Revision 1
December 2014
Physics
Student Guide
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ii
Table of Contents
INTRODUCTION .................................................................................................................1
TLO 1 UNITS OF MEASUREMENT ......................................................................................2
Overview ......................................................................................................................2
ELO 1.1 Fundamental Dimensions ..............................................................................3
ELO 1.2 Units of Measure ...........................................................................................4
ELO 1.3 Fundamental and Derived Measurements .....................................................8
ELO 1.4 Conversion between English and SI Systems .............................................10
ELO 1.5 Converting Time Units ................................................................................15
TLO 1 Summary ........................................................................................................17
TLO 2 VECTORS .............................................................................................................18
Overview ....................................................................................................................18
ELO 2.1 Vector Terminology ....................................................................................19
ELO 2.2 Identifying Vectors......................................................................................21
TLO 2 Summary ........................................................................................................22
TLO 3 SOLVING VECTOR PROBLEMS ..............................................................................23
Overview ....................................................................................................................23
ELO 3.1 Graphing Vectors ........................................................................................24
ELO 3.2 Determining Components of a Vector ........................................................29
ELO 3.3 Adding Vectors ...........................................................................................35
TLO 3 Summary ........................................................................................................45
TLO 4 FORCE .................................................................................................................46
Overview ....................................................................................................................46
ELO 4.1 Types of Forces ...........................................................................................46
ELO 4.2 Friction Force ..............................................................................................49
ELO 4.3 Force and Velocity ......................................................................................51
ELO 4.4 Weight .........................................................................................................53
ELO 4.5 Calculating Weight......................................................................................55
ELO 4.6 Free-Body Diagrams ...................................................................................57
ELO 4.7 Force Equilibrium .......................................................................................60
TLO 4 Summary ........................................................................................................66
TLO 5 MOTION ...............................................................................................................68
Overview ....................................................................................................................68
ELO 5.1 Newton's Laws of Motion ...........................................................................68
ELO 5.2 Universal Gravitation ..................................................................................70
TLO 5 Summary ........................................................................................................73
TLO 6 MOMENTUM ........................................................................................................73
Overview ....................................................................................................................73
ELO 6.1 Momentum and Conservation of Momentum .............................................74
ELO 6.2 Collisions and Velocity ...............................................................................78
TLO 6 Summary ........................................................................................................80
TLO 7 ENERGY, WORK, AND POWER..............................................................................81
Overview ....................................................................................................................81
ELO 7.1 Energy Definitions ......................................................................................81
ELO 7.2 Calculating Energy and Work .....................................................................85
ELO 7.3 Conservation of Energy...............................................................................89
ELO 7.4 Power...........................................................................................................91
ELO 7.5 Calculating Power .......................................................................................93
iii
TLO 7 Summary ........................................................................................................ 95
PHYSICS SUMMARY ........................................................................................................ 96
iv
Physics
Revision History
Revision
Date
Version
Number
Purpose for Revision
Performed
By
11/7/2014
0
New Module
OGF Team
12/11/2014
1
Added signature of OGF
Working Group Chair
OGF Team
Introduction
You will learn the concepts of motion, momentum, work, energy, and
power necessary to master later course material and to understand key
elements of the nuclear operator job in this module.
Physics
You must understand the concepts of classical physics to master reactor
physics and understand the workings of the nuclear power plant.
Rev 1
1
Objectives
At the completion of this training session, the trainee will demonstrate
mastery of this topic by passing a written exam with a grade of ≥ 80 percent
on the following Terminal Learning Objectives (TLOs):
1. Convert between units of measure associated with the English and
System Internationale (SI) measuring systems.
2. Describe vector quantities, and how they are represented.
3. Solve resultant vector problems.
4. Describe the measurement of force and its relationship to free body
diagrams.
5. Apply Newton's laws of motion to a body at rest.
6. Calculate the change in velocity when two objects collide, with
respect to the conservation of momentum.
7. Describe energy, work, and power in mechanical systems.
TLO 1 Units of Measurement
Overview
In this section, you will learn the units of measurement used in classical
physics, and learn how to convert between English and SI units.
Units of Measurement
Mastery of the units of measure used in physics is necessary to understand
later course content and to understand data used in the operator's job.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the three fundamental dimensions: length, mass, and time.
2. List standard units of the fundamental dimensions for the following
systems:
a. International System of Units (SI)
b. English system
3. Differentiate between fundamental and derived measurements.
4. Convert between English and SI units of mass and length.
5. Convert time measurements between the following:
a. Years
b. Weeks
c. Days
d. Hours
e. Minutes
f. Seconds
2
Rev 1
ELO 1.1 Fundamental Dimensions
Fundamental Dimensions
Physics is a science based upon exact measurement of physical quantities
that are dependent upon three fundamental dimensions. The three
fundamental or primary dimensions are mass, length, and time. It is
necessary that you understand these three fundamental units, as they form
the foundation for many concepts and principles presented later in this
lesson.
Fundamental Dimensions
Mass: Mass is the amount of material present in an object. This dimension
describes how much material makes up an object. Often, mass and weight
are confused as being the same because the units used to describe them are
similar. Weight is a derived unit, not a fundamental unit, and is a
measurement that describes the force of gravity on the "mass" of an object.
Length: Length is the distance between two points. We need the concept
of length to locate the position of a point in space and describe the size of a
physical object or system. For example, when measuring a length of pipe,
the ends of a pipe are the two points and the distance between the two
points is the length. A typical unit used to describe length is the meter.
Time: Time is the duration between two instants. Units of seconds,
minutes, or hours describe the measurement of time.
Units: A number alone is not sufficient to describe a physical quantity. For
example, to say that "a pipe must be 4 long to fit" has no meaning unless a
unit of measurement for length is also specified. It becomes clear by adding
units to the number "a pipe must be 4 meters long to fit."
The unit defines the magnitude of a measurement. The unit used to
describe the length could be a centimeter or meter if we have a
measurement of length, each of which describes a different magnitude of
length. It is especially important to specify the units of measurement of a
number when describing a physical quantity that it is possible to measure
using a variety of units. Common length units include meters, inches,
miles, furlongs, fathoms, kilometers, and a variety of other units.
Each of the fundamental dimensions mentioned previously has established
units of measurement for its use.
Knowledge Check
Which of the following is NOT a fundamental
dimension?
Rev 1
A.
Weight
B.
Length
C.
Mass
D.
Time
3
Knowledge Check
The fundamental dimensions are:
A.
time, force, and power
B.
mass, time, and force
C.
weight , length, and time
D.
mass, length, and time
Knowledge Check
Weight is:
A.
a fundamental unit
B.
a function of mass and length
C.
a measurement that describes the force of gravity on
the mass of an object.
D.
a constant, regardless of position with respect to other
bodies
ELO 1.2 Units of Measure
Introduction
There are two unit systems in use now, English Units and International
System of Units (SI).
English System: Units of Measure
In the United States, the English system is currently used. This system
consists of various units for each of the fundamental dimensions or
measurements. The table below shows these units. Throughout the United
States, the field of engineering uses the English system. The foot-poundsecond (FPS) system is the usual unit system used in the United States when
dealing with physics.
4
Rev 1
Length
Mass
Time
Inch
Ounce
Second*
Foot*
Pound*
Minute
Yard
Ton
Hour
Mile
Day
Month
Year
*denotes standard unit of measure
System Internationale
Over the years, there have been movements to standardize units so that all
countries, including the United States, will adopt the SI system. Two
related systems, the meter-kilogram-second (MKS) system, and the
centimeter-gram-second (CGS) system comprise the SI system.
The MKS and CGS systems are simpler to use than the English system
because they use a decimal-based system in which prefixes denote powers
of ten. For example, one kilometer is 1,000 meters and one centimeter is
one one-hundredth of a meter. Compared to the SI system, the English
system has odd units of conversion. For example, a mile is 5,280 feet, and
an inch is one twelfth of a foot.
In the field of physics, calculations primarily use the MKS system, while
the field of chemistry uses both the MKS and CGS systems. The two tables
below show the fundamental units for each of these systems.
The following prefixes denote powers of 10 of the base unit:




Rev 1
milli - denotes 1/1,000 of the base unit
centi - denotes 1/100 of the base unit
deci - denotes 1/10 of the base unit
kilo - denotes 1,000 of the base unit
5
MKS Units of Measure
Length
Mass
Time
Millimeter
Milligram
Second*
Meter*
Gram
Minute
Kilometer
Kilogram*
Hour
Day
Month
Year
* denotes standard unit of measure
CGS Units of Measure
Length
Mass
Time
Centimeter*
Milligram
Second*
Meter
Gram*
Minute
Kilometer
Kilogram
Hour
Day
Month
Year
Dimensions of Familiar Objects and Events
The following three tables show approximate lengths, masses, and times for
some familiar objects or events.
6
Rev 1
Approximate Lengths of Familiar Objects
Object
Length (meters)
Diameter of earth's orbit around the sun
2 x 1011
Football field
0.91 x 102 (100 yards)
Diameter of a dime
2 x 10-2
Thickness of a window pane
1 x 10-3
Thickness of paper
1 x 10-4
Approximate Masses of Familiar Objects
Object
Mass (kilograms)
Earth
6 x 1024
House
2 x 105
Car
2 x 103
Dime
3 x 10-3
Postage stamp
5 x 10-8
Approximate Time Durations of Familiar Events
Event
Time (seconds)
Age of the earth
2 x 1017
Human life span
2 x 109
Earth's rotation around the sun
3 x 107
Earth's rotation on its axis
8.64 x 104
Time between heart beats
1
Rev 1
7
Knowledge Check
The measurement system most used in the United
States when dealing with physics is:
A.
foot - pound - second
B.
mks
C.
cgs
D.
foot - pound - minute
Knowledge Check
The standard unit of measure for time is the
A.
second
B.
minute
C.
hour
D.
year
Knowledge Check
The standard unit of mass in the cgs system is the
A.
milligram
B.
gram
C.
kilogram
D.
lbm
ELO 1.3 Fundamental and Derived Measurements
Introduction
Units of measurement are vital to physics. This section explains those
fundamental units, and demonstrates how their combinations provide all of
the units of measure we need to work in the modern world.
Fundamental and Derived Measurements
Most physical quantities have units that are combinations of the three
fundamental dimensions of length, mass, and time. Derived units refer to
combinations of these dimensions or measurements. Therefore, these
combinations are derived from one or more fundamental measurements.
These combinations of fundamental measurements can be the same or
different units.
Fundamental and Derived Measurements Example
The following are examples of various derived units.
8
Rev 1
Area: Area is the product of two lengths (e.g., width multiplied by length
for a rectangle.) The units of length are squared (raised to the second
power), such as square inches (in2) or square meters (m2). For example,
1 𝑚 × 1 𝑚 = 1𝑚2 and 4 𝑖𝑛 × 2 𝑖𝑛 = 8 𝑖𝑛2 . You may see squared units
written as sq. ft for ft2, or sq. m for m2.
Volume: Volume is the product of three lengths (e.g., length multiplied by
width multiplied by depth for a rectangular solid.) The units of length are
cubed (raised to the third power), such as cubic inches (in3) or cubic meters
(m3). The liter is the specific unit for volume in the MKS and CGS unit
systems. For example, one liter (l) is equal to 1,000 cubic centimeters
(1 𝑙 = 1,000 𝑐𝑚3) or 2 𝑖𝑛 × 3 𝑖𝑛 × 5 𝑖𝑛 = 30 𝑖𝑛3 .
Density: Density is a measure of the mass of an object per unit volume. It
has units of mass divided by length cubed, such as kilograms per cubic
meter (kg/m3) or pounds-mass per cubic foot (lbm/ft3).
Velocity: Velocity is the change in length per unit time. Its units are
kilometers per hour (km/hr) or feet per second (ft/s).
Acceleration: Acceleration is a measure of the change in velocity per unit
time. Its units are centimeters per second per second (cm/s2) or feet per
second per second (ft/s2).
Knowledge Check
Match the following terms with the correct dimensions of
measurement.
1
A measure of the change in velocity
per unit time is centimeters per second
per second (cm/s2) or feet per second
per second (ft/s2)
A.
Acceleration
2
The product of three lengths (e.g.,
length x width x depth for a
rectangular solid) is (in3) or cubic
meters (m3)
B.
Volume
3
Mass of an object per unit volume is
kilograms per cubic meter (kg/m3) or
pounds per cubic foot (lbs/ft3)
C.
Density
4
Length per unit time is kilometers per
hour (km/hr) or feet per second (ft/s)
D.
Velocity
Rev 1
9
Knowledge Check
Derived measurements are
A.
combinations of the fundamental dimensions which are
useful in characterizing the physical property or
behavior of objects.
B.
conversion factors to relate the English system to the
mks system.
C.
always required to have all three fundamental
dimensions.
D.
conversion factors to relate the English system to the
cgs system.
Knowledge Check
Density is
A.
a measure of mass per unit volume and has dimensions
of mass per length cubed.
B.
a measure of mass per unit volume and has units of
mass per length squared.
C.
a measure of the gravitational force on a unit of mass.
D.
a measure of volume per unit mass and has units of
length cubed per unit mass.
ELO 1.4 Conversion between English and SI Systems
Introduction
Personnel at industrial facilities often use both the English and SI systems
of units in their work. In some cases, a measurement or an instrument will
provide a value in units that are different from the units required by a
procedure. This situation will require the conversion of measurements to
those required by the procedure. It is also possible that documentation
provided by industrial equipment vendors will use the English system of
measurement and may need conversion to SI measurements for use by
facility personnel or vice-versa.
In order to apply measurements from the SI system to the English system, it
is necessary to develop relationships of known equivalents (conversion
factors). Personnel can use these equivalents to convert units from the
given units of measure to the desired units of measure.
Converting Measuring Units
To convert measuring units, we use a systematic process to apply
conversion factors. The table below lists the process steps, and the tables
following that one list commonly used conversion factors.
10
Rev 1
Step
Action
1.
Select the appropriate relationship from the conversion table.
2.
Express the relationship as a ratio (desired units/present units).
3.
Multiply the quantity by the ratio.
4.
Repeat the steps until the value is in the desired units.
The following table lists many conversion factors fundamental and derived
units.
Converting Table
Measurement
Unit of
Measurement
Conversion Measurement
Time
60 seconds
60 minutes
= 1 minute
= 1 hour
Length
1 yard
12 inches
5,280 feet
1 meter
1 inch
= 0.9144 meter
= 1 foot
= 1 mile
= 3.281 feet
= 0.0254 meter
Mass
1 lbm
2.205 lbm
1 kg
= 0.4535 kg
= 1 kg
= 1,000 g
Area
1 ft2
10.764 ft2
1 yd2
1 mile2
= 144 in2
= 1 m2
= 9 ft2
= 3.098 x 106 yd2
Volume
7.48 gal
1 gal
1l
= 1 ft3
= 3.785 l
= 1,000 cm3
The following three tables list common conversion factors for mass, length,
and time, including a variety of different units that you may encounter as a
plant operator.
Rev 1
11
Common Conversion Factors for Units of Mass
Initial Unit
g
kg
t
lbm
1 gram
1
0.001
10-5
2.2046 x 10-3
1 kg
1,000
1
0.001
2.2046
1 metric ton (t)
106
1,000
1
2,204.6
1 pound-mass
(lbm)
453.59
0.45359
4.5359 x 10-4
1
1 slug
14,594
14.594
0.014594
32.174
Common Conversion Factors for Units of Length
Initial
Unit
cm
m
km
in
ft
mi
1
centimeter
1
0.01
10-5
0.3937
0.032808
6.2137 x
10-6
1 meter
100
1
0.001
39.370
3.2808
6.2137 x
10-4
1
kilometer
105
1,000
1
39,370
3,280.8
0.62137
1 inch
2.54
0.0254
2.54 x
10-5
1
0.0833
1.5783 x
10-5
1 foot
30.48
0.30480
3.0480
x 10-4
12.0
1
1.8939 x
10-4
1 mile
1.6093
1,609.3
x 105
1.6093
63,360
5,280
1
12
Rev 1
Common Conversion Factors for Units of Time
Initial
Unit
Sec
Min
Hr
Day
Year
1 second 1
0.017
2.7 x 10-4
1.16 x 10-5
3.1 x 10-8
1 minute 60
1
0.017
6.9 x 10-4
1.9 x 10-6
1 hour
3,600
60
1
4.16 x 10-2
1.14 x 10-4
1 day
86,400
1,440
24
1
2.74 x 10-3
1 year
3.15 x
107
5.26 x
105
8,760
365
1
The following table includes the steps, rationale, and results of an example
problem of converting units.
Conversion between Measurement Systems Demonstration
Step Convert 795 m to ft.
Result
1.
Select the equivalent
relationship from the
conversion table.
1 𝑚𝑒𝑡𝑒𝑟 (𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑢𝑛𝑖𝑡𝑠)
= 3.281 𝑓𝑡 (𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑢𝑛𝑖𝑡𝑠)
2.
Divide to obtain the factor
as a ratio.
3.281 𝑓𝑡
1𝑚
3.
Multiply the quantity by
the ratio.
(795 𝑚) × (3.281 𝑓𝑡/𝑚)
= 2.608 × 103 𝑓𝑡
Conversion between Measurement Systems Demonstration
You may need to use multiple conversion factors if an equivalent
relationship between the given units and you cannot find the desired units in
the conversion tables. Perform the conversion in several steps until the
measurement is in the desired units.
It is necessary to multiply the given measurement by each conversion factor
(ratio). After the units common to the numerator and the denominator
cancel each other out, the answer will be in the desired units.
Rev 1
13
Step:
Convert 2.91
square miles
to square
meters.
1.
Select the
equivalent
relationship
from the
conversion
table.
Multiple conversions are necessary because
there is no direct conversion shown for square
miles to square meters. For example, the
following conversions will be used in Step 2.
2.
Square miles
to square
yards to
square feet to
square meters
1 𝑚𝑖 2 = 3.098 × 106 𝑦𝑑 2
1 𝑦𝑑 2 = 9 𝑓𝑡 2
10.764 𝑓𝑡 2 = 1 𝑚2
3.
Express the
relationship as
a ratio
(desired
units/present
units).
4.
Multiply the
quantity by
the ratio.
Result
1 𝑚𝑖 2
9 𝑓𝑡 2
1 𝑚2
= (3.098 × 106 𝑦𝑑 2 ) (
)
(
)
1 𝑦𝑑 2 10.764 𝑓𝑡 2
= 2.590 × 106 𝑚2
106 𝑚2
(2.91 𝑚𝑖 (2.590 ×
)
1 𝑚𝑖 2
= 7.54 × 106 𝑚2
2)
Knowledge Check
Match the following dimensions to their equivalents.
1 3,280.8 feet
A.
1 mile per hour
2 453,590 grams
B.
1 lbm
3 0.017 hours
C.
1 minute
4 1.47 feet per second
D.
1 kilometer
14
Rev 1
Knowledge Check
Convert 55 miles per hour into feet per second.
A.
37 feet per second
B.
37.4 feet per second
C.
80.67 feet per second
D.
81 feet per second
Knowledge Check
Convert 1 cubic foot per second into liters per minute.
A. 55,742 liters per minute
B.
169,900,000 liters per minute
C.
1,699,000 liters per minute
D. 28,317 liters per minute
ELO 1.5 Converting Time Units
Introduction
The units of time must be consistent to solve problems in physics and build
understanding of physical concepts. This section will enable you to convert
time units so they are consistent.
Converting Time Units
To convert time units, use a similar systematic process to apply conversion
factors. The table below provides the systematic process steps, and the
table following gives commonly used conversion factors.
Step
Action
1.
Select the appropriate relationship from the conversion table.
2.
Express the relationship as a ratio (desired units/present units).
3.
Multiply the quantity by the ratio.
4.
Repeat the steps until the value is in the desired units.
Rev 1
15
Common Conversion Factors for Units of Time
Initial
Unit
Second
Minute
Hour
Day
Year
1
0.017
2.7 x 10-4
1.16 x 10-5
3.1 x 10-8
1
60
minute
1
0.017
6.9 x 10-4
1.9 x 10-6
1 hour
3,600
60
1
4.16 x 10-2
1.14 x 10-4
1 day
86,400
1,440
24
1
2.74 x 10-3
1 year
3.15 x 107
5.26 x 105
8,760
365
1
1
second
The table below includes a demonstration of time units conversion.
Converting time Units Demonstration
Step
Convert three years into
seconds.
Result
1.
Select the appropriate
relationship from the
conversion table.
The conversion factor goes
directly from years to seconds.
1 𝑦𝑒𝑎𝑟 = 3.15 × 107 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
2.
Express the relationship as a
ratio (desired units/present
units).
3.15 × 107 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
1 𝑦𝑒𝑎𝑟
Multiply the quantity by the
ratio.
3.15 × 107 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
(3 𝑦𝑒𝑎𝑟𝑠) (
)
1 𝑦𝑒𝑎𝑟
= 9.45 × 107 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
Repeat the steps until the
value is in the desired units.
Multiple steps are not necessary in
this case.
3.
4.
16
Rev 1
Knowledge Check
Calculate the number of minutes in 3.2 years.
A.
164,000
B.
168,000
C.
1,640,000
D.
1,680,000
Knowledge Check
Determine the number of hours in 46.5 days.
A.
1,016
B.
279
C.
2,790
D.
1,116
Knowledge Check
Determine the number of seconds in 8 hours.
A.
2,480
B.
24,800
C.
2,880
D.
28,800
TLO 1 Summary
Units of Measurement Summary
The three fundamental measurements are:



