PHYSICS 170 MECHANICS I Final Exam: 12 noon to 2:30 pm, Wednesday, April 17, 2013 SOLUTION TO QUESTION 1 a) Free-Body Diagram ? A . c (1,0, 2) m 1 Az 1 1 A - - - - A (20 marks) page 1 (4 marks) . D(-2, 0,1) m 1 15 1 A F BD FBC 10 $ Ay A Ax B (0, 2,0) m G ( 0 , 1 , - 0 , 5 ) m P 1 F G
PHYSICS 170 MECHANICS I Final Exam: 12 noon to 2:30 pm, Wednesday, April 17, 2013 SOLUTION TO QUESTION 1 b) Forces (20 marks) F = Ai+Aj+Ak Fpc=(i!2 j+2k)x ! ! ! page 2 X=F/ 2 + 2 + 22 F = (!21 ! 2 j +k)Y BD 12+ Y = F/ 22+22+12 BD F= ! 100(9.81) k Cartesian component force equations of equilibrium: ! F = 0: A+ X "2Y = 0 × X ! F = 0: y A "2X "2Y = 0 y ! F = 0: A +2X+Y " 981= 0 Z Z c) Couple moment M = ! 500i + 600 k Vector moment equation of equilibrium at point B: (Mp)p = ! M + ! (r " }) = ! 500i + 600 k ! ! ! i j k + 0 !2 0 Ax x AV A, ! ! ! i j k = 0 + 0 !1 !0.5 0 0 !981 (4 marks) (4 marks)
PHYSICS 170 MECHANICS I Final Exam: 12 noon to 2:30 pm, Wednesday, April 17, 2013 SOLUTION TO QUESTION 1 (20 marks) page 3 d) Cartesian component moment equations of equilibrium ! M = 0: X "500" 2A +981=0 Z ! M = 0: 0 = 0 y ! M = 0: Z 600+2A =0 X (4 marks) e) Solve the Cartesian component equations of equilibrium (4 marks) X = 356.2 FD=1.07kN BC A = ! 300N A = 769N Z A = 240N Y =28.1 BD F= 84.3N y