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# IM-1-Phy-121

```Module 1
Course Title
Course Number
Course Description
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Total Learning Time :
Pre-requisites
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Physics and Measurement
Physics for Engineers
Physics for Engineers 121
This course provides students with a clear and logical
presentation of the basic concepts and principles of
physics through a broad range of interesting real-world
application.
5 units (4 units lecture, 1 unit laboratory)
Calculus 1
Overview:
In this module, it will discuss the different kinds of physical quantities and some
preliminary concepts of measurements, units, modelling and estimation.
Learning Outcomes:
At the end of this module, the student should be able to:
1. Have a working knowledge to identify fundamental quantities in mechanics.
2. Have a working knowledge to express measurements in standard form, such as
SI -Standard International/Metric or English System.
3. Convert one units from one measurement system to another or convert within a
system.
Indicative Content:
This module discusses the following topics: Standards of Length, Mass and Time;
Modelling and Alternative Representations; Dimensional Analysis; Conversion of
Units; Estimates and Order-of-Magnitude Calculations; Significant Figures; and
Vectors.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
1
College of Engineering, Architecture and Technology
Name: ___________________________________ Date Submitted: __________
Course &amp; Year: ____________________________ Rating: _________________
Pre-Assessment
1. A proton, which is the nucleus of a hydrogen atom, can be modelled as a sphere
with a diameter of 2.4 m and a mass of 1.67 x 10 -27 kg. Determine the density
of the proton.
2. Two spheres are cut from a certain uniform rock. One has radius 4.5 cm. The
mass of the others is five times greater. Find its radius.
3. A solid piece of lead has a mass of 23. 94 g and a volume of 2.10 cm 3. From
these data, calculate the density of lead in SI units (kg/m 3)
4. One gallon of paint (volume = 3.78 x 10 -3 m3) covers an area of 25.0 m3. What
is the thickness of the fresh paint on the wall?
5. a) Compute the order of magnitude of the mass of a bathtub half full of water.
b) Compute the order of magnitude of the mass of a bathtub half full of copper
coins.
6. To an order of magnitude, how many piano tuners reside in NY City? The
physicist Enrico Fermi was famous for asking this question like this one on oral
PhD qualifying examination.
7. How many significant figures are in the following numbers?
a) 78.9 &plusmn; 0.2
b) 3.788 x 109
c) 2.46 x 10-6
d) 0.0053
8. Prove that one solution of the equation
2x4 – 3x3 + 5x = 70 if x = - 2.22
9. The consumption of natural gas by a company satisfies the empirical equation
V= 1.5t + 0.00800t2, where V is the volume of gas in millions of cubic feet and
t is the time in months. Express this equation in units of cubic feet and seconds.
Assume a moth is 30 days.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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10. A woman wishing to know the height of a mountain measures the angle of
elevation of the mountaintop as 120. After walking 1 km closer to the mountain
on level ground, she finds the angle to be 140.
a) Draw a picture of the problem, neglecting the height of the woman’s eyes
above the ground.
b) Using the symbol y to represent the mountain height and the symbol x to
represent the woman’s original distance from the mountain, label the
picture.
c) Using the labelled picture, write two trigonometric equations relating the
two selected variables.
d) Find the height y.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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Discussion:
Physics, the most fundamental physical science, is concerned with the
fundamental principles of the Universe. It is the foundation upon which the other
sciences- astronomy, biology, chemistry, and geology-are based. It is also the basis of
a large number of engineering applications. The beauty of physics lies in the simplicity
of its fundamental principles and in the manner in which just a small number of
concepts and models can alter and expand our view of the world around us.
Standards of Length, Mass and Time
To describe natural phenomena, we must make measurements of various aspects of
nature. Each measurements is associated with a physical quantity, such as the length
of an object. The laws of physics are expressed as mathematical relationships among
physical quantities. In mechanics, the three dimensional quantities are length, mass
and time. All other quantities in mechanics can be expressed in terms of these three.
In 1960, an international committee established a set of standards for the
fundamental quantities in science. It is called the SI (System International), and its
fundamental units of length, mass and time are the meters, kilogram, and second,
respectively. Other standards for SI fundamental units established by committee are
those for temperature (kelvin), electric current (ampere), luminous intensity (candela),
and the amount of substance (mole).
Length –is the distance between two points in space. In 1120, the king of England
decreed that the standard of length in his country would be named the yard and would
be precisely equal to distance from the tip of his nose to the end of his outstretched
arm. Similarly, the original standard for the foot adopted by the French was the length
of the royal foot of King Louis XIV. Neither of these standards is constant in time;
when a new king took the throne, length measurements changed! The French standard
prevailed until 1799, when the legal standard of length in France became the meter
(m), defined as one ten-millionth of the distance from the equator to the North Pole
along one particular longitudinal line that passes through Paris. Notice that this value
is an Earth –based standard that does not satisfy the requirement that it can be used
throughout the Universe.
