Question

Part B The dimension of the links in a four-bar mechanism shown in Figure 2a below are: 02A = 60 mm, AB = 110 mm, 04B = 90 mm, 0204 = 150 mm, AC = 80 mm, and <BAC = 30°. If 02 = 60° and the angular velocity of link 02A = 20 rad/s in clockwise direction. Using vectorial approach, 0,0, Figure 2 3. Determine the velocity of B, VB 4. Determine the velocity of C, Vc Hint: Figure 2b → 03 = 0 -ß and 04 = 180 - (8 +B)

Answer 1

X velocities of links AB, BO, respectively. let wzwy be angular given, 0, 5600 z = zoradisec cu ZBAC= 30°. step : using Geometry find all required angles to o o Aou Apply sine rule 150 sinß sin(180-(68+8)) a sin (120aß) - 150 0 - sinß ISO sin 1200 c0812 tang - B=23.410 ✓ nou 60 = Aly=130.784 mm I sin 60o - Sinß

now, apply cosine rule to AABOU, COSO - 1102 +130.7842 - 902 2 X 110 X 130 784 ::$ = 42.8207 An Also, by sinedule in DABOY, sinh sinu2.82 90 110 *2=56.178°. / >> G3 = $-$ = 19.414 → Oy = 190- (O+B) = 180 - (56-17 +23:41) = 100.42€ ✓ step(2) : find velocity comparing by vectorial approach, to Ws, Wu To = JA +UBIA = (60 x 20) (cos (90-03)i -sin (90-03)9) (10x esy) (cos (90-03) † - sin (90-03);) (131.25 + 365% so, + (- Goo - (03- 484,53 Lo

Also, To = Volon = (pow.) (cos (40+ Qubî +sin 690 7 Qu) ) = 98.515 wu i + 16.277W. ☺ ☺ 0 - © 1039.23 +36-5558W3 = 88.515 Wy .: Wy = 11.7407 + 0.913 Wz AISO, -600-103.748W3 = 16.277 Wy → Wy = -36,862 - 6.374 wg > 9 = > 11.7407 +0.413 W3 = -36.862 - 6.374 W2 --- = -7.161 Gadis - ④ = Wy = 8,783 rad's > Up = 777,427 î + 142.96 } -|- VB = 790,46 mm/s) ->2=180-(90+30-03) - =60603 1 now, uc = J + vara = 60K 20 Ccosso î - sin 20 ſ) Coox and (7.161)) (-cos (60+63) – sin C60403)] = 933.946 9 -36.877 (Vc = 934.67 mmis 1