14. Use the same diagram for problems 14 to 19.A force F = 800 N, has its line of action in the direction from point A to point B, as shown in the figure below.Express the force F as a Cartesian vector.2 mF- 800 NA ..3 m12 m•D800 (2,9.3)(2,-9,-3)(-2,9.5)800A)V94C)V94800B)800V94(-2,9,-3)D)V9415. Determine the components of the force F parallel to axis AD.-64A)V94-64C)V94(4,3.5)-64B)V94-4, 3, –5)D)16. Determine the components of the force F perpendicular to axis AD.A) (191.43, 722.82,280.55)B) (191.43, – 722.82, –280.55)(-191.43, – 722.82, –280.55)D) (-191.43,– 722.82, 280.55)C)17. What is the angle between OB and OD?A) 126.87°B) 127.65°C) 124.98°D) 125.84°18. Determine the moment of the force F about point O.800A) M, =(12,6,-10)800C) M, =V94:(-12,-6,-10)(-12,-6,10)V94800-10)800D) M, =194B) M, =12,-6,19419. What is the moment of the force F about the axis OC?990A) Molloc =-(4.8.0)960C) Molloc =V94890D) Molloc =V94880B) Mll oc = (4,8, 0)

Answered: Image /qna-images/answer/0eb67ffb-9bf2-4238-8a3e-68a7b7ef0c16.jpg

Answered: 4. Use the same diagram for problems 14…

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

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Q) A simplified model as shown in figure below of a planar robotic arm. You may assume that each segment has length 1.5 m, and the applied force is 250 N. 01 = 35 degree and 02 = 03 = 20 degree. Calculate the moment that the force, F, exerts .about the point A F C A M = -1042 M = -1055 M = -1050 M = -1046
Q4: The force in the Figure acts on the hook. Express it as a Cartesian vector. F = 675N (총)(S m) 2 m FB.uB (675)(0.7936i + 0.5151j+0.3282k) O FB.uB = (675)(-0.7936i + 0.5151j-0.3282k) FB.uB = (675)(-0.7936i - 0.5151j+0.3282k) FB-uB = (675)(-0.7936i + 0.5151j+0.3282k) O
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Q) A simplified model as shown in figure below of a planar robotic arm. You may assume that each segment has length 1.5 m, and the applied force is 250 N. 01 : 35° and 02 = 03 = 20°. Calculate the moment that the force, F, exerts about the point A. F В D 03 A
Q) A simplified model as shown in figure below of a planar robotic arm. You may assume that each segment has length 1.5 m, and the applied force is 250 N. e, = 35" and 02 0; 20°. Calculate the moment that the force, F, exerts about the point A. F B
Q) A simplified model as shown in figure below of a planar robotic arm. You may assume that each segment has length 1.5 m, and the applied force is 250 N. 01 = 35 degree and 02 = 03 = 20 degree. Calculate the %3D moment that the force, F, exerts .about the point A F A. M = -1042 O M = -1055 M = -1046 M = -1050 O
A:01 IE 0 & F1 F2 F3 b a F4 D E B FS Consider the following values: - a - 25 m; b - 5 m; a - 25 m; d - 5 m; F1 - 3 kN; F2 - 1 kN; F3 - 4 kN; F4 - 6 kN; F5 - 8 kN; a = 30°, 0 = 60°, B= 30° 1] What is the resultant moment of the five forces acting on the rod about point A? a) 85.8 kN.m b) - 21.1 kN.m c) - 13.5 kN.m d) 7.9 kN.m e) 24.1 kN.m f) - 57.8 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 188.8 kN.m b) - 153.5 kN.m c) 111.6 kN.m d) - 98.9 kN.m e) 85.4 kN.m f)- 244.2 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) 186 kN.m b) - 171 kN.m c) 155 kN.m d) 279 kN.m e) 423 KN.m 0 353 KN.m 4] What is the resultant moment of the five forces acting on the rod about point D? b) - 184.2 kN.m )-211.9 KN.m d) 126.9 kN.m e) - 27.1 KN.m ) 125.1 kN.m a) 117.8 kN.m Activate 5] What is the moment of the force F2 about point E? Go to Setting 1) 32.3 kN.m a) 11.7 KN.m b) 19.8 kN.m c) 54.6 kN.m d) 9.1 kN.m e) 41.1…