As recently as 1960, the length of the meter was defined as the distance between
two lines on a specific platinum-iridium bar stored under controlled conditions in
France. Current requirements of science and technology, however, necessitate more
accuracy than that with which the separation between the lines on the bar can be
determined. In the 1960s and 1970s, the meter was defined to be equal to 1650763.73
wavelength of orange-red emitted from a krypton-86 lamp.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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In October 1983, however, the meter was redefined as the distance travelled by
light in a vacuum during a time interval of 1/299 792 458 second. In effect, this
latest definition establishes that the speed of the light in vacuum is precisely 299 792
458 meters per second. This definition of the meter is valid throughout the Universe
based on the assumption that the light is the same everywhere. The speed of light also
allows us to define the light-year- the distance that light travels through empty
space in one year.
Mass – is related to the amount of materials that is present in the object, or to how
much that object resist changes in its motion. Mass is an inherent property of an
object and is dependent of the object’s surroundings and of the method used to
measure it. The SI fundamental unit of mass, the kilogram (kg), is defined as the
mass of a specific platinum-iridium alloy cylinder kept at the International
Bureau of Weights and Measures at Service, France. This mass standard was
established in 1887 and has not been changed since that time because platinumiridium is an unusually stable alloy.
Time – the standard of time was defined in terms of the mean solar day. (A solar day
is the time interval between successive appearances of the Sun at the highest point it
reaches in the sky each day). The fundamental unit of a second (s) was defined as
(1/60)(1/60)(1/24) of a mean solar day. This definition is based on the rotation of one
planet, the Earth. Therefore, this motion does not provide a time standard that is
universal. In 1967, the second was redefined to take advantage of the high precision
attainable in a device known as an atomic clock which measures vibrations of cesium
atoms.
One second is now defined as 9 192 631 770 times the period of vibration of
radiation from the cesium -133 atom. You should note that we will use the
notations time and time interval differently. A time is a description of an instant
relative to a reference time. For example, t = 10 s, refer to an instant 10 s after the
instants we have identified as t = 0. As another example, a time of 11:30 a.m., means
an instant 11.5 hours after our reference time of midnight.
On the other hand, a time interval refers to duration. He required 30 minutes to
finish to finish the task. It is common to hear a “time of 30 minutes” in this latter
example, but we will be careful to refer to measurements of duration as time intervals.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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Units and Quantities
In addition to SI, another system of units, the US customary system, is still used
in the United States despite acceptance of SI by the rest of the world. In this system,
the units of length, mass and time are the foot, slug and second respectively. We shall
use SI Units because they are almost universally accepted in science and industry. We
shall make some limited use of US customary units in the study of classical
mechanics.
In addition to the fundamental SI units of meter, kilogram, and second, we can
also use other units, such as millimetres and nanoseconds, where the prefixes milli
and nano denote multipliers of the basic units based on various power of ten.
The variable length, mass and time are examples of fundamental quantities. Most
other variables are derived quantities, those that can be expressed as mathematical
combination of fundamental quantities. Common examples are area (a product of two
length) and speed (a ratio of a length to a time interval).
Another example of a derived quantity is density. The density 𝜌 (Greek letterrho) of any substance is defined as its mass per unit volume.
𝜌 =
m/v
In terms of fundamental quantities, density is a ratio of mass to a product of three
lengths. Aluminum, for example, has a density of 2.70 x 103 kg/m3, and iron has a
density of 77.86 x 103 kg/m3. An extreme difference in density can be imagined by
thinking about a 10-cm cube of Styrofoam in one hand and a 10-cm cube of lead in
the other.
Modelling and Alternative Representations
One of the primary problems-solving methods in physics is to form an
appropriate model of the problem. A model is a simplified substitute for the real
problems that allows us to solve the problem in a relatively simple way. As long
as the predictions of the model agree to our satisfaction with the actual behaviour of
the real system, the model is valid. If the predictions do not agree, the model must be
refined or replaced with another model. The power of modelling is in its ability to
reduce a wide variety of very complex problems to a limited number of classes of
problems that can be approached in similar ways.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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The particle model is an example of the second category of models, which we
will call simplification models. In simplification model, details that are not
significant in determining the outcome of the problem are ignored. The third category
is that of analysis model, which are general types of problems that we have solved
before. An important technique in problem solving is to cast a new problem into a
form similar to one we have already solved and which can be used as model. The
fourth category is structural models, generally used to understand the behaviour of a
system that is that is far different in scale from our macroscopic world- either mush
smaller or much larger, so that we cannot interact with it directly. Intimately related to
the notion of modelling is that of forming alternative representation of the problem
that you are solving. A representation is a method of viewing or presenting the
information related to the problems.
