PHYSICS 170 MECHANICS I March Exam: 8:00 pm to 9:00 pm, Tuesday, March 5, 2019 SOLUTION TO QUESTION 1 a) Free-Body Diagram Z A Cx page 1 c (-1.75, 0, 2)m 182 A Cy Ax A (0-20)m 1 1 7 A 0 300 60° B (-100)m 12 By
PHYSICS 170 MECHANICS I March Exam: 8:00 pm to 9:00 pm, Tuesday, March 5, 2019 SOLUTION TO QUESTION 1 page 2 b) Cartesian component force equations of equilibrium Forces (suppressing units): F = 100 (cos 60° cos 30°[ + cos 60° sin 30° j - sin 60° k) F= Ai+Ak x Z r r F = B i + Bk z r B y T F = Ci + C j X y Equations of equilibrium: !F = 0: A + C + 100cos60° cos30° = 0 X X X (1) !F = 0: B + C + 100cos60° sin30° = 0 y (2) y y !F = 0: z A + B ! 100 sin60° = 0 Z (3) Z
PHYSICS 170 MECHANICS I Final Exam: 12 noon to 2:30 pm, Friday, April 22, 2016 SOLUTION TO QUESTION 1 page 3 c) Cartesian component moment equations of equilibrium Couple moment (suppressing units): 1 M = 200i ! 300 j + 400 k Vector moment equation of equilibrium at the origin: T (M) =! M+!(r"F) = 2001 ! 300 j + 400k r r i j r r k i r r j k r r r i j k + 0 !2 0 + !1 0 0 + !1.75 0 2 =0 A 0 A 0 B x z y B C CO x y z Equations of equilibrium: !M = 0: x 200!2A !2C =0 (4) Z y !M = 0: y !300+ B + 2C = 0 Z X (5) !M = 0: 400+ 2A ! B ! 1.75C = 0 z (6) X y y