In the Figure 6, the absolute force value in the bar AC is (in kN): Figure 6: 5 m 5 m A 20 KN 5 m B

An L-shaped bracket is supported by a frictionless pin at A and a cable between points B and D. If pin A is the origin, point C is at (-7, 0); and pin B is at (-7, -10). The cable is attached between point B and an anchor at point D, at (-2, -14). Units are in feet. Force F acts at point C at an angle of 60o from the positive x-axis. Determine the magnitude of the moment that the force F produces about point A. Determine the tension required to hold the bracket in position.

GO 321 3 in F1 F2 a F3 A d F4 F5 Consider the following values: - a = 25 m; b = 5 m; c = 25 m; d = 5 m; F1 = 7 kN; F2 = 8 kN; F3 = 9 kN; F4 = 10 kN; F5 = 11 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) - 158 kN.m b) 127 kN.m c) 146 kN.m d) - 103 kN.m e) 241 kN.m f) - 171 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) - 291.7 kN.m c) 311.6 kN.m d) – 264.6 kN.m e) 354.2 kN.m f) - 242.5 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) - 141.1 kN.m c) 14.6 kN.m d) - 272.1 kN.m e) 174.1 kN.m f) - 315.3 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) 82.7 kN.m c) 121.9 kN.m d) - 259.9 kN.m e) 127.1 kN.m f) 300.1 kN.m Activate 5] What is the moment of the force F2 about point E? a) 151.7 kN.m b) 118.1 kN.m c) 214.6 kN.m d) 158.6 kN.m e) 101.1 kN.m lo

Tire wrench moment, with 50lb force and 75 degree angle with length 2 feet

у- ахis F4 F5 X- axis BC F1 х- ахis y - axis F1 F2 Figure (1) Figure (2) 250 N 36.870 D F2 60° 0.5 m 1 m B 0.2 m 1 m 300 NE F1 1 m Figure (3) Figure (4) Q1] Answer the following questions: For figure (1): The object is subjected to two forces F1 = 25 N, and F2 = 50 N. Set x = 30°, 0 = 45°, and ß = 30°. 1- What is the magnitude of the resultant of these two forces? a) 39.4 N b) 80.1N c) 41.6 N 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? a) 14. 3° b) 392.6° c) 9.2° d) 322.9° e) 6.8° f) 344.2° g) None of them 500 N

2000Lbs 30° 60° y A' \ B The bar OA is fully constrained at O and attached to the cable AB. The bar length is 20', and it forms a 30° angle with the z-axis and its projection on the x– y plane forms a 60° angle with the x-axis. The point B is prescribed by the coordinates (8',16',0). If the force in the cable is 2000Lbs, what is the moment of this force about O? What is the moment of the 2000Lbs force about the axis OÁ.

Cartesian coordinat system; A(0,0,0) B(3,4,0) C(3,4,4) F=-20i +10 j+15k N is applied to point C Moment of force F about point A; M, Moment of force F about point B; M, Moment of force F about line AB; M AB Which following one is TRUE? ON MAB(-3i+4j)OM, O A) M = (3i – 4 j)®(-20i +10 j+15k) M, =(j)®(-20i +10 j +15k) O D Mg = 80i – 40j o M, = (3i+4 j+ 4k)® (-20i+10j+15k)

You need to find Components of all forces which is acting on Member ABC in the given figure [Must solve in handwritten format ]

vertica lues of x shown in the diagram. Even if a segment is 50 lb/ft 20 ft A V (lb) 600 500 400 300 *********** con 10 ft 80 lb 90000 I all 200 lb-ft F

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