The modelling approach recognizes that none of these processes affects the
motion of the Earth around the Sun to a measurable degree. Therefore these details
are all ignored. In addition, the size of the Earth does not affect the gravitational force
between the Earth and the Sun; only the masses of the Earth and Sun and the distance
between their centers determine this force. In simplified model, the Earth is imagined
to be particle, an object with mass but zero size. This replacement of an extended
object by a particle is called particle model, which is used extensively in physics.
The two primary conditions for using the particle model are as follows:


The size of the actual object is of no consequence in the analysis of its
motion.
Any internal processes occurring in the object are of no consequence in
the analysis of its motion.
Both of these conditions are in action in modelling the Earth as a particle. Its
radius is not a factor in determining its motion, and internal processes such as
thunderstorm, earthquakes, and manufacturing process can be ignored.
Four categories of models will help us understand and solve Physics problems.
1. Geometric model- we form a geometric construction that represents the real
condition. We then set aside the real problem and perform an analysis of the
geometric construction.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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Example #1:
You wish to find the height of a tree but cannot measure it directly. You stand 50.0
m from the tree and determine that the line of sight from the ground to the top of the
tree makes an angle of 25.00 with the ground. How tall is the tree?
The height of a tree can be found by measuring the distance from the tree
and the angle of sight to the top above the ground. This problem is a simple of
geometrically modelling the actual problem.
Solution:
Figure shows the tree and a right triangle corresponding to the information in the
problem superimposed over it. (We assume that the tree is exactly perpendicular to a
perfectly flat ground). In the triangle, we know the length of the horizontal leg. We
can find the height of the tree by calculating the length of the vertical leg. We do so
with the tangent function:
tan 𝜃 = opposite / adjacent = h/50.0 m
h = (50m) tan 𝜃 = (50m) tan 250
h = 23.3 m
2. Simplification model – details that are not significant in determining the
outcome of the problem are ignored. When we study rotation, objects will be
modelled as rigid objects. All molecules in a rigid object maintain their exact
positions with respect to forces to one another. We adopt this simplification
model because a spinning rock is much easier to analyze than a spinning block
of gelatin, which is not a rigid object. Other simplification models will assume
that quantities such as friction forces are negligible, remain constant, or are
proportional to some power of the object’s speed. We will assume uniform
metal beams, laminar flow of fluids, massless spring, symmetric distributions
of electric charge, resistance free wires, thin lenses and many more are
simplification models.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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3. Analysis model - an important technique in problem solving is to cast a new
problem into a form similar to one have already solved and which can be used
as a model. As we shall see, there are about two dozen analysis models that
can be used to solve most of the problems you will encounter. All of the
analysis models in classical physics will be based on four simplification
models: particles, system, rigid bodies and wave.
4. Structural model – these model are generally used to understand the
behaviour of a system that is far different in scale from our macroscopic
world- either much smaller or larger- so that we cannot interact with it
directly. As an example, the notion of hydrogen from an atom as an electron in
a circular orbit around a proton is a structural model of an atom. The ancient
geocentric model of the Universe, in which the Earth is theorized to be at the
center of the Universe, is an example of a structural model for something
larger in scale than our macroscopic world.
Types of representation can be of assistance to help you understand and solve
problems in different ways:
1. Mental representation – Imagine a scene that describes what is happening in
the word problem, then let time progress so that you understand the situation
and can predict what changes will occur in the situation. This step is critical in
approaching every problem.
2. Pictorial representation – Drawing a picture of the situation described in the
word problem can be great assistance in understanding the problem.
3. Simplified pictorial representation- It is often useful to redraw the pictorial
representation without completing details by applying a simplification model.
This process is similar to the discussion of the particle model described earlier.
4. Graphical representation- In some problems, drawing a graph that describes
the situation can be very helpful. A graphical representation is different from a
pictorial representation, which is also a two –dimensional display of
information but whose axes, if any, represent length coordinates. In a
graphical representation, the axes may represent any two related variables.
Therefore, a graphical representation is generally not something you would
see when observing the situation in the problem with your eyes.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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5. Tabular representation – It is sometime helpful to organize the information
in tabular form to help make it clearer. The periodic table of the elements is an
extremely useful tabular representation of information in chemistry and
physics.
6. Mathematical representation – The ultimate goal in solving a problem is
often the mathematical representation. You want to move from the
information contained that allow you to understand what is happening, to one
or more equations that represent the situation in the problem and that can be
solved mathematically for the desired result.
Fig 1A. A pictorial representation of
pop foul being hit by a baseball player.
