Thinking Like an Engineer 2e Instructor Slides SOLVEM Thinking Like an Engineer An Active Learning Approach, 2e Stephan, Bowman, Park, Sill, Ohland Copyright © 2013 Pearson Prentice-Hall, Inc. SOLVEM SOLVEM Thinking Like an Engineer 2e Instructor Slides Instructor Slides

Thinking Like an Engineer 2e SOLVEM SOLVEM Methodology S = Sketch O = Observations and / or Objectives L = List V = Variables and constants E = Equations M = Manipulate SOLVEM Thinking Like an Engineer 2e Instructor Slides

Thinking Like an Engineer 2e SOLVEM Example SOLVEM_1 What is the area of the largest equilateral triangle that can be cut from a circular piece of material 50 centimeters in diameter? SOLVEM Thinking Like an Engineer 2e Instructor Slides

Example, continued SOLVEM_1 Thinking Like an Engineer 2e SOLVEM Example, continued SOLVEM_1 What is the area of the largest equilateral triangle that can be cut from a circular piece of material 50 cm in diameter? SOLVEM H L 60° 120° 60° Thinking Like an Engineer 2e Instructor Slides

Example, continued SOLVEM_1 Thinking Like an Engineer 2e SOLVEM Example, continued SOLVEM_1 Objective: Determine the area of an equilateral triangle inscribed in a circle Observations: Each angle of the triangle is 60° Each side of the triangle is a chord of the circle The angle subtended by one of the chords is 120° (360°/3) 60° 120° H L SOLVEM Thinking Like an Engineer 2e Instructor Slides

Example, continued SOLVEM_1 Thinking Like an Engineer 2e SOLVEM Example, continued SOLVEM_1 D Diameter of circle 50 cm R Radius of circle φ Internal angle of equilateral triangle 60° θ Angle subtended by chord 120° L Length of chord (base of triangle) H Height of triangle A Area of triangle SOLVEM Thinking Like an Engineer 2e Instructor Slides

Example, continued SOLVEM_1 Thinking Like an Engineer 2e SOLVEM Example, continued SOLVEM_1 Radius of circle: R = 0.5 D Area of triangle: A = 0.5 L H Length of chord: L = 2 R sin(θ/2) tan(φ) = length of opposite side / adjacent side H/(L/2) for the right triangle defined by half of the equilateral triangle SOLVEM Thinking Like an Engineer 2e Instructor Slides

Example, continued SOLVEM_1 Thinking Like an Engineer 2e SOLVEM Example, continued SOLVEM_1 Length of chord: L = 2 R sin(60°) = 1.73 R Height of triangle: H = L tan(φ)/2 H = (1.73 R) tan(60°)/2 H = 1.5 R Area of triangle: A = 0.5 (1.73 R) (1.5 R) A = 1.30 R2 SOLVEM Thinking Like an Engineer 2e Note that by solving for the general case in terms of R (A = 1.30 R2), we can now easily calculate the area of an inscribed equilateral triangle in any size circle. Instructor Slides

Example, continued SOLVEM_1 Thinking Like an Engineer 2e SOLVEM Example, continued SOLVEM_1 NOW, plug in numbers! R = 0.5 D R = 0.5 (50 cm) = 25 cm A = 1.30 R2 A = 1.30 (25 cm)2 A = 811 cm2 SOLVEM Thinking Like an Engineer 2e Reasonableness check: Visual inspection leads one to think that the triangle area is about half the area of the circle. The area of the circle is 1960 cm2, so 811 seems reasonable. Instructor Slides

Example SOLVEM_2 You purchase a truck having wheels and tires such that each wheel-tire combination has a radius of 14 inches. The speedometer, which reads accurately, is calibrated for these tires. Inspired by a monster-truck event, you decide to jack up your truck such that the wheel-tire radius is 20 inches. Identify the effect this has on your speedometer reading. Specifically, if your speedo-meter indicates that you are traveling at 60 miles per hour, what is your actual speed? SOLVEM Thinking Like an Engineer 2e

Example, continued SOLVEM_2 Thinking Like an Engineer 2e

Example, continued SOLVEM_2 Observations: Assuming the tires do not slip, for one revolution, the tire circumference will be the distance traveled by the truck. The number of wheel revolutions per time determines the speedometer reading. Since more distance is covered per revolution for a bigger tire, the truck will be moving faster than the speedometer says it is in this case. Since we are comparing one tire to another, I should be able to solve this problem as a ratio, without having to worry about unit conversions. SOLVEM Thinking Like an Engineer 2e

Example, continued SOLVEM_2 R wheel radius 14 inches, 20 inches Vtire the truck velocity for a particular tire t time revtire revolution distance for a particular tire tire tire radius SOLVEM Thinking Like an Engineer 2e

Example, continued SOLVEM_2 Circumference of wheel = 2 π R V14 = (rev14 / t) V20 = (rev20 / t) rev14 = circumference of wheel14 rev20 = circumference of wheel20 SOLVEM Thinking Like an Engineer 2e

Example, continued SOLVEM_2 Thinking Like an Engineer 2e SOLVEM Example, continued SOLVEM_2 Step 1: Solve Equations 2 & 3 for time (t), which is unimportant since we are finding a rate. t = K (rev14 / V14) t = K (rev20 / V20) Step 2: Since unit time is the same in each case, equate the two expressions: (rev14 / V14) = (rev20 / V20) V20 = V14 (rev20 / rev14) SOLVEM Thinking Like an Engineer 2e Instructor Slides

Example, continued SOLVEM_2 Step 3: Substitute for circumference (rev20 /rev14) = (2 π R20) / (2 π R14) = (R20 / R14) Step 4: Now we can use ratio of radii instead of ratio of velocities V20 = V14 (rev20 / rev14) = V14 (R20 / R14) SOLVEM Thinking Like an Engineer 2e

Example, continued SOLVEM_2 NOW plug in number! V20 = V14 (R20 / R14) = (60 mph) (20 in / 14 in) V20 = 86 mph SOLVEM Thinking Like an Engineer 2e