Position Velocity and Acceleration Analysis: Slider-Crank Linkage AB R3 AA R2 R₁ ABA R4 ABA x AB (b) AA A Position analysis 031 = arcsin (asin02 b d = acos02-bcos03 asino2 03: =arcsin +π Velocity analysis a cose₂ d= =-a02 sine₂+b03 sin03 03 -002 b cosey 02 аз Acceleration Analysis aα₂ cosе₂-asin02 +bsin03 bcos03 d=-a2 sine₂-aw cos02 +bα3 sin03 +bwcos03 Acceleration Analysis: Inverted Slider-Crank Linkage R₂ R₁ b dot AAB AAB Corioli X α4= aacos(0-0)+sin(0-0)]+co sin(0,-03)-2003 b+ccos(03-04) amboos(0-0)+ccos(0-0)]+ax[sin(@2-03)-csin(04 −0₂)|| - • 26cm, sin(0, −0,)−m;[b² +c² +2bccos(0,−03)] b+ccos(03-04) " k=2 Energy Equation -,。, FV + 1 = a + k k=2 k=2 k=2 Consider the inverted slider-crank linkage as shown in the following figure. Given are: Link lengths: Link O₂A = 10 cm, Link O4B = 15 cm, Link 0204 = 20 cm. Positions: y=90, 02 = 140°, 03198.8", 04 = 108.8, b = AB = 24.113 cm. Angular velocities: (02-15 rad/s (CW), 003 004 = -3.22 rad/s (CW) Velocity of slip: b = -80 cm/s = ⚫ Link O₂A rotates with a constant angular velocity. a) Calculate the angular acceleration of links 4 and 3 and the acceleration of slip at point B. b) Calculate the magnitude and direction of the acceleration of point B. YA 02 Ꮎ, 03 B X 04 04 Position, Velocity and Acceleration Analysis: Pin-jointed Fourbar linkage Y VBA R3 R4 R2 d R₁ -B±√B²-4AC 0412 =2arctan 2A Position Analysis x 04 VBA -E±√E²-4DF 03,2 = 2 arctan 2D D A cose₂-K₁ - K₂cos02 + K3 B = -2sin02 ⚫C=K₁- (K₂+1)cos0₂+K3 cose₂-K₁ - K₁cose₂+ Ks E = -2sin02 F • K₁ = K₁+(K-1)cos02 + K5 . K₁₁ = • K₂ a²-b²+c²+d² • K3 2ac • K₁₁ = c2-d²-a²-b2 ⚫ K5 2ab PS: if 0 is calculated before you can use one of the equations below to solve for 03 bcos03-acos02+ccos04+d bsin03=-asin02 + csin04 Velocity Analysis aw2 sin(02-03) 004 c sin(04-03) a 003 sin(04-0₂) b sin(03-04)