Fig. 1B. A simplified pictorial 1A
Representation for the situation
Fig. 1C. A graphical representation of the position as a function of time of a
block shanging from a spring and oscillating.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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Dimension Analysis
The word dimension denotes the physical nature quantity. The distance between
two points, for example, can be measured in feet, meters, or furlongs, which are all
different units for expressing the dimension of length. The symbol we use to specify
the dimension of length, mass and time are L, M, and T. respectively. We shall often
use brackets [ ] to denote the dimension of a physical quantity. For example, the
symbol we use for speed is v, and in our notation, the dimensions of speed are written
[v] = L/T. As another example, the dimensions of area A are[A ] = L2. The
dimensions and units of area, volume, speed, and acceleration are listed below.
In many situations, you may have to check a specific equation to see if it matches
your expectations. A useful procedure for doing that, called dimensional analysis,
can be used because dimensions can be treated as algebraic quantities. For example,
quantities can be added or subtracted only if they have the same dimensions.
Furthermore, the terms on both sides of an equation must have the same dimensions.
By following these simple rules, you can use dimensional analysis to determine
whether an expression has the correct form. Any relationship can be correct only if
the dimensions on both sides of the equation are the same.
To illustrate this procedure, suppose you are interested in an equation for the
position x of a car at a time t if the car starts from rest at x = 0 and moves with
constant acceleration a. The correct expression for this situation is x= 1/2at2 as we
show in the next topic. The quantity x on the left side has the dimension of length. For
the equation to be dimensionally correct, the quantity on the right side must also have
the dimension of length. We can perform a dimensional check by substituting the
dimensions for acceleration. L/T2 and time, T, into the equation. That is, the
dimensional form of the equation x=1/2at2 is:
L = L/ T2 . T2 = L
The dimensions of time cancel shown, leaving the dimension of length on the righthand side to match that on left. A more general procedure using dimensional analysis
is to set up an expression of the form
x ∝ a nt n
where n and m are exponents that must be determined and the symbol ∝ indicates a
proportionality. This relationship is correct only if the dimensions of both sides are the
same. Because the dimension of the left side is length, the dimension of the right side
must also be length. That is,
[ a n t n ] = L = L1 T0
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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Table 1.1 Dimensions and Units of Four Derived Quantities
Quantity
Area
Volume
Speed
Acceleration
(A)
(V)
(s)
(a)
Dimensions
L2
L3
L/T
L/T2
SI Units
m2
m3
m/s
m/s2
US customary units
ft2
ft3
ft/s
ft/s2
Because the dimensions of acceleration are L/T2 and the dimension of time is T,
we have
(L/T2) n T m = L1 T 0
(Ln T m -2n ) = L1 T 0
The exponents of L, and T must be the same on both sides of the equation. From the
exponents of L, we see immediately that n =1. From the exponents of T, we see that
m - 2n = 0, which, once we substitute for n, gives us m = 2. Returning to our original
expression x ∝ a n t m, we conclude that x ∝ at2.
Example # 2. Analysis of an Equation
Show that the expression v = at, where v represents speed, a acceleration, and t an
instant of time, is dimensionally correct.
Identify the dimensions of v from Table 1.1
[ 𝑣 ] = L/T
Identify the dimensions of a from Table 1.1 and multiply by dimension of t
[ 𝑎𝑡 ] = L/T2 T = L/T
Therefore, v = at is dimensionally correct because we have the same dimensions on
both sides. (If the expression were given a v = at2, it would be dimensionally in
correct.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
12
Example # 3. Analysis of a Power Law
Suppose we are told that the acceleration of a particle moving with uniform speed
v in a circle of radius r is proportional to some power of r, say r n , and some power
of v, say v m. Determine the values of n and m and write the simplest form of an
equation for the acceleration.
Write an expression for a with a dimensionless constant of proportionality k:
a = k r nv m
Substitute the dimensions of a, r, and v
L/T2 = Ln (L/T) m = Ln + m / T m
Equate the exponents of L and T so that the dimensional equation is balanced:
n + m = 1 and m = 2
Solve the two equations for n:
n=-1
Write the acceleration expression:
a = k r -1 v 2 = k v 2/r
Conversion of Units
Sometimes, it is necessary to convert units from one measurement system to
another or convert within a system. Conversion factors between SI and US customary
units of length are follows.
1 mile = 1609 m = 1.609 km
1 ft. = 0.3048 m = 30.48 cm
1m=
1 in. = 0.0254 m = 2.54 cm
39.37 in. = 3.281 ft.
A more complete list of conversion factors see Conversion Table.
Like dimensions, units can be treated as algebraic quantities that can cancel each
other. For example, suppose we wish to convert 15 in to cm.
15 in =
(15 in) (2.54 cm / 1 in) = 38.1 cm
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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Where the ratio in parentheses is equal to 1. We express 1 as 2.54 cm/1 in. (rather
than 1 in/2.54 cm) so that the unit “inch” in the denominator cancels with unit in the
original quantity. The remaining unit is the centimeter, our desired result.
