Question

2. The crank AB rotates counter-clockwise with the angular velocity of 10 rad/s. The piston is displaced with da = 7.803 in from home position Co. The home position of the piston is yco = 43.803”. The unit vector of the sliding direction is: u Determine: (50 pts) u © - = 7.803 in DO-+c 2 in 24 in в (о 12 in А

Answer 1

solution Co DG3 26.56 de = 7.803 in DA 115 in 2 in gin 124 in (2 B 12 B B 9.- 4.50 12 in 12 45° А دم G2 G2 =D. D G 'B с A O x O 0 2 36 36 6 36 21 y O 12.00 CO Now , Displacement of biston from Point Co is given by BD BD -Sinzo R = AB [11-coso) + ( AB VA BD and l Let AB n 7 AB [(-c050) + Hence, x = (n²_ sin o) dx at dx do do dt. Velocity of piston, V

NOW, 4. = V = sing sino + z ſm²- sinzo Sin90° ته 12.2 (+10) Sin-45° + 26.83 1212 ہی वं.-V +2016/ in/ dv aa वv व . at at Cos20 wwcoso %3D -- → : d do h Cos-90 a 1212 (10) ८०545 + ofcos 12n 2 ने व %3D 120 in 24 12 10 Now, 4.5 wi = -V/(241) 22. 20161136 483+4 . --5.59_rad/g. "B

Now, Velocity of point B is given by WAB (AB) = 10 (1252) = 169,70 inle Ve Now, Linear accleration of point B ab - a A + AB/A • Relative acceleration = a B O + XABX (8B/A) - WAB (VB/A) => a = 10²(1252) =) _Qg = 1697.0 in /8 tony 0 + O OB for unK BD ao a + AB/D + => 1637 (coses î - sinas ģ) = 120 ſ 7-5.55*(= 19 i 424 ġ) + Kxf(-12 € -24 j) Comparing component we get - TG97 12 (5.59) + 480 (24) ABD 65.6 2 rad/82.

Linear acceleration of 67, an w? ( A ) = 10 (2) 848.52 ins? -> acon For linear acceleration off na Gore Wh (BP) =-5.59944 (aaa) विक.) 41979 inte (962) 280 (BEJE 65:62 793.73 2:19) in /s* (Ama) + = 879.96 2 ana 419.19 + $79.96€ 974.70 in/s For linear accleration of G₂ Le ac = 120 in/8