In (Figure 1), F1 = 80 lb and F₂ = 30 lb. Figure M₁ = 5 ft 30 Submit 0 4 ft IVE] ΑΣΦ | 41 vec 4 6 ft 60 120 Determine the resultant moment of the two forces about the a axis Express the result as a Cartesian vector Enter the components of the moment separated by commas. Express your answer in pound-feet to three significant figures. Previous Answers Request Answer K 1 of 1 3 d C QWP ? X Incorrect; Try Again; 5 attempts remaining D ilb-ft j lb-ft, k lb-ft

In (Figure 1), F1 = 80 lb and F₂ = 30 lb. Figure M₁ = 5 ft 30 Submit 0 4 ft IVE] ΑΣΦ | 41 vec 4 6 ft 60 120 Determine the resultant moment of the two forces about the a axis Express the result as a Cartesian vector Enter the components of the moment separated by commas. Express your answer in pound-feet to three significant figures. Previous Answers Request Answer K 1 of 1 3 d C QWP ? X Incorrect; Try Again; 5 attempts remaining D ilb-ft j lb-ft, k lb-ft

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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In, F₁ = 80 lb and F₂ = 40 lb. Part A M = Submit Provide Feedback Request Answer 5 ft ? 30 0 4 ft B. Determine the resultant moment of the two forces about the a axis. Express the result as a Cartesian vector. Enter the components of the moment separated by commas. Express your answer in pound-feet to three significant figures. IVE ΑΣΦΗ. If Ivec 30 45 120 6 ft ilb ft. j lb ft, k lb-ft Next >
A cleat on a boat is used to support forces from three ropes as shown. Determine the resultant force vector R, using Cartesian representation, and determine the magnitude R. Also express R in polar vector representation. Given: F = 600 N. y F2 = 200 N F1 F3 = 100 N 12 %3D 40° 30° The Cartesian representation of the resultant force is as follows: The magnitude of the resultant force is |N. The polar representation of the resultant force is as follows: R = N@ (Express the angle in the positive counterclockwise direction from the positive x axis.).
Determine the magnitude of the vector difference V' = V2 -2V, and the angle 0, which V' makes counterclockwise from the positive x-axis. The angle 0, will be a positive number between 0 and 360°. Complete both graphical and algebraic solutions. You will have to draw an accurate figure for the graphical solution. y V = 28 unit$ V = 31 units 4 2 29° Answers: %3D V' = i units
If F1 = 10N, F2 = 15N, F3 = 8N, F4 = 10N, F5 = 18N, and the length of the side of a red square is 1m. Solve resultant force (in degrees), magnitude of resultant moment (in N-m), magnitude of resultant force (N).
QI) At point A, the four forces Fs, Fc, Fp and Fe act, the lines of action for these forces are given by the straight lines AB, AC, AD and AF, as shown in Figure 1. Coordinate the points are: A 3 B 2. D 4 12 cm Figure 1 The magnitudes of the forces are: F7 KN, F= 3 KN, F=9 KN and F= 13 KN. Represent this force Facting on the point 4 as a Cartesian vector and determine its direction.
The magnitude of the resultant force FR of the 2 tractors pulling on the lazy cow is 11650 N and is directed along the positive r-axis. y, 02 F2 Values for the figure are given in the following table. Note the figure may not be to scale. Variable Value 35 degrees 6200 N F Using the law of sines and/or cosines, a. Determine the required magnitude of F2 b. Determine the direction of F2, 01 relative to the positive r-axis in degrees. Answer must be between 0 ° s0 s 90° c. In which quadrant is F2 located? Round your final answers to 3 significant digits/figures. F N degrees In which quadrant is F2 located? O fourth O third O first O second
*Q2: Resolve the force (450) to its components along coordinates AC & AB (nun perpendicular coordinate) 45 450 Ib %23 %23
he maghitude and direction of the resultant according to the chosen scale. = 246 N, 255° Test yourself 3: 1. Two forces of 60 N and 8O N respectively act simultaneously at a point. F = 60 N Sketch and determine their resultant by using the following methods: a) tail-to-head 60° b) tail-to-tail (122 N 25 2. Two forces, F, and F, act at a point as indicated. a) Draw a sketch diagram to show the resultant force when angle 6 = 120. The angle is now decreased from 120° to 20°. F = 80 N F (The magnitudes of the forces remain the same.) b) Draw a new sketch diagram to show the new resultant. Compare the sizes of the resultants and write your conclusion. Forces in equilibrium A stationary object might have no forces acting on it. Forces on the object can also be balanced so there is a zero resultant In
C h kN y bkN a kN In the accompanying sketch, triangle OAB lies in the ry-plane. The component of F represented by line segment AC is parallel to the z-axis. If a = 4.8, b = 7.8, h = 3.9 and B= 27°, express the force F in cartesian vector form. %3D Note that A, B and C do NOT denote points measured with respect to distances; they relate to forces. F = k) kN j+
Q Avecar A f Length 10 units.ma Kes an angle of 30 with avector Bof benath 6unuts. Find the wagnitude of the Vector difference A-B and the angle itmakes with vector A- O by the para llelagram method. O by the triangle methad. © bay the rectamqular vesalutien.
In your response, please display the full computation and all FBDs that were used. Only round off at the final answer with 4 decimal places. P=2kN, F=3kN, and moment M=0.5kN-m are acting on the structure in the illustration. A pin at Z and a cable RQ connected by a connecting pin R support it. Im ● F ● P S TO M Im R lu W OZ a' 0.8m 3m If the pin support Z's factor of safety against shear stress is 3, determine whether a single or double shear should be provided for an economical and safe design. 4 m The mas allowable shear stress in the pin is 50MPa. Consider the following specifications. Single shear design: 25mm pin diameter Double shear design: 17mm pin diameter
6 m! 1.0 m TS 2 m 3 m 2m 2 m The magnitude of one of the forces is T2 = 80 kN. Find the rectangular representation of T2. * The magnitude of one of the forces is T2 = 80KN. Find the rec tan gular representation of T2. T2 = (-22.86 i + 34.29 / + 68.57 k) N T2 = (-22.86 i – 34.29 j + 68.57 k) N T2 = (22.86 i - 34.29 j + 68.57 k) N T2 = (22.86 i- 34.29 /-68.57 k) N