Example # 4:
a) On an interstate highway in a rural region of Wyoming, a car is traveling at a
speed of 38 m/s. Is the driver exceeding the speed limit of 75 mi/h?
Solution:
Convert meters to miles and second to hours:
(38 m/s) (1 mi/1609 m) (60 s/ min) (60 min / 1 h) = 85 mi/h
The driver is indeed exceeding the speed limit and should slow down.
b) What if the driver were from outside the United States and is familiar with
speeds measured in kilometre per hour? What is the speed of the car in kph?
We can convert our final answer to the appropriate units:
(85 mi/h) (1.609km/1 mi) = 137 km/h
Estimates and Order -of-Magnetic Calculations
Suppose someone asks you the number of bits of data on a typical Blu-ray Disc. In
response, it is not generally expected that you would provide the exact number but
rather an estimate, which may be expressed in scientific notation. The estimate may
be made even more approximate by expressing it as an order of magnitude, which is a
power of 10 determined as follows:
1. Express the number in scientific notation, with the multiplier of the power of
10 between 1 and 10 and a unit.
2. If the multiplier is less than 3.162 (the square root of 10), the order of
magnitude of the number is the power of 10 in the scientific notation. If the
multiplier is greater than 3.162, the order of magnitude is one larger than the
power of 10 in the scientific notation.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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We use the symbol – for “is on the order of”. Use the procedure above to verify the
orders of magnitude for the following lengths:
0.0086 m ~ 10-2 m
0.0021 ~ 10-3 m
720 m ~ 103 m
Usually, when an order-of-magnitude estimate is made, the results are reliable to
within about a factor of 10.
Inaccuracies caused by guessing too low for one number are often cancelled by
other guesses that are too high. You will find that with practice your guesstimates
become better and better. Estimation problems can be fun to work because you freely
drop digits, venture reasonable approximations for unknown numbers, making
simplifying assumptions, and turn the question around into something you can answer
in your head or with minimal mathematical manipulation on paper. Because of the
simplicity of these types of calculation, they can be performed on a small scrap of
paper and are often called back-of-the-envelope calculations.
Example # 5:
Estimate the number of breaths taken during an average human lifetime.
Solution:
We start by guessing that the typical human lifetime is about 70 years. Think about
the average number of breaths that a person takes 1 min. This number varies
depending on whether the person is exercising, sleeping, angry, serene, and so forth.
To the nearest order of magnitude, we shall choose 10 breaths per minute as our
estimate. (This estimate is certainly closer to the true average value than an estimate
of 1 breath per minute or 100 breaths per minute).
Find the approximate number of minutes in a year:
1 yr. (400 days/1 yr.) (25 h/1 day) (60 min/1 h) = 6 x 105 min
Find the approximate number of minutes in a 70-year lifetime:
number of minutes = (70 yr.) (6 x 105 min/yr.) = 4 x 107 min
Find the approximate number of breaths in a lifetime
Number of breaths = (10 breaths/min) (4 x 107 min) = 4 x 108 breaths
Therefore, a person takes on the order of 109 breaths in a lifetime. Notice how much
simpler it is in the first calculation above to multiply 400 x 25 than it is to work with
the more accurate 365 x 24.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
15
Significant Figures
When certain quantities are measured, the measured values are known only to
within the limits of the experimental uncertainty. The value of this uncertainty can
depend on various factors, such as the quality of the apparatus, the skill of the
experimenter, and the number of measurements performed. The number of significant
figures in a measurement can be used to express something about the uncertainty. The
number of significant figures related to the number of numerical digits used to express
the measurement.
As an example of significant figures, suppose we are asked to measure the radius
of a Blu-ray Disc using a meter stick as a measuring instrument. Let assume the
accuracy to which we can measure the radius of the disc is &plusmn; 0.1 cm. Because of the
uncertainty of &plusmn; 0.1 cm, if the radius is measured to be 6.0 cm, we can claim only that
its radius lies somewhere between 5.9 cm and 6.1 cm. In this case, we say that the
measured value of 6.0 cm has two significant figures. Note that the significant figures
include the first estimated digit. Therefore, we could write the radius as (6.0 &plusmn; 0.1)
cm.
Zeros may or may not be significant figures. Those used to position the decimal
point in such numbers as 0.03 and 0.0075 are not significant. Therefore, there are one
and two significant figures, respectively, in these two values. When the zeros come
after other digits, however, there is a possibility of misinterpretation. For example,
suppose the mass of an object is given as 1500 g. This value is ambiguous because we
do not know whether the last two zeros are being used to locate the decimal point on
whether they represent significant figures in the measurements.
