A rigid system DA is subjected to two forces at point A as shown in the figure below. Given FAB = -0.47i +0.47j - 1.884k (KN) and FAC = -2.34i +1.872k (KN). D 4m A FAB 4m. 4m FAC B(4,5,0) 3) The magnitude of moment caused by force FAB about y-axis is ) The magnitude of moment caused by force FAC about OB is The resultant moment due to FAB and FAC about origin O is KN.m and its direction is KN.m and its direction is KN.m
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- Q4 ) Determine the moment of each of the forces about point (A) shown in figure M = ( 50.7 Nm , 55.7 Nm , 60 Nm ) * %3D Q4 F1 = 10 N F2 = 5 N 3. 4m F3 = 20 N AHOME WORK 3 Determine the resultant force at A. for the figure below Hint : 1- Find the position of all points A,B,C (x, y, z) 2- Find FAR and FAC 3- Find |raR| and Iracl 4- Find FAR = F UAB and FAC = F UAC 5- Finally find R and a, B,Y 3m Fc= 500N FB= 500N 3m 5m 3m B 4m 2m 5mSituation 1- For the forces shown in the Figure AP-4.1: 1. Determine the X-component of P= 300 N A the resultant force in Newtons. Answer: 185.12 N 2. Determine the Y-component of the resultant force in Newtons. Answer: 117.37 N 3. Determine the moment about point O due to forces, in N-m. Answer: 107.87 N-m 0.2 m BI R= 100 N Q = 150 N 0.2 m Figure AP-4.1
- - Two different ropes connected to the A ring, FB and FC as in the figure. pulling forces are applied. (ring A is in the x-y plane and its z coordinate is zA =0.) Accordingly, a-) Express the FB force as vector. b-) Vectorial moment of FC force with respect to point B Calculate as. FB=(26)kN, FC = (12)kN, hB = (7)m, hC=(8)mENGINEERING 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 foroA member is loaded with the given system of forces as shown in figure. Find the moment about A & B. Where AB = AD = 80 m, F1=30 N F2=10N, F3=36N, F4=36N, 01=40° , 02= 50°, 03=35° & 0 4=70°. F1 F2 A B D F3 F4 The Moment about "A" (in Nm) = The Moment about "B" (in Nm) = The Moment about "C" (in Nm) = The Moment about "D" (in Nm) = %3D
- Untitled Section F1 F2 a F3 d F4 Consider the following values: - F5 a = 16 m; b = 11 m; c = 13 m; d = 8 m; F1 = 3 kN; F2 = 11 kN; F3 = 2 kN; F4 = 20 kN; F5 = 2 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) 88.8 kN.m b) 76.3 kN.m c) 19.6 kN.m d) 55.5 kN.m e) 59.1 kN.m f) 81.3 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 88.8 kN.m b) 120.8 kN.m c) 31.6 kN.m d) 98.9 kN.m e) 41.1 kN.m f) 24.2 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) 125.7 kN.m b) 178.1 kN.m c) 254.6 kN.m d) 439.1 kN.m e) 144.1 kN.m f) 215.3 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) 77.8 kN.m b) 98.3 kN.m c) 100.6 kN.m d) 18.9 kN.m e) 97.1 kN.m f) 60.1 kN.m 5] What is the moment of the force F2 about point E? Activate V ) 151.3 kN.m Go to Setting! a) 122.7 kN.m b) 108.1 kN.m c) 66.6 kN.m d) 83.1 kN.m e)…8:05 png.صباحي q2 Compute the moment of these forces as shown in the figure with respect to point A. 6 kN 30° 130° 1.5 m 4 kN -1.5 m- -1.5 m-Ke-45 Homework-3- 1. Calculate the moment at point (A) of the forces shown in figure. T:-10 2.5 m E- -- 1-25m-1 T-3 N T T3=6 N M=5 N.m T-4 N T;=2 N 0-45° 3 m 5 m 2. Determine the moment of a 90 N force about point C for the condition 0 = 15°. 600mm B A 90N 3. Find the forces in member AB and BC by component method. TAK Tc 45 30 45 30° Ring Pulley 60 200 N Pulley 200 N 200 N 200 N 800mm
- (B Page view A Read 1-3 3) Find the magnitude and direction of the resultant force and moment at point A 250lb 500lb/ft 1.5 30° 20 342lb 250lb 5 1.5 FORCE e F =F.Cos(e) F=F.Sin( e) (Ib) M, =Fxd FORCES COUPLES R. = R= M.= RESULTANT FORCE AND MOMENT DIAGRAM +M -M, +Y R=2 / 5 78% + 1-2 2) Find the magnitude and direction of the resultant force and moment at point A 250lb 150lb A JA ++ 350lb 400lb 1" 1 \150lb FORCE e F, =F.Cos(e) F, =F.Sin( e) (Ib) 5M, =Fxd FORCES COUPLES R= R = M= +M, -M, RESULTANT FORCE AND MOMENT DIAGRAM R= OR = ++ M.= -Y AY 6. étv A Aa20 KN 2 kN /m 40 kN m 150 kN m 450 mm- 600 mm 8m 3m 300 mm 5KN 3 KN /m +15 m--1.5 m- 3m S kip t 15 kip ft 15 kip ft 6 ft 10 t 6 ft Determine the moment of the force system of Figure below with respect to point A and B