1 = the 0.18LONG M. 2.FMJCS.GFind10.1m0410to Tambo Eli Revo do (^~^-h^)5.8m1YZ6.3mAB7.3mLaOrqueсFLXCriven:posimon Vector for Force D, E, F0 = (0₁5.8, 07E = L 6.3 15,9,0)F = L13.6 15.9, 07fD = 4.19KNE = 5.36 KNmagnitude of the moment about org.in due to this force?F = 10.2 KN

Answered: Image /qna-images/answer/68e2220a-3dd3-4e0b-a8ff-88bce78450ce.jpg

+ ovellus 0.18 =18 6.3m mu 22.20 10.1m Find DE 5.8mp J ZA Zoodo (^-^) SN A Z O B VE 7.3m c FL Triven: for Force D, E, F Posimon Vector 0 = (0₁.5.8₁ 07 E = ( 6.3, 15,9,0) F = 13.6 15.9, 07 D=4.19KN E = 5.36 KN magnitude of the moment about orgin due to this force. F = 10.2 KN

+ ovellus 0.18 =18 6.3m mu 22.20 10.1m Find DE 5.8mp J ZA Zoodo (^-^) SN A Z O B VE 7.3m c FL Triven: for Force D, E, F Posimon Vector 0 = (0₁.5.8₁ 07 E = ( 6.3, 15,9,0) F = 13.6 15.9, 07 D=4.19KN E = 5.36 KN magnitude of the moment about orgin due to this force. F = 10.2 KN