Length: distance between two points
Mass: amount of material in an object
Time: duration between two instants
The English system of units consists of the following standard units:



Foot
Pound
Second
The SI system of measurement consists of the following standard units:





Rev 1
Meter
Kilogram
Second
Centimeter
Gram
17

Second
Combinations of units comprise derived units to describe various physical
quantities. For example:



Mass - pounds-mass (lbm) or kilograms (kg)
Volume - cubic inches (in3) or liters (l)
Density - mass per unit volume (lbm/in3 or kg/l)
Unit Conversion:


Conversion tables list equivalent relationships.
Obtain factors by dividing to get a multiplying factor (1).
Unit Conversion Steps:



Step 1: Select the equivalent relationship from the conversion table.
Step 2: Express the relationship as a conversion factor.
Step 3: Multiply the given quantity by the conversion factor.
Units of Measurement Summary
Now that you have completed this lesson, you should be able to:
1. Describe the three fundamental dimensions: length, mass, and time.
2. List standard units of the fundamental dimensions for the following
systems:
a. International System of Units (SI)
b. English system
3. Differentiate between fundamental and derived measurements.
4. Convert between English and SI units of mass and length.
5. Convert time measurements between the following:
a. Years
b. Weeks
c. Days
d. Hours
e. Minutes
f. Seconds
TLO 2 Vectors
Overview
Vectors are quantities with both magnitude and direction. They are useful
in understanding physical concepts and solving various problems involving
motion.
Vectors are a necessary tool for understanding the interrelationship of force
and motion.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Define the following as they relate to vectors:
a. Scalar quantity
b. Vector quantity
18
Rev 1
c. Vector component
d. Resultant
2. Describe methods for identifying vectors in written material.
ELO 2.1 Vector Terminology
Introduction
In this section, you will learn what vectors are and the terms used in
describing them.
Scalar and Vector Quantities
Scalars are quantities that have magnitude only; they are independent of
direction. Vectors have both magnitude and direction. Graphically, a line
with an arrow end indicates a vector. The length of a vector (arrow)
represents the magnitude of the vector. The arrow shows direction of the
vector.
Scalar Quantities
Most of the physical quantities encountered in physics are either scalar or
vector quantities. A scalar quantity is a quantity that has magnitude only.
Typical examples of scalar quantities are time, speed, temperature, and
volume. A scalar quantity or parameter has no directional component, only
magnitude. For example, the units for time (minutes, days, hours, etc.)
represent an amount of time only and tell nothing of direction. Additional
examples of scalar quantities are density, mass, and energy.
Vector Quantities
We define a vector quantity as a quantity that has both magnitude and
direction. To work with vector quantities, one must know the method for
representing these quantities. Magnitude, or size of a vector, is also referred
to as the vector's displacement. You can think of this as the scalar portion
of the vector; the length of the vector represents this attribute. By
definition, a vector has both magnitude and direction. Direction indicates
how the vector is oriented relative to some reference as shown in the figure
below.
Rev 1
19
Figure: Vector Reference Axis
Using north/south and east/west reference axes, vector A is oriented in the
NE quadrant with a direction of 45° north of the E-W axis. Giving direction
to scalar A makes it a vector. The length of A is representative of its
magnitude or displacement.
Scalar versus Vector Example
To help distinguish between a scalar and a vector, consider an example
where the only information known is that a car is moving at 80 kilometers
per hour. The information given (80 km/hour) only refers to the car's speed,
which is a scalar quantity. It does not indicate the direction the car is
moving. However, the same car traveling at 80 km/hour due east indicates
the velocity of the car because it has magnitude (80 km/hour) and direction
(due east); therefore, a vector is indicated.
Diagramming a Vector Quantity
When we diagram a vector, a straight line shows the vector length. Add an
arrow on one end of the line. The length of the line represents the
magnitude of the vector, and the arrow represents the vector direction.
Description of a Simple Vector
Vectors are simple straight lines used to illustrate the direction and
magnitude of certain quantities. Vectors have a starting point at one end
(tail) and an arrow at the opposite end (head), as shown in the figure below.
Figure: Simple Vector
Examples of Vector Quantities
Displacement, velocity, acceleration, and force are examples of vector
quantities. Momentum and magnetic field strength are also good examples
of vector quantities, although somewhat more difficult to understand. In
20
Rev 1
each of these examples, the main ingredients of magnitude and direction are
present.
Vector Component
A vector component is a vector that is part of an overall result, or resultant
vector. You can determine the vector comprising a single vector
component either graphically or by using trigonometry.
Resultant
A resultant vector is a single vector that results from combining two or
more vector components.
Knowledge Check
Select all that are true:
A.
Vectors have direction and magnitude.
B.
Scalars have direction and magnitude.
C.
Temperature is a vector quantity.
D.
Force is a vector quantity.
Knowledge Check
Which of the following is a scalar quantity?
A.
Velocity
B.
Displacement
C.
Force
D.
Density
Knowledge Check
Which of the following is a vector quantity?
A.
Acceleration
B.
Energy
C.
Heat rate
D.
Mass
ELO 2.2 Identifying Vectors
Introduction
There are specific ways to symbolize vectors in texts and on graphs, using
letters or rectangular coordinates.
Rev 1
21
Vector Identification
In textbooks, vector quantities are often represented by simply using a
boldfaced letter (e.g. A, B, C, R). Particular quantities are predefined (F force, V - velocity, and A - acceleration). Sometimes, a bold capital letter
with an arrow above it represents vector quantities, as shown below:
⃗𝑨 ⃗𝑩
⃗ ⃗𝑪 ⃗𝑹
⃗
Regardless of the convention used, specific vector quantities must include
magnitude and direction (for example, 50 km/hour due north, or 50 lbf at
90°).
We represent vector quantities graphically using a rectangular coordinate
system.



Two-dimensional system that uses an x-axis and a y-axis
x-axis is horizontal straight line
y-axis is a vertical straight line, perpendicular to x-axis
The following sections will provide examples and practice using the
rectangular coordinate system for graphic representation of vectors.
Knowledge Check
Which of the following is not a way to represent a
vector?
A.
As a quantity with both magnitude and direction (50
km/hr, due north)
B.
As a quantity (30 mi/hr)
C.
As a bold, capital letter with an arrow over it
D.
As a bold capital letter for predefined items, such as F
for force
Knowledge Check
Vectors are sometimes represented by:
A.
ampersands followed by a letter designation.
B.
terms that imply motion, such as velocity, with speed but
not necessarily direction.
C.
numerical quantities such as mass.
D.
bold capital letters with arrow over them.
TLO 2 Summary
Vectors Summary
Scalar quantities:

22
Magnitude only
Rev 1


Independent of direction
Examples of scalars include time, volume, and temperature.
Vector quantities:




Both magnitude and direction
Length of arrow signifies magnitude
Arrow shows direction
Examples of vectors include force, velocity, and acceleration.
Vector identification:



Boldfaced capital letters (A, F, R) with arrows over them
Graphically, or (x, y) coordinates
Directional Coordinates, or Degrees
Important facts:


A resultant is a single vector that can replace two or more vectors.
You can obtain components for any two non-parallel directions if the
vectors are in the same plane.
 Restricting the treatment to perpendicular directions and twodimensional space, the components of a vector are the two vectors in
the x and y (or east-west and north-south) directions which produce
the same effect as the original vector (or add to produce the original
vector).
 Components are determined from data, graphically or analytically.
Now that you have completed this lesson, you should be able to:

Define the following as they relate to vectors: a. scalar quantity b.
vector quantity c. vector component d. resultant
 Describe methods for identifying vectors in written material.
TLO 3 Solving Vector Problems
Overview
Now that you know how to distinguish between a scalar and a vector, you
can solve vector problems. These problems range from graphing vectors,
determining a vector’s components, and adding vectors by various methods.
Vectors are a necessary tool for understanding the interrelationship of force
and motion.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Given a magnitude and direction, graph a vector using the rectangular
coordinate system.
2. Determine components of a vector from a resultant vector.
3. Add vectors using the following methods:
a. Graphical
b. Component addition
c. Analytical
Rev 1
23
ELO 3.1 Graphing Vectors
Introduction
In this section, you will learn to graph vectors.
Graphing Vectors
Vector quantities are graphically represented using the rectangular
coordinate system, a two-dimensional system that uses an x-axis and a yaxis. The x-axis is a horizontal straight line. The y-axis is a vertical
straight line, perpendicular to the x-axis. The figure below shows an
example of a rectangular system. The intersection of the axes is called the
point of origin. Each axis is marked off in equal divisions in all four
directions from the point of origin. On the horizontal axis (x), values to the
right of the origin are positive (+). Values to the left of the origin are
negative (-).
Figure: Rectangular Coordinate System
On the vertical axis (y), values above the point of origin are positive (+).
Values below the origin are negative (-). It is very important to use the
same units (divisions) on both axes.
The rectangular coordinate system creates four infinite quadrants, as shown
in the figure above.




