PHYSICS 170 MECHANICS I Midterm Exam: 7:30 pm to 8:30 pm, Wednesday, February 12, 2014 SOLUTION TO QUESTION 1 a) Diagram: 13=1600 ?-700g+900g)N·m 1 (20 marks) AZ M Fg = 500 ` N page 1 (3 marks) 1 Fg = 800 R N 1- A 3 B(0, 4,0)m × · P (320)m = (6, 40) m FC = 300 JN
PHYSICS 170 MECHANICS I Midterm Exam: 7:30 pm to 8:30 pm, Wednesday, February 12, 2014 SOLUTION TO QUESTION 1 (20 marks) b) Resultant force: F2 = > F=(5007 +300]+800k)N Resultant couple moment: .. (MR)p = ! M + ! (r " }) = 600i ! 700 j + 900 k i + ! ! ! j k !3 !2 0 500 0 0 ! ! ! i j k + !32 0 0 0 800 ! ! ! i j k + 3 2 0 0 300 0 = 2200i +1700 j + 2800 k (MR)p = (2.20i +1.70 j +2.80k)kN!m c) Magnitude and coordinate direction angles: F = 990N (! , ", #) = (59.7!, 72.4', 36.1 !) (3 marks) d) Magnitude and coordinate direction angles: (Mp)p = 3.95 kN!m (! , ", #) = (56.1,64.5',44.8!) (3 marks) e) Wrench? : (Mp), is not parallel to Fp : FR and (Mp), do not form a wrench. page 2 (3 marks) (6 marks) (2 marks)
PHYSICS 170 MECHANICS I Midterm Exam: 7:30 pm to 8:30 pm, Wednesday, February 12, 2014 SOLUTION TO QUESTION 2 a) Free-Body Diagram: Z A À Az è (20 marks) page 1 (4 marks) 1 Cz 450N Cy C 45 ( - 0 . 6 , 1.2 0. 4 ) m DIO, 1.2, 0. 4 ) m BZ A 300N.m Ax X 1ª 10 A (0,0,0) m Bx B (0, 0. 8, 0) m y