Replace the force and couple moment system acting on the beam in Fig. 4-44a by an equivalent resultant force, and find where its line of action intersects the beam, measured from point O. 8 kN SOLUTION 4 kN Force Summation. Summing the force components, (FR)x= EF (FRx = s KN() = 4.80 kN- 15 kN m 0.5 m %3D 1.5 m--1.5 m--1.5 m--1.5 m-

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

I need quick solution includes Free body diagram. [Moment]

ENGINEERING MECHANICS STATICS Example: Determine the resultant couple moment of the three couples acting on the plate in the figure. F = 200 lb %3D F3 = 300 lb -d = 4 ft- F2 = 450 lb A %3D d3 = 5 ft d2 = 3 ft %3D F = 450 lb B. %3D F = 200 lb F3 = 300 lb Solution: 2och nair of counle foro

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Staticsols X 2 *moments H.W.pdf Google zilai x + O File | F:/Physics%20videos/moments%20H.W.pdf E A Road aloud V Draw V Highlight v O Erase 3 The force P is applied perpendicular ty the portion BC of the bent bar. Find the moment of P about point C, B, and A. M- Pad --P,* (D %3D Q4 P 30 N CD: 15+1-6Singo- 58 40 1.5 m II D

Can you answer d through e on this question? thank you

A bent tube is attached to a wall with brackets as shown. . A force of F = 970 lb is applied to the end of the tube with direction indicated by the dimensions in the figure. a.) Determine the force vector F in Cartesian components. b.) Resolve the force vector F into vector components parallel and perpendicular to the position vector I DA- Express each of these vectors in Cartesian components. NOTE: The parallel component of the force is the same as the force along that direction. The perpendicular component of the force can be found from the relationship: Fparallel + Fperpendicular = Ftotal 2013 Michael Swanbom BY NC SA d A FK values for dimensions on the figure are given in the table below. Note the figure may not be to scale. Be sure to align your cartesian unit vectors with the coordinate axes shown in the figure. d 78 in h 57 in g 60 in Variable value 13 in 36 in 45 in F|DA = i. k) lb and b. FL DA =

A bent tube is attached to a wall with brackets as shown. A force of F = 785 lb is applied to the end of the tube with direction indicated by the dimensions in the figure. a.) Determine the moment about point D due to the force F Enter your answer in Cartesian components with units of ft- lbs. b.) Determine the moment about a line (i.e. axis) running from D to C due to the force F. Enter your answer in Cartesian components with units of ft-lbs. 2013 Michael Swanbom x BY NC SA g Z h A с FK kaz Values for dimensions on the figure are given in the table below. Note the figure may not be to scale. Be sure to align your cartesian unit vectors with the coordinate axes shown in the figure. Variable Value α 4.84 in b 13.2 in с 12.5 in d 30.8 in h 18.7 in 22.0 in →> a. MD=( i+ k) ft- lb →> b. MDC = î + k) ft- lb

Plz solve correctly all parts Will definitely