Quadrant I - located above and to the right of the origin.
Quadrant II - located above and to the left of the origin.
Quadrant III - located to the left and below the origin.
Quadrant IV - located below and to the right of the origin.
The table below lists the steps required to graph a vector.
24
Rev 1
Step
Action
1.
Label the X and Y axes.
2.
Mark off equal divisions in all 4 directions on the graph.
3.
Label the values above and to the right of the origin positive,
below and to the left negative.
4.
Draw the vector as follows: If the vector lies along either of the
axes (vertical or horizontal), then simply draw the line along the
axis.
If the vector is given in graph coordinates (x and y coordinates)
count off the x coordinate horizontally, and then the y coordinate
vertically. This will give you the position of the vector head.
Draw a line from the origin to the head to represent the vector.
If the vector is given in terms of a magnitude and angle, then use
a protractor to draw the angle, and dividers or a scale to measure
the magnitude on the axis. Then, using the angle drawn and the
magnitude determined by the dividers (or scale), mark the
position of the vector head.
Beginning at the point of origin (intersection of the axes), draw a line
segment of the proper length along the x-axis, in the positive direction.
This line segment represents the vector magnitude, or displacement. Place
an arrow at the head of the vector to indicate direction. The tail of the
vector is located at the point of origin (see figure below).
Figure: Displaying Vectors Graphically - Magnitude
Rev 1
25
When drawing vectors that do not fall on the x- or y-axes, the tail is located
at the point of origin. Depending on the vector description, there are two
methods of locating the head of the vector. If coordinates (x, y) are given,
you can plot these values to locate the vector head. If degrees describe the
vector, rotate the line segment counterclockwise from the x-axis about the
origin to the proper orientation, as shown in the figure below.
Figure: Displaying Vectors Graphically - Direction
Because the x- and y-axes define direction, conventional directional
coordinates and degrees also serve to identify the x- and y-axes (see figure
below).
Figure: Degree Coordinates
26
Rev 1
Graphing Vectors Demonstration
You can graph vectors by using either Cartesian coordinates or angle for
direction. An example of a vector beginning at the origin and pointing to
coordinate (2,5) is given in the figure below.
Figure: Vector from Origin to (2,5)
Another example of a vector originating at the origin and extending a
magnitude of 7 units at 30° is shown in the following figure.
Figure: Vector, Magnitude 7, 30°
Rev 1
27
Knowledge Check
The graph of a vector is:
A.
a straight line with an arrow is drawn on one end. The
length of the line represents the magnitude of the
vector, and the arrow represents the direction of the
vector.
B.
intended to convey direction only. Notes are needed to
convey magnitude.
C.
indicative of a quantity with magnitude but no
direction.
D.
a means of showing the direction of the force, but is
not drawn to scale.
Knowledge Check
Which of the following is not a way to represent a
vector?
A.
As a quantity with both magnitude and direction (50
km/hr, due north )
B.
As a quantity (30 mi/hr)
C.
As a bold, capital letter with an arrow over it
D.
As a bold capital letter for predefined items, such as F
for force
Knowledge Check
Which of the following is an appropriate way to
represent a vector?
28
A.
As a capital letter with a subscript "v"
B.
As a quantity with both magnitude and direction (50
km/hr, due north)
C.
As a bold, capital letter in quotation marks
D.
As a quantity (30 mi/hr)
Rev 1
ELO 3.2 Determining Components of a Vector
Introduction
A resultant is a single vector that represents the combined effect of two or
more other vectors (called components). The components can be
determined either graphically or by using trigonometry.
Determining Components of a Vector by Graphing
Components of a vector are vectors, which when added, yield the resultant
vector. For example traveling 3 kilometers north and then 4 kilometers east
yields a resultant displacement of 5 kilometers, in a direction 37° north of
east.
This example shows how to determine the component vectors of any two
non-parallel vectors from any resultant vector in the same plane. The table
below lists the individual steps in the process. For the purposes of this
lesson, the discussion of vectors is limited to two-dimensional space. The
student should realize however, that vectors can and do exist in threedimensional space.
For the example given above, an alternate problem might be, "If the final
displacement of the individual is 5 kilometers from the starting point,
northeast along a line 37° north of east, how far north and how far east did
the individual travel from his original position?"
Step
Action
1.
Graph the vector, magnitude, and direction, beginning at the
origin.
2.
Draw a vertical line from the head of the vector to the x-axis.
Note the x coordinate.
3.
This information provides one component, with magnitude equal
to the x coordinate and direction 0° (or 180° if the x coordinate
is negative).
4.
Draw a horizontal line from the head of the vector to the y-axis.
Note the y coordinate.
5.
This information provides the second component, with
magnitude equal to the y coordinate and direction 90° (or 270° if
the y coordinate is negative).
Determining Components of a Vector Using Trigonometry
The table below provides instructions for determining vector components
using trigonometry.
Rev 1
29
Step
Action
1.
Graph the vector, magnitude, and direction, beginning at the
origin.
2.
Determine the cosine of the vector's angle.
3.
The vector cosine multiplied by the magnitude provides the x
component vector.
4.
Determine the trigonometric sine of the vector angle.
5.
The vector sine multiplied by the magnitude provides the y
component vector.
Plotting Component Vectors
Component vectors can be determined by plotting them on a rectangular
coordinate system. For example, a resultant vector of 5 units at 53° can be
broken down into its respective x and y magnitudes. The x value of 3 and
the y value of 4 can be determined graphically or using trigonometry. One
of several conventions can express their magnitudes and position, including
(3, 4), (x = 3, y = 4), (3 at 0°, 4 at 90°), and (5 at 53°). In the first
expression, the first term is the x-component (Fx), and the second term is the
y-component (Fy) of the associated resultant vector. As in the previous
example, if given only the resultant, instead of component coordinates, one
can determine the vector components as illustrated in the figure below.
First, plot the resultant on rectangular coordinates and then project the
vector coordinates to the axis. The length along the x-axis is Fx, and the
length along the y-axis is Fy. The following example demonstrates this
method.
30
Rev 1
Figure: Vector Components
Example: For the resultant vector shown in the figure below, determine the
component vectors given FR = 50 Newtons (N) at 53°.
Figure: Graphing Vector Demonstration
Solution: First, project a line from the head of FR perpendicular to the xaxis, and a similar line perpendicular to the y-axis. Where the projected
lines meet the axes determines the magnitude (size) of the component
vectors. In this example, the component vectors are 30 N at 0° (Fx) and 40
N at 90° (Fy). If the resultant vector FR had not already been drawn, the
first step would have been to draw the vector.
Rev 1
31
Determining Components of Vectors by Graphing
Demonstration
You can also use trigonometry to determine vector components. Before
explaining this method, it may be helpful to review the fundamental
trigonometric functions. Trigonometry is a branch of mathematics that
deals with the relationships between angles and the length of the sides of
triangles. Three ratios define the relationship between an acute angle of a
right triangle, shown as θ in the figure below, and the triangle sides.
For the triangle shown above, these defining relationships are:
Sine θ = opposite/hypotenuse = a/c
Cosine θ = adjacent/hypotenuse = b/c
Tangent θ = opposite/adjacent = a/b
Most equations abbreviate sine, cosine, and tangent to sin, cos, and tan,
respectively.
Determining Components of Vectors by Graphing
Demonstration
First, make a rough sketch that shows the approximate relationship of the
vector angle and the sides. Determine the component vectors, Fx and Fy,
for FR = 50 N at 53° in the figure below. Use trigonometric functions.
Figure: Graphing Vector Demonstration
32
Rev 1
Step
Fx is calculated as follows:
1.
2.
cos 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝐹
cos 𝜃 = 𝐹𝑥 or 𝑭𝑥 = 𝑭𝑅 cos 𝜃
𝑅
3.
𝐹𝑥 = (50 𝑁)(cos 53°)
4.
𝐹𝑥 = (50 𝑁)(0.6018)
5.
𝐹𝑥 = 30 𝑁 𝑜𝑛 𝑥– 𝑎𝑥𝑖𝑠
Step
1.
2.
Fy is calculated as follows:
sin 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝐹𝑦
sin 𝜃 = 𝐹 or 𝐹𝑦 = 𝐹𝑅 sin 𝜃
𝑅
3.
𝐹𝑦 = (𝐹𝑅 )(sin 𝜃)
4.
𝐹𝑦 = (50 𝑁)(sin 53°)
5.
𝐹𝑦 = (50 𝑁)(0.7986)
6.
𝐹𝑦 = 40 𝑁 𝑜𝑛 𝑦– 𝑎𝑥𝑖𝑠
Therefore, the components for FR are Fx = 30 N at 0° and Fy = 40 N at 90°.
Note that this result is identical to the result obtained using the graphic
method.
Rev 1
33
Knowledge Check
A vector begins at (2, 3) and ends at (-1, 7). Select all
of the statements about this vector that are true.
A.
The vector has a magnitude of 5.
B.
The vector has a horizontal component of magnitude 3,
direction 180°, and a vertical component of magnitude
4, direction 90°.
C.
The vector has a direction of 127°.
D.
The vector has a direction of 120°.
Knowledge Check
What is the magnitude and direction of the vector
below?
34
A.
Magnitude 7.2, direction 56.4°
B.
Magnitude 6.6, direction 56.4°
C.
Magnitude 7.2, direction 60°
D.
Magnitude 6.6, direction 60°
Rev 1
ELO 3.3 Adding Vectors
Introduction
It is possible to add component vectors together to determine the resultant
vector. For example, when two or more forces are acting on a single object,
we use vector addition to determine the direction and magnitude of the net
(resultant) force on the object. Consider an airplane that travels due east for
100 kilometers at 500 km/hr, then NE for 50 kilometers at 400 km/hr, and
finally north for 500 kilometers at 500 km/hr. We use vector addition to
determine the net distance the airplane is from its point of origin or to
predict when it will arrive at its destination.
Methods Used to Add Vectors
There are several methods to add vectors. This chapter explains the
graphical, component-addition, and analytical methods. Either the
graphical method or the component addition method will provide a fairly
accurate result. If a higher degree of accuracy is required, use an analytical
method with geometric and trigonometric functions to add the vectors.
Adding Vectors
The graphic method utilizes a five-step process, listed in the table below.
You will need a protractor and a scale, as well as pencil and paper to
perform graphical vector addition.
Step
Action
1.
Plot the first vector on the rectangular (x-y) axes. Ensure that
both axes use the same scale. Place the tail (beginning) of the
first vector at the origin of the axes as shown in the figure below.
2.
Draw the second vector connected to the end of the first vector.
Start the tail of the second vector at the head of the first vector.
Ensure that you draw the second vector to scale. Ensure proper
angular orientation of the second vector with respect to the axes
of the graph (see second figure below).
Rev 1
35
Step
Action
3.
Add other vectors sequentially. Add one vector at a time.
Always start the tail of the new vector at the head of the previous
vector. Draw all vectors to scale and with proper angular
orientation.
4.
When all given vectors have been drawn, construct and label a
resultant vector, FR, from the point of origin of the axes to the
head of the final vector. The tail of the resultant is the tail of the
first vector drawn as shown in the third figure below. The head
of the resultant is at the head of the last vector drawn.
36
Rev 1
Step
Action
5.
Determine the magnitude and direction of the resultant.
Measure the displacement and angle directly from the graph
using a scale and a protractor. Determine the components of the
resultant by projection onto the x- and y-axes.
Adding Vectors
To add vectors using the component-addition method, use the following
four-step method. This method does not require a scale or protractor.
Step
Action
1.
Determine x- and y-axes components of all original vectors.
2.
Mathematically combine all x-axis components.
Note: When combining, recognize that positive-x components at
180° are equivalent to negative-x components at 0° (+x at 180° =
-x at 0° ).
3.
Mathematically combine all y-axis components (+y at 270° = -y
at 90°).
4.
Resulting (x, y) components are the (x, y) components of the
resulting vector.
Math Review for Adding Vectors
Calculations using trigonometric functions are the most accurate method for
adding vector components to determine the magnitude and direction of the
resultant. The graphic and component addition methods of obtaining the
resultant of several vectors described previously can be hard to use and time
consuming. In addition, accuracy is a function of the scale used in making
the diagram, and the accuracy of the vector drawings. The analytical
method is simpler and far more accurate than these previous methods.
In earlier mathematics lessons, you learned how to use the Pythagorean
Theorem to relate the lengths of the sides of right triangles such as in the
figure below. The Pythagorean Theorem states that in any right triangle,
the square of the length of the hypotenuse equals the sum of the squares of
the lengths of the other two sides. For the right triangle shown in the figure
below, this expression is:
𝑎2 + 𝑏 2 = 𝑐 2 or 𝑐 = √𝑎2 + 𝑏 2
Rev 1
37
Figure: Right Triangle – Pythagorean Theorem
Also, recall the three trigonometric functions reviewed in an earlier chapter
and listed below:
sin 𝜃 =
𝑎
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
=
𝑐 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝜃 =
𝑏
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
=
𝑐 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
tan 𝜃 =
𝑎 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
=
𝑐 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
The cosine will be used to solve for Fx. Sine will be used to solve for Fy.
Tangent will normally be used to solve for θ, although sine and cosine may
also be used, depending on which vector characteristics are known.
On a rectangular coordinate system, the sine values of θ are positive (+) in
quadrants I and II and negative (-) in quadrants III and IV. The cosine
values of θ are positive (+) in quadrants I and IV and negative (-) in
quadrants II and III. Tangent values are positive (+) in quadrants I and III
and negative (-) in quadrants II and IV.
When mathematically solving for tan θ, calculators will specify angles in
quadrants I and IV only. Actual angles may be in quadrants II and III.
Analyze each problem graphically in order to ensure a realistic solution. To
obtain angles in Quadrants II and III, add or subtract 180° from the value
calculated.
Analytical Method
Use the analytical method in the table below to solve this example:
A man walks 3 kilometers in one direction, then turns left 90° and continues
to walk for an additional 4 kilometers. In what direction and how far is he
from his starting point?
38
Rev 1
Step
Approach
1.
Result
The first step in
solving this
problem is to
draw a simple
sketch as
shown:
2.
His net
displacement is
found as
follows:
3.
𝑅 = √𝑎2 + 𝑏 2
𝑅 = 5 𝑚𝑖𝑙𝑒𝑠
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝑎
tan 𝜃 =
𝑏
4
tan 𝜃 =
3
tan 𝜃 = 1.33
𝜃 = tan−1 1.33
tan 𝜃 =
His direction
(angle of
displacement)
is found using
the tangent
function.
𝜃 = 53°
4.
Determine
direction and
distance.
Therefore, his new location is 5
kilometers at 53° from his starting point.
Adding Vectors
The table below provides instructions for adding vectors by the analytical
method.
Step
Action
1.
Draw x and y coordinates for all vectors from the point of origin
or the center of the object.
2.
Resolve each vector into its rectangular components.
3.
Sum the x and y components.
4.
Calculate the magnitude of FR.
Rev 1
39
Step
Action
5.
Calculate the angle of displacement.
To carry this approach a step further, consider the following model
developed for finding the resultant of several vectors.
To demonstrate the model, consider three forces (F1, F2, and F3) acting on
an object as shown below. The goal is to find the resultant force (FR).
Step 1: Draw x and y coordinates and the three forces from the point of
origin or the center of the object, as shown in the figure below. The
drawing includes component vectors and angles to aid in the discussion.
Figure: Force Acting on an Atomic Nucleus
Step 2: Resolve each vector into its rectangular components:
Vector
Angle
x component
y component
F1
θ1
𝐹1𝑥 = 𝐹1 cos 𝜃1
𝐹1𝑦 = 𝐹1 sin 𝜃1
F2
θ2
𝐹2𝑥 = 𝐹2 cos 𝜃2
𝐹2𝑦 = 𝐹2 sin 𝜃2
F3
θ3
𝐹3𝑥 = 𝐹3 cos 𝜃3
𝐹3𝑦 = 𝐹3 sin 𝜃3
Step 3: Sum the x and y components.
𝐹𝑅𝑥 = 𝛴𝐹𝑥 = 𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥
𝐹𝑅𝑦 = 𝛴𝐹𝑦 = 𝐹1𝑦 + 𝐹2𝑦 + 𝐹3𝑦
40
Rev 1
Step 4: Calculate the magnitude of FR.
2
2
𝐹𝑅 = √𝐹𝑅𝑥
+ 𝐹𝑅𝑦
Step 5: Calculate the angle of displacement.
tan 𝜃 =
𝐹𝑅𝑦
𝐹𝑅𝑥
𝜃 = tan−1
𝐹𝑅𝑦
𝐹𝑅𝑥
Adding Vectors Demonstration
Determine the magnitude and direction of the resultant for the following:
F1 = 3 units at 300°
F2 = 4 units at 60°
F3 = 8 units at 180°
FR = 4 units at 150°
The figure below shows construction of the three vectors and their resultant:
Figure: Addition of Vectors
Adding Vectors Demonstration
Use the analytical method in the table below to solve this example:
Given the following vectors what are the coordinates of the resultant vector,
that is, the sum of the vectors?
F1 = (4, 10)
F2 = (-6, 4)
F3 = (2, -4)
F4 = (10, -2)
Rev 1
41
Step
Approach
1.
Determine the x-axis and
y-axis components of all
four original vectors.
2.
3.
Result
𝑥– 𝑎𝑥𝑖𝑠 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠
= 4, −6, 2, 10
𝑦– 𝑎𝑥𝑖𝑠 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 =
10, 4, −4, −2
Mathematically combine
all x-axis components.
𝑥 = 4 + (−6) + 2 + 10
𝑥 = 4 − 6 + 2 + 10
𝑥 = 10
Mathematically combine
all y-axis components.
𝑦 = 10 + 4 + (−4) + (−2)
𝑦 = 10 + 4 − 4 − 2
𝑦=8
Express the resultant
vector.
The resultant components from the
previous additions are the
coordinates of the resultant, that is,
FR = (10, 8)
4.
Adding Vectors Demonstration
Given three forces acting on an object, determine the magnitude and
direction of the resultant force FR.
F1 = 90 N at 39°
F2 = 50 N at 120°
F3 = 125 N at 250°
Step 1: First, draw x and y coordinate axes on a sheet of paper. Then, draw
F1, F2, and F3 from the point of origin. It is not necessary to be very
accurate in placing the vectors in the drawing. The approximate location in
the right quadrant is all that is necessary. Label the drawing as below.
42
Rev 1
Step 2: Resolve each force into its rectangular coordinates.
Force
F1
Magnitude
90 N
Angle
39°
X Component
𝐹1𝑥 = (90) cos 39°
𝐹1𝑦 = (90) sin 39°
𝐹1𝑥
= (90) (0.777)
𝐹1𝑦 = (90)(0.629)
𝐹1𝑥 = 69.9 𝑁
𝐹2𝑥
= (50) cos 120°
F2
50 N
120°
𝐹2𝑥 = (50)(−0.5)
𝐹2𝑥 = −25 𝑁
F3
125 N
250°
Y Component
𝐹1𝑦 = 56.6 𝑁
𝐹2𝑦 = (50) sin 120°
𝐹2𝑦 = (50)(0.866)
𝐹2𝑦 = 43.3 𝑁
𝐹3𝑥
= (125) cos 250°
𝐹3𝑦
= (125) sin 250°
𝐹3𝑥
= 125)(−0.342)
𝐹3𝑦
= (125)(−0.94)
𝐹3𝑥 = −42.8 𝑁
𝐹3𝑦 = −117.5 𝑁
Step 3: Sum the x and y components.
𝐹𝑅𝑥 = 𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥
𝐹𝑅𝑥 = (69.9 𝑁) + (−25 𝑁) + (−42.8 𝑁)
𝐹𝑅𝑥 = 2.1𝑁
𝐹𝑅𝑦 = 𝐹1𝑦 + 𝐹2𝑦 + 𝐹3𝑦
𝐹𝑅𝑦 = (56.6 𝑁) + (43.3 𝑁) + (−117.5 𝑁)
𝐹𝑅𝑦 = −17.6 𝑁
Step 4: Calculate the magnitude of FR.
𝐹𝑅 = √𝐹𝑥2 + 𝐹𝑦2
𝐹𝑅 = √(2.1)2 + (−17.6)2
𝐹𝑅 = √314.2
𝐹𝑅 = 17.7 𝑁
Step 5: Calculate the angle of displacement.
𝐹𝑅𝑦
𝐹𝑅𝑥
−17.6
tan 𝜃 =
2.1
tan 𝜃 = −8.381
tan 𝜃 =
Rev 1
43
𝜃 = tan−1 (−8.381)
𝜃 = −83.2°
Therefore:
𝐹𝑅 = 17.7 𝑁 𝑎𝑡 – 83.2° 𝑜𝑟 276.8°.
Note: A negative angle means a clockwise rotation from the zero degree
axis.
Knowledge Check
Determine the resultant of the following vectors:
F1, magnitude 6, angle 150°
F2, magnitude 11, angle 240°
F3, magnitude 5, 0°
A.
Magnitude 8.7, 228°
B.
Magnitude 7.9, 210°
C.
Magnitude 11.1, 196°
D.
Magnitude 7.9, 196°
Knowledge Check
Given XL = 50 ohms at 90°, R = 50 ohms at 0°, and XC
= 50 ohms at 270°, what is the resultant Z?
A.
Z = 70.7 ohms at 45°
B.
Z = 70.7 ohms at -45°
C.
Z = 50 ohms at 0°
D.
Z = 50 ohms at 90°
Knowledge Check
Determine the resultant, FR given: F1 = 30 N at 0°, 10
N at 90°; F2 = 50 N at 0°, 50 N at 90°; F3 = 45 N at
180°, 30 N at 90°; F4 = 15 N at 0°, 50 N at 270°.
44
A.
FR = 50 N at 0°, 140 N at 90°
B.
FR = 140 N at 0°, 40 N at 90°
C.
FR = 140 N at 0°, 140 N at 90°
D.
FR = 50 N at 0°, 40 N at 90°
Rev 1
Knowledge Check
Determine the sum of the following vectors: F1,
magnitude 6 at 45°; F2, magnitude 8 at 30°; F3,
magnitude 5 at 120°.
A.
Magnitude 18.5 at 44°
B.
Magnitude 15.3 at 55.4°
C.
Magnitude 11.6 at 39.7°
D.
Magnitude 21.6 at 44.3°
TLO 3 Summary
Vector Problems Summary
Graphic Method Summary:



