For a short interval of motion, the input gear pivoted at O rotates with a constant angular velocity = 4 rad/s in clockwise sense (as shown in figure below). In the position shown, the force applied to the piston is 2F where F = 200 N; The diameter of the gear (G) is 30 cm, link AB is 70 cm and y₁ = 20 cm. a) Draw the vector loop for the given mechanism. b) Find the position of the piston. c) Find the velocity and acceleration of the piston. d) Using energy method, determine the torque T and the power required from the motor to operate the mechanism. e) Find the force in the member (BA). The piston has a mass mp = 0.65 kg. Neglect the mass of other links. All dimensions are in cm. A Y^ "(G) Ут B 2F X Position, Velocity and Acceleration Analysis: Pin-jointed Fourbar linkage VBA 041,2 002 R4 -B±√B2-4AC R1 Position Analysis = Zarctan 2A 004 VBA 03,2 = 2 arctan -E±√E2-4DF 2D A cos02-K₁ - K₂cos02 + K3 • B = -2sin02 C K₁ (K2 + 1)cos62 + K3 ⚫D=cose₂-K₁ - K4cos62 + K5 ⚫ E = -2sin02 F K₁+(K4-1)cos02 + K5 K₁₁ = 1 • K₁ = . K₂ -b²+c²+d² • K3 = 2ac c2-d²-a²-b2 ⚫ Ks 2ab PS: if is calculated before you can use one of the equations below to solve for 03 bcos03-acos02 + ccos04+d bsin03=-asin02 + csin04 aw2 sin(02-03) 004 c sin(04-03) Velocity Analysis as sin(04-82) 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

Mechanical Engineering

For a short interval of motion, the input gear pivoted at O rotates with a constant angular velocity = 4 rad/s in clockwise sense (as shown in figure below). In the position shown, the force applied to the piston is 2F where F = 200 N; The diameter of the gear (G) is 30 cm, link AB is 70 cm and y₁ = 20 cm. a) Draw the vector loop for the given mechanism. b) Find the position of the piston. c) Find the velocity and acceleration of the piston. d) Using energy method, determine the torque T and the power required from the motor to operate the mechanism. e) Find the force in the member (BA). The piston has a mass mp = 0.65 kg. Neglect the mass of other links. All dimensions are in cm. A Y^ "(G) Ут B 2F X Position, Velocity and Acceleration Analysis: Pin-jointed Fourbar linkage VBA 041,2 002 R4 -B±√B2-4AC R1 Position Analysis = Zarctan 2A 004 VBA 03,2 = 2 arctan -E±√E2-4DF 2D A cos02-K₁ - K₂cos02 + K3 • B = -2sin02 C K₁ (K2 + 1)cos62 + K3 ⚫D=cose₂-K₁ - K4cos62 + K5 ⚫ E = -2sin02 F K₁+(K4-1)cos02 + K5 K₁₁ = 1 • K₁ = . K₂ -b²+c²+d² • K3 = 2ac c2-d²-a²-b2 ⚫ Ks 2ab PS: if is calculated before you can use one of the equations below to solve for 03 bcos03-acos02 + ccos04+d bsin03=-asin02 + csin04 aw2 sin(02-03) 004 c sin(04-03) Velocity Analysis as sin(04-82) 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 a short interval of motion, the input gear pivoted at O rotates with a constant angular velocity = 4 rad/s in clockwise sense (as shown in figure below). In the position shown, the force applied to the piston is 2F where F = 200 N; The diameter of the gear (G) is 30 cm, link AB is 70 cm and y₁ = 20 cm. a) Draw the vector loop for the given mechanism. b) Find the position of the piston. c) Find the velocity and acceleration of the piston. d) Using energy method, determine the torque T and the power required from the motor to operate the mechanism. e) Find the force in the member (BA). The piston has a mass mp = 0.65 kg. Neglect the mass of other links. All dimensions are in cm. A Y^ "(G) Ут B 2F X Position, Velocity and Acceleration Analysis: Pin-jointed Fourbar linkage VBA 041,2 002 R4 -B±√B2-4AC R1 Position Analysis = Zarctan 2A 004 VBA 03,2 = 2 arctan -E±√E2-4DF 2D A cos02-K₁ - K₂cos02 + K3 • B = -2sin02 C K₁ (K2 + 1)cos62 + K3 ⚫D=cose₂-K₁ - K4cos62 + K5 ⚫ E = -2sin02 F K₁+(K4-1)cos02 + K5 K₁₁ = 1 • K₁ = . K₂ -b²+c²+d² • K3 = 2ac c2-d²-a²-b2 ⚫ Ks 2ab PS: if is calculated before you can use one of the equations below to solve for 03 bcos03-acos02 + ccos04+d bsin03=-asin02 + csin04 aw2 sin(02-03) 004 c sin(04-03) Velocity Analysis as sin(04-82) 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

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

pls very urgent using the formulas required in the picture

For a short interval of motion, the input gear pivoted at O rotates with a constant angular velocity = 4
rad/s in clockwise sense (as shown in figure below). In the position shown, the force applied to the piston is
2F where F = 200 N; The diameter of the gear (G) is 30 cm, link AB is 70 cm and y₁ = 20 cm.
a) Draw the vector loop for the given mechanism.
b) Find the position of the piston.
c) Find the velocity and acceleration of the piston.
d) Using energy method, determine the torque T and the power required from the motor to
operate the mechanism.
e) Find the force in the member (BA).
The piston has a mass mp = 0.65 kg. Neglect the mass of other links. All dimensions are in cm.
A
Y^
"(G)
Ут
B
2F
X
Position, Velocity and Acceleration Analysis: Pin-jointed Fourbar linkage
VBA
041,2
002
R4
-B±√B2-4AC
R1
Position Analysis
= Zarctan
2A
004
VBA
03,2
= 2 arctan
-E±√E2-4DF
2D
A
cos02-K₁ - K₂cos02 + K3
• B = -2sin02
C K₁ (K2 + 1)cos62 + K3
⚫D=cose₂-K₁ - K4cos62 + K5
⚫ E = -2sin02
F
K₁+(K4-1)cos02 + K5
K₁₁ = 1
•
K₁ =
.
K₂
-b²+c²+d²
• K3 =
2ac
c2-d²-a²-b2
⚫ Ks
2ab
PS: if is calculated before you can use one of the equations below to solve for 03
bcos03-acos02 + ccos04+d
bsin03=-asin02 + csin04
aw2 sin(02-03)
004
c sin(04-03)
Velocity Analysis
as sin(04-82)
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

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