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- The cable AB supports a uniformly distributed load of 12 lb/ft. Determine the angles A and B, and the cable tension at A.The end of a water hose weighing 0.5 lb/ft is pulled with a 40-lb force that is inclined at 14 to the horizontal. Determine the length s of the hose that is lifted off the ground and the corresponding horizontal distance L.The properties of the unequal angle section are Ix=80.9in.4,Iy=38.8in.4, and Iu=21.3in.4. Determine Ixy.
- Resolve the 360-lb force into components along the cables AB and AC. Use =60 and =40.The center of gravity of the 850-N man is at G. If the man pulls on the rope with a 388-N force, determine the horizontal distance b between the man's feet and G.The 110-lb traffic light is suspended from two identical cables AB and BC, each weighing 0.80 lb/ft. If the maximum allowable horizontal force exerted by a cable on a vertical post is 170 lb, determine the shortest possible length of each cable and the corresponding vertical distance h.
- The string attached to the kite weighs 0.4 oz/ft. If the tension in the string is 2.8 lb at O and 3.2 lb at B, determine the length s of the string and the height H of the kite.The 30-lb block is held in place on the smooth inclined plane by the force P. (a) Knowing that =20, determine P. (b) Given that P=40lb, find the angle .If the intensity of the line loading is w=[(40xx2)/40]lb/in., where x is measured in inches, use integration to find the resultant.
- Determine the location of the resultant force acting on the shaft shown from left. options are as follows: a) 4.5m b) 4m c) 3.5m d) 2mQ3 / Determine and locate the resultant of the system shown in fig (3) If you know that the values of Ax = 240 N. Ay = 2000 N. F = 160 N, N = 2400 N. W = 400 N and M = 2N.mShow the complete solution, for this eq.