Draw rectangular coordinates.
Draw first vector.
Draw second vector connected to the end (head) of first vector with
proper angular orientation.
 Draw remaining vectors, starting at the head of the preceding vector.
 Draw resultant vector from the origin of axes to head of final vector.
 Measure the length of the resultant vector.
 Measure the angle of the resultant vector addition.
Component-Addition Method Summary:




Determine the x- and y- axes of all original vectors.
Mathematically combine all x-axis components.
Mathematically combine all y-axis components.
The results are the components of the resultant vector.
Analytical Method Summary:






Draw x and y coordinate axes.
Draw component vectors from point of origin.
Resolve each vector into rectangular components.
Sum the x and the y components.
Calculate magnitude of FR.
Calculate angle of displacement.
Summary
Now that you have completed this lesson, you should be able to:

Given a magnitude and direction, graph a vector using the rectangular
coordinate system.
 Determine components of a vector from a resultant vector.
 Add vectors using the following methods:
a. Graphical
b. Component addition
c. Analytical
Rev 1
45
TLO 4 Force
Overview
You will learn the types of forces and their effects on objects at rest or in
motion in this section.
It is vital to understand how different forces act on objects, and the effect
the application of force has on those objects.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the following types of forces:
a. Tensile
b. Compressive
c. Frictional
d. Centripetal
e. Centrifugal
2. Describe the factors that affect the magnitude of the static and kinetic
friction force.
3. Describe force as it applies to an object and its velocity.
4. Describe weight as it applies to an object and its velocity.
5. Given the mass of an object and a value for gravity, calculate the
weight of the object.
6. Describe the purpose of a free-body diagram, and given all necessary
information, construct a free-body diagram.
7. Describe the conditions necessary for a body to be in force
equilibrium.
ELO 4.1 Types of Forces
Introduction
It is important to understand the types of forces that act on an object when
determining how an object reacts to a force or forces.
Types of Forces
Tensile and Compressive Forces
In discussing the types of forces, we use a simple rule to determine if the
force is a tensile or a compressive force. If an applied force on a member
tends to pull the member apart, the force is a tensile force. If a force tends
to compress the member, the force is a compressive force. Ropes, cables,
etc., attached to bodies can only support tensile loads, and therefore such
objects are in tension when placed on the free-body diagram. In addition,
when a fluid is involved, note that fluid forces are usually compressive
forces.
Friction
Friction is a resistance force when two surfaces are in contact and one of the
surfaces is attempting to move parallel to or over the other surface. The two
types of friction forces are coulomb and fluid.
46
Rev 1
Centripetal Force
An object moving at constant speed in a circle is not in equilibrium. The
direction of the velocity is continually changing but the magnitude of the
linear velocity is not changing. Acceleration requires an object moving in a
circular path to have a constant acceleration towards the center.
Newton's second law of motion, 𝐹 = 𝑚𝑎, dictates that a force is required to
cause acceleration. Centripetal force must have constant acceleration
towards the center of the circular path and must be a net force acting
towards the center. An object will move in a straight line without this force.
The figure below illustrates the centripetal force.
Figure: Centripetal Force
Centrifugal Force
Centrifugal force appears to be opposite the direction of motion acting on
an object that follows a curved path. This force is a fictitious force that
appears to be directed away from the center of the circular path. It is an
apparent force that describes the force’s presence due to an object's rotation.
Imagine a string attached to the plane in the above figure to understand this
force better. The string places a centrifugal force as the plane rotates about
the center and changes the direction causing it to travel in a circle.
Rev 1
47
Figure: Apparent Centrifugal Force
The apparent outward force (apparent centrifugal force) seems to pull the
plane away from the center, as shown above. This is the same force one
feels when riding in a car that is traveling in circles. Cutting the string
proves that centrifugal force is not an actual force and the plane will fly off
in a straight line that is tangent to the circle at the velocity it had the
moment the string was cut. If there were an actual centrifugal force present,
the plane would not fly away in a line tangent to the circle, but would fly
directly away from the circle in the direction of the apparent centrifugal
force.
Knowledge Check
Match the following terms to the appropriate definitions.
1 tensile forces only
A.
ropes and
cables
2 cause of circular motion
B.
fluid forces
3 contact between objects
C.
friction
4 compressive forces only
D.
centripetal
48
Rev 1
Knowledge Check
Which of the following is not a real force?
A.
centripetal
B.
tensile force
C.
centrifugal
D.
friction
Knowledge Check
Fluid forces are:
A.
generally insignificant
B.
compressive
C.
dominated by friction forces
D.
tensile
ELO 4.2 Friction Force
Introduction
Friction occurs when two surfaces are in contact and one surface is
attempting to move parallel to or over the other surface when two surfaces
are in contact.
Friction Force
The two friction forces are coulomb and fluid frictions.
Fluid Friction
Fluid friction develops between layers of fluid moving at different
velocities. The flow of fluids through pipes is an example of this friction.
Dry Friction
The following experiment demonstrates the laws of dry friction.
Consider a block of weight W placed on a horizontal plane surface (as
shown in the figure below). The forces acting on the block are its weight
(W) and the equal and opposing normal force (N) of the surface, as shown
in part (a) below. The weight and normal force have no horizontal
components, since their orientations are purely vertical. Suppose that we
apply a horizontal force (P) to the block, as shown in part (b) below. If the
magnitude of P is small, the block will not move. Some other horizontal
force must therefore exist, that opposes P. This opposing force is the
friction force (F), which is actually the resultant of a great number of forces
acting over the entire contact surface between the block and the plane.
These friction forces result from two aspects: irregularities of the surfaces in
contact and the attractive coulombic forces between the atoms/molecules of
the block and plane surface.
Rev 1
49
Figure: Frictional Forces
When the force P is increased, the friction force F initially increases and
continues to oppose P. This occurs until the magnitude of F reaches a value
of Fm, as shown in part (c) above. At this point, the friction force F is no
longer sufficient to oppose P and the block starts sliding.
The magnitude of F drops from Fm to a lower value Fk, when force P
overcomes the static -friction force and the block has been set in motion
(note: the term static implies no motion). From this point, (if we maintain
force P), the block will keep sliding, while the magnitude of the kineticfriction force, denoted by Fk, remains approximately constant (note: the
term kinetic implies movement).
Static-Friction Force
Experimental evidence shows that the maximum static value of the friction
force (Fm) is directly proportional to the normal force N (N is equivalent in
magnitude to the object's weight W). The constant of proportionality, µs, is
called the coefficient of static friction.
𝐹𝑚 = µ𝑠 𝑁
Kinetic-Friction Force
The kinetic-friction force (Fk), which is also directly proportional to N, may
be similarly expressed. The constant of proportionality, µk, is the
coefficient of kinetic friction.
𝐹𝑘 = µ𝑘 𝑁
Magnitudes of Coefficients of Friction
The coefficients of friction (µs and µk) are independent of the surfacecontact area, but depend strongly on the nature of the surfaces.
Frictional forces are always opposite in direction to the motion (or
impending motion) of an object.
50
Rev 1
Knowledge Check
Select all statements about the friction force that are
true.
A.
The direction of the friction force is opposite the
direction of motion.
B.
The coefficient of kinetic friction between two surfaces
is greater than the coefficient of static friction for the
same two surfaces.
C.
Friction force is proportional to the normal force.
D.
When an object is not moving, the friction force is
zero.
Knowledge Check
Friction is:
A.
insignificant in most real world applications
B.
higher when the objects are in relative motion than
when they are static with respect to each other
C.
caused by motion
D.
Caused by interaction between two surfaces in contact
Knowledge Check
Select the statement which is NOT true.
A.
The friction between two surfaces is dependent on
whether they are in motion or at rest with respect to
each other.
B.
The magnitude of the friction force is dependent on the
surface area in contact.
C.
The friction between two surfaces is dependent on the
roughness of the surface.
D.
The coefficient of friction between two materials is
dependent on the surface area in contact.
ELO 4.3 Force and Velocity
Introduction
We define force as any action on a body that tends to change the velocity of
the body, which could be either a pushing or a pulling force.
Rev 1
51
Force and Velocity
Force is a vector quantity that tends to produce an acceleration of a body in
the direction of its application. Changing the body's velocity causes the
body to accelerate. Therefore, the mathematical definition of force is as
given by Newton's second law of motion:
𝐹 = 𝑚𝑎
Where:
F = force on object (Newton (kg-m/s2) or lbf)
m = mass of object (kg or lbm)
a = acceleration of object (m/s2 or ft/s2)
Its point of application, magnitude, and direction characterizes a force. A
force that is distributed over a small area of the body upon which it acts
may be considered a concentrated force if the dimensions of the area
involved are small compared with other pertinent dimensions.
Two or more forces can act upon an object without affecting its state of
motion. A book resting on a table has a downward force acting on it
(gravity) and an upward force exerted on it from the tabletop (reaction).
These two forces cancel and the net force on the book is zero. We can
verify this fact by observing that the book is not in motion; there has been
no change in the book’s state of motion.
Knowledge Check
Select all the true statements.
A.
Application of force always causes acceleration.
B.
Force is a vector quantity.
C.
When an object remains at rest, there is no force
applied to the object.
D.
When an object remains at rest, the net force on the
object is zero.
Knowledge Check
In the equation F = ma
52
A.
"a" refers to the acceleration, which has dimensions of
length per unit of time ( e.g., feet per second.)
B.
The equation clearly shows that an object at rest must
have no forces applied to it.
C.
"m" refers to mass, which is essentially the same as
weight.
D.
"F" refers to the total of all forces acting on the object.
Rev 1
Knowledge Check
When an object is at rest,
A.
no force is applied to the object.
B.
the net force applied is not large enough to move the
object.
C.
gravitational force on the object is zero.
D.
the net force, sum of all forces applied to the object, is
zero.
ELO 4.4 Weight
Introduction
Weight is a force exerted on an object due to the object's position in a
gravitational field.
Weight
Weight is a special application of Newton's second law. We define weight
as the force exerted on an object by the gravitational field of the earth, or
more specifically the pull of the earth on the body.
𝑚𝑔
𝑊 = 𝑚𝒈 or 𝑊 =
𝑔𝑐
Where:
W = weight (N or lbf)
m = mass (kg)
g = the local acceleration due to gravity (on earth = 9.8 m/sec2 or 32.17
ft/sec2)
gc = a conversion constant employed to facilitate the use of Newton's
second law of motion with the English system of units, equal to 32.17 ftlbm/lbf-sec2
*Note that gc has the same numerical value as the acceleration due to
gravity at sea level.
The mass of a body is the same, wherever the body is located. The weight
of a body, however, depends upon the local acceleration due to gravity.
Thus, the weight of an object is less on the moon than on earth, because the
local acceleration due to gravity on the moon is less.
Example 1: Calculate the weight of a person with a mass of 84 kg.
𝑾 = 𝑚𝒈
𝑊 = (84 𝑘𝑔) [9.81
𝑚
]
𝑠𝑒𝑐 2
𝑊 = 8.24 × 102 𝑁
Example 2: Calculate the weight of a person with a mass of 84 kg on the
moon. The acceleration due to gravity on the moon is 1.63 m/sec2.
Rev 1
53
𝑾 = 𝑚𝒈
𝑊 = (84 𝑘𝑔) [1.63
𝑚
]
𝑠𝑒𝑐 2
𝑊 = 1.37 × 103 𝑁
If we drop an object, it will accelerate as it falls, even though it is not in
physical contact with any other body. Scientists developed the theory of
gravitational force to explain this concept. The gravitational attraction of
