Answer the question.Thank you. 3-43-2) Find the TRUE magnitude and direction of the RESULTANT FORCE and MOMENTAT POINT A1/200 IbFORCE AND COMPONENT DIAGRAMForcelbs0200 60%100100 45° 70.71Rx=30°e FF.Cos(0) F=F.Sin(0)A100 lb/ftA173.2170.71+M₁M=FxdForcesRy=RESULTANT FORCE AND MOMENT DIAGRAM173.21*6=1639,26'10.71 x 12 =४५४,521MR 1887.78 FT.45Couples-MA OVIO VA100 lb

3-4 3-2) Find the TRUE magnitude and direction of the RESULTANT FORCE and MOMENT AT POINT A TTTTT Force lbs 6 30° 200 lb 0 200 60% 100 45° FORCE AND COMPONENT DIAGRAM 100 lb/ft e Fx=F.Cos(0) F, =F.Sin(6) 100 70.71 173.21 -70.71 M=Fxd Forces 173.21*6= -2- 1039.26 10.71 * 12' = 848.521 45 Couples 100 lb

3-4 3-2) Find the TRUE magnitude and direction of the RESULTANT FORCE and MOMENT AT POINT A TTTTT Force lbs 6 30° 200 lb 0 200 60% 100 45° FORCE AND COMPONENT DIAGRAM 100 lb/ft e Fx=F.Cos(0) F, =F.Sin(6) 100 70.71 173.21 -70.71 M=Fxd Forces 173.216= -2- 1039.26 10.71 12' = 848.521 45 Couples 100 lb

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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3 / 5 73% 1-3 3) Find the magnitude and direction of the resultant force and moment at point A 250lb 500lb/ft 4.5 342lb 250lb FORCE e F =F.Cos(e)F, =F.Sin( e) (Ib) ) M =Fxd FORCES COUPLES R= M = RESULTANT FORCE AND MOMENT DIAGRAM A +M -MA R= +X OR = M= étv A Aa W
F1 F2 a F3 A d. F4 E Consider the following values: - a = 11 m; b = 6 m; c = 8 m; d = 3 m; F1 = 3 kN; F2 = 11 kN; F3 = 2 KN; F4 = 20 kN; F5 = 2 kN; F5 a = 30° , 0 = 60° ,B= 30° 1] What is the resultant moment of the five forces acting on the rod about point A? a) -1.2 kN.m b) 12.7 kN.m c) - 4.6 kN.m d) 22.9 kN.m e) - 14.1 kN.m f) 21.3 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 28.8 kN.m b) 42.7 kN.m c) - 1.6 kN.m d) 10.9 kN.m e) - 1.1 kN.m f) 24.2 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) 63.6 kN.m b) -11.1 kN.m c) 77.4 kN.m d) - 9.1 kN.m e) 24.1 kN.m f) - 85.3 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) 33.6 kN.m b) 83.4 kN.m c) - 1.6 kN.m d) 52.1 kN.m e) 21.1 kN.m f) - 40.1 kN.m 5] What is the moment of the force F2 about point E? a) 80.6 kN.m b) - 8.1 kN.m c) 54.6 kN.m d) 33.1 kN.m e) - 22.1 kN.m f) 28.3 kN.m
10 / 36 110% Q22](problem 2/28: p37) (¿ãt2) Determine the resultant `R" of the two forces Shown in fiqure, and find the angle o" between R and A- 10 800 N the x-axis . 900 N 25 4" Ane R=10379 N . 0 =66-8 Q23]( problem 2/25 's p37) Exemple201-2o At What angle x must the Type here to search
Answer for the 1st part of the problem. magnitude of force A = 74 N direction of force A = 25o north of west second force = P direction of force P = α south of west arrow_forward The resultant of two forces is horizontal , it means that the y component of resultant = 0 Ry = ASin(25o) - PSin(α) 0 = (74×0.42) - PSin(α) PSin(α) = 31.08 NFor, the smallest force P , the value of Sinα must be maximum which is 1 when the value of αis 90 degree Thus, P×Sin(90o) = 31.08 N P = 31.08 N The magnitude of smallest force P = 31.08 N the direction of force P = 90 degree south of west = along - y axis= arrow pointing vertically downw
%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)…
The curved pipe has a radius of 5 ft. Figure 60 1001 > Part A If a force of F=75 lb acts at its end as shown in (Egure 1), determine the moment of this force about point C Enter the components of the moment separated by commas. Express your answer in pound-feet to three significant figures. VAX i vec Me 322,247,47.0 Submit Previous Amers Bequest Ac X Incorrect: Try Again Provide Feedback ? Review Ilb-ft, Jlb-ft, klb-ft Next >
Problem 3. What is the moment of the 1.4 kN force along the direction of the 1.0 kN force? Express your answer as a magnitude and sketch the direction of rotation about the axis defined by the 1.0 kN force TIGBAO TK 1.0KN X 539 indi 21 150 mm 195 mm 2 4 160 mm 21% ARD 19TO 1.4 kN MEBIRT 0 DAW ITOM 00 1400
F1 F2 a F3 A B d b F4 E F5 Consider the following values: - a = 25 m; b = 5 m; c = 25 m; d = 5 m; F1 = 3 kN; F2 = 1 kN; F3 = 4 kN; F4 = 6 kN; F5 = 8 kN; a = 30°, e = 60° %3D B = 30° %3D 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 f) 353 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) - 184.2 kN.m c)– 211.9 kN.m d) 126.9 kN.m e) - 27.1 kN.m ) 125.1 kN.m Activate 5] What is the moment of the force F2 about point E? Go to Setting f) 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)…
F1 F2 a C F3 A B b F4 D E Consider the following values: - F5 a = 25 m; b = 5 m; c = 25 m; d = 5 m; F1 = 4 kN; F2 = 2 kN; F3 - a = 30° , 0 = 60°, B = 30° 5 kN; F4 = 7 kN; F5 = 3 kN; %3D 1] What is the resultant moment of the five forces acting on the rod about point A?
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 m F- 800 N A .. 3 m 12 m •D 800 (2,9.3) (2,-9,-3) (-2,9.5) 800 A) V94 C) V94 800 B) 800 V94 (-2,9,-3) D) V94 15. Determine the components of the force F parallel to axis AD. -64 A) V94 -64 C) V94 (4,3.5) -64 B) 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. 800 A) M, = (12,6,-10) 800 C) M, = V94 :(-12,-6,-10) (-12,-6,10) V94 800 -10) 800 D) M, = 194 B) M, = 12,-6, 194 19. What is the moment of the force F about the axis OC? 990 A) Molloc =- (4.8.0) 960 C) Molloc = V94 890 D) Molloc = V94…
4:48 68 O E B18 Statics_Moment.. o 240 N 2. A prybar is used to remove a nail as 15° shown. Determine 350 mm the moment of the 240 N force about the point o of 65° contact between the prybar and the small support block. 30 mm PROBLEMS 240cos10=236.35 N Mo=? 240 N 15° 65 240sin 10=41.68 N 350 mm M=236.35(0.35) +41.68(0.03)=83.97 N -m 65 (clockwise direction) 30 mm PROBLEMS 900 mm 700 mm 3. An experimental device imparts a force of magnitude F = 225 N to the front edge of the rim at A to simulate the F 300 mm effect of a slam dunk. Determine the moments of the force F about point O and about point B. Finally, locate, 3050 mm
touch & OL166% 00:27 MENG250 - Practice Assi... Not saved yet •2-77. Determine the magnitude and coordinate direction angles of F, so that the resultant of the two forces is zero. Y2 B2 a2 y 60° 15°. F= 180 N O