To remove this ambiguity, it is common to use scientific notation to indicate the
number of significant. In this case, we would express the mass as 1.5 x 10 3 g if there
are two significant figures in the measurement value, 1.5 x 10 3 g if there are three
significant figures and 1.500 x 103 g if there are four. The same rules holds for
number less tahn1, so 2.3 x 10-4 has two significant figures (and therefore could be
written 0.00023) and 2.50 x 10-4 has three significant figures (and therefore written as
0.000230).
s
In problem solving, we often combine quantities mathematically through
multiplication, division, addition, subtraction, and so forth. When doing so, you must
make sure that the result has the appropriate number of significant figures. A good
rule of thumb to use in determining the number of significant figures that can be
claimed in a multiplication or a division is as follows:
When multiplying several quantities, the number of significant figures in the
final answer is the same as the number of significant figures in the quantity
having the smallest number of significant figures. The same rule applies to
division.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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Let’s apply the rule to find the area of the Blu-ray Disc whose radius we measured
above. Using the equation for the area of a circle,
A = 𝜋r2 = 𝜋(66666.0 cm) 2 = 1.1 x 102 cm2
If you perform this calculation on your calculator, you will likely see 113 097
335.5. It should be clear that you don’t want to keep all of these digits, but you might
be tempted to report the result as 113 cm2. This result is not justified because it has
three significant figures, whereas the radius only has two. Therefore, we must report
the result with only two significant figures as shown above.
For addition and subtraction, you must consider the number of decimal places
when you are determining how many significant figures to report:
When numbers are added or subtracted, the number of decimal places in the result
should equal the smallest number of decimal places of any term in the sum or
difference.
As an example of this rule, consider the sum
23.2
+
5.174
= 28.4
Notice that we do not report the answer as 28.374 because the lowest number of
decimal places is one, for 23.2. Therefore, our answer must have only one decimal
place.
The rule for addition and subtraction can often result in answers that have a
different number of significant figures than the quantities with which you start. For
example, consider these operations that satisfy the rule:
1.0001
+
0.0003 = 1.0004
1.0002 - 0.998
= 0.004
In the first example, the result has five significant figures even though one of the
terms, 0.0003, has only one significant figure. Similarly, in the second calculation, the
result has only one significant figure even though the numbers being subtracted have
four and three, respectively.
Example # 6.
A carpet is to be installed in a rectangular room whose length is measured to be
12.71m and whose width is measured to be 3.36 m. Find the area of the room.
Solution:
If you multiply 12.71 by 3.46 m on your calculator, you will see an answer of
43.9766 m2. How many of these numbers should you claim? Our rule of thumb for
multiplication tells us that you can claim only the number of significant figures in
your answer as are present in the measured quantity having the lowest number of
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
17
significant figures. In this example, the lowest number significant figures is three in
3.46 m, so we should express our final answer as 44.0 m2.
Vectors
Physical quantities fall into two categories. One type, known is scalar quantity can
be completely described by a single number (with appropriate units) giving its
magnitude or size. Some common scalars are mass, temperature, volume and speed.
For example, a car’s speed can be completely described by noting the number on its
speedometer. A basketball’s mass can be specified by measuring a single number with
a scale.
The other type of quantity known as a vector quantity, has both magnitude and a
direction. Velocity is a common vector quantity with magnitude specifying how fast
an object is moving and a direction specifying the direction of travel.
For example, a car’s velocity could specified by noting it was traveling 60 miles
per hour in the direction due north. A car traveling 60 miles per hour in an eastward
direction would have different velocity vector but the same scalar speed.
In general, a vector quantity is characterized by having both a magnitude and
direction. By contrast, a scalar quantity has magnitude, but no direction.
Equality of Two Vectors
Two vectors A and B are equal if they have the same magnitude and the same
direction. They need not be located at the same point in space. The four vectors are all
equal to each other. Moving a vector from one point in space to another doesn’t
change its magnitude or its direction.
A + B
=
C differ significantly from A + B = C
The first is a vector sum, which must be handled graphically or with components,
whereas the second is a simple arithmetic sum of numbers.
When two or more vectors are added, they must all have the same units. For example,
it doesn’t make sense to add a velocity vector, carrying units of meters per second, to
a displacement vector, carrying units of meters. Scalar obey the same rule: It would be
similarly meaningless to add temperatures to volumes or masses to time intervals.
Vectors can be added geometrically or algebraically. To add vector B to vector A
geometrically, first draw A on a piece of graph paper to some scale, such as 1 cm
= 1 m, so that its direction is specified relative to a coordinate system.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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Then draw vector B to the same scale with the tail of B starting at the tip of A, as
shown in Fig.1.12a. Vector B must be drawn along the direction that makes the proper
angle relative to vector A. The resultant vector R = A + B is the vector drawn from
the tail of A to the tip of B. This procedure is known as the triangle method of
When two vectors are added, their sum is independent of the order of the
addition: A + B = B + A. This relationship can be seen from the geometric
construction in Fig. 1.12b, and is called the commutative law of addition.