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

ull JAWWAL ? 6:20 PM @ 50% A moodle.najah.edu МOODLE Recent - For the following system, answer the following assuming each tension in cable = 10 kN 1) Express tensions in Cartesian coordinate system 2) Find moment due to both cables and weight of door= 1 kN about point A 3) Find moment due to both cables and weight of door= 1 kN about z-axis 960 mm 450 mm 90 mm 675 mm B 90 mm 690 mm 270 mm Fig. P4.113 II
QI5: Delermihe The resultant of The forces shoan in figure () 40, KN 6.0 48N 4 KN CAKN 2 KN 56 KN Fy-(15)
N 1 91% 1 07:00 elearning.nust.na/eleal 23 Determine the magnitude of the resultant force (in N) of the coplanar force system shown in the figure below where the horizontal force P = 210 N and the couple M = 527 Nm. M N m A 150 N 300 250 mm 400 mm 90 N Answer: 81.6 Leave a message P.
1. ! SHOW COMPLETE SOLUTION ! *pls draw FBD *pls use data give properly : rate will be low if not explained properly Problem 1: The pulleys are frictionless and of negligible weight Find: 1.1 Tension (N) in the cord supporting block B. 1.2 Acceleration at Block A (m/s^2) 1.3 Acceleration at Block B (m/s^2) 1.4 Acceleration at Block C (m/s^2)
%A I. Untitled Section F1 F2 a C F3 A b a F4 D E B Consider the following values: - F5 a = 16 m; b = 11 m; c = 13 m; d = 8 m; F1 - 3 kN; F2 - 11 kN; F3 - 2 kN; F4 - 20 kN; F5 - 2 KN; a = 30°, 0 = 60° , B= 30° 11 What is the resultant moment of the five forces acting on the rod about point A? a) 88.8 kN.m b) 76.3 kN.m c) 19.6 kN.m d) 55.5 kN.m e) 59.1 kN.m f) 81.3 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 88.8 kN.m b) 120.8 kN.m c) 31.6 kN.m d) 98.9 kN.m e) 41.1 kN.m f) 24.2 kN.m 31 What is the resultant moment of the five forces acting on the rod about point C? a) 125.7 kN.m b) 178.1 kN.m c) 254.6 kN.m d) 439.1 kN.m e) 144.1 kN.m ) 215.3 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) 77.8 kN.m b) 98.3 kN.m c) 100.6 kN.m d) 18.9 kN.m e) 97.1 kN.m H 60.1 kN.m 5] What is the moment of the force F2 about point E? c) 66.6 kN.m Activate W f) 151.3 kN.m a) 122.7 kN.m b) 108.1 kN.m d) 83.1 kN.m e)…
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…
3) A point mass is attached to one end of a beam levered against a spike which acts as the pivot. The point mass weighs 250N and the beam has a uniform mass of 10kg and is 8m long. Parts a and b Part c Fapplied 3 of 6mov 1m 8m Krist 0 = 60° Scale 250 N Scale a. If a scale placed beneath the point mass reads 250N, what is the magnitude and direction for the force at the pivot point? Direction: North, 90° from x-axis Magnitude to k* 250N = 2500 N b. For the same scenario, how far must the pivot point be located relative to the left edge of the beam? (Hint, both the boxes weight and the force from the scale act at the farthest right edge of the beam.) www.acta 0672nos onl c. The pivot is then moved to be 6m from the left edge of the beam which now makes a 60 angle with the floor. How much downward force would be needed at the left-most edge of the beam to reduce the reading on the scale to zero? (Hint: the beam is still in static eq. when the scale reads zero.)
A:YV 36 ll 46.ll Test 3.pdf F1 F2 a F3 A d. C a F4 E Consider the following values: - F5 a = 12 m; b = 10 m; c = 10 m; d = 8 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) - 111.6 kN.m b) 112.6 kN.m c) - 184.6 kN.m d) - 109.6 kN.m e) - 104.1 kN.m f) None of them 2] What is the resultant moment of the five forces acting on the rod about point B? a) -95.8 kN.m b) 42.7 kN.m c) 80.9 kN.m d) 10.9 kN.m e) 51.8 kN.m f) None of them 3] What is the resultant moment of the five forces acting on the rod about point C? a) 32.6 kN.m b) - 21.1 kN.m c) 17.4 kN.m d) - 16.5 kN.m e) 15.6 kN.m f) None of them 4] What is the resultant moment of the five forces acting on the rod about point D? a) 37.6 kN.m b) 73.8 kN.m c) 71.6 kN.m d) 52.1 kN.m e) 21.1 kN.m f) None of them 5] What is the moment of the force F2 about point E? a) 90.7 kN.m b) - 88.1 kN.m c) 54.6 kN.m d) 103.1 kN.m e)…
P 4:01 A moodle.nct.edu.om 480% * 4G OMANTEL l. Find the magnitude and direction of the resultant force for the system of forces shown in the figure. Where F1 =15 N, F2 =35N, F3 =5 N, F4 =15N. The angles 01 =15°,02 =25° , 03 = 30°, 04 = 20 ° (Enter only the values in the boxes by referring the units given in bracket. Also upload the hand written copy in the link provided) F2 F1 02 F3 F4 II
Choose the following systems that are equivalent to the one shown. Choose all that apply. (Remember conditions for equivalency require the sum of forces to be equal and the sum of moments about on arbitrary point to be equal) 50 N DA OB. OC OD. 1m 1 m 1 m 50 N 1 m 50 N 1 m 100 N 1 m 50 N 1m 50 N 1 m 50 N-m 1 m 100 N-m 1 m
MEC107-Lectur... Prashant Kumar Pan... A lever AB = 2 m long is inclined at 35⁰ with horizontal A force of 180 N is acting at B vertically downwards. The moment about 'A' is (a) 29.5 Nm B (b) 295 Nm -2m (c) 2950 Nm 180N (d) 2.950 Nm 6/9/2022 35°
Sy Step-by-Step Calculator - Syn x b Change in length of strut DC X b The displacements at points I x A WebAssign - ENGR 2331 Su2 x -> A webassign.net/web/Student/Assignment-Responses/last?dep=26956849#Q6 W 24 Statically Indeterminate S x E Apps M Gmail D YouTube Maps O Reading list A rigid bar of weight W= 810 N hangs from three equally spaced vertical wires (length L- 150 mm, spacing a= S0 mm): two of steel and one of aluminum. The wires also support a load Pacting on the bar. The diameter of the steel wires is d-2 mm, and the diameter of the wire is d,=4 mm. Assume E, - 210 GPa and E, = 70 GPa. Rigid bar of weight W (a) L. Rigid bar of weight W (b) (a) What load P (in N) can be supported at the midpoint of the bar (x = a) if the allowable stress in the steel wires is 200 MPa and in the aluminum wire is 65 MPa? (Se figure part (a).) (b) What is P (in N) if the load is positioned at x = a/2? N (c) Repeat part (b) if the second and third wires are switched as shown in the figure part…