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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 002 R4 -B±√B2-4AC R1 Position Analysis 004 VBA 041,2 A =2arctan 2A cos02-K₁ - K₂cos02 + K3 • B = -2sin02 C K₁ (K2 + 1)cos62 + K3 03,2 = 2 arctan -E±√E2-4DF 2D ⚫D=cose₂-K₁ - K4cos62 + K5 ⚫ E = -2sin02 •F K₁+(K4 - 1)cos02 + K5 K₁₁ = 1 • K₁ = . K₂ 2-b²+c²+d² • K3 = 2ac 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 bsin03=-asino₂ + csin04 aw2 sin(02-03) 004 c sin(04-03) Velocity Analysis as sin(04-0₂) 003 = 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

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 002 R4 -B±√B2-4AC R1 Position Analysis 004 VBA 041,2 A =2arctan 2A cos02-K₁ - K₂cos02 + K3 • B = -2sin02 C K₁ (K2 + 1)cos62 + K3 03,2 = 2 arctan -E±√E2-4DF 2D ⚫D=cose₂-K₁ - K4cos62 + K5 ⚫ E = -2sin02 •F K₁+(K4 - 1)cos02 + K5 K₁₁ = 1 • K₁ = . K₂ 2-b²+c²+d² • K3 = 2ac 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 bsin03=-asino₂ + csin04 aw2 sin(02-03) 004 c sin(04-03) Velocity Analysis as sin(04-0₂) 003 = 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

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
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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 002 R4 -B±√B2-4AC R1 Position Analysis 004 VBA 041,2 A =2arctan 2A cos02-K₁ - K₂cos02 + K3 • B = -2sin02 C K₁ (K2 + 1)cos62 + K3 03,2 = 2 arctan -E±√E2-4DF 2D ⚫D=cose₂-K₁ - K4cos62 + K5 ⚫ E = -2sin02 •F K₁+(K4 - 1)cos02 + K5 K₁₁ = 1 • K₁ = . K₂ 2-b²+c²+d² • K3 = 2ac 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 bsin03=-asino₂ + csin04 aw2 sin(02-03) 004 c sin(04-03) Velocity Analysis as sin(04-0₂) 003 = b sin(03-04)
Transcribed Image Text: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 002 R4 -B±√B2-4AC R1 Position Analysis 004 VBA 041,2 A =2arctan 2A cos02-K₁ - K₂cos02 + K3 • B = -2sin02 C K₁ (K2 + 1)cos62 + K3 03,2 = 2 arctan -E±√E2-4DF 2D ⚫D=cose₂-K₁ - K4cos62 + K5 ⚫ E = -2sin02 •F K₁+(K4 - 1)cos02 + K5 K₁₁ = 1 • K₁ = . K₂ 2-b²+c²+d² • K3 = 2ac 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 bsin03=-asino₂ + csin04 aw2 sin(02-03) 004 c sin(04-03) Velocity Analysis as sin(04-0₂) 003 = 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
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
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