This same general approach can also be used to add more than two vectors, as is
done in Fig 1.13 for four vectors. The resultant vector sum R = A + B + C + D is
the vector sum drawn from the tail of the first vector to the tip of the last. Again, the
order in which the vectors are added is unimportant.
Fig. 1.11 These four vectors are equal because they have equal lengths and point
in the same direction
(a)
(b)
Fig. 1.12 (a) When vector B is added to vector A, the vector sum R is the vector
that runs from the tail of A to the tip of B. (b) Here the resultant runs from the
tail of B to the tip of A. These constructions prove that A + B = B + A
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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Fig. 1.13 A geometric construction for summing four vectors. The resultant
vector R is the vector that complete the polygon.
Negative of a Vector
The negative of the vector A is defined as the vector that gives zero when added to
A. This means that A and –A have the same magnitude but opposite in directions.
Subtracting Vectors
Vector subtraction makes use of the definition of the negative of a vector. We
define the operation A –B as the vector –B added to the vector A:
A – B = A + (-B)
Vector subtraction is really a special case of vector addition. The geometric
construction for subtracting two vectors is shown in Fig. 1.14
Fig. 1.14 This construction shows how to subtract vector B from vector A. The
vector –B has the same magnitude as vector B but points in the
opposite direction.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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Multiplying or Dividing a Vector by a Scalar
Multiplying or dividing a vector by a scalar gives a vector. For example, if vector A is
multiplied by the scalar number 3, the result, written 3A, is a vector with a magnitude
three times that of A and pointing in the same direction. If we multiply vector A by
the scalar -3, the result is -3A, a vector with a magnitude three times that of A and
pointing in the opposite direction (because of the negative sign).
Example:
A car travels 20 km due north and then 35 km in the direction 60 0 west of north,
as in fig. 1.15. Using a graph, find the magnitude and direction of a single vector that
gives the net effect of the car’s trip. This vector is called the car’s resultant
displacement.
Fig. 1.15 A graphical method for finding the resultant displacement vector
R=A+B
Solution:
Let A the first displacement vector, 20 km north, and B the second displacement
vector, extending west of north. Carefully graph two vectors, drawing a resultant vector
R with its base touching the base of A and extending to the tip of B. Measure the length
of this vector, which turns out to be about 48 km. The angle 𝛽, measured with a
protractor, is about 390 west of north.
Components of a Vector
One method of adding vector make use of the projections of a vector along the axes of
a rectangular coordinate system. These projections are called components. Any vectors
can be completely described by its components.
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Consider a vector A in a rectangular coordinate system, as shown in fig. 1.16. A
can be expressed as the sum of two vectors Ax parallel to the x-axis, and Ay, parallel to
the y-axis. Mathematically,
A = Ax + Ay
Fig. 1.16 Any vector A lying in the xy-plane can be represented by its
rectangular components Ax and Ay.
where Ax and Ay are component vectors of A. The projection of A along the x-axis, Ax,
is called the x-component of A, and the projection of A along the y-axis, Ay, is called
the y-component of A. These components can be either positive or negative numbers
with units. From the definition of sine and cosine, we see that cos 𝜃 = Ax/A and sin 𝜃
= Ay/A, so the components of A are:
Ax = A cos 𝜽
(1.4a)
Ay = A sin 𝜽
(1.4b)
These components form two sides of a right triangle having a hypotenuse with
magnitude A. It follows that A’s magnitude and direction are related to its components
through the Pythagorean theorem and the definition of the tangent:
A = √ Ax2 + Ay2
Tan 𝜃 = Ax/Ay
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
(1.5)
(1.6)
22
Fig. 1.17 The angle 𝜽 need not always be defined from the positive x-axis.
To solve for the angle 𝜃, which is measured counterclockwise from the positive xaxis by convention, the inverse tangent can be taken of both sides of equation
𝜃 = tan-1 (Ax /Ay)
This formula gives the right answer for 𝜃 only half the time! The inverse tangent
function returns values only from -900 to + 900, so the answer in your calculator window
will only be correct if the vector happens to lie in the first or fourth quadrant. If it lies
in the second or third quadrant, adding 180 0 to the number in the calculator window
will always give the right answer. The angle must be measured from positive x –axis.
Other choices of reference line are possible, but certain adjustments must then be made.
If a coordinate system other than the one shown in Fig. 1.16 is chosen, the
components of the vector must be modified accordingly. In many applications it’s more
convenient to express the components of a vector in a coordinate system having axes
that are not horizontal and vertical but are still perpendicular to each other. Suppose a
vector B makes an angle 𝜃’ with the x’-axis defined in Fig. 1.18. The rectangular
components of B along the axes of the figure are given by Bx’ = B cos 𝜃’ and By’ = B
sin 𝜃’. The magnitude and direction of B are then obtained from expressions equivalent
to equations 1.5 &amp; 1.6.