two objects depends upon the mass of each and the distance between them.
Newton's law of gravitation addresses this concept.
Knowledge Check
Select all the true statements.
A.
The mass of a body is the same, wherever the body is
located. The weight of a body, however, depends upon
the local acceleration due to gravity.
B.
Weight is a function of mass only, and does not vary
based on location.
C.
Weight is a vector quantity.
D.
Weight is a scalar quantity.
Knowledge Check
Weight is defined as:
A.
a measure of the size of an object
B.
a measure of the mass of an object
C.
a function of how fast an object will fall to earth
D.
the force exerted on an object by the gravitational field
of the earth
Knowledge Check
Weight is:
54
A.
a vector quantity
B.
a force with no direction
C.
a measure of mass
D.
a scalar quantity
Rev 1
ELO 4.5 Calculating Weight
Introduction
You will learn to calculate the weight of an object in this section.
Calculating Weight
The table below gives instructions for calculating the weight of an object.
Step
Action
1.
Determine which system is being used (English or SI) and choose
the correct formula.
2.
Determine the mass of the object.
3.
Determine the local acceleration due to gravity.
4.
Complete the formula and calculate the weight of the object.
𝑊 = 𝑚𝑔 or 𝑊 =
𝑚𝑔
𝑔𝑐
Where:
W = weight (N or lbf)
m = mass (kg or lbm)
g = the local acceleration of gravity (on earth = 9.8 m/sec2 or 32.17 ft/sec2)
gc = a conversion constant employed to facilitate the use of Newton's
second law of motion with the English system of units, equal to 32.17 ftlbm/lbf-sec2.
*Note that gc has the same numerical value as the acceleration of gravity at
sea level.
Calculating Weight Demonstration
Step
Calculate the weight of an
object with mass of 262 kg.
Result
1.
Determine which system is
being used (English or SI) and
choose the correct formula.
Since the mass is given in kg,
the SI system is being used, and
the formula is 𝑊 = 𝑚𝑔.
2.
Determine the mass of the
object.
The mass is given as 262 kg.
3.
Determine the local
acceleration due to gravity.
Local acceleration due to gravity
is 9.8 m/sec2.
Rev 1
55
Step
Calculate the weight of an
object with mass of 262 kg.
Result
𝑊 = 𝑚𝑔
4.
Complete the formula and
calculate the weight of the
object.
𝑚
)
𝑠𝑒𝑐 2
𝑘𝑔– 𝑚
= 2.6 × 103
𝑠𝑒𝑐 2
3
= 2.2 × 10 𝑁
= (262 𝑘𝑔) (9.8
Knowledge Check
An object weighs 144 pounds-mass on earth. What
would it weigh on a planet with acceleration due to
gravity of 14.9 ft/sec2?
A.
66.6 pounds-mass
B.
66.6 kilograms
C.
2.19 x 102 pounds-mass
D.
2.19 x 102 kilograms
Knowledge Check
An object weighs 230 pounds-mass on earth. What
does it weigh on the moon?
Note: Acceleration due to gravity on the moon is 5.35
ft/sec2
A.
38.3 pounds-mass
B.
43.0 pounds-mass
C.
38.3 kg
D.
43.0 kg
Knowledge Check
Calculate the weight of an object with mass of 2.60 x
101 grams.
56
A.
0.254 pounds-mass
B.
2.54 x 102 N
C.
2.54 x 102 pounds-mass
D.
0.254 N
Rev 1
ELO 4.6 Free-Body Diagrams
Introduction
In studying the effect of forces on a body, it is necessary to isolate the body
and determine all forces acting upon it. This is referred to as the free-body
method and is essential in understanding basic and complex force problems.
Isolate the body in question from all other bodies to ensure a complete and
accurate account of all forces.
The diagram of such an isolated body with the representation of all external
forces acting on it is called a free-body diagram, as shown in the figure
below.
Free-Body Diagrams
Figure: Book on a Table
The figure above shows a book sitting stationary on a table. The book has
two forces acting on it to keep it in a stationary position. One force is the
weight (W) of the book exerting a force downward on the table. The
second force is the normal force (N) exerted upward by the table to hold the
book in place. A normal force is defined as any force that acts
perpendicular to a surface (and could be the normal component of an angled
force). The right side of the figure displays the free-body diagram of the
isolated book and presents the forces acting on it.
Free-Body Diagrams
The table below gives instructions for developing a free-body diagram.
Step
Action
1.
Determine which body is to be isolated. The body chosen will
usually involve one or more of the desired unknown quantities.
2.
Isolate the body chosen with a diagram that represents its complete
external boundaries.
3.
Represent all forces that act on the isolated body in their proper
Rev 1
57
Step
Action
positions in the diagram. Do not show the forces that the object
exerts on anything else, since these forces do not affect the object
itself.
4.
Indicate the choice of coordinate axes directly on the diagram;
include pertinent dimensions for convenience.
The free-body diagram serves the purpose of focusing attention on the
action of the external forces; therefore, do not clutter the diagram with
excessive information. Force arrows should be clearly distinguished from
other arrows to avoid confusion.
When the above steps are completed, a correct free-body diagram will
result, and you will be able to apply the appropriate equations to the
diagram to find the desired information.
Free-Body Diagram Demonstration
Figure: Free-Body Diagram
Consider that a force of some magnitude is towing the car illustrated above
backwards. The diagram above is a free-body diagram showing all the
forces acting on the car. Use the following steps to develop the free-body
diagram:
1. Determine which body is to be isolated. The body chosen will
usually involve one or more of the desired unknown quantities given
by the problem statement.
2. Isolate the body chosen with a diagram that represents its complete
external boundaries. Note that the car is the only body in the diagram
and there are no other connections to the car.
3. Represent all forces that act on the isolated body in their proper
positions in the diagram of the isolated body. Do not show the forces
that the object exerts on anything else, since these forces do not affect
the object itself. Note that arrows indicate all forces acting on the car;
arrows show the direction of each force.Those forces are:
 Fapp - The force applied to tow the car.
58
Rev 1

- The frictional force that opposes the applied force due
to the weight of the car and the nature of the surfaces (the
car's tires and the road surface).
 W - The weight of the car.
 N - The normal force acting on the car.
4. Indicate the choice of coordinate axes directly on the diagram; show
pertinent dimensions for convenience.
5. To solve for the net force acting on the object, assign known values
for each individual force as determined by data given in the problem.
After assigning a sign convention (e.g., positive (+) for forces upward
and to the right, negative (-) for forces downward and to the left), all
forces would be summed to find the net force acting on the body.
Using this net force information and appropriate equations, obtain
solutions for requested unknown values.
Fk
Knowledge Check
Select all the true statements.
Rev 1
A.
If the hanging object is at rest, the net forces applied to
it must be zero.
B.
Cable T2 has no force applied to it.
C.
Cable T2 has a horizontal force pulling the hanging
object to the left but no vertical force applied.
D.
Cable T1 has a vertical force to match the weight of the
hanging object, and a horizontal force pulling the
hanging object to the right.
59
Knowledge Check
A free-body diagram must contain:
A.
the object of interest and all forces acting on the object.
B.
a detailed description of the object in question.
C.
any other information of interest about the object,
whether pertinent to the problem or not.
D.
all physical connections to the object.
Knowledge Check
Assume the crate shown below is sitting on a level
concrete floor. Which of the following correctly
describes the forces applied to the crate?
A.
A horizontal friction force exists which always
precisely matches any other horizontal force on the
crate.
B.
A horizontal friction force is present at all times.
C.
The upward force is the normal force. The downward
force is the weight of the crate. They are equal.
D.
There is no force on the crate. If there was, it would be
in motion.
ELO 4.7 Force Equilibrium
Introduction
Knowledge of the forces required to maintain an object in equilibrium is
essential in understanding the nature of bodies at rest and in motion.
Net Force
When multiple forces act on an object, the result may change the object's
state of motion. If certain conditions are satisfied, however, the forces may
combine to maintain a state of equilibrium or balance. It is necessary to
60
Rev 1
assess the overall effect of all the forces acting on a body to determine if a
body is in equilibrium. All of the forces that act on an object combine into
a single resultant or net force that influences the object's motion. All net
forces are vector quantities. When analyzing various forces you must
account for both the magnitude of the force and the direction in which the
force is applied. As described in the previous chapter, a free-body diagram
is the best analysis tool. To understand this concept more clearly, consider
the book resting on a table in the figure below:
Figure: Net Force Acting on Book on Table
The net force on the book in A above is zero, because the upward (normal)
force exactly balances the weight of the book; hence, with no net force, the
book remains at rest. If a horizontal force acts on the book (see B above)
and we neglect the effect of friction, the net force will be equal to the
horizontal applied force and the book will move in the direction of the
applied force. The free-body diagram in C above shows that the normal
force (N) cancels the weight (W) of the book, since they are equal in
magnitude and opposite in direction. The resultant (net) force is therefore
equal to the applied force (FAPP).
Force Equilibrium
Equilibrium
We consider an object in equilibrium to be in a state of balance, therefore,
the net force on the object is equal to zero. This means that if the vector
sum of all the forces acting on an object is equal to zero, then the object is
in equilibrium. Newton's first law of motion describes equilibrium and the
effect of force on a body that is in equilibrium. The law states that, "An
object at rest will remain at rest or an object moving in a straight line with a
constant velocity will remain so, if the net force on it is zero."
Inertia
Inertia is the tendency of a body to resist a change in its state of motion.
Newton's first law of motion is also called the Law of Inertia.
The First Condition of Equilibrium
The First Condition of Equilibrium, which is a consequence of Newton's
first law, states that, "A body will be in translational equilibrium if and only
if the vector sum of the forces exerted on the body by its environment
equals zero." For example, if three forces act on a body, it is necessary for
the following to be true for the body to be in equilibrium:
Rev 1
61
𝐹1 + 𝐹2 + 𝐹3 = 0
We can simplify this equation as follows:
𝛴𝐹 = 0
This sum includes all forces exerted on the body by its environment. The
vanishing of this vector sum is a necessary condition that must be satisfied,
to ensure translational equilibrium. In three dimensions (x, y, and z), the
component equations of the First Condition of Equilibrium are:
𝛴𝐹𝑥 = 0
𝛴𝐹𝑦 = 0
𝛴𝐹𝑧 = 0
This condition applies to objects in motion with constant velocity and
bodies at rest or in static equilibrium.
Equilibrant
If the sum of all forces acting upon a body is equal to zero; we refer to that
body as in force equilibrium. If the sum of all the forces is not equal to
zero, any force or forces capable of balancing the system is defined as an
equilibrant.
Force Equilibrium Example
The hanging object in the figure below has a weight of 125 kg. Assume
that the object is suspended by cables T1, T2, and T3 as shown. Calculate
the tension (T1) in the cable at 30° with the horizontal.
Figure: Hanging Object
The tension in a cable is the force transmitted by the cable. We can
measure the tension at any point in the cable by removing a suitable length
of the cable and inserting a spring scale:
62
Rev 1
Figure: Free-Body Diagram: Hanging Object
Solution 1:
Since the object and its supporting cables are motionless (i.e., in
equilibrium), the net force acting on the intersection of the cables is zero.
Since the net force is zero, the sum of the x-components of T1, T2, and T3 is
zero and the sum of the y-components of T1, T2, and T3 is zero.
𝛴𝐹𝑥 = 𝑇1𝑥 + 𝑇2𝑥 + 𝑇3𝑥 = 0
𝛴𝐹𝑦 = 𝑇1𝑦 + 𝑇2𝑦 + 𝑇3𝑦 = 0
The tension T3 is equal to the weight of the object, 125 N. The x and y
components of the tensions can be found using trigonometry. Substituting
known values into the second equation above yields the following:
𝛴𝐹𝑦 = (𝑇1 )(sin 30°) + (𝑇2 )(sin 180°) + (𝑇3 )(sin 270°) = 0
𝛴(𝑇1 )(0.5) + (𝑇2 )(0) + (125 𝑁)(−1) = 0
(0.5)(𝑇1 ) − 125 𝑁 = 0
(0.5)(𝑇1 ) = 125 𝑁
𝑇1 = 250 𝑁
Solution 2:
A simpler method to solve this problem involves assigning a sign
convention to the free-body diagram and examining the direction of the
forces. By choosing (+) for the upward direction and (-) for the downward
direction:



