Problem Statement Based on Problem 7-14 from the textbook. Determine the normal force, shear force, and moment at points F and E. -3 ft- E B 2 ft W b FD 2 ft
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- 3-4 Sketch a free-body diagram of each element in the figure. Compute the magnitude and direction of each force using an algebraic or vector method, as specified. F=400 N. 30° B 60° 2 1.9 m 0 9 m C D 60° E x? ReviewIn the following armor find the forces in the members CD, KD and JK of section 1-1 indicated in the figure.
- (٢٦+ * Find the two forces Fbc & Fac as :shown in the figure below Fab=30 N 110 30/Fbc (Fbc=45.18 N and Fac= 58.8 N) O (Fbc=38.56 N and Fac= 56.38N) O (Fbc=75.18 N and Fac= 78.8 N) O ! FacThe pipe is pulled by three cables as shown on the figure below. All three cables are subjected to known forces. Another cable is planned to be added on the system. This additional cable is designed to pull the pipe along the positive y-axis. The proposed cable, Fy is twice as much as the only force which tends to pull the pipe below the x-axis. (Fa= 7.5kN; Fb= 1.25kN; Fc=5kN) Find the summation of forces in x axis and y axis.4.Three members of a frame on a knot plate can be seen in the picture below. Determine the size and direction of the Result of Style on the knot plate analytically and graphically F1 Pelat Buhul F2 F3 F1 = 15 N F2 = 13 N F3 = 10 N Teta = 30 ^o
- The shelf bracket is subjected to the force F = 336 Newtons at an angle 0 = 25.5°. Compute the moment (in N-m) that this force exerts about each of the two attachment points (screw locations in the figure). Take counterclockwise moments to be positive. a F 2013 cc CC Michael Swanbom BY NC SA Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. Variable Value a 48.0 cm 45.6 cm 4.10 cm The moment about the upper attachment point is N-m. The moment about the lower attachment point is N-m. luk-The force P3 to make it to equilibrium in kN is The Stress in section 1 in N/mm2 is The Compressive Stress in section 2 in N/mm2 is The stress in section 3 in N/mm2 is The total change in length in x10-3 mm isExample 1- determining the components of the specified forces in the direction of the x and y axes numerically + y F1x = F1 cos a, = 400N cos 45 = 282,8N F2x = F2 cos a, = 200N cos 130 = -128,6N F3x = F3 cos oaz = 300N cos 220 = - 229,8N Fax = F4 cos a= 350N cos 45 = 0 N %3D F: F1 + X F3 = F1 sin a, = 400N sin 45 = 282,8 N F4 %3D %3D 1y F = F2 sin a, = 200N sin 130 = 153,2 N %3D 2y F3y = F3 sin a, = 300N sin 220 = - 192,8 N Fay = F4 sin a, = 350N sin 270 = - 350 N %3D -y
- with the following modifications: Add an additional force P = 20kN acting downwards at point BENTER YOUR ANSWER THE WAY I SPECIFY HERE IN ADDITION TO DOING IT ON SCRATCH PAPER 400 N 800 N B D A E C 2 m Use the method of sections to calculate the axial force in members AC and BC. Enter your answer in the following way rounded to the nearest whole number. Example If your answer is TAC 2 N tension TBC =3.1 N compression. enter 2Tension;3Compression Enter TAC then TBC (no space)Answer for the 1st part of the problem. magnitude of force A = 74 N direction of force A = 25o north of west second force = P direction of force P = α south of west arrow_forward The resultant of two forces is horizontal , it means that the y component of resultant = 0 Ry = ASin(25o) - PSin(α) 0 = (74×0.42) - PSin(α) PSin(α) = 31.08 NFor, the smallest force P , the value of Sinα must be maximum which is 1 when the value of αis 90 degree Thus, P×Sin(90o) = 31.08 N P = 31.08 N The magnitude of smallest force P = 31.08 N the direction of force P = 90 degree south of west = along - y axis= arrow pointing vertically downw