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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- h1= 1.05 m h2=0.25 m w=0.50 mConsider the pipe assembly shown in (Figure 1). Neglect the weight of the pipe. Take F₁ = (200i 200j - 250k) N and F2 = (2001+ 500j) N. Figure ▸ ▼ Part A Part B Part C 1.5 m Submit Mr, My, Mz = 525,- 250,0 B Previous Answers Im Determine the x, y, and z components of internal moment acting at a section passing through point B on the left part of the body relative to the cross-section. Express your answers in newton-meters to three significant figures separated by commas. ▸ View Available Hint(s) LIVE ΑΣΦ 41 vec 1 m X Incorrect; Try Again; 4 attempts remaining 1 of 1 ? N.mThe problem I am solving shows a point B on the OD of the pipe that lies on the z-axis. When using the equation tau =(Mr)/J, I get 15.5 MPa, not 124 MPa
- The state of 2D stress at a point is given by a matrix [100 30] MPa MPа 30 20 Ox Txy Tyx Oyy The maximum shear stress in MPa isUse the method of SECTIONS to find the true magnitude and direction in bar HG of the truss shown below in Fig 3.5 -500lb -100lb A 45° -200lb H -100lb 2 B -200lb -200lb MAGNITUDE AND DIRECTION OF SUPPORTS -500lb H -100lb 45° 2¹ C2D 2¹ Fig 3.5. -200lb -100lb LI 2 B 2₁ 2₁ D 2₁ EP₂ 7+1 * P₁ ART 2 P (P₁ - P₂- Zgp) 21 / Prove that :- m² a cda A₂ == * Cde actual mip Solution
- 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 downwAn aluminum rod is hanging from one end. The rod is 1 m long and has a square cross section 20 mm by 20 mm. Find the extension of the rod resulting from its own weight. Take E = 70 GPa and the unit weight = 27 kN/m3.
- NoneThe cross section of a concrete dam is shown in the figure. Take the weight-density (= pg) of water to be 10 kN/m³ and that of concrete to be 25 kN/m³. For the given design of the cross-section, find the ratio h/H that is safe enough for the dam to not tip over (about the downstream edge E). Here you will enter the numerical answer. In the handworked file you must show your worked solution. H E 60° H/10 4H/10 hPlease find the question attached.