Question

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Write out a hand solution for the following problem. Given: L1 2 feet · L2 = 1 foot . L3 = 2 feet . The length of rope BD is 2.5 feet . The unstretched length of the spring AC is 2.5 feet . The spring constant for the spring is k- 10 lb/in. Find . The values of the weights hung and A & B (in pounds) . The tension in the spring and in each rope. Remember: you must draw all relevant FBDs. Be sure to include o the name of the point that your FBD is depicting o the magnitudes of known forces, or a label of unknown forces (you can pick the names) o All internal forces drawn as tension. o All relevant angle information

Answer 1

Let correct in S.I unit L_1 = 2 feet = 2 times 0.305 == 0.61m L_2 = 1 feet = 0.305 m L_3 = 2 feet = 0.61 m BD = 2.5 feet = 0.7625 m AC = 2.5 feet = 0.7625 m Spring k = 10 lb/in = 4.54 kg/0.0254 m = 178.74 kg/m \r\n Now calculating theta, & CA^2 = L^2_3 + L^2_1 CA' = 0.8626 Now as theta = L_c/CA = 0.61/0.8626 = 0.707 = > theta almostequalto 45 degree sin theta = 0.707 similarly sin = L_3/BD = 0.61/0.7625 = 0.8 cos = 0.6 Apply Theorem At point A Kx/.sin 90 degree = T_2/sin (90 degree + theta) = T_1/sin (180 degree - theta) = T_1/sin theta \r\n Now 'x' [] = CA' - CA = 0.8626 - 0.7625 almostequalto 0.1 m thus (178.74) times 0.1 = T_2/0.707 = T_1/0.707 T_1 = 12.637 N T_2 = 12.637 N. Apply Theorem at point B. T_3/sin 90 degree = T_2/sin (90 degree