For the figure show to the right, neglect all self-weight and compute the following: a) The moment of force F about point B. Give your 3 m answer as a Cartesian vector. b) The magnitude of the tension in cables AB and AC. C The support reactions at D. Give your answers as 6 m Cartesian vectors. d) The magnitude of the internal bending moment about the y-axis at the midpoint of strut AD. F = (-1201 + 180j – 340k )N C. 20

A P = 10 kip 1.5 ft B C 2 ft The bars in the structure to the left are pinned and have circular cross-sections. The diameter of the cross- section is 1.5 inches. The bars are made of steel, which has an elastic modulus of 29,000 ksi. A point force of 10 kips is applied at point C, as shown. What is the change in length of member BC? Is the bar getting longer or shorter?

The frame shown supports a load W DE 1000 lb through the cable that passes through 3 pulleys and at the end is attached to a frame, also supports the distributed loads ω1 = 100 lb / ft; ω2 = 60 lb / ft and ω3 = 50 lb / ft. Determine: 1. Reactions at supports A and E. 2. Forces acting on bar ABCDE, show the results in graphical form of the beam. That is, you must make the simplified free-body diagram of the loads that act only on the horizontal element. 3. Make the shear force and bending-moment diagrams for the entire beam. The geometry and load data are as follows: see the geometry and load in the images. Clarification: THE CABLE GOES BEHIND THE FRAME AND THE CONTACT IS ONLY WITH THE PULLEYS AND THE GRIP TO THE FRAME (WALL).All pulleys have the same radius.