The upward component of T1 is +(T1)(sin 30°)
The tension T3 is -125 N
T2 has no y-component
Therefore, using the same equation as before, we obtain the following.
𝛴𝐹𝑦 = (𝑇1 )(sin 30°) − 125 𝑁 = 0
(0.5)(𝑇1 ) = 125 𝑁
𝑇1 = 250 𝑁
Rev 1
63
Knowledge Check
A 900-kg car is accelerating (on a frictionless surface)
at a rate of 2 m/sec2. What force must be applied to the
car to act as an equilibrant for this system?
64
A.
1,800 N
B.
1,800 foot-pounds
C.
8,820 N
D.
8,820 foot-pounds
Rev 1
Knowledge Check
If the hanging object, supported by cables T1, T2 and
T3, has a mass of 200 kg and is in force equilibrium,
what is the force applied by cable T2?
Figure: Hanging Object
Rev 1
A.
200 N
B.
200 foot-pounds
C.
346.4 foot-pounds
D.
346.4 N
65
Knowledge Check
Assuming the crate below is at rest, which of the
following statements about forces on the crate is false?
A.
Any horizontal force applied must be less than the
force caused by static friction, or the crate would move.
B.
The crate is in force equilibrium.
C.
No horizontal force can be applied, or the crate would
not be at rest.
D.
The normal force applied is equal to the force due to
gravity.
TLO 4 Summary
Measurements of Force



A tensile force is a force that tends to pull an object apart.
A compressive force is a force that tends to compress an object.
Frictional force is the force resulting from two surfaces in contact,
where one of the surfaces is attempting to move with respect to the
other surface. The magnitude of the frictional force is affected by the
following:
— Weight of the object being moved
— Type of surface on the object being moved
— Type of surface on which the object is moving
 Static-frictional forces are those frictional forces present when an
object is stationary, whereas kinetic-frictional forces are those
frictional forces present between two objects that are moving.
 Centripetal force is the force on an object moving in a circular path
that is directed towards the center of the path.
 Centrifugal force is the fictitious force that appears to be directed
away from the center of the circular path.
Definition of Force
Force is a vector quantity that tends to produce an acceleration of a body in
the direction of its application.
66
Rev 1
𝐹 = 𝑚𝒂
Weight is the force exerted on an object due to gravity. (On the earth, it is
the gravitational pull of the earth on the body.)
𝑚𝒈
𝑊 = 𝑚𝒈 𝑜𝑟 𝑊 =
𝑔𝑐
(English system)
Free-Body Diagrams
A free-body diagram isolates a body and illustrates all the forces that act on
the body so that a complete and accurate account of all of those forces may
be considered. Constructing a free-body diagram requires the following
four steps:
1. Determine the body or combination of bodies to be isolated.
2. Isolate the body or combination of bodies with a diagram that
represents the complete external boundaries.
3. Represent all forces that act on the isolated body in their proper
positions within the diagram.
4. Indicate the choice of coordinate axes directly on the diagram.
Net Force, Inertia, and Equilibrium





The force that is the resultant force of all forces acting on a body is
defined as the net force.
If the vector sum of all the forces acting on an object is equal to zero,
then the object is in equilibrium.
Inertia is the tendency of a body to resist a change in its state of
motion.
The First Condition of Equilibrium is stated as follows: "A body will
be in translational equilibrium if and only if the vector sum of forces
exerted on a body by the environment equals zero,” or 𝐹1 + 𝐹2 +
𝐹3 = 0 or, 𝛴𝐹 = 0
Any force or system of forces capable of balancing a system so that
the net force is zero is defined as an equilibrant.
Summary
Now that you have completed this lesson, you should be able to:
1.
2.
3.
4.
5.
Rev 1
Describe the following types of forces:
a. Tensile
b. Compressive
c. Frictional
d. Centripetal
e. Centrifugal
Describe the factors that affect the magnitude of the static and kinetic
friction force.
Describe force as it applies to an object and its velocity.
Describe weight as it applies to an object and its velocity.
Given the mass of an object and a value for gravity, calculate the
weight of the object.
67
Describe the purpose of a free-body diagram, and given all necessary
information, construct a free-body diagram.
7. Describe the conditions necessary for a body to be in force
equilibrium.
6.
TLO 5 Motion
Overview
You will learn the physical laws that govern interactions between forces and
moving objects in this section.
Motion
Learning the laws governing moving objects will enable you to understand
the interactions of particles and atoms in the fission process, the way
turbines turn steam expansion into mechanical energy, the causes of water
hammer, and many other important physical phenomena important to power
plant operation.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe Newton's 1st, 2nd, and 3rd laws of motion.
2. Describe Newton's law of universal gravitation.
ELO 5.1 Newton's Laws of Motion
Introduction
Sir Isaac Newton developed the basis for modern mechanics in the
seventeenth century. He formulated three fundamental laws on objects in
motion. The study of Newton's laws of motion allows us to understand and
accurately describe the motion of objects and the forces that act on those
objects.
Newton's Laws of Motion
Newton's First Law
Newton's first law of motion states that, "An object at rest will remain at
rest or an object moving in a straight line with a constant velocity will
remain so, if the net force on it is zero."
Newton's Second Law
Newton's second law states that, "The acceleration of a body is inversely
proportional to its mass and directly proportional to the net (i.e., resultant)
force acting on it; the acceleration acts in the direction of the net force."
This law establishes the relationship between force, mass, and acceleration,
expressed mathematically as:
𝐹 = 𝑚𝑎
Where:
F = force (Newton = 1 kg-m/sec2, or lbf)
68
Rev 1
m = mass (kg or lbm)
a = acceleration (m/sec2 or ft/sec2)
This law defines force (and its units) and is one of the most important laws
in physics. Newton's first law is a consequence of the second law, since an
object experiences no acceleration if there is no net force and it will be
either at rest, or moving at a constant velocity.
Newton's second law (𝐹 = 𝑚𝑎) can be used to calculate an object’s weight
at the surface of the earth. In this special case, weight (W) is the force
caused by the gravitational field of the earth acting on the mass (m) of the
object.
For this case, acceleration is represented by g, which equals 9.8 m/sec2 or
32.17 ft/sec2 (g is referred to as the acceleration due to gravity). Thus, the
equation becomes 𝑊 = 𝑚𝑔.
Newton's Third Law
Newton's third law of motion states, "If a body exerts a force on a second
body, the second body exerts an equal and opposite force on the first." This
law implies that for every action there is an equal and opposite reaction.
Thus, the downward force exerted on a desk by a book is accompanied by
an opposing upward force of equal magnitude and in the opposite direction
exerted on the book by the desk. This principle holds for all forces, variable
or constant, regardless of their source.
Knowledge Check
State Newton's Second Law of Motion.
Rev 1
A.
A particle with a force acting on it has an acceleration
proportional to the magnitude of the force and in the
direction of that force.
B.
For every action, there is an equal but opposite
reaction.
C.
An object remains at rest (if originally at rest) or moves
in a straight line with constant velocity if the net force
on it is zero.
D.
Motion of an object is determined by the size and
shape of the object, not the mass of the object.
69
Knowledge Check
An object is moving at a constant velocity in a straight
line. The net force acting on the object is zero. Which
of the following statements regarding the object is
true?
A.
The object will continue to move in a straight line at a
constant velocity regardless of the net force acting on
the object.
B.
The object will eventually come to a stop, regardless of
the net force acting on the object.
C.
The object will never come to a stop, regardless of the
net force acting on the object.
D.
As long as the net force acting on the object remains
zero, the object will continue to move in a straight line
at a constant velocity.
Knowledge Check
A force of 2,000 N is acting on a car causing it to
accelerate at a constant rate of 2 m/sec2. Determine
the mass of the car.
A.
2,000 kg
B.
1,000 kg.
C.
2,000 pounds
D.
1,000 pounds
ELO 5.2 Universal Gravitation
Introduction
One additional law attributed to Newton concerns the mutual attractive
forces between two bodies. It is known as the universal law of gravitation.
Universal Gravitation
"Each and every mass in the universe exerts a mutual, attractive
gravitational force on every other mass in the universe. For any two
masses, the force is directly proportional to the product of the two masses
and is inversely proportional to the square of the distance between them."
Newton expressed the universal law of gravitation as follows:
𝑚1 𝑚2
𝐹 = 𝐺( 2 )
𝑟
70
Rev 1
Where:
F = force of attraction (Newton [kg-m/sec2] or lbf)
G = universal constant of gravitation (6.67 x 10-11 m3/kg-sec2 or 3.44 x 10-8
ft3/slug-sec2 [note: 1 slug = 32.2 lbm])
m1 = mass of the first object (kg or lbm)
m2 = mass of the second object (kg or lbm)
r = distance between the centers of the two objects (m or ft)
Calculating the Gravitational Acceleration Constant
The value of g (acceleration due to gravity) at the surface of the earth can be
determined, using this universal law of gravitation. First, assume that the
earth is much larger than a given object and the object resides on the surface
of the earth; hence, the value of r will be equal to the radius of the earth.
Second, the force of attraction (F) for the object is equal to the object's
weight (W), as described by Newton's second law (𝑊 = 𝑚𝑔). Setting these
two equations equal to each other yields the following:
𝑊 = 𝐺(
𝑀𝑒 𝑚1
) = 𝑚1 𝑔
𝑟2
Where:
Me = mass of the earth (5.97 x 1024 kg)
m1 = mass of the object
r = radius of the earth (6.371 x 106 m)
The mass of the object (m1) cancels, and the value of g can be determined as
follows:
𝑔 = 𝐺(
𝑀𝑒
)
𝑟2
𝑔 = (6.67 × 10−11
𝑚3
5.97 × 1024 𝑘𝑔
)
𝑘𝑔– 𝑠𝑒𝑐 2 (6.371 × 106 𝑚)2
𝑚
𝑠𝑒𝑐 2
The acceleration due to gravity, g, is not a constant value, but varies with
distance from the earth (i.e., altitude). If the object is at an altitude of 30
km (18.63 mi), then we can calculate the value of g as follows:
𝑔 = 9.81
𝑟 = 30,000 𝑚 + 6.371 × 106 𝑚 = 6.401 × 106 𝑚
𝑔 = (6.67 × 10
−11
𝑚3
5.97 × 1024 𝑘𝑔
)
𝑘𝑔– 𝑠𝑒𝑐 2 (6.401 × 106 𝑚)2
𝑚
𝑠𝑒𝑐 2
An altitude of 30 km (18.63 mi) only changes g from 9.81 m/sec2 to 9.72
m/sec2. There will be an even smaller change for objects closer to the
earth's surface; hence, we normally consider g to be a constant value.
𝑔 = 9.72
Rev 1
71
Knowledge Check
Select all statements about the Law of Universal
Gravitation that are true.
A.
The law applies to every mass in the universe.
B.
The force due to gravity is mutual.
C.
The force due to gravity is directly proportional to the
distance between the bodies.
D.
The force due to gravity is attractive.
Knowledge Check
Two objects are located 10 meters apart. The first
object has a mass of 5.000 x 103 kg. The second object
has a mass of 2.500 x 103 kg. Determine the
gravitational force between these two objects.
A.
8.341 x 10-6 kg-m/sec2
B.
8.341 x 10-5 kg-m/sec2
C.
8.341 x 10-5 m3/kg-sec2
D.
8.341 x 10-6 m3/kg-sec2
Knowledge Check
The Law of Universal Gravitation states that:
72
A.
All masses repel each other with a force that is
inversely proportional to the distance between the
masses.
B.
All masses attract each other with a force inversely
proportional to the product of their masses.
C.
All masses attract each other with a force inversely
proportional to the distance between the masses.
D.
All masses repel each other with a force directly
proportional to the product of their masses.
Rev 1
TLO 5 Summary
Newton's First Law of Motion

An object at rest will remain at rest or an object moving in a straight
line with a constant velocity will remain so, if the net force on it is
zero.
Newton's Second Law of Motion

The acceleration of a body is inversely proportional to its mass and
directly proportional to the net (i.e., resultant) force acting on it; the
acceleration acts in the direction of the net force.
 𝐹 = 𝑚𝑎
Newton's Third Law of Motion

The forces of action and reaction between interacting bodies are equal
in magnitude and opposite in direction.
 For every action there is an equal and opposite reaction.
Newton's Universal Law of Gravitation

