Question

2- In the following mechanism find a4 , AA, AB, Ac where w2=2 rad/s [CW] all dimensions are in mm and az 8 rad/s [CW] A 01A = 30 02 = 70° AB=70 AC=29 BC=53 04B = 25 01 17 64 04 B Solutions must include both graphical and analytical methods. Explain the step-by- step procedure for graphically developing the velocity/acceleration image. Detailed analytical procedure including all trigonometric functions used must be presented. The consistency between the graphical and analytical results must be confirmed.

Answer 1

= ABX WAB VALE No,A 2 configuration diagram lcm = lomm = Ca xo, А Velocity A link with seespect to = Vo, A O A = 30mm 2rad/sec 2 2 0 and on are fixed and there is not relative velocity in a and Celockwise) Gomm/sec ou No,a I 2 x 30 r 2.9 Velocity diagram scale rem = 20 mun/sec 15 By measuring diagram 3 3x15 = 0,53cm Vab = 2.3x15234-5mm/sec z 15mm/sec Ubc = 31.5 mm [sec Voub 45 mm/sec 3 3cm Yaoi 15.3 70° Vac 1.7 C 이, Ou Vab le Veja LA 1 Ou B 2.5 B o, ou -3 meter. 1 mm lo 45 mmlsec Noub wy I l. 8 rad/sec oub 25 mm زياع)

We accelaration diagram scale Icon=15mm (seca oaca t a AC 2 For Draw the accelatation diag reun need to Calcalabe both radial aand taugeutical component of each each link. ang = radial Vas component a AB XABX AB VO, A (6o)" 120mm/sec? aoa = 0 (Because we can Constant). ta O AB AB "AB a BC a Ebap AB с“, А 2 qAB OA 30 latoub le 10 Coub (posapeudicule to) lle (panelled to ar 그 V2 llink a Varu (mm) (cum/sec) at=&exr (mm/seca) O. OB that link) 10, A 30 Go 120 CHITOB) O, ou 34.5 ? 17 TAB 24 70 AC ? 15 29 7.75 atout = 3.4cm 31.5 18.72 ܟܢ 51 = xu (29) Bc53 -5lman (sec" Xu=2.04 rad /sect AA = CAK 25.5mm/sec2 1,2 X 15 18 mm/sec² Aca Aca aca = 1x15 18 am/seca = 1.7x15 a us 81 JOuB 25 AB (mm/sec)