Question

How do I solve for the angle for the link BC? The angle BC I'm referring to is using the horizontal axis as a reference. Thank you.

Problem 1: (10 Marks), (Figure from Reference 1. Problem 3.46, modified For the mechanism in the posture shown, e2 = 100, and 42 = 15 dCW constant. Consider 02A = 0.5 m, AB = 1 m, and BC = 1.2 m. Link BC is pinned to link AD. a) Determine the velocity vector of the slider C on the inclined surface with the method of relative kinematics. b) Determine the acceleration vector of point B with the method of relative kinematics. Note: DO NOT solve any part of the problem with the loop closure method. Mechanism Dimensions 1.4 m -------- 1.7 m --- --

Answer 1

solution:- Y we have given in the question that D2 = 1000 8 = 15 radsec, cw OZA=0.5m, AB: 1m), BC= 1-2m - -- we will draw velocity diagram for the given Mechanism using Relative velocity VAG= 6 x 0₂A = 15X0.5 VAQ=smys now taking scale of smys = 5cm, let take faxed pointe as Ozo yer 5 cm - - - - - - - - - - - - - - -

omitially we take the perpendicular to the on and draw it. Then we take line parallel to Ab since it is slotted mechanism. Take perpendicular line, to AD and join it with D. Take a midolle on line at AD which will be point to Draw a line parallel to slider motion of c, from point in velocity diogram. & Draw a perpendicular linet BC links from point b. ① Measure the line oc and we will get so = 3.1cm so from scale 5 m/2 - 5cm. 1m = 1 cm 3. 1cm = 3.1 m/s Thereforeo!. 3.1 m/s Answer ül for accelelalóon xector keink B?- a = artat when ars radial acceleration als tangential aceelevation are war since w is same as w of link o, A that is a = to 15 radser length of ob can be a mealmeed from velocity diagram. we get Ob = 3,8 cm this Ozb is velocity 50 To = 3.8 m/sec a = y2 = (3.82 & 16.04 m/ 22 0.9 AD will only reciprate there will be no at. sine e link Tar 16.04 / 2 | Answer