PHYSICS 170 MECHANICS I Midterm Exam: 2:00 pm to 2:50 pm, Wednesday, October 24, 2012 QUESTION 1 (20 marks) SOLUTION a) Free-Body Diagram Z . C ( 0 , 8 6 ) ft 1 1 1 m2 A My Ay 17 A1000)ft - 2 A2 (4 marks) m B (12, 4,0 )ft (12,0,0) ft F x
PHYSICS 170 MECHANICS I Midterm Exam: 2:00 pm to 2:50 pm, Wednesday, October 24, 2012 QUESTION 1 (20 marks) SOLUTION Moments about B b) Reactions at A and force in the cable (suppressing units) (4 marks) F = Ai+Aj ! ! x* y ! M = M i + M j +M k ! ! Z ! ! Fp=(!12 i + 4j + 6k)X BC X = F / 122+42+62 c) Cartesian component equations of equilibrium (8 marks) Cartesian component force equations of equilibrium: ! F = 0: `x A !12X + 20=0 X ! F = 0: y A + 4X ! 40= 0 y ! F = 0: Z 6X!90=0 Vector moment equation of equilibrium at point B: (Mp)p= ! M + ! (r " }) ! ! ! X = M i + M j + M k + Z y ! ! i j k !12 !4 0 ! A, A, 0 ! ! ! i j k = 0 + 0 !4 0 20 !40 !90
PHYSICS 170 MECHANICS I Midterm Exam: 2:00 pm to 2:50 pm, Wednesday, October 24, 2012 QUESTION 1 SOLUTION (20 marks) Moments about B Cartesian component moment equations of equilibrium: ! M = 0: M +360=0 X ! M = 0: y ! M = 0: M = 0 y Z M +4A !12A +80= 0 X y d) Solution to the six Cartesian equations of equilibrium (4 marks) X=15 F = 2101b BC X A = 160lb A = ! 20.0lb y y M = 0 N M = ! 960 lbft X M = ! 360 lbft The negative signs indicate that the directions are along the negative coordinate axes.