PHYSICS 170 MECHANICS I Midterm Exam: 7:30 pm to 8:30 pm, Wednesday, February 11, 2015 SOLUTION TO QUESTION 1 (20 marks) page 1 a) Free-body diagram (2 marks) x 14- 300 Z 4 0 60 0 135 100 600 0 60 1,1 - y t V F ( 4 , 4 - 2 ) m
PHYSICS 170 MECHANICS I Midterm Exam: 7:30 pm to 8:30 pm, Wednesday, February 11, 2015 SOLUTION TO QUESTION 1 (20 marks) b) Forces page 2 (8 marks) F = F (cos60° cos30° i - cos60° sin 30° j + sin 60° k) ! F2 = F2 (cos135° i + cos60° j + cos60° k) ! F= (4i + 4j ! 2 k)X X = F /V42+42+22 F = (19 i ! 8 j ! 5 *) c) Cartesian component force equations of equilibrium ! F = 0: F, cos60° cos30° + F, cos135° + 4X ! 9 = 0 (1) x (8 marks) ! F = 0: ! F cos60° sin 30° + F, cos60° + 4X ! 8 = 0 (2) y ! F = 0: F sin60° + F2 cos60° ! 2X ! 5 = 0 (3) N d) Solution to the equations of equilibrium (2 marks) F = 8.26kN 1 F2 = 3.84kN F2 = 12.2kN
PHYSICS 170 MECHANICS I Midterm Exam: 7:30 pm to 8:30 pm, Wednesday, February 11, 2015 SOLUTION TO QUESTION 1 (20 marks) page 3 SOLVING THE EQUATIONS: REDUCED ROW ECHELON FORM METHOD · Equations (1) to (3) can be solved using the reduced row echelon form matrix program rref([A]) on a calculator where [A] is the 3! 4 matrix 0 cos60 cos30 " o $ [A]= $ $ O cos135 o o % 4 9 ' ' ! cos60 sin30 o o # sin60 cos60 0 cos60 4 8 ' ' !2 5 & This yields ! 1 0 0 F # # 010 F # 001X % $ 1 0 0 8.256 $ & & = 0 1 0 3.843 & & # 0 0 1 2.036 % ???…