Question

2. (20 points) Figure 2 shows a four-bar linkage. Body 1 is the fixed link or the ground, body 2 is the crankshaft OA, body 3 is the coupler AB, and body 4 is the rocker BC. Using the complex-algebra approach, obtain an expression for the angular velocities of the coupler and the rocker in terms of the angular orientation and angular velocity of the crankshaft. Assume that the angular velocity of the crankshaft is constant and is equal to 30 rad/s, and the lengths of the crankshaft, coupler, and the rocker are 0.2, 0.4, and 0.5 m, respectively, and the distance OC is 0.4 m. y3 D C Figure 2 Four-bar mechanism

Answer 1

04 2m 4B 2 4 m B c2.5m O c = 4m 84 9t is also cla Av nechor loop approach- y rector i = n(cose tisino) Breio close loop! - 827 83 - 84-8izo reld + 83 83 rue ou 8,680 - 0 => 02 ( cos Oz ti sin02) + Y3 (costs + isin83) - ruccosout i sinow) - Mi Cos 8tisingi) = 0 geperating real and imaginaey parts Real: 82 coso2 +83 Cosoz - ry cosOy - ricoso, = 0 - ② quaqinay? 82 sind + 83 sino 3 - ru sindy - risind, zo - ③ Now are know a so differentiate eq? ☺ we get 820zi e d2 + 8z dzi i redui idu o=0 - expard equation ☺ and seperate real and imaginary parts constants Real! _ 82 82 sin Oz - 83 8z singz + ryby sin du = 0 - ② { ixi=-1} quaginary: 82 or coso 2 + 83 di cosez - ry by cosoy = 0 - 6 from eq ② and ③ if 81, 82, 83, ry are given then W3 = 68 = -82 sin (02-84) wo --.2 sin ( 02-04) x 30 8z sin (03-04) P4 sin ( 03-9u wy- ou = -82 sin (02-03) w - - •2 ein (02-03) x 30 ry sin(03-04) 5 sin (03-04) de 191 2 De Wy= - 15 wys - 19 & in (02-04) sin (03-04) sin (02-03) ein (03-04)