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DEXAMPLE 7-3 Acceleration Analysis of a Fourbar Crank-Slider Linkage with a Vector Loop Method. Problem: Solution: Given a fourbar crank-slider linkage with the link lengths L₂ = a = 40 mm, L3 = b = 120 mm, offset = c = -20 mm. For 02 = 60°, 002 = −30 rad/sec, and o₂ = 20 rad/sec², find α3 and linear acceleration of the slider for the open circuit. Use the angles, positions, and angular velocities found for the same linkage in Examples 4-2 and 6-8. (See Figure 7-6 for nomenclature.) 1 Example 4-2 found angle 03 = 152.91° and slider position d = 126.84 mm for the open circuit. Example 6-8 found the the coupler angular velocity 003 to be 5.616 rad/sec. 2 Using equation 7.16d and the data from step 1, calculate the coupler angular acceleration α.3. α3 aα₂ cose₂-asin02 +bw sin03 bcos03 40(20) cos 60°-40(-30)² sin 60° +120(5.616)² sin 152.91° 120 cos 152.91° =271.94 rad/sec² (a) 3 Using equation 7.16e and the data from steps 1 and 3, calculate the slider acceleration ä. ä=-aα₂ sine₂-aw cosе2 +bα3 sin03 +b² cos03 =-40(20) sin 60°-40(-30)² cos 60° +120(271.94)sin 152.91° +120(5.616)² cos152.91° =-7.203 m/sec² (b)

DEXAMPLE 7-3 Acceleration Analysis of a Fourbar Crank-Slider Linkage with a Vector Loop Method. Problem: Solution: Given a fourbar crank-slider linkage with the link lengths L₂ = a = 40 mm, L3 = b = 120 mm, offset = c = -20 mm. For 02 = 60°, 002 = −30 rad/sec, and o₂ = 20 rad/sec², find α3 and linear acceleration of the slider for the open circuit. Use the angles, positions, and angular velocities found for the same linkage in Examples 4-2 and 6-8. (See Figure 7-6 for nomenclature.) 1 Example 4-2 found angle 03 = 152.91° and slider position d = 126.84 mm for the open circuit. Example 6-8 found the the coupler angular velocity 003 to be 5.616 rad/sec. 2 Using equation 7.16d and the data from step 1, calculate the coupler angular acceleration α.3. α3 aα₂ cose₂-asin02 +bw sin03 bcos03 40(20) cos 60°-40(-30)² sin 60° +120(5.616)² sin 152.91° 120 cos 152.91° =271.94 rad/sec² (a) 3 Using equation 7.16e and the data from steps 1 and 3, calculate the slider acceleration ä. ä=-aα₂ sine₂-aw cosе2 +bα3 sin03 +b² cos03 =-40(20) sin 60°-40(-30)² cos 60° +120(271.94)sin 152.91° +120(5.616)² cos152.91° =-7.203 m/sec² (b)

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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DEXAMPLE 7-3 Acceleration Analysis of a Fourbar Crank-Slider Linkage with a Vector Loop Method. Problem: Solution: Given a fourbar crank-slider linkage with the link lengths L₂ = a = 40 mm, L3 = b = 120 mm, offset = c = -20 mm. For 02 = 60°, 002 = −30 rad/sec, and o₂ = 20 rad/sec², find α3 and linear acceleration of the slider for the open circuit. Use the angles, positions, and angular velocities found for the same linkage in Examples 4-2 and 6-8. (See Figure 7-6 for nomenclature.) 1 Example 4-2 found angle 03 = 152.91° and slider position d = 126.84 mm for the open circuit. Example 6-8 found the the coupler angular velocity 003 to be 5.616 rad/sec. 2 Using equation 7.16d and the data from step 1, calculate the coupler angular acceleration α.3. α3 aα₂ cose₂-asin02 +bw sin03 bcos03 40(20) cos 60°-40(-30)² sin 60° +120(5.616)² sin 152.91° 120 cos 152.91° =271.94 rad/sec² (a) 3 Using equation 7.16e and the data from steps 1 and 3, calculate the slider acceleration ä. ä=-aα₂ sine₂-aw cosе2 +bα3 sin03 +b² cos03 =-40(20) sin 60°-40(-30)² cos 60° +120(271.94)sin 152.91° +120(5.616)² cos152.91° =-7.203 m/sec² (b)
Transcribed Image Text:DEXAMPLE 7-3 Acceleration Analysis of a Fourbar Crank-Slider Linkage with a Vector Loop Method. Problem: Solution: Given a fourbar crank-slider linkage with the link lengths L₂ = a = 40 mm, L3 = b = 120 mm, offset = c = -20 mm. For 02 = 60°, 002 = −30 rad/sec, and o₂ = 20 rad/sec², find α3 and linear acceleration of the slider for the open circuit. Use the angles, positions, and angular velocities found for the same linkage in Examples 4-2 and 6-8. (See Figure 7-6 for nomenclature.) 1 Example 4-2 found angle 03 = 152.91° and slider position d = 126.84 mm for the open circuit. Example 6-8 found the the coupler angular velocity 003 to be 5.616 rad/sec. 2 Using equation 7.16d and the data from step 1, calculate the coupler angular acceleration α.3. α3 aα₂ cose₂-asin02 +bw sin03 bcos03 40(20) cos 60°-40(-30)² sin 60° +120(5.616)² sin 152.91° 120 cos 152.91° =271.94 rad/sec² (a) 3 Using equation 7.16e and the data from steps 1 and 3, calculate the slider acceleration ä. ä=-aα₂ sine₂-aw cosе2 +bα3 sin03 +b² cos03 =-40(20) sin 60°-40(-30)² cos 60° +120(271.94)sin 152.91° +120(5.616)² cos152.91° =-7.203 m/sec² (b)
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