Practice Test 1 with Solutions - Introduction to Engineering Analysis | ENGR 1100, Exams of Engineering

Download Practice Test 1 with Solutions - Introduction to Engineering Analysis | ENGR 1100 and more Exams Engineering in PDF only on Docsity! 1 Intro to Engineering Analysis, Studio/Laptop Version Test #1, Sept. 18, 1998 NAME_ Solution________________ If you wish, you can use Maple to do the problems below, but you must show your methodology on the test pages. Remember that clarity is important. Problem 1 (48 points, 8 points each) Use Cartesian vectors to solve the following problems based on the attached sketch. Boldface indicates a vector quantity. a) Write the position vector from O to A. Also write a position vector along OA that has half the magnitude of the vector from O to A. Solution: OA = (2i + 3j + 7k) m HOA = (i + 1.5j + 3.5k) m b) Write a force vector of magnitude 10 N that acts along direction OA. Solution: OA = (2i + 3j + 7k) m eOA = {(2i + 3j + 7k)/(2 2 + 3 2 + 7 2 )} eOA = {(2i + 3j + 7k)/7.87 F = 10 N (0.254i + 0.381j + 0.889k) F = (2.54i + 3.81j + 8.89k) N c) Write a force vector F of magnitude F that acts along direction AO. Solution: eAO = -eOA = -0.254i - 0.381j - 0.889k F = F (-0.254i - 0.381j - 0.889k) d) Write the position vector OB, when the length of vector OB is 10 m, x = 90 degrees, y = 45 degrees, and OB is in the y-z plane. Solution: OB = 10 m (cos x i + cos Y j + cos Z k) Where x = 90 , Y = 45 , Z = 45 OB = 10 m (0i + 0.707j + 0.707k) Y = 45 y z x C (3,-2,5)m • • • A (2,3,7)m B X = 90 O 2 OB = (7.07j + 7.07k) m e) What is the cosine of the angle between position vector OC and the positive x-axis? Solution: eOC  eX = (eOC)(ex) cos  (by the definition of the dot product) cos  = (eOC  eX) *Note that eX is the unit vector along the x-axis (eX = i). OC = (3i - 2j + 5k) m; eOC = {(3i - 2j + 5k)/(3 2 + -2 2 + 5 2 )} cos  = (eOC  i) cos  = (0.487i – 0.325j + 0.812k)  (i) cos  = 0.487 f) Write a position vector BA that goes from point B to A. Solution: BA = OA – OB = {(2 - 0)i + (3 - 7.07)j + (7 – 7.07)k)} m BA = (2i – 4.07j - 0.07k) m 5 Problem 3 (32 points) Answer questions a, b and c (below) for the following problem (from Engineering Mechanics - Statics by Hibbeler): Romeo is trying to reach Juliet by climbing with a constant velocity up a rope that is knotted at point A. Any of the three segments of the rope can sustain a maximum force of 2000 N before breaking. Romeo has a mass of 65 kg and Juliet has a mass of 60 kg. Juliet joins Romeo and they are at rest on the rope at Romeo’s position shown in the diagram. What is the tension in each rope, and will any of the ropes break? a) (10 points) Draw an appropriate free body diagram for this problem. Solution: y x W = (65+60)(9.81)N W TAC TAB 60 6 b. (10 points) Write out the relevant vectors and identify the unknown quantities. Solution: TAB = TAB (-cos 60i + sin 60j) TAC = TAC (i) W = 1225 (-j) N Unknowns: TAB and TAC b) (12 points) Set up, but don’t solve, the scalar equations that would have to be solved for the unknown quantities in this problem. NOTE: If you wish to solve the equations, correct numerical answers will be worth up to 5 extra bonus points for the test. Solution: For static equilibrium, F = 0 FX = -TAB cos 60 + TAC = 0 (1) FY = TAB sin 60 – 1225 N = 0 (2) Extra Credit: Solving equation (2) directly: TAB = 1414.5 N Substituting into equation (1) and solving: TAC = 707.3 N Both of these tensions are less than the limit (2000 N). Therefore, Romeo and Juliet will climb away to safety and live happily ever after…