The figure below shows a pin-jointed four-bar mechanism. The input link (EC) has a constant angular velocity of 15 rad/s CCW. (DB) is the output link and (CB) is the coupler link. At the instant shown: a) Draw the vector loop of the mechanism b) If 0: 85, prove that the input angle in (X,Y) the global coordinate system is 02, global = 10.52' c) Determine the angular velocity of output link (DB) and coupler link (CB). d) Determine the absolute velocity of point A in cartesian form (also find its magnitude and direction); if 8=15.8° and CA = a = 1.84 m. Y' 0.9 m 0.5 m C 03 0.6 m 0.25 m B 2.1 m D- B 04 2.0 m X Position, Velocity and Acceleration Analysis: Pin-jointed Fourbar linkage VBA R3 R4 R2 004 d R1 -B±√B²-4AC 041,2 = 2arctan 2A Position Analysis VBA (-E±√√E²-4DF 031.2 = 2arctan 2D A = cos 2-K₁ - K₂cos02 + K3 B = -2sin02 C K₁ (K₂+ 1)cos02 + K3 - D= cose₂-K₁- K₁cos02 + K5 ⚫ E = -2sin02 ⚫F K₁+(K4-1)cos82 + K5 K₁ == K₂ +d2 • K3 2ac d K₁ = • K4 c2-d2-a2-b2 ⚫ K5 2ab PS: if is calculated before you can use one of the equations below to solve for 03 bcos03=-acos02+ ccos04+d bsin03 = −asin02 + csin04 aw2 sin(02-03) 004= c sin(04-03) Velocity Analysis aw2 sin(04-02) 003 = b sin(03-04) Position Velocity and Acceleration Analysis: Slider-Crank Linkage AA R2 AB R3 R₁ ABA R4 ABA x AB AA A Position analysis 031 = arcsin (asin02 b (b) d = acos02-bcos03 asino2 03: =arcsin +π Velocity analysis a cos02 d= =-a02 sine₂+b03 sin03 03 -002 b cose 02 аз Acceleration Analysis aα₂ cosе₂-asin02 +bsin03 bcos03 d=-aa2 sine₂-awcos02 +bα3 sin 03 +bwcose Acceleration Analysis: Inverted Slider-Crank Linkage B AABoriali R₂ b dot AAB R1 X A X α4= aacos(0-0)+sin(03-03)]+co sin(0,-03)-2003 b+ccos(03-04) Jamboos(0-0)+cos(0-0)]+ax[sin(0-0)-csin(04 −0₂)|| - • 26cm, sin(0, −0,)−m;[b² +c² +2bccos(0, −03)] b+ccos(03-04) " k=2 Energy Equation Fk Vk + kk = mak • Vk + Σkαk ·@k k=2 -,。, k=2 k=2