45° FR F 75° 60 200 lb Using the sine rule, we have: O a F- 200lb Ob F- 273lb Oc F=189lb O d. F= 245lb

1. Note: for part a), you can do this without a calculator; it might be easier to work out the problem symbolically, and then substitute the numbers at the end of the problem. В 2L The axial rod is fixed to the two walls, as shown above. Portion AB is composed of one material (EAB = E), while portion BD is composed of a different material (EBD = 2E). Portion AB has cross sectional area AAB = A, while portion BD has cross sectional area ABD = 2A. EAB = E = 15×106 psi AAB = A = 1 in² EBD = 2E = 30×10° psi ABD = 2A = 2 in² LAB = L = 10 in. LBC = 2L = 20 in. LcD = L = 10 in. An external force, P = 100 kips, is applied at point C as shown below. a) Determine the reaction forces at wall A and D, Ra and Rp. Specify the direction, (→) or (+). b) Deflection at point C. Specify the direction, (→) or (+). (Ans: RA = 100/7 kips →, Rp = 600/7 kips →, & = 1.429x10² in –)

A cylindrical buoy 0.60 m in diameter and 1.80m high weighs 205kg. It is moored in salt water to a 12m length of chain weighing 12 kg/m of length. At high tide the height of the buoy protruding above water is 0.84m. What could be the height pf protrusion of the buoy if the tide dropped 2.10m? Specific weight of the steel chain is 7,790 kg/m^3. 206kg 0.84m W.S. 0.96m h,

Problem 3 Compute the moment about point D.