Exploring Engineering

Exploring Engineering Chapter 2 Key elements in Engineering Analysis

What we are going to learn • Maybe the most important single lecture in this course (which you should have already read ahead). • Engineering is about units as well as numbers. • How to deal with units and dimensions • Newton’s 2nd law of motion • SI and Engineering English units • “gc” and “g” • Significant figures

Units and dimensions • All engineers will have to understand this material irrespective of sub discipline • Let’s start with my favorite superhero…it’s Superman! • Why is this relevant?

Superman – Engineering Hero • Superman represents the many of the qualities that engineers must master: Consider his qualities… • Faster than a speeding bullet • More powerful than a mighty locomotive • Can leap tall buildings with a single bound • Keeps falling into kryptonite traps

Superman – Engineering Hero • Superman embodies many engineering concepts! • Faster … speeding bullet velocity/speed* • More powerful … mighty locomotive power • Can leap …tall buildings force and energy • …kryptonite traps Information (or lack ofit!) • Yes! There is a difference. We will see later!

Superman – Engineering Hero • Suppose I asked you what is ? • Pretty good answer is 3.14, or 3.142, or 3.141593 • Suppose now I ask what is Superman’s speed? • Is 800 an answer? • No! Not unless we add something - i.e., 800 m/s • The units, meters/second, really adds some new information..had we said 800 inches/hr Superman would be called “Supermolasses”!

Variable Units Number Velocity/speed m/s(miles per hour - mph, furlongs per fortnight) 800(1789, 4.81 x 106) Power hp (kW) 2,000 (1491) Energy N-m (ft lbf) 9.81 x 104 (72,300) Informa-tion Bits Need enough to dodge kryptonite!

Units and Dimensions • Did you see that we converted from one set of units to another as in m/s converted to furlongs/fortnight? • There is a “fail-safe” method of converting*. • Example: What’s the volume of a 1 ft cube in m3 if 1 m = 3.28 ft (or 3.28 [ft/m])? • V = 1 ft3, V = 1/3.283[ft3][m/ft]3= 0.028 m3 • In simple cases the free web program Convert.exe is pretty good too!

Units and Dimensions • What’s the acceleration of a rocket in mph/s if you know it in SI units, a = 55 m/s2? • 1 mile = 1609 m [i.e., 1609 m/mile] & 1 hour = 3,600 s [i.e 3600 s/hr]. • a = 55  3600/1609[m/s2] [s/hr][mile/m] = 120 mph/s (to 2 significant* figures) • … of which more later

Units and Dimensions • In this course we will require the units to be manipulated in square brackets […] in each problem. • While easy to get the previous solutions without this method, many engineering problems are much harder than this & need this apparently clumsy methodology. • Computerized unit conversions are available in free software on the Internet (for example at: http://joshmadison.com/software/convert-for-windows)

More Conversion Examples: • These use conversion factors you can paste from Convert.exe • 800 m/s to mph • 800 [m/s][3.28ft/m][1/5280 miles/ft][3600 s/hr] • 800 x 2.236 = 1790 [mph] • 2,000 hp to kW • 2,000 [hp][0.7457 kW/hp] = 1492 kW • 9.81 x 104 N mto ft lbf • 9.81 x 104[N m][1/4.448lbf/N][3.28 ft/m] •  9.81 x 104 x 0.737 = 7.23 x 104 ft lbf

Newton’s 2nd Law and Units • What Newton discovered was not “may the force be with you”, nor “may the mass  acceleration be with you” but that force is proportional to the acceleration that it produces on a given mass.

Force, Weight, and Mass • In high school you learned F = ma but there’s more to it • Newton said that force was proportional to mass x acceleration (not equal to it) because the equation also defines force • So an undefined force is given by Fma and in some also undefined unit system F1m1a1 (e.g., Force in units of wiggles, mass in carats and acceleration in furlongs/fortnight2) • Eliminate the proportionality,

Force, Weight, and Mass • The ratio (F1/m1a1) is arbitrary. Picking it defines the unit of force. • SI system: F1  1 Newton whenm1 = 1 kgand a1 = 1 m/s2 • Then you can use F = ma • English system: F1  1lb force whenm1 = 1 lb massand a1 = 32.174 ft/s2

Example 1 • What would the SI force on a body if its mass were 856 grams? • Need: Force on a body of mass 856 g (= 0.856 kg) accelerated at 9.81 m/s2 • Know: Newton’s Law of Motion, F = ma • How: F in N, m in kg and a in m/s2. • Solve: F = ma = 0.856  9.81 [kg] [m/s2 ] = 8.397 =8.40 N

Example 2 • What would the lbf force on a body if its mass were 3.25 lb mass? • Need: lbf on a body of 3.25 lbm accelerated at 32.2 ft/s2 • Know: Newton’s Law of Motion, F = ma/gc • How: gc = 32.2 lbm ft/lbf s2 • Solve: F = ma/gc = 3.25  32.2 /32.2[lbm] [ft/s2 ][lbf s2]/[lbm ft] = 3.25 lbf • Weight is W = mg/gc – a special familiar force.

Example 3 • What would the lbf force on a body located on the moon (g = 5.37 ft/s2) if its mass were 3.25 lbm? • Know: Newton’s Law of Motion, F = ma/gc • How: gc = 32.2 lbm ft/lbf s2 unchanged • Solve: F = ma/gc = 3.25  5.37 /32.2[lbm] [ft/s2 ][lbf s2]/[lbm ft] = 0.542 lbf

Newton’s 2nd Law and Units • It bears repeating: SI system is far superior and simpler: • Example: How many N to accelerate 3.51 kg by 2.25 m/s2? • Ans: F = 3.51 x 2.25 [kg][m/s2] = 7.88 N

Significant figures • Arithmetic cannot improve the accuracy of a result • 10 meters, 10. meters, 10.0 meters and 10.00 meters are not identical • 10 meters implies you have used a 10 meters scale; 10. meters implies you have used a 1 meter scale; 10.0 meters implies you have used a 0.1 meter scale and 10.00 meters implies you have used a 0.01 meter scale

Significant figures • Thus 10/6 = 2 and not 1.66666667 etc. as displayed in your calculator • A significant figure is any one of the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0. Note that zero is a significant figure except when it is used simply to fix the decimal point or to fill the places of unknown or discarded digits.

Significant figures • 1.23 has 3 sig. figs. • 4567 has 4 sig. figs. • 0.0123 has three sig. figs. • 12,300 has three sig.figs. (The trailing zeroes are place holders only) • 1.23 x 103, 1.230 x 103, 1.2300 x 103 have 3, 4, and 5 sig. figs. respectively

Significant figures – example • Round off 123.456 − 123.0 • 123.456 has 6 sig. figs. • 123.0 has 4 sig. figs. • But 123.0 is the least precise of these numbers with just 1 figure to right of decimal place • Thus 123.456 − 123.0 = 0.456 = 0.46 = 0.5 • The moral: In this course you will be graded on significant figures – read your text for all the relevant rules of round-off!

Summary Engineering problems need precise mathematics • But not more precise than can be justified (see text, Chapter 1) • Units must be consistent • […] method is very helpful in maintaining correct units • Newton’s 2nd law defines force and gives rise to different sets of units • In SI, force = ma and wt = mg • In English units, force = ma/gc and wt = mg/gc • gc is a universal constant that defines force in lbf and g is merely the acceleration due to gravity on Earth Significant Figures are important in engineering calculations.

Exploring Engineering