Forces and Precision. Exploring Engineering Chapter 2 Key elements in Engineering Analysis.
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Transcript of Forces and Precision. Exploring Engineering Chapter 2 Key elements in Engineering Analysis.
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Forces and Precision
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Exploring Engineering
Chapter 2Key elements in
Engineering Analysis
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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 dimensionsNewton’s 2nd law of motion
• SI and Engineering English units
• “gc” and “g”
Significant figures
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Units and dimensions
All engineers will have to understand this material irrespective of sub disciplineLet’s start with my favorite superhero…it’s
Superman! Why is this relevant?
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• Superman represents the many of the qualities that engineers must master: Consider his qualities…
1) Faster than a speeding bullet
2) More powerful than a mighty locomotive
3) Can leap tall buildings with a single bound
4) Keeps falling into kryptonite traps
Superman – Engineering Hero
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• 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 of it!)
* Yes! There is a difference. We will see later!
Superman – Engineering Hero
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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/sThe units, meters/second, really adds some new
information..had we said 800 inches/hr Superman would be called “Supermolasses”!
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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!
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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!
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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
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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)
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More Conversion Examples:
These use conversion factors you can paste from Convert.exe
800 m/s to mph800 [m/s][3.28 ft/m][1/5280 miles/ft][3600 s/hr]800 x 2.236 = 1790 [mph]
2,000 hp to kW2,000 [hp][0.7457 kW/hp] = 1492 kW
9.81 x 104 N m to ft lbf 9.81 x 104 [N m][1/4.448 lbf/N][3.28 ft/m] 9.81 x 104 x 0.737 = 7.23 x 104 ft lbf
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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.
F mass acceleration
or F ma
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Force, Weight, and Mass
In high school you learned F = ma but there’s more to itNewton 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,
maam
FF
am
ma
F
F
11
1
111
and
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Force, Weight, and Mass
The ratio (F1/m1a1) is arbitrary. Picking it defines the unit of force.
SI system: F1 1 Newton when m1 = 1 kg
and a1 = 1 m/s2
Then you can use F = ma English system: F1 1 lb force when m1 =
1 lb mass and a1 = 32.174 ft/s2
lbfandslbf
ftlbm174.32 Define
2c
c g
maFg
c 2
lbm×ft 32.174 g
lbf ×s
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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 = maHow: 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
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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.
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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
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Newton’s 2nd Law and Units
It bears repeating: SI system is far superior and simpler:
2 provided m in kg and a in m/ sF ma
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
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Significant figures
Arithmetic cannot improve the accuracy of a result10 meters, 10. meters, 10.0 meters and 10.00
meters are not identical10 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
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Significant figures
Thus 10/6 = 2 and not 1.66666667 etc. as displayed in your calculatorA 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.
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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
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Significant figures – example
Round off 123.456 − 123.0123.456 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!
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Summary
Engineering problems need precise mathematicsBut not more precise than can be justified (see text,
Chapter 1)Units must be consistent
• […] method is very helpful in maintaining correct unitsNewton’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 engineeringcalculations.