PHYSICS 170 MECHANICS I March Exam: 8:00 pm to 9:00 pm, Tuesday, March 6, 2018 SOLUTION TO QUESTION 1 a) Free-Body Diagram 12 (10-23) my 1 1 1 Az FBC 12 FBD Az 10 17 E (300)m 1 x B(60,00 m 600N page 1 D (0, 2, 3 ) m A (0,0,0)m - Ay y
PHYSICS 170 MECHANICS I March Exam: 8:00 pm to 9:00 pm, Tuesday, March 6, 2018 SOLUTION TO QUESTION 1 page 2 b) Cartesian component force equations of equilibrium Forces (suppressing units): F = A? +A ]+Ak F .= (-67-2]+ 3k) X F. = (-67 +2] + 3k)Y F ==- 600 k Force equations of equilibrium ?? =0: x X = F_| V62 + 22 + 32 BC Y = F / V62 + 2 + 32 BD A-6X-6Y=0 (1) EF =0: y A-2X+2Y =0 (2) y ?F=0: A +3X + 3Y - 600 = 0 Z z (3)
PHYSICS 170 MECHANICS I March Exam: 8:00 pm to 9:00 pm, Tuesday, March 6, 2018 SOLUTION TO QUESTION 1 page 3 c) Vector moment equation of equilibrium (about point A) (M ) =??+ ?(F x F)=600j +1200k i i j k + 6 0 0X+6 0 0Y+3 0 0 =0 -6 -2 3 -6 2 3 0 0 -600 Vector moment equation of equilibrium (about point B) (M.) =?? + ?(F x F)=600j +1200k i + -6 0 0 +-3 A A x k i j 0 0 =0 A y 2 0 0 -600