Question

3- ( 20% ) Figure 3 shows a linkage (not to scale) and its dimensions, where LOA = 0.8 in@241", Loub= 2.97 in, Luc = 2.61 in, L0204 =1.85in @278.5º. Graphically find the accelerations of points A, B, and C for the position shown if w, = 40 rad/min and a, = -1500 rad/min? as shown. (Tip: please redraw the linkage according to the actual scale before solving the problem.) 25 2.97 261 - 0,0,185278 Figure 3

Answer 1

trinen 0 Loup = 2.97in = 7.5438 cm LOLA = 0.8 in 0 241 = 2.032 can Loc=2.61 in = 6.6294 cm LOLA = 3.3 cm 20402 = 1.85ine 278° = 4.699 cm W22 40 rad/pec (cow) L2 = -1500 rad/m² = -1500 radh 3600 & 2 = 0.4166 rad /s2 B. No Totion used I perpendicular 11 parallel Wirite with respect to General Du 0 acc. diagram Vay relocity diagram relocity of a with respect to y In actual relocity of A Scale 2:1 aay لعنلممود acc , wirt SA means I am=o.sun/s scale 2:1 lan=0.5cm/? A IB 02 VAJA oul VAOR A atay = tangential are ole n wirit y B

60 so I 3.3 - teps to draw relocity triangle o first I draw WA Oq = W2X (02A) O2 A - 40 x 21032 = 1.35 46 anls so I take scale 2:1 and draw 2.709 cm on diagram. and this velocity will be I to link 2. 2 now relocity of A' (point on link 4) wirrt A will be parallel to link 4. so ļ draw A A (magnitude unknown) DVA ou = "Aloz [beth O2 & ou are on afined link] must be I to link yo so g got the point A. measure O2 A' = 14 an = Q. 7an/s so I calculate 2.8 -0.8484 red Is O4 A row VB = WX ou B = 0.84848 7.5438 = 6.4 cm runts I so q draw = 12.80 cm on diagram Wit VA = 0.7 = 0.2121 (onsular vel of link 4) ou A 3.3 VB = W X ou B = 0.2121 X 7 5 438 = 1.6cnls go g draw 3.2cm on diagram 9 VCB = mas unknown but direction known i. e I to link 5 and Vc = mag. unknown but direction known je horizantal' By was intersection of these two lines I got poida and I measure it 3.3 cm i e e = 1.65cm/s now coming to acceleration diagram steps :- 1 ano;=anor tanor ad A2 = wax 0₂A = (40 x 2.032 = 0.903 11 cm/s2 ho 8 N.

(3 3600 30 g draw how 11 to linku) 2 = 0.148 con 12 but I take ail scale and draw 1.806 cm on diagrams. 02A1 = 1.806cm 0903) can/h a a TAO2 = &R=-1500 rad read/ X 2.032 = 0.8466 cen/s² 1.69 an on diagram A, A = 1.69am = 0.8466 an/s? now acc. of A with respect to A must be Il to link 4 (I draw a line with which is how radial acc of A Worot 4 wyt (Aon) = 10.2121) X (3:3) so g draw 0.296 cm on diagram and then I draw at Ao which is t to link 4, and I got point A I measure it o. bem so at Al = 0.3 m/s² a=a=0.3 - o.ogo rad/s2 R a Boy = % (046) = 0.0909x 7.5438 = 0.68 socm/s so I draw 137cm on diagram (t to ou B) a boy = wux (BO4) = 6-21215x7.5438 = 0.3393 so I draw 0.6787 an on diagram (11 to 4B) so & reached at point B. g measure it from O2 = 1.8cm ag=0.9 anls & Reza w 2 x BC = NCBI 2 Vox =0.9 BC = 0.12 21 20 cm/s2 6.6294 -0.3 ou A 33 2 now 2 X BC Bc

now I draw at co sie t to link be and then foon of a wirit O2 is horizartal § got point c. 9 measure it g measure it = 02 C=1.5cm accal point e=0.75 anls? ac = 0.75 cm/s2