02-03) 04 c sin(04-03) W3 = b sin(03-04)
Consider the washing mechanism within an agricultural production line as shown in the following figure. Link lengths are shown in the figure in mm. The input link (OA) has a constant angular velocity of 10 rad/s CCW. At the instant shown, if OD = 551.77 mm: a) Draw the vector loop of the mechanism and establish the local coordinate system (x-y). b) Determine the local angular positions of output link (DB) and coupler link (AB). c) Determine the angular velocities of output link (DB) and coupler link (AB). d) Deduce, depending on part (b), the transmission angle. 100 mm 1. 80 mm -240 mm- B. 360 mm D Position, Velocity and Accleration Analysis: Pin-jointed Fourbar linkage 041,2 A R3 03 VB VBA VBA VB VA (b) 002 a VA R2 02 R1 02 b R4 x C 04 004 - X 04 Position Analysis = 2 arctan -B± √√B² - 4AC 2A -E±√E²-4DF 031,2 = 2 arctan 2D cos02 K₁ - K₂cos02 + K3 Dcos02 K₁ + K4cos02 + K5 E = -2sin02 FK₁+(K4-1)cos02 + K5 • A • B = -2sin02 • C K₁ (K2 + 1)cos02 + K3 d • K₁ = • K₂ = C a²-b²+c²+d2 • K3 = 2ac K₁ = K₁ = c²-d²-a²-b² K5 = 2ab PS: if is calculated before you can use one of the equations below to solve for 03 bcos03=-acos02 + ccos04 +d Velocity Analysis bsin03=-asin02 + csin04 aw2 sin(04-02) aw2 sin(02-03) 04 c sin(04-03) W3 = 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
I want to solve it based on the pin joint 4 bar linkage rule, which is given in an image. Please, I do not want to solve it at all using the geometric method.
Transcribed Image Text:Consider the washing mechanism within an agricultural production line as shown in the following figure.
Link lengths are shown in the figure in mm. The input link (OA) has a constant angular velocity of 10 rad/s
CCW. At the instant shown, if OD = 551.77 mm:
a) Draw the vector loop of the mechanism and establish the local coordinate system (x-y).
b) Determine the local angular positions of output link (DB) and coupler link (AB).
c) Determine the angular velocities of output link (DB) and coupler link (AB).
d) Deduce, depending on part (b), the transmission angle.
100 mm
1.
80 mm
-240 mm-
B.
360 mm
D
Transcribed Image Text:Position, Velocity and Accleration Analysis: Pin-jointed Fourbar
linkage
041,2
A
R3
03
VB
VBA
VBA
VB
VA
(b)
002
a
VA
R2
02
R1
02
b
R4
x
C
04
004
- X
04
Position Analysis
= 2 arctan
-B± √√B² - 4AC
2A
-E±√E²-4DF
031,2
= 2 arctan
2D
cos02 K₁ - K₂cos02 + K3
Dcos02 K₁ + K4cos02 + K5
E = -2sin02
FK₁+(K4-1)cos02 + K5
•
A
•
B = -2sin02
•
C K₁ (K2 + 1)cos02 + K3
d
•
K₁ =
•
K₂ =
C
a²-b²+c²+d2
•
K3
=
2ac
K₁ =
K₁ =
c²-d²-a²-b²
K5 =
2ab
PS: if is calculated before you can use one of the equations below to solve for 03
bcos03=-acos02 + ccos04 +d
Velocity Analysis
bsin03=-asin02 + csin04
aw2
sin(04-02)
aw2 sin(02-03)
04
c sin(04-03)
W3
=
b sin(03-04)
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