Question

Find the acceleration using both the graphical and analytical method

5 m P#4 L1= 6m L2 3 m La= 5 m L4= 4 m e2= 65" W2= 1.5 rad/s (CCW) alphag = 0 rad/s O3= ?? W3 = ??, direction ?? ??, direction ?? W4 VAB ?? B04=?? alphag ??, direction = ?? alpha, ??, direction ?? 6 m 3 m 65.00

Answer 1

Seln 5m 4m w1.5m Sm t to seal 2 65D 6m we knaw- (V= 2 NAD2 0AO2 A02 2 - (5)x 3 6.75 ms aBA Since do mot haure hane VAS? we BA BA dthese, NBot Bo4 to do B0A analya O HOHACKs (AD). d Ap

Veloaity analye we w(Ao) 15 x 3 4.5 m A 65 02 ceniudien, 65 ding Vel. VADA 4,5ml& e4 129 are knaw D Vho2 C,0,) ab 0.48 CC 2.4m IS.; w om R 0.85 c 3.4 m. ;w

Noo- 6.75 x mls t a 1.152 m)st BA 8A BA 1 Bo4 2.83 m/s BO4 Bot

Pugcedure acceluration ω.R.t. Paint = 6.15 m/s2 O 2 A a apA 1-152 mls (BA) () BA - 2.89 m/s2 B 4 unten make the dingram Steps to stationmy peints 1) Plot the lank paper ine 15), paxallee to 2) Daw a abselute ngen hal empenunt is zerg aA acch 3) frem (a) nw a ine (15); ve (b 8inee line de we dent KaDw 4.) Praw a 5 kem (0, ) do ine (2.8) porallel B04;(b). he pandiuloy Point 6) rom daw a intursut Jine which will to,oa) p6tain abanlt acc" 7) Fein apos to obtain absolute accl" &) Join

Aceleration iggram B 15s 65 O2 04

Now (ram g) 3-4 (BA) (SA) t a A3-4/5 (8a) () 0-68 xad/s & 5.1 1-275 xnds Other answers 03 129 0.48 xadls (Ccw) (ccs) い4= 0.85 ad/s NTIBOA) 3.4 m)s VT (AB) ネチ mys.