L=5m ·x- T B 1.2 m -1.6 m→
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The 90-kg man, whose center of gravity is at G, is climbing a uniform ladder. The length of the
ladder is 5 m, and its mass is 20 kg. Friction may be neglected. Compute for the following given that x = 1.5 m:
[1] the normal reaction at B (Answer: 858.375 N) and the total force at A (Answer: 647.479 N)
[2] the vertical reaction at A (Answer: 392.400 N) and the horizontal reaction at A (Answer: 515.025 N)


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- Figure Part A Fuc= 0.97 5m HA 2m KN Determine the magnitude of the projected component of the F-5 kN force acting along the axis BC of the pipe. Suppose that a 3.1 m and b=4.4 m (Figure 1) Express your answer to three significant figures and include the appropriate units. Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining B ? a 1m 1 of 1QI/ The rigid pole and cross-arm assembly is supported by the three cables shown. A turnbuckle at D is tightened until it induces a tension T in CD of 1.2 kN. Express T as a vector. Does it make any difference in the result which coordinate system is used? Ans. T = 0.321i + 0.641j – 0.962k kN, 1.5 m /C B 1m 1.5 m T = 1.2 kN 3m E 01.5 m 3 m 2 m mthe horizontal and vertical components of several forces are: (a) Ph = -200lb and Pv = 100lb : (b) Fh = 300lb and Fv = -200lb . determine each force and its angle with respect to the hozirontal (x-axis)
- At a point in an elastic material under strain, the stresses on the three mutually perpendicular planes are as follows:A normal tensile stress of 60 N/mm^2and shear stress of 40 N/mm2 on one plane and a normal tensile force of 40 N/mm^2and a complimentary shear stress of 40 N/mm^2 on another plane. Find the following using Mohr circle only (take 5 N/mm2 = 1 cm)a. The principal stresses and principal planes.b. The maximum shear stress and its plane.c. The normal and shear stress on a plane inclined at an angle of 30Oto major principal plane.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 downwThe inner diameter of the AB compressed air tank is 160, the wall thickness is 1.Since the manometer pressure in the tank is 1,6 ,at point a at the top of the tank maximum normal stress and determine the maximum plane shear stress.(c=16 , d=38, e =32, g=160, h=1, k =1.6 , F=160)
- Please find the question attached.F2 Q1) Axial displacement of point C in the system shown on the left is 0.01 cm. Find the maximum elongation of the bar and the maximum normal stress. A F1 A. GIVEN: F2 = 100 kN, (1 = 240 cm, €2= 160 cm, bi = 5 cm, bz=10 cm, h=5 cm, E=2.107 N/em², a=20 cm Note: neglect stress concentration. h bị b2 A-A sectionQ6

