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- 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 foroThe force vector F = (ai + bj + ck) kN acts on point A of the system shown in the figure. Bar OB and point B with coordinates B (x, y, 0) m is on the x-y plane. By using the values given to your name in the data table;a) Find the moment at point O in vectorial form,b) Calculate the magnitude of the moment with respect to the rod OB F = {ai + bj + ck} kN A = 60 B = 70 C = 80 Meter L1 = 16 L2 = 6 B (x, y, 0) coordinates (m) x = 16 y = 3Example: The resultant of the concurrent forces shown in figure is 300 N pointing up along the y-axis. Compute the value of F and e required to give this resultant. 50ON 30 240N o Since the resultant is up along y- axis then EF = 0 o HW. o F = Fcos 0 + 240 cos 30 – 500 = 0 o Fcos 0- 292.15 = 0 o Repeat this problem if the resultant is 40ON down to the right at 60° with the x-axis. o Also repeat the problem if the body is in equilibrium. o Fcos 0 = 292.15... .... (1) o F, = R = Fsin 0 - 240 sin 30 = 300 Fsin 0 = 300+ 120 = 420 ..... (2) Divide eqn. (2) by (1) yields F sin 0 420 %3D F cos e 292.15 o tan 0 = 1.4376, 0 = tan 1.4376 = 55.18° 420 F = sin 55.18 = 511.6N
- The moment of the force F=960-lb about the point O in Cartesian vector form is. 1 ft 4 ft 2ft Select one: O a. Mo = 1600 – 3200A O b. Mo = -200j + 400k Oc. Mo =-200; – 400k O d. Mo= 200; + 400k 1:15 PMProblem 4 From the figure, w1 =13kN/m, w2=10kN/m, F1 =7kN 1. Find the resultant of the forces 2. Find the location of the resultant from point B. 7m 4m 2mA force couple acting on the rod. Find: The couple moment acting on the rod in Cartesian vector notation. F= (-14i + 8j + 6k)N 0.5 m 0.8 m 1.5 m 0.5 m F = (14i - 8j - 6k )N
- w1 Problem 4 From the figure, wi =15KN/m, w2=12kN/m, F1 =7kN 1. Find the resultant of the forces 2. Find the location of the resultant from point B. to 7m 4m 2mUntitled Section F1 F2 a C F3 A d F4 Consider the following values: - F5 a 12 m;b 10 m; c 12 m; d = 10 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) 110.8 kN.m b) 88.7 kN.m c) 214.6 kN.m d) 75.9 kN.m e) 25.1 kN.m f) 142.5 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 78.8 kN.m b) 21.7 kN.m c) 341.6 kN.m d) 87.9 kN.m e) 161.1 kN.m f) 175.4 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) 425.7 kN.m b) 711.1 kN.m c) 414.6 kN.m d) 139.1 kN.m e) 144.1 kN.m f) 220.4 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) 174.8 kN.m b) 822.7 kN.m c) 120.6 kN.m d) 142.9 kN.m e) 217.1 kN.m f) 120.6 kN.m Activate V 5] What is the moment of the force F2 about point E? 69 te Setting ) 74.1 kN.m a) 25.7 kN.m b) 18.1 kN.m c) 54.6 kN.m d) 43.1 kN.m…a = 4 mb = 3.2 mc = 2 mF = {-7i+4j+12k} NSpecified axis: AF Find the following: Find the z-component of the moment of F about the specified axis. Find the x-component of the moment of F about the specified axis. Find the magnitude of the moment of F about the specified axis. Find the y-component of the moment of F about the specified axis.
- y M C F1 L1 F2 L2 B a A a D ASituation 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-4.1The Force shown in the figure is F=30i +40 j-120k N D 4 m 2 m 1.5 m 2.5 m 2 m ní 0.5 m B Which following equation is used to find the moment magnitude of the force F about AC line?