PHYSICS 170 MECHANICS I Final Exam: 12 noon to 2:30 pm, Monday, April 8, 2019 SOLUTION TO QUESTION 1 a) Free-body diagram c (1,0, 2) m € A 1- Az 4 ? A Z 1 1 - - - FBC - Ay A page 1 D(-2, 0,1) m 1 1 1 15 1 1 F BD B (0, 2,0) m G (0, 1, -0,5) m Dy F G
PHYSICS 170 MECHANICS I Final Exam: 12 noon to 2:30 pm Monday April 8 2019 SOLUTION TO QUESTION 1 page 2 b) Cartesian component force equations of equilibrium Forces: F = A? + A ]+Ak y Z Fp=(1-2]+2k)X F. = (-27-2]+])Y F ==- 80(9.81) =- 784.8k Equations of equilibrium: ??=0: A +X-2Y =0 x x BC X=F /V12+22 +22 Y=F /122+22 +12 BD (1) ?F =0: y (2) x A -2X-2Y =0 ?F=0: A +2X+Y-784.8 =0 Z Z (3)
PHYSICS 170 MECHANICS I Final Exam: 12 noon to 2:30 pm Monday April 8 2019 SOLUTION TO QUESTION 1 page 3 c) Cartesian component moment equations of equilibrium Couple moment: M =- 400?+200k Vector moment equation of equilibrium at point B: (M ). = ??+?(FxF)= i -400? +200k + 0 -2 i j 0+ 0 -1 0.5 =0 A A A x y z? 0 0 -784.8 Equations of equilibrium: ?? =0: -400-2A +784.8 = 0 x Z (4) ?? =0: y 0 =0 ??=0: 200+2A =0 x (5)