PHYSICS 170 MECHANICS 1: Final Exam: 3:30 PM to 6:30 PM, Saturday, Dec. 16, 2017 SOLUTION TO QUESTION 1 (25 MARKS) a) Free-Body Diagram Z 1 E14,0,27ft A(0,0,0) ft L 1 Ay Ax TIL Mx DE D(3,2,0) ft 1 × G(0,5,2,-1)ft page 1 (7 marks) C(-3,0,2)ft 1 1 1 711 BC B(-2, 2,0)ft 3 15 < A Y 13
PHYSICS 170 MECHANICS 1: Final Exam: 3:30 PM to 6:30 PM, Saturday, Dec. 16, 2017 SOLUTION TO QUESTION 1 (25 MARKS) page 2 b) Cartesian component equations of equilibrium (15 marks) Forces and moments (suppressing units): (5 marks) FA = Ai + A,j FBC=(-?-2]+2k) X X = FRc / V12 + 22 + 22 = FRC / 3 FDE =(i-2] +2k) Y W =- 50k Y = FPE / V12 + 22 + 22 = FOR / 3 MA=M? + My j Cartesian component force equations of equilibrium: (3 marks) x ?FR =0: ?Fy =0 : EF2=0: A - X +Y = 0 (1) AV-2X -2Y = 0 (2) 2X+2Y-50=0 (3) Vector moment equation of equilibrium at point A: (4 marks) TAB =- 27 +2] TAD = 37 +2] TAG =0.51 +2j-k (MR)A=?M + ?(FXF)= M.i +M ]+ i jk li jk i j +-2 2 0 X+3 2 0Y + 0.5 2 -1 -2 2 1 -2 2 0 0 -50 k -1 =0
PHYSICS 170 MECHANICS 1: Final Exam: 3:30 PM to 6:30 PM, Saturday, Dec. 16, 2017 SOLUTION TO QUESTION 1 (25 MARKS) page 3 Cartesian component moment equations of equilibrium: (3 marks) M +4X +4Y -100 =0 My +4X -6Y +25 =0 6X-8Y=0 (4) (5) (6) c) Solution to the equations of equilibrium: (3 marks) X =14.286 lb FBC = 42.9 lb Ax = 3.57 lb M =0.00 lb -ft Y = 10.714 lb FDE =32.1 lb A = 50.0 lb M =- 17.9 lb .ft The negative sign indicates that the direction is along the negative y axis.