Problem 3. As shown in the Figure below, please determine the internal force in the middle of AB, with the given data: OA-15 cm, OB-30 cm, OC-60 cm. W=40N, Wo=60N. F=1214N, B-10¹, FM = 1236 N, and 8=15%. Fj B FM W Wo 4
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- Solve the preceding problem for the following data: b = 8.0 in., k = 16 lb/in., a = 45°, and P = 10 lb.A cargo ship is tied down to marine boll arts at a number of points along its length while its cargo is unloaded by a container handling crane. Each bollard is fastened to the wharf using anchor bolts. Three cables having known tension force magnitudes F, = ll0 kN.F, = 85kN.and F, 9OkNare secured to one bollard at a point A with coordinates (0.0.45 m. 0) in the x-r-: coordinate system shown in the figure part b. Each cable force is directed at an attachment point on the ship. Force F, is directed from point A to a point on the ship having coordinates (3 m, 9 m. 0) force F, is directed at a point with coordinates (6.5 m. 8.5 m. 2 m) and force F, is directed at a point with coordinates (8 m. 9 m. S m). The diameter of each anchor bolts is 4 24 mm. (a) Find the reaction forces and reaction moments at the base of the bollard. (b) Calculate the average shear stress in the anchor bolts (in the x-: plane). Assume each bolt cart ics an equal share of the total force.Solve the preceding problem for the following data:P = 160 kN,JV = 200 tN,L = 2 m,b = 95 mm, h = 300 mm, and d = 200 mm
- F = F cos # F = F + F F F sin 8. # tan Exampie (1):Find the two components of the force (100 X) if: 0 = 30", 120 270° as shown in figure. F = 100 N F = 100 N 6=30 0 =120° 0 =270 0 = 30 0 =60 F = 100 N lution:tol F = Fe F= F + F COs # F F sin a. # tan Example (L):Find the two components of the force (100 X) if: 0 = 30, 120 270° as shown in figure. F = 100 N F = 100 N 6%330 6 =120° 0 =270° 0 = 300 0 60 F = 100 N ution:ENTER 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)
- 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 x1) Given the figure below, answer the following questions: a) What are the x and y components offorce F? (use Fx and Fy)b) What are the x and y components offorce P? (use Px and Py)(Hint: The x-y plane is differentfrom the true horizontal and verticalaxes)c) What are the vertical and horizontalcomponents of force F? (use Fh & Fv)d) What are the vertical and horizontalcomponents of force P? (use Ph & Pv)e) What is the magnitude of the resultant?f) What angle does the resultant make withthe horizontal axisg) What angle does the resultant makewith the inclined x-axis.QUESTION 8 The y-components results of the force vectors that has been shown in Figure. F2 = 80 N 20 %3D F = 150 N %3D 30° 15° F = 100 N A F3 = 110 N %3D Fly=75 N, F2y=-75.2 N, F3y=-110 OA. "N, F4y=-25.9 N Fly=75 N, F2y=75.2 N, F3y=110 N, OB. 'F4y=-25.9 N Fly=75 N, F2y-75.2 N, F3y=-110 'N, F4y-25.9 N Fly=-75 N, F2y-75.2 N, F3y=-110 'N, F4y=-25.9 Ñ OC. OD.
- Given values: m1 = 10kg, r1=75 cm, m3=20kg, r3=25cm, m4=5kg, had to find the value of r4=? I got r4=50cm m1 = 10 kg at r1 = 75cm m3 = 20 kg at r3 = 25cm m4 = 5 kg at r4 = 50cm 1. Compute the theoretical value of r3 using the following equation: r1m1 = r3m3 + r4m4 2. Compute the % error of r3 using the following equation: % error = |theoretical value - experimental value| / theoretical value x 100%The figure shows a rigid bar that is supported by a pin at A and two rods, one made of steel and the other of bronze. Neglecting the weight of the bar, compute the force (in N) in the bronze rod caused by the 45846-Nload, using the following data: length of steel = 11 m, length of bronze = 2.3 m Area of steel = 562 mm, Area of bronze = 335 mm Modulus of elasticity of steel = 195 Gpa, Modulus of elasticity of bronze = 80 Gpa x= 0.55 m, y = 1.12 m, and z = 0.78 m. Round off the final answer to two decimal places. ... Bronze SteelA utility hook was formed from a round rod of diameter d= 0.75 in into the geometry shown in the figure. Given: F= 1500 lbf, L = 7 in, and D;= 2.5 in. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. L.