Fig. 1.18 The components of vector B in a tilted coordinate system.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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Sample Problem:
a) Find horizontal and vertical components of the d = 1 x 102 m displacement of
a superhero who flies from the top of a tall building along the path shown in
the fig.1.20a
b) Suppose instead the superhero leaps in the other direction along a displacement
vector B to the top of a flagpole where the displacement components given by
Bx = - 25 m and By = 10 m. Find the magnitude and direction of the displacement
vector.
Fig. 1.20
a) Find the vector components of A from its magnitude and direction
Ax = A cos 𝜃 = (1 x 10 2 m) cos (300) = 86.6 m
Ay = A sin 𝜃 = (1 x 10 2 m) sin (-300) = -50.0 m
b) Find the magnitude and direction of the displacement vector B from its
components.
B = √ Bx2 + By2
= √(−25 m)2 + (10m)2 = 26.9 m
𝜃 = tan-1 (By/Bx) = tan-1 (10/-25)
𝜃 = -21.80
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
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The graphical method of adding vectors is valuable in understanding how vectors
can be manipulated, but most of the time vectors are added algebraically in terms of
their components. Suppose R = A + B. Then the components of the resultant vector R
are given by:
Rx = Ax + Bx
Ry
= Ay
+ By
So x- components are added only to x-components, and y-components only to
y- components. This magnitude and direction of R can be consequently be found with
equations 1.5 and 1.6.
Subtracting two vectors works the same way because it’s a matter of adding the
negative of one vector to another vector. You should make a rough sketch when adding
or subtracting vectors, in order to get an approximate geometric solution as a check.
Sample Problem:
A hiker begins a trip by first walking 25 km 450 south of east from her base camp. On
the second day she walks 40 km in a direction 60 0 north of east, at which point she
discovers a forest ranger’s tower.
a) Determine the components of the hiker’s displacement in the first and second
days.
b) Determine the components of the hiker’s total displacement for the trip.
c) Find the magnitude and direction of the displacement from base camp.
(a)
(b)
Fig. 1.21 a) Hiker’s path and the resultant vector. b) Components of the
hiker’s total displacement from camp.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
25
Solution:
a) Find the components of A
Ax = A cos (450) = (25 km) (0.0707) = 17.7 km
Ay = A sin (-450) = (25km) (0.0707) = - 17.7 km
Find the components of B
Bx = B cos 600 = (40 km) (0.50) = 20 km
By = B sin 600 = (40 km) (0.806) = 34.6 km
b) Find the components of the resultant vector, R = A + B
To find Rx add the x-components of A &amp; B:
Rx = Ax + Bx
= 17.7 km + 20 km = 37.7 km
To find Ry , add the y –components of A &amp; B:
Ry = Ay + By = - 17.7 km + 34.6 km = 16.9 km
c) Find the magnitude and direction of R
Use the Pythagorean theorem to get the magnitude:
R = √ Rx2 + Ry2 = √ (37.7 km)2 + (16.9 km)2 = 41.3 km
Calculate the direction of R using the inverse tangent function:
𝜃 = tan-1 (16.9 km/37.7 km) = 21.40 NE
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College of Engineering, Architecture and Technology
Name: ___________________________________Date Submitted: __________
Course &amp; Year: ____________________________ Rating: _________________
Evaluation:
1. a) Estimate the number of times your heart beats in a month.
b) Estimate the number of human heartbeats in an average lifetime.
2. Answer each question yes or no. Must two quantities have the same dimensions
a)
b)
c)
d)
e)
If you are adding them?
If you are multiplying them?
If you are subtracting them?
If you are dividing them?
If you are equating them?
3. Under what circumstances would a vector have components that are equal?
4. How many significant figures are there in
a)
b)
c)
d)
78.9 &plusmn; 0.2
3.788 x 109
2.46 x 1026
0.0032
5. A firkin is an old British unit of volume equal to 9 gallons. How many cubic
meters are there in 6 firkins?
6. The speed of light is about 3 x 108 m/s. Convert this figure to miles per hour.
7. A ladder 9 m long leans against the side of a building. If the ladder is inclined
at an angle of 750 to the horizontal, what is the horizontal distance from the
bottom of the ladder to the building?
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
27
8. In fig. P1.49, find (a) the side opposite 𝜃 (b) the side adjacent to ∅ (c) cos 𝜃,
and (d) tan ∅.
9. A person walks 250 north of east for 3.10 km. How much far due to north and
how far due east would she have to walk to arrive at the same location?
10. A commuter plane starts from an airport and takes the route shown in fig. P
1.71. The plane first flies to city A, located 175 km 20 0 west of north, to city B.
Finally, the plane flies 190 km due to west, to city C. Find the location of city
C relative to the location of the starting point.
ENGR. MA. INEZ STEPHANY UVAS, CHE, MEE
28
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