e. What is the moment caused by the force, F, -9641bf, on around POINT A on the block if 0 = 42°, a 4.6 ft., and b = 8.1ft.? F 0 b A f. What is the moment caused by the weight of the block, W=6731bf, (applied at the center of mass, which is in the middle of the block) around point A? g. What is the magnitude of the force, F2, necessary to create a resultant moment of zero from all forces on the block around A. h. Is the block in equilibrium (consider forces and moments)?
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- Find the moment about A, B, & C, where AB = AD = 19 m, F₁= 3 N, F₂ = 7 N, F3 = 11 N & F4 = 19 N. Each force is acting at the mid-point of each side. --(3 Marks) F4 F₁ F3 B F2 (NOTE: ENTER ONLY THE VALUES BY REFERRING TO THE UNITS GIVEN IN THE BRACKET) The moment of force about "A" (in Nm) = The moment of force about "B" (in Nm) = The moment of force about "C" (in Nm) =Find the moment about A & B, where AB = AD = 12 m, F1= 4 N, F2 = 8 N, F3 = 13 N & F4 = 19 N. Each force is acting at the mid-point of each side. -- 4 Marks %3D %3D F1 F2 F4 D F3 (NOTE: ENTER ONLY THE VALUES BY REFERRING TO THE UNITS GIVEN IN THE BRACKET. ALSO, SUBMIT THE HANDWRITTEN ANSWER SHEET IN THE LINK PROVIDED) The moment of force about "A" (in Nm) (2 Marks) %3D The moment of force about "B" (in Nm) = (2 Marks) %3DFind the moment about C & D, where AB = AD = 16 m, F,= 3 N, F2 = 6 N, F3 = 12 N & F4 = 16 N. Each force is acting at the mid-point of each side. F1 F2 F4 Fs (NOTE: ENTER ONLY THE VALUES BY REFERRING TO THE UNITS GIVEN IN THE BRACKET. ALSO, SUBMIT THE HANDWRITTEN ANSWER SHEET IN THE LINK PROVIDED) The moment of force about "C" (in Nm) = The moment of force about "D" (in Nm)
- Question 2 Find the moment of the force about each coordinate axis (x, y, and z). Then determine the moment of the force about line AO. Mx = 72 Nm 2; My = 108 Nm G; Mz=96 Nm 6 A (0,6,4)m - OR - M. A MA - { l19.1je 79.4 è 3 Nm B(8,0.6)mIf you know that the moment of the force F about point A should look like the formula: M, i+ M,j• M, k, what is the value of the M,? knowing that the force vector equals to: F= -160.11 + -624.2) + 867Ak lb Note: Your answer should contain only numbers, not units. F B 8 ft 1.5 ft 0.25 ftThe rigid pipe is subject to the following loads. Neglecting the weight of the pipe, determine the moment at O. SOLVE THE FF 1. Vector component form of 50-lb force (ANSWER: 25.000i +0j -43.30k lb) and the Moment at point O, in vector component form. (ANSWER: 1153.01i -829.52j +250.00k lb-in) 2. Magnitude of the resultant force (ANSWER: 78.10 lb) 3. Position vector of the 50-lb force from point O to its point of application on the pipe. (ANSWER: -15i -10j +12k inch)
- A flag pole has a staked line to keep it from tipping over. The height of the flagpole is 17 ft. The cable can hold a maximum tension of 1500 lb. Determine the moment created at the support Q by the tension in cable. Enter your answer in Cartesian components with units of kip-ft (1 kip = 1000 lb). Round to 3 significant digits. top T a y Values for dimensions on the figure are given in the table below. R. Variable Value a 13 ft b 18 ft M i + а. A = (| k) kip-ftThe connected bar BC is used to increase the lever arm of the crescent wrench as shown. If a clockwise moment of MA = 100 Nm is needed to tighten the nut at A and the extension d = 235 mm, determine the required force F %3D C 15 F 33° 200 mm B. in order to develop this moment. A Question 1.1 Calculate the magnitude of force F by finding the moment arm D (# d) of F about A and using MA = FD. Expected answer: 360.8 Moment arm, D = 353.0 mm Expected answer: 277.1 Force, F = 285.8 N %3D Moment arm DQ1: Determine the moment of 50N force shown in Fig. about point O by at least two ways? 200 mm 100 mm-
- 7 The solid frame is subjected to the forces shown, calculate the distance (d) if the moment at A equal (MA=-11 N.m) (clockwise + and counter clockwise -) 50N 45° 2m 1m 40N 27 kN 3m 30 2m 100N4 6m 6mFind the moment about A, C, and F for the forces acting on the link shown below. Point C is at the center of B and D. Take F1 = 49 N, F2 = 16 N, F3 = 21 N, F4 = 28 N, F5 = 41 N and F6 = 24 N. F1 F5 4m 7 F6 6m 4m A B F2 F3 Solution: (i) Moment about A in Nm is (ii) Moment about C in Nm is (iii) Moment about F in Nm is C I 10m E F4 60°Find the moment about point A due to the forces shown in the Figure below. HINT: Choose clockwise as negative and counterclockwise as positive. a. C. 10 N b. M M e. d. M 30 N 30 N = 1 m = -10 N. 1.5 m - 30 N. 1 = 2 m = -10 N 1.5 m + 30 N. 1 M +10 N 1.5 m - 30 N.3 1.5 m = -10 N. 1.5 m - 30 N. 2 M 10 N 2 m + 30 N 1.5 n O O O O