Each and every mass in the universe exerts a mutual, attractive
gravitational force on every other mass in the universe. For any two
masses, the force is directly proportional to the product of the two
masses and is inversely proportional to the square of the distance
between them.
𝑚1 𝑚2
 𝐹 = 𝐺( 2 )
𝑟
TLO 6 Momentum
Overview
You will learn the physical laws that govern interactions between forces and
moving objects in this section.
Learning the laws governing moving objects will enable you to understand
the interactions of particles and atoms in the fission process, the way
turbines turn steam expansion into mechanical energy, the causes of water
hammer, and many other important physical phenomena important to power
plant operation.
Objectives
Upon completion of this lesson, you will be able to do the following:
1
2
Rev 1
Describe momentum and conservation of momentum.
Using the conservation of momentum, calculate the velocity of an
object (or objects) following a collision of two objects.
73
ELO 6.1 Momentum and Conservation of Momentum
Introduction
Momentum describes the motion of a moving body. An understanding of
momentum and the conservation of momentum provides essential tools in
solving physics problems.
Momentum
Momentum is the measure of the motion of a moving body. It is the
product of the body's mass and the velocity at which it is moving.
𝑝 = 𝑚𝑣
Where:
p = momentum of the object (kg-m/s or ft-lbm/s)
m = mass of the object (kg or lbm)
v = velocity of the object (m/s or ft/s)
Momentum is a vector quantity, since it results from the velocity of the
object. If we want to add different momentum quantities, we must take into
account the direction of each momentum vector.
Momentum Example
Step
Calculate the momentum for a 7.0-kg bowling ball rolling
down a lane at 7.0 m/s.
1.
𝒑 = 𝑚𝒗
2.
𝑝 = (7.0 𝑘𝑔) (7.0
3.
𝑝 = 49
𝑚
)
𝑠
𝑘𝑔– 𝑚
𝑠
Force and Momentum
There is a direct relationship between force and momentum. Specifically,
the rate at which momentum changes with time (∆p/∆t) is equal to the net
force (F) applied to an object. From Newton's second law of motion, a net
force is also equal to the product of an object's mass (m) and its acceleration
(a), which is the rate of change of its velocity (∆v/∆t).
∆𝑣
∆𝑣
Given that 𝐹 = 𝑚𝑎 and 𝑎 = ( ∆𝑡 ), then, 𝐹 = 𝑚 ( ∆𝑡 ). As 𝑝 = 𝑚𝑣 (and
∆𝑝
∆𝑝 = 𝑚∆𝑣, for a constant-mass object), then 𝐹 = ∆𝑡 . Hence, as stated
above, the net force on an object is equal to the rate of change of its
momentum.
Force and Momentum Example
Problem statement:
74
Rev 1
The velocity of a rocket must reach 35 m/s to achieve proper orbit around
the earth. If the rocket has a mass of 5.00 x 103 kg and it takes 9 seconds to
reach orbit, what is the required thrust (force) to achieve this orbit?
Follow the steps in the table below to solve the problem.
Step
Approach
1.
Assuming that the initial velocity
∆𝑣
is zero, the change in velocity
𝐹 = 𝑚( )
∆𝑡
(∆v) is 35 m/s.
2.
3.
4.
Result
Enter all known quantities.
𝑚
35 𝑠
𝐹 = (5.00 × 103 𝑘𝑔)
9𝑠
Calculate.
𝐹 = 1.94 × 104
Convert units.
𝐹 = 1.94 × 104 𝑁
𝑘𝑔– 𝑚
𝑠2
Conservation of Momentum
The momentum of an object that experiences no net force is always
conserved. In other words, if no net external force acts upon an object, the
momentum of the object remains constant (i.e., ∆𝑝 = 0).
The conservation of momentum is important, when solving problems
involving collisions, explosions, etc., where the external force is negligible
and the problem states that the momentum before and after the event are
equal.
For example, the conservation of momentum applies when firing a bullet
from a gun. Prior to firing the gun, both the gun and the bullet are at rest
(i.e., vG and vB are zero) and, therefore, the total momentum is zero. This
can be written as follows:
𝑚𝐺 𝑣𝐺 + 𝑚𝐵 𝑣𝐵 = 0
or
𝑚𝐺 𝑣𝐺 = −𝑚𝐵 𝑣𝐵
Upon firing the gun, the momentum of the recoiling gun is equal and
opposite to that of the bullet; i.e., the momentum of the bullet (mBvB) is
equal to the momentum of the gun (mGvG), but opposite in direction.
Elastic and Inelastic Collisions
The conservation of momentum is independent of whether or not a collision
is elastic or inelastic. In an elastic collision, both momentum and kinetic
energy (i.e., the energy due to an object's velocity; see below) are
conserved. A common example of an elastic collision is the head-on
collision of two billiard balls.
Rev 1
75
In an inelastic collision, momentum is conserved, but system kinetic energy
is not. An example of an inelastic collision is the head-on collision of two
automobiles, in which part of the initial kinetic energy is lost (as the metal
distorts on impact). Although kinetic energy is not conserved in an inelastic
collision, the total system energy is invariably conserved; this concept is
captured in the law of conservation of energy.
Law of Conservation of Momentum
Mathematically, we can express the law of conservation of momentum in
several different ways. The general statement is that the sum of a system's
initial momentum is equal to the sum of a system's final momentum.
𝛴𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝛴𝑝𝑓𝑖𝑛𝑎𝑙
In the case where a collision of two objects occurs, the equation expressing
the conservation of momentum is as follows:
𝑝1 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑝2 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝1 𝑓𝑖𝑛𝑎𝑙 + 𝑝2 𝑓𝑖𝑛𝑎𝑙
or
(𝑚1 𝑣1 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + (𝑚2 𝑣2 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑚1 𝑣1 )𝑓𝑖𝑛𝑎𝑙 + (𝑚2 𝑣2 )𝑓𝑖𝑛𝑎𝑙
In the case where two bodies collide and have identical final velocities, the
equation below applies:
(m1 v1 + m2 v2 )initial = (m1 + m2 )vfinal
For example, consider two railroad cars rolling on a level, frictionless track
(see figure below). The cars collide, become coupled, and roll together at a
final velocity (vfinal). We express the momentum before and after the
collision as:
(𝑚1 𝑣1 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + (𝑚2 𝑣2 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑚1 𝑣1 )𝑓𝑖𝑛𝑎𝑙 + (𝑚2 𝑣2 )𝑓𝑖𝑛𝑎𝑙
If we know the initial velocities of the two objects (v1 and v2), we can
calculate the final velocity (vfinal) by rearranging the above equation:
𝑣𝑓𝑖𝑛𝑎𝑙 =
(𝑚1 𝑣1 + 𝑚2 𝑣2 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑚1 + 𝑚2
Figure: Momentum
76
Rev 1
Knowledge Check
Define momentum.
A.
A measure of the motion of a moving body. It is the
result of the product of the body's mass and the
velocity at which it is moving.
B.
A measure of the velocity of a moving body.
C.
A measure of the mass of a moving body.
D.
A measure of the density of a moving body.
Knowledge Check
Calculate the momentum of a 3,500-pound car
traveling at 55 miles per hour.
A.
621.133 kg-ft/sec.
B.
621,133 kg-ft/sec.
C.
3,106 kg-ft/sec.
D.
677,600 kg-ft/sec.
Knowledge Check
Select all the true statements.
A.
Momentum is conserved in all collisions.
B.
Momentum is conserved in elastic collisions, but not in
inelastic collisions.
C.
Momentum of an object may be changed by applying
force to the object.
D.
Momentum is a vector quantity.
Knowledge Check
Which of the following statements describes the
conservation of momentum?
Rev 1
A.
The sum of a system's initial momentum is equal to the
sum of a system's final momentum.
B.
The force acting on an object is a product of the
object's mass and its acceleration.
C.
For every action, there is an equal and direct reaction.
D.
The sum of a system's initial momentum is greater than
the sum of a system’s final momentum.
77
Knowledge Check
Select the equation that correctly describes the
conservation of momentum, in the case where a
collision of two objects occurs.
B.
𝛥𝑃
𝛥𝑡
p = mv
C.
F = ma
D.
(m1v1)initial + (m2v2)initial = (m1v1)final + (m2v2)final
A.
𝐹=
ELO 6.2 Collisions and Velocity
Introduction
In this section, you will learn to calculate the final velocities of objects after
collisions.
Collisions and Velocity
The table below provides instructions for calculating velocities of objects
after elastic collisions.
Step
Action
1.
Draw the interaction, including known masses and velocities of all
objects, before and after the collision.
2.
Determine which directions are positive (e.g., up and to the right)
in the problem.
3.
Organize the known information, including masses and velocities
of all objects, before and after the collision.
4.
Determine the unknown that you must find.
5.
Use the formulas given below to determine the solution.
Formula to use if objects are together after the collision:
(𝑚1 𝑣1 + 𝑚2 𝑣2 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑚1 + 𝑚2
Formula to use if objects are separate after the collision:
𝑣𝑓𝑖𝑛𝑎𝑙 =
(𝑚1 𝑣1 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + (𝑚2 𝑣2 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑚1 𝑣1 )𝑓𝑖𝑛𝑎𝑙 + (𝑚2 𝑣2 )𝑓𝑖𝑛𝑎𝑙
Collisions and Velocity Demonstration
Consider the two railroad cars shown in the figure below.
78
Rev 1
Figure: Collision of Railroad Cars
Problem statement:
The railroad cars above have masses m1 = 1,000 kg and m2 = 1,300 kg. The
first car is moving at a velocity of 9.0 m/s and the second car is moving at a
velocity of 4.0 m/s. The first car overtakes the second car and couples with
it. Calculate the final velocity of the coupled cars.
Step
Approach
Result
1.
The final velocity
(vfinal) can be
calculated, using
the appropriate
formula.
(𝑚1 𝑣1 + 𝑚2 𝑣2 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑚1 + 𝑚2
𝑣𝑓𝑖𝑛𝑎𝑙
2.
Enter the given
quantities.
3.
𝑣𝑓𝑖𝑛𝑎𝑙 =
Calculate the final
velocity.
𝑚
𝑚
[(1,000 𝑘𝑔) (9.0 𝑠 ) + (1,300 𝑘𝑔) (4 𝑠 )]
=
1,000 𝑘𝑔 + 1,300 𝑘𝑔
𝑣𝑓𝑖𝑛𝑎𝑙 = 6.2
𝑚
𝑠
Knowledge Check
An object with mass of 1.0 x 102 grams is traveling at
50 m/s when it strikes an object with mass of 10 kg,
which is at rest. Assuming that the first object is at rest
after the elastic collision, what is the velocity and
direction of the second object?
Rev 1
A.
0.50 m/s, in the same direction that the first object was
moving
B.
0.50 m/s, in the opposite direction that the first object
was moving
C.
5.0 m/s, in the same direction that the first object was
moving
D.
5.0 m/s in the opposite direction that the first object
was moving
79
Knowledge Check
Two moving objects collide. The first object was
traveling at 5.00 m/s and has a mass of 5 kg. The
second object has a mass of 2 kg. At what velocity and
direction must the second object have been traveling,
for the combined objects to be at rest after the
collision?
A.
25.0 m/s , in the same direction as the first object.
B.
12.5 m/s, perpendicular to the first object.
C.
25.0 m/sec, in the opposite direction of the first object.
D.
12.5 m/s, in the opposite direction of the first object.
Knowledge Check
Two railroad cars have masses of m1 = 1,500 kg and
m2 = 1,250 kg. The first car is moving at a velocity of
7.0 m/s and the second car is moving at a velocity of
5.0 m/s. The first car overtakes the second car and
couples with it. Calculate the final velocity of the two
cars.
A.
6.1 m/s
B.
5.9 m/s
C.
6.2 m/s
D.
6.0 m/s
TLO 6 Summary
Momentum

Momentum is the measure of the motion of a moving body. It is
equal to the body's mass multiplied by the velocity at which it is
moving.
 Therefore, momentum can be defined as: 𝑝 = 𝑚𝑣
The Law of Conservation of Momentum

