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
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- Solve the preceding problem for the following data: b = 8.0 in., k = 16 lb/in., a = 45°, and P = 10 lb.The following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. d 2 Pl CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m С. 62,5 N*m D. Object 1 doesn't exert a torque.Example ( 2-1) : The ring in the figure below is subjected to two forces ( F1) and ( F2). Determine the magnitude and direction of the resultant force. 10° F, = 150 N F = 100 N %3D 15° Solution : Parallelogram Law. The parallelogram is formed by drawing a line from the head of F1 that is parallel to F2 , and another line from the head of F2 that is parallel to F1 , the resultant force FR extends to where these lines intersect at point A. The two unknowns are the magnitude of ( FR) and the angle ( 0) (theta).
- Example 15: The man shown in Figure pulls on the cord with a force of 70 lb. Represent this force acting on the support A as a Cartesian vector and determine its direction. 30ft -S t- 6 ft 12 itThe following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m C. 62,5 N*m D. Object 1 doesn't exert a torque. 2. How much is the torque done by the force of gravity of the beam around point P? A. 20,8 N*m N-m ך177 .B C. 156 N*m D. 313 N*m 3. If you needed to cancel the nomal forces of the two objects, where you should place object 2? The axis of rotation is point P. A. 3,30 m from point B. 9,18 m from point P C. 5,69 m from point P D. 3,62 m from point PAn 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.
- The concurrent force system in space is composed of three forces as shown in the figure: Where: P1 = 182 kN at (3,4,2) P2 = 68 kN at (4,1,-2) P3 = 86 kN at (2,-3,3) Problem: a) Solve for the x-component of the resultant force in kN.b) Solve for the y-component of the resultant force in kN.c) Solve for the z-component of the resultant force in kN.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 loFind the resultant of force Y 141.4 N 60 300 N 240 N MATHalino.com 260 N 30 1 m 12 1m Figure P-264
- F1 F2 F3 A B d b F4 D F5 Consider the following values: - a = 25 m; b = 5 m; c = 25 m; d = 5 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) - 154.8 kN.m b) 111.9 KN.m c) - 174.6 kN.m d) 162.9 kN.m e) 214.1 kN.m 0-47.1 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? | a) - 209.3 kN.m b) 215.7 kN.m c) 311.6 kN.m d) - 98.9 kN.m e) 125.4 kN.m f) - 124.2 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) - 204 kN.m b) 121 kN.m c) 146 kN.m d) - 279 kN.m e) 241 kN.m D- 353 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) – 212.7 kN.m b) 182.7 kN.m e) - 21.9 kN.m d) 36.9 kN.m e) - 27.1 kN.m ) 30.1 kN.m 5] What is the moment of the force F2 about point E? a) 100.3 kN.m b) 118.1 kN.m c) 224.6 kN.m d) 193.1 kN.m e) 151.1 kN.m ) 221.3 kN.m E.From the given figure, what is the y-component of vector BC? F = 120 lb 1 ft B 4 ft 2 ft A Figure: 4PO O +4ft -4ft +2ft -2ftA: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…