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Consider the inverted slider-crank linkage as shown in the following figure. Given are: • • Link lengths: Link O2A = 10 cm, Link O4B = 15 cm, Link 0204 = 20 cm. Positions: y=90°, 02=140, 03 = 198.8", 04 = 108.8", b = AB = 24.113 cm. Angular velocities: 002=-15 rad/s (CW), 003 004 = -3.22 rad/s (CW) • Velocity of slip: b = -80 cm/s • Link O2A rotates with a constant angular velocity. a) Calculate the angular acceleration of links 4 and 3 and the acceleration of slip at point B. b) Calculate the magnitude and direction of the acceleration of point B. A 0, TITITIT 02 B 03 04 ᎾᏊ X X Question 1: 9=70 b=24,113; c = 15; d=20 8=90°; 02=140°; 3 = 198,8°; 04 = 10818 α2=0 w₁ = -15 (d/s); 3 = 4 = -3,22 (rd(s); b = -80 (cm/s). 24=x3 = 57,33 (ra/15²)) A= 56973,77. -515,2. b = (A+B+C+D) 6+269 (2509) B= 0 C= 204 jou D= 8361,458 b=133 465 ((m/s) b = 19,34 (315)

Consider the inverted slider-crank linkage as shown in the following figure. Given are: • • Link lengths: Link O2A = 10 cm, Link O4B = 15 cm, Link 0204 = 20 cm. Positions: y=90°, 02=140, 03 = 198.8", 04 = 108.8", b = AB = 24.113 cm. Angular velocities: 002=-15 rad/s (CW), 003 004 = -3.22 rad/s (CW) • Velocity of slip: b = -80 cm/s • Link O2A rotates with a constant angular velocity. a) Calculate the angular acceleration of links 4 and 3 and the acceleration of slip at point B. b) Calculate the magnitude and direction of the acceleration of point B. A 0, TITITIT 02 B 03 04 ᎾᏊ X X Question 1: 9=70 b=24,113; c = 15; d=20 8=90°; 02=140°; 3 = 198,8°; 04 = 10818 α2=0 w₁ = -15 (d/s); 3 = 4 = -3,22 (rd(s); b = -80 (cm/s). 24=x3 = 57,33 (ra/15²)) A= 56973,77. -515,2. b = (A+B+C+D) 6+269 (2509) B= 0 C= 204 jou D= 8361,458 b=133 465 ((m/s) b = 19,34 (315)

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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i need to solve part b pls quickk very urgent part (a) is solved
Consider the inverted slider-crank linkage as shown in the following figure. Given are: • • Link lengths: Link O2A = 10 cm, Link O4B = 15 cm, Link 0204 = 20 cm. Positions: y=90°, 02=140, 03 = 198.8", 04 = 108.8", b = AB = 24.113 cm. Angular velocities: 002=-15 rad/s (CW), 003 004 = -3.22 rad/s (CW) • Velocity of slip: b = -80 cm/s • Link O2A rotates with a constant angular velocity. a) Calculate the angular acceleration of links 4 and 3 and the acceleration of slip at point B. b) Calculate the magnitude and direction of the acceleration of point B. A 0, TITITIT 02 B 03 04 ᎾᏊ X X Question 1: 9=70 b=24,113; c = 15; d=20 8=90°; 02=140°; 3 = 198,8°; 04 = 10818 α2=0 w₁ = -15 (d/s); 3 = 4 = -3,22 (rd(s); b = -80 (cm/s). 24=x3 = 57,33 (ra/15²)) A= 56973,77. -515,2. b = (A+B+C+D) 6+269 (2509) B= 0 C= 204 jou D= 8361,458 b=133 465 ((m/s) b = 19,34 (315)
Transcribed Image Text:Consider the inverted slider-crank linkage as shown in the following figure. Given are: • • Link lengths: Link O2A = 10 cm, Link O4B = 15 cm, Link 0204 = 20 cm. Positions: y=90°, 02=140, 03 = 198.8", 04 = 108.8", b = AB = 24.113 cm. Angular velocities: 002=-15 rad/s (CW), 003 004 = -3.22 rad/s (CW) • Velocity of slip: b = -80 cm/s • Link O2A rotates with a constant angular velocity. a) Calculate the angular acceleration of links 4 and 3 and the acceleration of slip at point B. b) Calculate the magnitude and direction of the acceleration of point B. A 0, TITITIT 02 B 03 04 ᎾᏊ X X Question 1: 9=70 b=24,113; c = 15; d=20 8=90°; 02=140°; 3 = 198,8°; 04 = 10818 α2=0 w₁ = -15 (d/s); 3 = 4 = -3,22 (rd(s); b = -80 (cm/s). 24=x3 = 57,33 (ra/15²)) A= 56973,77. -515,2. b = (A+B+C+D) 6+269 (2509) B= 0 C= 204 jou D= 8361,458 b=133 465 ((m/s) b = 19,34 (315)
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