Question

Use the non-rectangular component approach. The screw eye shown is subjected to two forces F1 and F2. Draw and determine the magnitude of the resultant of the two forces using the Parallelogram Law. Find its orientation theta with respect to the plus x-axis? Test Problem 3 Use the non-rectangular component approach. The engine block shown is suspended by a system of cables. The mass of the engine block is 200 kg. For equilibrium, determine the tension in cable AD and the tension in cables AB and AC.

Answer 1

Solution Problem 2 -F cos - 40lbxcos 30° 34.641b F) = F sine. - 501b xsin 20° 17.101b F 1199.96+292.41 1492.37 38.631b Therefore the magnitude the resultant force is 38.63lb tan 6 17.101b 34.641b 0.49 = tan-1 (0.49 26.270 Therefore the orientation is 26.27

Fi θ 120° 30°

Problem 3 Solution Tension along AD T = mg = 200kg × 9.8m/s* = 1 960N Tension in equation of equilibrium Tc cos 45°-TB coS 60°-0(1) TAc sin 45° COS 60°-1960N-0 ....(2) TAc cos 45°- cos 60° 0 Tc sin 45° COS 60 1960 TAc cos 45°+Tc sin 45° -1960 TCOS45+sn 451960 Tc [0.70+0.70] 1960 c [1.4]-1960 = 1400N Substitute Tscin equation 1 14 00Ncos4 5°-TAB cos 60°-0 TAB cos 60°-1400Ncos45° 1400Ncos45° AB cos 60° 1979.88N Therefore the tension along AD is 1960N Therefore the tension along AB is 1979.88N Therefore the tension along AC is 1400N