PHYSICS 170 MECHANICS I March Exam: 6:00 pm to 7:00 pm, Thursday, March 12, 2020 SOLUTION TO QUESTION 1 a) Free-Body Diagram x 4 8(0,-3, 2) m 1E AB - 1 AC Dx A(60,0)m 30 400N page 1 Az c (0,0,2) m A Dz è D 10,0,00 m J Dy y
PHYSICS 170 MECHANICS I March Exam: 6:00 pm to 7:00 pm Thursday March 12, 2020 SOLUTION TO QUESTION 1 Forces and couple moments (suppressing units) F = 400 (sin 30° ] - cos 30°k) FD= D?+D ]+Dk Fp=(-6i- 3]+ 2k) X F =(-6i + 2k)Y page 2 M =- 600 j+ 720k X = F / V62 + 32 + 22 AB Y = F_/ V62 + 22 AC b) Cartesian component force equations of equilibrium Equations of equilibrium: ??=0: D -6X-6Y =0 X (4 marks) (1) x ?F =0: y y D - 3X + 400sin 30° = 0 (2) ?F=0: D +2X +2Y - 400 cos 30° = 0 Z Z (3)
PHYSICS 170 MECHANICS I March Exam: 6:00 pm to 7:00 pm Thursday March 12 2020 SOLUTION TO QUESTION 1 page 3 c) Cartesian component moment equations of equilibrium (5 marks) Vector moment equation of equilibrium at point A: (M.) =?? + ?(Fx F) i j ? 0=0 = - 600 } +720k + -6 0 x D D D y Z Equations of equilibrium: ?? =0: 0 = 0 x ?? =0: 6D - 600 = 0 y Z (4) ?? =0: 2 -6D +720 = 0 y (5)