F1 F2 a C F3 A d F4 D Consider the following values: - F5 a = 25 m; b = 5 m; c = 25 m; d = 5 m; F1 = 4 kN; F2 = 2 kN; F3 = 5 kN; F4 = 7 kN; F5 = 3 kN; a = 30° , 0 = 60° , B = 30° 1] What is the resultant moment of the five forces acting on the rod about point A?
F1 F2 a B C. F4 D FS Consider the following values: a - 25 m; b - 5 m; c - 25 m; d - 5 m; F1 = 3 kN; F2 - 4 kN; F3 - 5 kN; P4 - 6 kN; F5 - 7 kN; a = 30°, 0 = 60°, B= 30° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 65.8 kN.m )-74.6 kN.m d) 42.9 KN.m e) - 53.7 kN.m f) - 97.1 kN.m b) 82.7 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 98.8 kN.m b) 112.7 KN.m c) 321.6 kN.m d) 88.9 kN.m e) - 138.7 kN.m f) 214.2 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) - $6.7 kN.m b) 111.1 KN.m e) - 254.6 kN.m d) 217.9 kN.m e) - 146.2 kN.m ) 135.3 KN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) - 117.8 kN.m b) 182.7 KN.m c)- 281.9 KN.m d) 316.9 kN.m e) - 141.4 kN.m ) 210.1 kN.m 5] What is the moment of the force F2 about point E? a) 111.7 kN.m b) 28.1 kN.m c) 54.6 kN.m d) 93.1 kN.m e) 79.3 kN.m f) 21.3 kN.m
Q) A simplified model as shown in figure below of a planar robotic arm. You may assume that each segment has length 1.5 m, and the applied force is 250 N. 01 = 35 degree and 02 = 03 = 20 degree. Calculate the moment that the force, F, exerts * .about the point A F В C B.
F1 F2 a F3 A d. F4 Consider the following values: F5 a = 25 m;b = 5 m; c = 25 m; d = 5 m; F1 = 3 kN; F2 = 11 kN; F3 = 2 kN; F4 = 20 kN; F5 = 2 kN; a = 30°, 0 = 60° , B = 30° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 15.8 kN.m b) 12.7 kN.m c) - 74.6 kN.m d) 52.9 kN.m e) 24.1 kN.m f)- 47.1 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 8.8 kN.m b) 2.7 kN.m c) 31.6 kN.m d) 8.9 kN.m e) 5.4 kN.m ) 24.2 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) 86.7 kN.m b) 51.1 kN.m c) 14.6 kN.m d) 27.9 kN.m e) 74.1 kN.m ) 35.3 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) - 17.8 kN.m b) 82.7 kN.m c) - 21.9 kN.m d) 36.9 kN.m e) - 27.1 kN.m f) 30.1 kN.m Activate V 5] What is the moment of the force F2 about point E? Go to Setting ) 321.3 kN.m a) 111.7 kN.m b) 218.1 kN.m c) 254.6 kN.m d) 93.1 kN.m e) 101.1 kN.m bo
F1 F2 F3 A B d b F4 D F5 Consider the following values: - a = 25 m; b = 5 m; c = 25 m; d = 5 m; F1 - 6 kN; F2 - 7 kN; F3 = 8 kN; F4 - 9 kN; F5 - 10 kN; a = 30°, 0 = 45° , B = 60° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 154.8 kN.m b) 111.9 KN.m c) - 174.6 kN.m d) 162.9 kN.m e) 214.1 kN.m 0-47.1 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? | a) - 209.3 kN.m b) 215.7 kN.m c) 311.6 kN.m d) - 98.9 kN.m e) 125.4 kN.m f) - 124.2 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) - 204 kN.m b) 121 kN.m c) 146 kN.m d) - 279 kN.m e) 241 kN.m D- 353 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) – 212.7 kN.m b) 182.7 kN.m e) - 21.9 kN.m d) 36.9 kN.m e) - 27.1 kN.m ) 30.1 kN.m 5] What is the moment of the force F2 about point E? a) 100.3 kN.m b) 118.1 kN.m c) 224.6 kN.m d) 193.1 kN.m e) 151.1 kN.m ) 221.3 kN.m E.

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4. Use the same diagram for problems 14 to 19. A force F = 800 N, has its line of action in the direction from point A to point B, as shown in the figure below. Express the force F as a Cartesian vector.