Question

4713 guide pulle mnm 200 mm The linkage mechanism shown at right is driven by pulling the flexible band F, which is attached to the rotating sector at E. a) (8 points) If the band is being pulled at a constant speed of u = 4 m/s past the guide pulley, find the angular velocities of links AD and BD at the instant shown. W WBD 125 ro semmo geach wag sha

Answer 1

ANSWER Given ahta: * The linkage mechanism shows is driven by pulling the flexible band 'F' is a hatched at point 'E' of the circular body and leads over the guide publey * Band is being pulled at a constant speed, 4n1s 4m/sec 75 Guide Pulley MF 200 mm VA 425 Rotating Sector V * In the above figure Shows, at point I' is located the intersection and perspordicular to VA and VD Here, I is the Instantaneous center of Velocity

(a) * Instantaneous Center ode 2eso Velocity as link 'Ad' Given, velocity of constant speed us 4 m/s we have → Angular velocity at of WOE = 4 mis 200 mm * WE = 4mis 0.2001 - → WOE - 20 godians / sec At point 'OE' dotates about the point ó velocity, VA: WO E 1251m - 12517 we have) 0.125m) We = 20 radians / sec Substitute WOE value VA - ( 20 radlians/sec ) x (0.125m) Here 125mm is converted into 0.1257) VA" 2.5ms |

- Angulis velocity at_line AD WAD - VA AI we have solved above YA = 2.5 m/s 7 → WAD - 225 mls : goom 200mm is converted into 0. 200 meters >> WAD - 125 rodians /sec DI velocity , VO - OI. WAD → WAD - VD VD - DI, WAD (0.150 m) x (12.5 rod/sec) 150 mm is converted into Orisom We have already solved, WAD = 12.5 roplians/sec - VD = 1.875 M/S 7

* Angular velocity at link BD WBD = VD BD Here, VO = 1.875 Mis BD = 250mm > 0.250m = . won 1:875 ms 0.250m » TuoD = 7:5. radians /sec ]