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- The bar ABC is supported by three identical, ideal springs. Note that the springs are always vertical because the collars to which they are attached are free to slide on the horizontal rail. Find the angle at equilibrium if W = kL. Neglect the weight of the bar.The bent rod of negligible weight is supported by the ball-and-socket joint at B and the cables attached to points A and C. Find the forces in the cables and the magnitude of the reaction at B. Dimensions Figure P.5.39The steel cylinder with a cylindrical hole is connected to the copper cone. Find the center of gravity of the assembly. The weight densities of steel and copper are 489lb/ft3 and 556lb/ft3, respectively.
- The figure shows the Russel fracture traction device and a mechanical model of the leg. The leg is held in balance in the position indicated by the two weights attached to the two cables. The combined weight of the leg and cast is W=180 N. The distance between the points A and B where the cables are attached to the leg is given as L=100 cm and the angle of the leg with the horizontal is given as γ=8°. Point C is the center of gravity of the cast and leg at three quarters of the L measured from point A (3L/4= 75 cm). The angle that cable 2 makes with the horizontal is measured as β=50°. Accordingly, in order for the leg to remain in balance in the shown position; a) Find the tensile force T1 in cable 1. (Write your result in N)b) Find the tensile force T2 in cable 2. (Write your result in N)c) Find the angle α of cable 1 with the horizontal.The figure shows the Russel fracture traction device and a mechanical model of the leg. The leg is held in balance in the position indicated by the two weights attached to the two cables. The total weight of the leg and the cast is W=250 N. The distance between the points A and B where the cables are attached to the leg is given as L=100 cm and the angle of the leg with the horizontal is γ=6°. Point C is the center of gravity of the cast and leg at three quarters of the L measured from point A (3L/4= 75 cm). The angle that cable 2 makes with the horizontal is measured as β=40°. Accordingly, in order for the leg to remain in balance in the shown position; a) Find the tensile force T1 in cable 1. (Write your result in N) Answerb) Find the tensile force T2 in cable 2. (Write your result in N) Answerc) Find the angle α of cable 1 with the horizontal. Responsegive the complete solution to the given problem and show the free body diagram
- Question 2) The figure shows a mechanical model of the Russel fracture traction device and the leg. The leg is held in balance in the position indicated by the two weights attached to the two cables. The combined weight of the leg and the cast is W=210 N. The horizontal distance between points A and B where the cables are attached to the leg is L=100 cm and the vertical distance is d=6 cm. Point C is the center of gravity of the cast and leg at three quarters of the L measured from point A (3L/4= 75 cm). The angle that cable 2 makes with the horizontal is measured as β=33°. Accordingly, in order for the leg to remain in balance in the shown position; a) Find the tensile force T1 in cable 1. (Write your result in N) b) Find the tensile force T2 in cable 2. (Write your result in N) c) Find the angle α of cable 1 with the horizontal. ResponseShow complete solution with free body diagram.PLEASE ANSWER IN COMPLETE AND CLEAR SOLUTIONS. INCLUDE NECESSARY FREE-BODY DIAGRAMS. A 6-m long ladder has a mass of 16 kg, and its center of gravity is 2.4 m from the bottom along the ladder. The ladder is resting on a 5° upward sloping floor at A and on a wall inclined to the right at 5° from the vertical at B, so that it makes an angle of 60° with the horizontal. How far up along the ladder can a 77-kg man climb before the ladder is on the verge of slipping? The angle of friction at all contact surfaces is 15°. Also determine the total reaction at A and B in Newton. Note: Point A is at the left of point B.
- show solution with a free body diagramThe uniform bar AB weighing 240 lb is mounted as shown in the figure upon a carriage weighing 480 lb. The center of gravity of the carriage is at C midway between the wheels. If P = 180 lb and there is no frictional resistance at the wheels, find the value of R11. What’s the difference between the center of gravity and the center of mass? 2. How is the specific weight of a body defined? 3. What does it mean when we say a body is homogeneous? 4. What are the theorems of Papus and Guldinus used for?