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 B/0,-3,27m 1E A8 page 1 (3 MARKS) c (0,0,2) m 1 0 x A (600)m 1 30 400N 2 AC Dx è D (0,0,00 m 17 Dy y
PHYSICS 170 MECHANICS I March Exam: 6:00 pm to 7:00 pm Thursday March 12, 2020 SOLUTION TO QUESTION 1 page 2 Forces and couple moments (suppressing units) F = 400 (sin 30° ] - cos 30° k) M =- 600}+720k F = D? + D ]+D? F .= (-61- 3] + 2k) X F =(-6i + 2k)Y AC X = F. / V62 + 32 + 22 AB Y = F / V62 + 22 b) Cartesian component force equations of equilibrium (4 marks) Equations of equilibrium: ?F =0: D -6X-6Y =0 x (1) EF =0: y D - 3X + 400 sin 30° = 0 y (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 ) . = ??+ ?(7 x F) =- 600 j+720k + -6 0 0 =0 ? j D. Dy z D Equations of equilibrium: ?? =0: 0 = 0 x ??=0: y 6D - 600 = 0 (4) Z ?? =0: -6D +720 = 0 Z y (5)