i need to solve part b pls quickk very urgent part (a) is solved 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, 03 = 198.8", 04 = 108.8", b = AB = 24.113 cm.Angular velocities: 002=-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.A0,TITITIT02B0304ᎾᏊXXQuestion 1:9=70b=24,113; c = 15; d=208=90°; 02=140°; 3 = 198,8°; 04 = 10818α2=0w₁ = -15 (d/s); 3 = 4 = -3,22 (rd(s); b = -80 (cm/s).24=x3 = 57,33 (ra/15²))A= 56973,77.-515,2.b =(A+B+C+D)6+269 (2509)B=0C=204jouD= 8361,458b=133 465 ((m/s)b= 19,34 (315)

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Answered: Consider the inverted slider-crank…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

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theta O = 55 The link length and value of O2 for some four bar linkage are defined below, (drive link =4.2 cm , coupler link =5.9 cm , follower link = 5.9 cm, fixed link= 6.85 cm 1. draw the linkage to scale and graphically find all possible solution (both open and cross) for 03 and O4 2. If w = 2 rad /s (CCW) find the angular velocity for bar 3 and 4 3.if a= 5 rad/s? (Ccw) find the normal and tangential acceleration for link 3 and 4
A 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/s
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 4
A 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 %3D
A 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 0, = 0, 02 = 60°, and 04 = 119°. The angular velocity of link2 w2 = 10 rad/s CCW. L4 L2 02= 60° @2 B. Ao O = 0° Figure 5.32 Problems 5.3 and 5.4 The angular velocities of link 3 w3 and and link 4 w4 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 5.29 rad/s CW, w4 = 4.80 rad/s CCW O d. w3 = 3.94 rad/s CCW, W4 = 6.14 rad/s CCW
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 = 04-03
The linkage in the following figure has O2A = 0.8, AB = 1.93, AC = 1.33, and offset = 0.38 in. The crank angle in the position shown is 34.3° and angle BAC = 38.6°. Find a 3, A4, Ag and Ac for the position shown for 02 = 15 rad/s, a2= 10 rad/sec² in direction shown. 5. a) Using the acceleration difference graphical method b) Using an analytical method. A 3 B 02 4 1
Problem 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.
Qestion-1: A slider-crank mechanism is shown in figure below. Position vectors for various linkages are drawn as shown in figure. Length of the rotating Link-1 (L1) is continuously changing due to the slider placed on it. Link-3 is fixed and aligned along y-axis. Note: All angles are measured anti-clockwise from x-axis and Link-1 is rotating with uniform velocity Take: L2 = 20mm, 02 =45°, L3= 30mm, o=10rad/sec, V2= 1000mm/s. a) Formulate the vector loop, position, velocity and acceleration equations b) Solve to find 01, L1, linear velocity of slider and angular acceleration of link-1. c) Identify if the mechanism can be declared as spatial or planar mechanism also identify total number of full and half joints.
1. For the fourbar linkage shown (Fig. 1) [5]: 25 cm A 34 10 cm (crank) 02 3 20 cm 25 cm 25 cm P 04 FIGURE 1. Hoeken's linkage (a) Solve for 03 and 04 if the crank angle 02 = 120°. Show your work. (You may want to use the scripts to verify your answers.) (b) Find the position of point P when the crank angle is at 60 degrees.
A 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
According to the below figure, if the rotational velocity of link AB is constant and equal to 2 rad/s (wAB = 2rad/s), find the velocity of point E. This problem should be solved based on the concept of the 400 mm instantaneous center of rotation. Other methods are not acceptable. 700 mm 400 700 mm mm

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