Q2. Determine all forces acting on body A for the system shown in figure 2. Body A weighs 200 N and the cylinder B weighs 500 N. assume all the contact surfaces are smooth. 24 ring Cable 05m 30 0.5 m A 1 m 30
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Q: Q2. Determine all forces acting on body A for the system shown in figure 2. Body A weighs 200 N and…
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Q: Q2. Determine all forces acting on body A for the system shown in figure 2. Body A weighs 200 N and…
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- The homogeneous bar AB weighs 25 lb. Determine the magnitudes of the forces acting on the bar at A and B. Neglect friction.In the equilibrium position shown, the resultant of the three forces acting on the bell crank passes through the bearing O. Determine the vertical force P. Does the result depend on 0? 77 lb 8" 24° $ 46 lb Answer: P = i O 10" lbQ2 The 80-lb block C rests on the uniform 15-lb bar AB. The cable connecting C to B passes over a pulley at D. Find the magnitude of the force acting between the block and the bar. C. 30
- 2. The homogeneous bar AB weighs 25 lb. Determine the forces acting on the bar at A and B. Neglect friction. 60° AQ2. Determine all forces acting on body A for the system shown in figure 2. Body A weighs 200 N and the cylinder B weighs 500 N. assume ail the contact surfaces are smooth. 24 ring Cable A r= 0.5 m 30 0.5 m B. 2 m 1 m 1m 30Statics of Rigid Bodies (Round off to two (2) decimal places. Include negative sign (-) if the answer is negative.) The line of action of the 3000 lb force runs through te points A and B as shown in the figure. Determine: y scalar component of F
- 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 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=5 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=30°. 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. ResponseStatics of Rigid Bodies (Round off to two (2) decimal places. Include negative sign (-) if the answer is negative.) The line of action of the 3000 lb force runs through te points A and B as shown in the figure. Determine the x scalar component of F?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 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.
- In the equilibrium position shown, the resultant of the three forces acting on the bell crank passes through the bearing O. Determine the vertical force P. Does the result depend on 6? 55 lb 78 lb Answer: P = $ 25+30 cm 18 cm Hom Scm Hom Sum 2. The 30 cm handle is connected rigidly to the 5cm bar at a right angle. Determine the force developed at A if a force P = 100 N is applied perpendicular to the handle when the handle is at the position indicated.300 lb NOT TO SCALE B. E D 4 ft 650 lb · ft 3 ft 2 ft -2 ft→ The pin at point C fits into a smooth slot cut into member AC. Since there is no friction between the pin and the slot, the force there is normal to the slot. For component directions, use x positive to the right and y positive upward.