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 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 O2O4 = 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 VBA R2 d x R₁ 04 Position Analysis -B±√B²-4AC -E±√E²-4DF 0412 =2arctan 03,2 = 2 arctan 2D 2A A cos02-K₁ - K₂cos02 + K3 B = -2sin02 ⚫ C=K₁- (K₂+ 1)cos02 + K3 ⚫D = cose₂-K₁ - K₁cose₂ + K5 E = -2sin02 F • K₁ = K₁+(K-1)cos02 + K5 . K₁ = • 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)