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 La = 3.5 in.Using the current position, 02 110°, the position analysis has determined that 0 = 209.21°.Currently, link 2 is traveling at 10 rad/s CWB.8, =0Figure 6.7 Four-bar linkage acceleration analysisThe angular velocities of link 3 wz and and link 4 wa are:Select one:O a. w3 = -4.29 rad/s, wa = 4.81 rad/sO b. w3 = 2.7 rad/s, w4 = 3.428 rad/sO c. W3 = -4.29 rad/s, wa = -4.38 rad/sO d. w3 = 3.589 rad/s, wa = 4.81 rad/s%3!%3D

Answered: Image /qna-images/answer/3929a9d6-9103-4b6e-b15c-156d4e2c2a1b.jpg

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 La = 3.5 in. Using the current position, 02 110°, the position analysis has determined that 0 = 209.21°. Currently, link 2 is traveling at 10 rad/s CW CO B. 8, =0 L, Figure 6.7 Four-bar linkage acceleration analysis The angular velocities of link 3 wz and and link 4 wa are: Select one: O a. w3 = -4.29 rad/s, wa = 4.81 rad/s O b. w3 = 2.7 rad/s, w4 = 3.428 rad/s O c. W3 = -4.29 rad/s, wa = -4.38 rad/s O d. wz = 3.589 rad/s, w = 4.81 rad/s %3!

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 La = 3.5 in. Using the current position, 02 110°, the position analysis has determined that 0 = 209.21°. Currently, link 2 is traveling at 10 rad/s CW CO B. 8, =0 L, Figure 6.7 Four-bar linkage acceleration analysis The angular velocities of link 3 wz and and link 4 wa are: Select one: O a. w3 = -4.29 rad/s, wa = 4.81 rad/s O b. w3 = 2.7 rad/s, w4 = 3.428 rad/s O c. W3 = -4.29 rad/s, wa = -4.38 rad/s O d. wz = 3.589 rad/s, w = 4.81 rad/s %3!

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

Show how you would find the velocity of each point (no values given just the equation) for an overlapping/superimpose 4 bar linkage. Find the equation for 100 degree.
Hand written plz asap please... Solve 1 hourr plzzzzz hand written plz asap ..i'll rate plz asap
Which of the following is the correct velocity vector equation for the linkage below? OA OB. je OD. je @ je jo 44 je 4+ @ -r@ je OPT+re+-= trwe 34 3 je Le 3=r₁₂@ jo je ³=r₁₂@₁₂e २० 2 2e 2 je 12 V3 1₁ 14
For the four-bar- linkage shown in the following figure. BC=68mm, CD=100mm, AD=120mm. Determine the range of AB to make it a crank-rocker mechanism.
Please solve question A6.3 attached below, Thank You!
Please all choose with explain
5. Estimate the sliding velocity d and the angular velocity @3 using instant center velocity analysis for the open configuration of the crank-slider linkage in Table P6-2 row g by drawing the linkage to scale and measuring lengths using a ruler and angles using a protractor. Link 2 Link 3 Offset 1.4 4 1 2 6 -3 3 8 2 3.5 10 1 20 -5 f 13 0 g 25 10 *Drawings of these linkages are in the PDF Problem Workbook folder. Row TDC04 a b с d e 002 Link 2 02 Y A LO 5 3 7 A 0₂ Link 3 Offset Slider position d 0₂ 45 60 -30 120 225 100 330 YA B 4 002 10 -12 -15 24 -50 -45 100 X 04= = 90° X
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
Find the number of cycles of all elements in the planetary mechanism in the figure. z2 = 16 z3 = 30 z5 = 60 z6 = 18 z7 = 24 nentry= 4400 d/dk
Consider an offset inverted four-bar slider crank mechanism as shown in the following figure. Determine the angular acceleration of link 3 given L2 = 6 cm, 0₂ = 60°, and other dimensions shown in the figure. 02 = = 70 rad/sec² CCW 0₂ = 90 rpm CW A 3 02 3.0 cm 10.0 cm 4 B3 on 3 B4 on 4
I want to explain where the numbers 6.2772 and -1.274 come from(are marked in red) just calculation for the two nb
P1 00:00 A 650.00 F 420.00 P B 450.00 P2 A four-bar mechanism is given in the below figure. Link P₁A and P₂B are made by steel and have uniform cross sections. The PAB section, which is a right triangle, made by aluminium. Link P₁A has rotational velocity and acceleration of 8 rad/s and 20 rad/s² (both counterclockwise), respectively. There is a horizontal force at P of F= 200 N. Given G₂, G3 and G4 are the centre of mass of the links P₁A, PAB and P₂B, respectively. The masses of the links P₁A, PAB and P₂B are in turn m₂ = 1 kg: m3 = 2 kg and m4 = 1.2 kg. The moment of inertia of the links P1A, PAB and P₂B are in turn IG2 = 0.08 kgm²; IG3 = 0.15 kgm² and IG4 = 0.1 kgm². Find all the forces acting on the joints at this instant.