**Mechanics Problems: Applying Vector Concepts**Use the figure provided for both problems.**Problem 4: Determining the Angle with the x-axis**The magnitude of α₁, the angle F1 makes with the x-axis, is nearest to:- a) 35°- b) 50°- c) 71°- d) 66°- e) None of the above**Problem 5: Calculating the x-component of F2 Force**The x-component of the F2 force can be expressed mathematically as:- a) 750 cos 50° sin 40°- b) 550 sin 45° cos 45°- c) −550 cos 45° sin 45°- d) 310 N- e) 2.75 kN**Explanation of the Diagram:**The diagram shows a structure with two forces, F1 and F2, acting on it. Specific angles are given:- F1 is a 750 N force shown with angles 40° and 45° relative to some axes.- F2 is a 550 N force with angles 45° and 120°.- The forces and angles are illustrated along with Cartesian coordinate axes (x, y, z).These angles and forces are crucial for solving the problems related to vector components and magnitudes.

Answered: Image /qna-images/answer/7ecf900f-d74a-41ea-b0ef-673102551ced.jpg

4. The magnitude of aa1, the angle F1 makes with the x-axis, is nearest a) 35⁰ b) 50° c) 71° d) 66° F₂=550N 120° 45° 40° F₁ = 750N e) None of the above 5. The x-component of the F2 force can be expressed mathematically as a) 750 cos 50° sin 40° b) 550 sin 45° cos 45º c) -550 cos 45° sin 45° d) 310 N e) 2.75 kN

4. The magnitude of aa1, the angle F1 makes with the x-axis, is nearest a) 35⁰ b) 50° c) 71° d) 66° F₂=550N 120° 45° 40° F₁ = 750N e) None of the above 5. The x-component of the F2 force can be expressed mathematically as a) 750 cos 50° sin 40° b) 550 sin 45° cos 45º c) -550 cos 45° sin 45° d) 310 N e) 2.75 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

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a docs.google.com/forms/d/e/1FAlpQLSel : Untitled Section F1- 10 kN F2- 26 kN -45 F3 = 8 kN BI 4 m 8m F4- 44 KN a-45 F5 = 8 kN 1] What is the resultant moment of the five forces acting on the rod about point A? a) 77.5 kN.m b) 108.1 kN.m e) 59.8 kN.m d) 153.1 kN.m e) 23.1 kN.m ) S6.4 kN.m 21 What is the resultant moment of the five forces acting on the rod about point C? a) 4.5 KN.m b) 18S7.1 AN. e) 59.8 kN.m d) 153.4 kN.m e) 99.1 KN.m n 56.4 kN.m 31 What is the moment of the force F1 about point E? ) - 14.6 KN.m b) - 106.1 kN.m )- 48.S KN.m d)-2158 EN.m ) - 99.7 kN.m 9- 188.1 kN. 4] What is the moment of the force F2 about point D? a) 19.5 kN.m b) 122.1 AN.m e) 88.S kN.m d) 147.1 kN.m •) 58.1 kN. m ) 135.4 kN.m 5] What is the moment of the force F5 about point A? a) 11.8 KN.m b) 5,7 kN.m e) 44.8 kN.m d) 19.1 kN.m e) 69.4 kN.m 0 1.4 kN.m * :The answer is [1 a O ->
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x- axis y-axis Figure (1) For figure (1): The object is subjected to two forces F1 = 25 N, and F2 50 N. Set oc = 30°, 0 = 45°, and ß = 30°. 1 - What is the magnitude of the resultant of these two forces? c) 41.6 N a) 39.4 N b) 80.IN d) 69.8 N e) 71.4 N f) 46.1 N g) None of them 2 - What is the direction of the resultant force measured counterclockwise from the positive x-axis? ) 344.2° a) 14.3° b) 392.6° c) 9.2° d) 322.9° e) 6.8° g) None of them
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Two equal forces are acting at a point with an angle of 60° between them. If the resultant force is + 20/3 N, themagnitude of each force is equal to 1) 40N 2) 20N 3) 10/3 N 4) 20/3 cos 60
Solve with diagram as fast as.
6- The adjacent figure shows a collision with a body at rest. What is the velocity and angle of scattering of the second body after the collision? 4.33 m/s 5.00 m/s .** 30° Before the collision After the collision. A) 8-60 V = 2.5 B) 0=60 V = 3.5 C) @=-30 V = 4,33 D) 8= -50 V 1.5 A В C D
F1 = 10 kN F2 = 26 kN 0 = 45 B = 60° F3 = 8 kN A B 4 m 8 m F4 = 44 kN D E lJa = 30° F5 = 8 kN 1] What is the resultant moment of the five forces acting on the rod about point A? a) 32.4 kN.m b) 22.5 kN.m c) 44.1 kN.m d) 19.5 kN.m e) 9.1 kN.m f) 21.1 kN.m 2] What is the resultant moment of the five forces acting on the rod about point C? a) 149.5 kN.m b) 33.5 kN.m c) 123.9 kN.m d) 111.5 kN.m e) 42.1 kN.m ) 29.1 kN.m 3] What is the moment of the force F1 about point E? а) - 98.9 kN.m b) - 59.8 kN.m c) - 114.8 kN.m d) - 77.1 kN.m e) - 29.7 kN.m f) – 66.9 kN.m 4] What is the moment of the force F2 about point D? a) 112.5 kN.m b) 33.5 kN.m c) 129.1 kN.m d) 99.5 kN.m e) 55.1 kN.m f) 29.1 kN.m 5] What is the moment of the force F5 about point A? а) - 9.6 kN.m b) - 18.8 kN.m c) – 39.8 kN.m d) - 2.1 kN.m e) - 24.7 kN.m f) – 44.9 kN.m
6 - at point O in the figure F1 = 32i + 37j F2 = -27i + 14j F3 = -16i - 21j F4 = -12i - 11j Find the sum of the vectors. What is the strength of the total vector? A) 29.13B) 29.83C) 30,26D) 28.93E) 23.19
17 - In the figure at point O F1 = 119i + 124j F2 = -113i + 117j F3 = -115i - 108j F4 = -137i - 111j Find the sum of the vectors. What is the strength of the total vector? A) 256,56B) 245.62C) 246.98D) -243.96E) -241,26