Find the velocities of points C and D.
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- The link length and value of O2 for some four bar linkages are defined below, (fixed link =12 cm, drive link =5 cm, coupler link = 10 cm, follower link= 8 cm) 1. draw the linkage to scale and graphically find all possible solution (both open and cross) for 03 and O4, analytical . 2. If w = 2 rad /s (CCW) find the angular velocity for bar 3 and 4 3.if a= 5 rad/s² (CCW) find the normal and tangential acceleration for link 3 and 4A pinjointed fourbar linkage is shown in the figure below. The link lengths are L1 = 7.8 in., L2 = 2.9 in., L3 = 7.25 in. and L4= 3.5 in. Using the current position, 02-100°, the position analysis has determined that 04 = 214.25°. Currently, link 2 is traveling at 5.50 rad/s CCW Lep B. Figure 6.7 Four-bar linkage acceleration analysis The angular velocities of link 3 w3 and and link 4 Wa are: Select one: O a. w3 = 3.589 rad/s, Wa = -2.91 rad/s O b. w3 = 2.07rad/s, wa = -2.91 rad/s C. W3 = 2.7 rad/s, wa 3.428 rad/s O d. w3 = 2.07 rad/s, w4 = -3.38 rad/sGiven a fourbar linkage (not drawn in scale) with the link lengths L2 = a = 6 cm, L3 = b = 25 cm, L4 = c = 30 cm, L1 = d = 12 cm and θ2 = 450. For RP = 4 cm, θ3 = 87.80, θ4 = 103.10, δ3 = 450, ω2 = 10 rad/s, ω3 = -7.7 rad/s, ω4 = -5.2 rad/s and α2 = 10 rad/sec2, the angular acceleration of link 3, α3 is equal to
- 1. Find a combination of link lengths where motion of a point on output link is one quarter of a circle. 2. Find the value of all 0, 0, 0, and y in open and close configuration Read the value of link lengths and the input angle 8., then use the formulae given below to calculate the value of unknowns 03, 0, and y K₁ = = K₂= d K2 K3 = a²-b²+c²+d² 2ac A = cos 0₂ - K₁ - K₂ cos 0₂ + K3 B = -2 sin 0₂ C = K₁ (K₂ + 1) cos 02 + K3 -B± √B²-4AC 2A 0412 = 2tan-1 d K₁ = — K5 = c²d²a²-6² 2ab D = cos 0₂ - K₁ - K4 cos 0₂ + K5 E = -2 sin 0₂ FK₁+ (K₁ - 1) cos 02 +K5 0312 2 tan-1 (-E± -E± √E²4DF 2D Y = 04-031 of 1 3. A four-bar linkage is shown below. (A) the directions of each. (B) E on link 3. Use the method of velocity vector polygons to calculate w, and wq. Report Find and report the magnitude and direction of the absolute velocity of point RAD = 10 cm, Rap = 15 cm, Ra = 60 cm; R = 65 cm; Rp = 15 cm W2 = 2 rad/s cw (Note: ADB and FEC are right triangles.) link 3 A. link 2A general pinjointed fourbar linkage is shown in the figure below. It has the followings: The link lengths are L1 = 8.50 in., L2 = 3.00 in., L3 = 5.00 in. and L4 = 4.50 in. The values of θ1 = 0, θ2 = 60°, and θ4 = 119°. The angular velocity of link2 ω2 = 10 rad/s CCW. The angular velocities of link 3 ω3 and and link 4 ω4 are: Select one: a. ω3 = 5.29 rad/s CW, ω4 = 4.80 rad/s CCW b. ω3 = 5.29rad/s CW, ω4 = 6.14 rad/s CW c. ω3 = 3.94 rad/s CCW, ω4 = 4.8 rad/s CCW d. ω3 = 3.94 rad/s CCW, ω4 = 6.14 rad/s CCW
- For the mechanism shown in figure. Obtain the angular velocity of link 3 and the slipping velocity between link 3 and link = 5 inches RAO 4 = 4 Using loop closure method. The lengths of links RPA 5î in/sec. LAPO = 90 degrees. 10 inches and VA = 0& TA r₂ 1 "A pinjointed fourbar linkage is shown in the figure below. • The link lengths are L, = 7.8 in., L2 = 2.9 in., L3 = 7.25 in. and L4 = 3.5 in. Using the current position, 82 =110°, the position analysis has determined that 04 = 209.21°. Currently, link 2 is traveling at 10 rad/s CW Le B. O, =0 Figure 6.7 Four-har linkage acceleration analysis The angular velocities of link 3 w3 and and link 4 wa are: Select one: a. W3 = 3.589 rad/s, wa = 4.81 rad/s O b. w3 = -4.29 rad/s, w4 = -4.38 rad/s O c. W3 = -4.29 rad/s, wa = 4.81 rad/s d. w3 = 2.7 rad/s, wa- 3.428 rad/sQ2: The four-bar linkage mechanism O₂ABO4 shown in Figure 2 is driven by link O₂A 604 at w₂ = 45 rad/s (ccw). Link AO₂= 101.6 mm, Link AB = 254 mm, Link Be₂ 304.8 mm and the fixed link O₂O4= 304.8 mm, link AO2= 101.6 mm, link AB = 254 mm, link BQ₂ = 304.8 mm and the fixed link O₂O4= 304.8 mm. and = 120⁰. BO4 Find the followings: 1. Velocity of point C located at the middle of the link AB. 2. Angular velocities of links AB and BO4 A W2 Ө B Figure 2
- Motor is driving a gear attached to the linkage in the mechanism as shown: Let r₁ r₂ = 10mm; c= 22mm; b= 100mm; a = 3mm; Determine the functions , of velocity of the slider with parameters (a,b,c) using the vector approach. (Hint: gears do not slip, link o-D is attached rigidly to the gear, pins at E and D allow free rotation) Schematic of kinematics: Velocity vector of point D: Velocity vector of point E: Equation of Velocity for the link E to F: b r₁ 12 WmA general pinjointed fourbar linkage is shown in the figure below. It has the followings: The link lengths are L1 = 8.50 in., L2 = 3.00 in., L3 = 5.00 in. and L4 = 4.50 in. The values of e1 = 0, 02 = 60°, and 04 = 119°. The angular velocity of link2 w2 = 10 rad/s CCW. L4 2 = 60° Ao B. O, = 0° Figure 5.32 Problems 5.3 and 5.4 The angular velocities of link 3 wz and and link 4 wa are: Select one: O a. w3 = 5.29rad/s CW, w4 = 6.14 rad/s CW O b. w3 = 3.94 rad/s CCW, W4 = 4.8 rad/s CCW O c. W3 = 3.94 rad/s CCW, wa = 6.14 rad/s CCW %3D O d. w3 = 5.29 rad/s CW, w4 = 4.80 rad/s CCW %3D %3DProblem 4-6a The link lengths (a, b, c, d) and the value of 2 for a crank-rocker linkage are defined as 2, 7, 9, 6, 30°, respectively. Draw the scaled linkage. Find all possible solutions (both open and crossed) for angles 03 and 04 graphically. Орen B A LNCS 4 a GCS र 4 4" Crossed (This is not the scaled kinematic diagram.) Problem 4-7a Repeat Problem 4-6a except solve by the vector loop method.