The Law of Conservation of Momentum states that if no net external
force acts upon a system, the momentum of the system remains
constant. If the net force (F) is equal to zero, then ∆p = 0.
 The momentum before and after a collision can be calculated using
the following equation:
(𝑚1 𝑣1 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + (𝑚2 𝑣2 )𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑚1 𝑣1 )𝑓𝑖𝑛𝑎𝑙 + (𝑚2 𝑣2 )𝑓𝑖𝑛𝑎𝑙
Summary
Now that you have completed this lesson, you should be able to:
1. Describe momentum and conservation of momentum.
80
Rev 1
2. Using the conservation of momentum, calculate the velocity of an
object (or objects) following a collision of two objects.
TLO 7 Energy, Work, and Power
Overview
In this section, you will learn the concepts of energy, work, and power.
The concepts of energy, work, and power are vital to understanding the
operation of power plant components and phenomena that occur in power
plant operation. Understanding these concepts enables the operator to
foresee consequences of different actions.
Objectives
Upon completion of this lesson, you will be able to do the following:
1
2
3
4
5
Describe the following terms:
a. Energy
b. Potential energy
c. Kinetic energy
d. Work
Calculate energy and work for a mechanical system.
State the First Law of Thermodynamics, "Conservation of
Energy."
State the mathematical expression for power.
Calculate power in a mechanical system.
ELO 7.1 Energy Definitions
Introduction
Energy is the measure of a system's (or object's) ability to do work or cause
a change. It is not possible to create or destroy energy. It is possible to
transfer energy from one system (or object) to another or transform energy
from one type of energy to another. Work is a measure of the amount of
energy required to move an object.
Energy Definitions
Energy is the capacity of a system or object to perform work and may be
stored in various forms, such as potential energy and/or kinetic energy
(otherwise referred to as mechanical forms of energy).
Complex systems may include other types of energy, such as chemical,
electromagnetic, thermal, acoustic, and/or nuclear. A pile-driver hammer
performs work by virtue of its motion. Coal burned in a fossil-fueled power
plant releases chemical energy. Fuel elements in a nuclear reactor produce
energy by nuclear reactions. For the purposes of this lesson, the discussion
will be limited to mechanical and thermal (e.g., heat) forms of energy.
However, the principles involved with energy calculations are similar for all
types of energy.
Rev 1
81
Transient and Stored Energy
We separate both thermal and mechanical energies into two categories:
transient and stored. Transient energy is energy in motion (i.e., energy that
moves from one place to another). Stored energy is the energy contained
within a substance or object.
Potential Energy
Potential energy is the energy stored in an object, due to its position within
a force field. An example is the potential energy of an object above the
surface of the earth, situated within the earth's gravitational force field.
Potential energy also applies to energy originating from the separation of
electrical charges and that stored in a spring. Upon burning, the potential
energy stored in hydrogen and oxygen atoms is released to form water.
When quantifying potential energy, the position of an object (relative to
other objects or a reference point) must be determined. Typically, the
measure of an object's position is its vertical distance above a reference
point, which is normally the earth's surface.
The potential energy of an object represents the work that would be
required to elevate the object to some position above the reference point.
The equation below presents the mathematical expression for potential
energy:
𝑚𝑔𝑧
𝑃𝐸 = 𝑤𝑒𝑖𝑔ℎ𝑡 × ℎ𝑒𝑖𝑔ℎ𝑡 = (
)
𝑔𝑐
Where:
PE = potential energy in N-m (Joule) or ft-lbf
m = mass in kg or lbm
g = 9.81 m/s2 (32.2 ft/s2)
gc = 32.2 (lbm-ft)/(lbf-s2)
z = height above the reference point in m or ft
gc is only used in the English system of measurement.
Note
Kinetic Energy
Kinetic energy is the energy stored in an object, due to its motion. For
example, a stationary baseball has no kinetic energy, because it is not
moving. When someone throws the ball, it will acquire kinetic energy and
will exhibit some velocity.
Velocity is a measure of motion; we can calculate kinetic energy in terms of
its velocity, as shown below:
82
Rev 1
𝐾𝐸 =
𝑚𝑣 2
𝑚𝑣 2
or 𝐾𝐸 =
2
2𝑔𝑐
Where:
KE = kinetic energy in N-m (Joule) or ft-lbf)
m = mass in kg or lbm
v = velocity in m/s or ft/s
gc = 32.2 lbm-ft/lbf-s2
The kinetic energy of an object represents the amount of energy that is
required to increase its velocity from rest (v = 0) to its final velocity.
Thermal and Mechanical Energy
Thermal Energy
Thermal energy is related to temperature; the higher an object's temperature,
the greater its molecular movement, and the greater its thermal energy. If
one object has more thermal energy than an adjacent one, the higher
temperature object will transfer thermal energy (at an atomic or molecular
level) to the other. Because the energy is in motion, we refer to the energy
as transient energy (or, in this case, heat).
Internal Energy
Internal energy is another term for thermal energy; i.e., it is the energy
stored in a substance, due to the motion of its composite particles (atoms or
molecules).
Mechanical Energy
Mechanical energy is the energy related to an object's motion or position.
We commonly refer to transient mechanical energy as work. As discussed
previously, stored mechanical energy exists in one of two forms: kinetic or
potential.
Work
Most people think of work as any activity that requires exertion. In physics,
however, the definition is more specific; work results from a force acting on
an object, and only if the object exhibits motion in the direction of the
applied force. For example, if a force acts on a box (such as pushing it) and
it moves some distance, the force has performed work on the box. If a force
acts on the box and the box does not move, then the force has not performed
work (as defined in physics) on the box.
The equation below provides the mathematical definition of work:
𝑊 = 𝐹𝑑
Where:
W = work done in Joules (J) or ft-lbf
F = force applied to the object in N or lbf
d = distance the object is moved (in the direction of the applied force) in m
or ft
Rev 1
83
Knowledge Check
Match the following terms with the appropriate definitions.
1 Energy stored in an object because of
its motion
A.
Work
2 The energy stored in an object because
of its position
B.
Thermal energy
3 Force applied through a distance
C.
Kinetic energy
4 That energy related to temperature
D.
Potential energy
Knowledge Check
The ability to do work is known as:
A.
power
B.
kinetic energy
C.
energy
D.
potential energy
Knowledge Check
Potential energy is:
84
A.
energy due to position or elevation.
B.
the capability of an object or substance to do work if
utilized
C.
a possible future energy source
D.
energy that could be applied to perform work
Rev 1
ELO 7.2 Calculating Energy and Work
Introduction
In this section, you will learn to calculate energy and work.
Calculating Potential Energy
The table below provides instructions for calculating potential energy.
Step
Action
1.
Determine whether the problem uses the English or SI system.
2.
Determine the height above the reference point and the mass of the
object in question.
3.
Ensure all units are consistent. Convert to appropriate units, if
necessary.
4.
Apply the appropriate formula: 𝑃𝐸 = 𝑚𝑔𝑧 (SI) or 𝑃𝐸 =
𝑚𝑔𝑧
𝑔𝑐
(English)
5.
Calculate the potential energy.
Calculating Kinetic Energy
The table below provides instructions for calculating kinetic energy.
Step Action
1.
Determine whether the problem uses the English or SI system.
2.
Determine the velocity and mass of the object in question.
3.
Ensure all units are consistent. Convert to appropriate units, if
necessary.
4.
Apply the appropriate formula: 𝐾𝐸 =
𝑚𝑣 2
2𝑔𝑐
(English) or 𝐾𝐸 =
𝑚𝑣 2
2
(SI)
5.
Rev 1
Calculate the kinetic energy.
85
Calculating Work
The table below provides instructions for calculating work.
Step
Action
1.
Determine whether the problem uses the English or SI system.
2.
Determine the force applied and the distance through which the
force acts on the object.
3.
Ensure all units are consistent. Convert to appropriate units, if
necessary.
4.
Use the formula to calculate the work done: 𝑊 = 𝐹𝑑
Calculating Potential Energy Demonstration
The table below shows the steps in calculating potential energy for the
following:
What is the potential energy of a 50-kg object suspended 10.0 m above the
ground?
Step
Approach
Result
1.
Determine the appropriate
formula.
𝑃𝐸 = 𝑚𝑔𝑧
2.
Replace the variables with
the appropriate values.
𝑃𝐸 = (50 𝑘𝑔)(9.81
𝑚
)(10.0 𝑚)
𝑠2
𝑃𝐸 = 4.90 × 102 𝑁– 𝑚
3.
Using the order of
operations, calculate the PE.
or,
4.90 × 102 𝐽
Calculating Kinetic Energy Demonstration
The table below shows the steps in calculating kinetic energy for the
following:
What is the kinetic energy of a 10-kg object that has a velocity of 8.0 m/s?
Step
Approach
Result
1.
Determine the appropriate
formula.
𝑚𝑣 2
𝐾𝐸 =
2
86
Rev 1
Step
Approach
2.
Replace the variables with
the appropriate values.
3.
Using the order of
operations, calculate the
KE.
Result
𝐾𝐸 = (10
𝑘𝑔
𝑚 2
) (8.0 )
2
𝑠
𝐾𝐸 = 3.2 × 102 𝐽
Calculating Work Demonstration
The table below shows the steps in calculating work for the following
example:
An individual pushes a large box for three minutes. During that time, the
person exerts a constant force of 200 N on the box, but it does not move.
How much work has the person accomplished?
Step
Approach
Result
1.
Determine the appropriate
formula.
𝑊 = 𝐹𝑑
2.
Replace the variables with the
appropriate values.
𝑊 = (200 𝑁)(0 𝑚)
3.
Using the order of operations,
calculate the work.
𝑊 = 0 𝑁– 𝑚 or 𝑊 = 0 𝐽
If the force does not produce movement, the force accomplishes no work.
The table below shows the steps in calculating work for the following
example:
An individual pushes a large box for three minutes. During that time, the
person applies a horizontal force of 200 N to the box, and the box moves
5.0 m horizontally. How much work has the person accomplished?
Step
Approach
Result
1.
Determine the appropriate
formula.
𝑊 = 𝐹𝑑
2.
Replace the variables with the
appropriate values.
𝑊 = (200 𝑁)(5.0 𝑚)
3.
Using the order of operations,
calculate the work.
𝑊 = 1.0 × 103 𝑁– 𝑚
or,
1.0 × 103 𝐽
Rev 1
87
Knowledge Check
Match the terms below with the appropriate formula.
1 1
𝑚𝑣 2
2
A.
Work
2 mgz
B.
Momentum
3 Fd
C.
Kinetic energy
4 mv
D.
Potential energy
Knowledge Check
Calculate the kinetic energy of a 500-kg car traveling at
8.00 x 101 km/hr.
88
A.
1.23 x 105 N
B.
1.23 x 105 J
C.
3.20 x 106 J
D.
3.20 x 106 N
Rev 1
Knowledge Check
Calculate the potential energy of 2,000 kg of water
which is being stored in a water tank 15 meters above
ground level.
A.
30,000 Joules
B.
294,000 Joules
C.
30,000 Newtons
D.
294,000 Newtons
Knowledge Check
A force of 5.00 x 101 N is applied to an object through
a distance of 16 feet. Calculate the amount of work
done.
A.
8.00 x 102 J
B.
8.00 x 102 ft-lbf
C.
2.44 x 102 ft-lbf
D.
2.44 x 102 J
ELO 7.3 Conservation of Energy
Introduction
In this section, you will learn the law of conservation of energy.
Conservation of Energy
The first law of thermodynamics, or the law of conservation of energy,
states, "Energy cannot be created or destroyed, only altered in form."
As discussed previously, potential energy is a measure of the force that we
must apply to an object, to raise it from a reference point to another height.
The energy (or work) expended in raising the object is equivalent to the
potential energy gained by the object, due to its height. This is an example
of a transfer of energy and a conversion of the type of energy.
Another example is a person throwing a baseball. While the ball remains in
the person’s hand, it contains no kinetic energy. When the person throws
the ball, the person applies a force to it and the ball acquires kinetic energy;
the amount of kinetic energy the ball contains is equivalent to the work the
person expended in throwing it.
The following simplified equation describes this transfer of energy (E)
mathematically:
𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝐸𝑎𝑑𝑑𝑒𝑑 − 𝐸𝑟𝑒𝑚𝑜𝑣𝑒𝑑 = 𝐸𝑓𝑖𝑛𝑎𝑙
Rev 1
89
Where:

Einitial is the energy initially stored in an object/system. This energy
generally exists in a combination of kinetic energy and potential
energy.
 Eadded is the energy added to the object/system. Heat may be added to
or work done on the object/system.
 Eremoved is energy removed from an object/system. Heat may be
rejected from or work done by the object/system.
 Efinal is the energy remaining within the object/system after all energy
transfers and transformations occur.
Simplified Energy Balance
To expand upon the above equation, each component can be broken down
as follows:




𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐾𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑃𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙
Eadded = work done on and/or heat added to the object/system
Eremoved = work done by and/or heat removed from the object/system
𝐸𝑓𝑖𝑛𝑎𝑙 = 𝐾𝐸𝑓𝑖𝑛𝑎𝑙 + 𝑃𝐸𝑓𝑖𝑛𝑎𝑙
The equation below shows the resulting energy balance:
𝐾𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑃𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝐸𝑎𝑑𝑑𝑒𝑑 − 𝐸𝑟𝑒𝑚𝑜𝑣𝑒𝑑 = 𝐾𝐸𝑓𝑖𝑛𝑎𝑙 + 𝑃𝐸𝑓𝑖𝑛𝑎𝑙
Neglecting any heat removed from or added to the object/system, we obtain
the following equation:
𝐾𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑃𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑊𝑜𝑛 = 𝐾𝐸𝑓𝑖𝑛𝑎𝑙 + 𝑃𝐸𝑓𝑖𝑛𝑎𝑙 + 𝑊𝑏𝑦
Many sources refer to the above equation as the simplified energy balance
and the equation is an expression the law of conservation of energy. The
simplified equation applies only to problems involving work and
mechanical energy (i.e., kinetic and potential energies), since heat is
neglected.
Knowledge Check
Match the terms with the appropriate definitions.
1 Energy cannot be created or destroyed,
only altered in form.
A.
Kinetic energy
2 𝐾𝐸1 + 𝑃𝐸1 + 𝐸𝑎𝑑𝑑𝑒𝑑
= 𝐾𝐸2 + 𝑃𝐸2 + 𝐸𝑟𝑒𝑚𝑜𝑣𝑒𝑑
B.
Conservation of
energy
3 energy due to position
C.
Potential energy
4 energy due to motion
D.
Energy balance
90
Rev 1
Knowledge Check
The law of conservation of energy states:
A.
Energy should not be wasted
B.
The sum of potential energy and kinetic energy is
always the same
C.
Energy cannot be altered in form.
D.
Energy cannot be created or destroyed
Knowledge Check
Which of the following is not a term in the simplified
energy balance?
A.
Energy added
B.
Wasted energy
C.
Kinetic energy
D.
Potential energy
ELO 7.4 Power
Introduction
Power is a measure of the rate of application or consumption of energy.
Thermal power refers to the transfer of heat; mechanical power refers to
work being performed on or by an object/system.
Power
Power is defined as the amount of heat transferred per unit time or the rate
of doing work and is measured in units of Joules/sec (also known as Watts;
1 J/s = 1 watt (W), British Thermal Units (BTU), horsepower (hp), or ftlbf/s.
Thermal power is the measure of thermal energy transferred per unit time
(i.e., the rate of heat flow). Typical thermal power units include BTU and
kilowatts (kW).
Calculate thermal power using the following expression:
𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 =
ℎ𝑒𝑎𝑡 𝑢𝑠𝑒𝑑
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
Mechanical energy transferred per unit time (i.e., the rate at which work is
done) is referred to as mechanical power. Typical mechanical power units
are units of J/s or W (in the MKS system) and ft-lbf/s or hp (in the English
system).
We can calculate mechanical power using the following expression:
Rev 1
91
𝑃𝑜𝑤𝑒𝑟 =
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
We can define work as the product of force and distance, therefore, the
following equation applies:
𝐹𝑑
𝑡
Where:
𝑃=
P = power in W or ft-lbf/s
F = force in N or lbf
d = distance in m or ft
t = time in seconds
In the English system of measurement, horsepower is a commonly used
term for describing the power ratings of pumps, motors, and engines (1 hp =
550 ft-lbf/s or 745.7 W).
Since distance travelled per unit time is equivalent to velocity, an alternate
description of power is as follows:
𝐹𝑣
550
Where:
𝑃=
P = power in hp
F = force in lbf
v = velocity in ft/s
The above equations for power assume that force and velocity are constant;
otherwise, you must use average values of force and velocity.
Knowledge Check
Match the following terms to the appropriate definitions.
1 energy used per unit time
A.
Horsepower
2 J/s
B.
Thermal power
3 550 ft-lbf/s
C.
Watt
4 heat used per unit time
D.
Power
92
Rev 1
Knowledge Check
Which of the following is NOT an accurate definition
of power?
A.
Force times distance
B.
Rate of doing work
C.
Energy used per unit of time
D.
Heat used per unit of time
Knowledge Check
Which of the following is NOT a unit used to express
power in mechanical systems?
A.
ft-lbf/s
B.
Horsepower
C.
Newtons
D.
Watts
ELO 7.5 Calculating Power
Introduction
In this section, you will learn to calculate power in mechanical systems.
Calculating Power
The table below provides instructions for calculating power.
Step
Action
1.
Determine the information known and information needed.
2.
Ensure that units are consistent. Convert units as necessary.
3.
Using the appropriate formula below, calculate power.
𝑃=
𝐹𝑑
𝑡
𝑃𝑜𝑤𝑒𝑟 =
𝑃=
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
𝐹𝑣
550
Rev 1
93
Calculating Power Demonstration
A pump provides a flow rate of 10,000 liters per minute and performs 1.5 x
108 J of work every 100 minutes. What is the power of the pump?
Solution:
𝑃𝑜𝑤𝑒𝑟 =
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
1.5 × 108 𝐽𝑜𝑢𝑙𝑒𝑠 1 𝑚𝑖𝑛
𝑃=(
)(
)
100 𝑚𝑖𝑛
60 𝑠𝑒𝑐
𝑃 = 2.5 × 104 𝑊𝑎𝑡𝑡𝑠
Knowledge Check
A man constantly exerts 50.0 lbf to move a crate 50
feet across a level floor. The move requires 10
seconds. What is the power expended by the man
pushing the crate?
A.
0.450 horsepower
B.
2.50 x 103 foot-pounds force
C.
4.50 horsepower
D.
2.50 x 103 watts
Knowledge Check
A boy rolls a ball with a steady force of 1.0 lbf, giving
the ball a constant velocity of 5 ft/s. What is the power
expended by the boy in rolling the ball?
94
A.
9.0 x 10-3 watts
B.
9.0 x 10-2 watts
C.
9.0 x 10-3 horsepower
D.
9.0 x 10-2 horsepower
Rev 1
Knowledge Check
A race car is traveling at a constant velocity and goes
one-quarter mile (1,320 feet) in 5 seconds. If the motor
is generating a forward force of 1.89 x 103 lbf on the
car, what is the power of the motor in hp?
A.
1.00 x 103 watts
B.
5.00 x 103 watts
C.
9.07 x 102 horsepower
D.
5.00 x 103 horsepower
TLO 7 Summary
Energy, Work, and Power
 Energy is the ability to do work. The work done on an object is the
product of the force on the object and the distance the object moves
(in the direction of the force).
 Kinetic energy is the energy an object has, due to its motion.
𝑚𝑣 2
𝑚𝑣 2

𝐾𝐸 =

Potential energy is the energy an object has, due to its position.
𝑚𝑔𝑧
 𝑃𝐸 = 𝑚𝑔𝑧 or 𝑃𝐸 = 𝑔

2
or 𝐾𝐸 =
2𝑔𝑐
𝑐
The law of conservation of energy: Energy cannot be created or
destroyed, only altered in form.
 Simplified energy balance:
 𝐾𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑃𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝐸𝑎𝑑𝑑𝑒𝑑 = 𝐾𝐸𝑓𝑖𝑛𝑎𝑙 + 𝑃𝐸𝑓𝑖𝑛𝑎𝑙 + 𝐸𝑟𝑒𝑚𝑜𝑣𝑒𝑑
 Power is the amount of energy used per unit time.
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
 𝑃𝑜𝑤𝑒𝑟 =
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
Summary
Now that you have completed this lesson, you should be able to:
1. Describe the following terms:
a. Energy
b. Potential energy
c. Kinetic energy
d. Work
2. Calculate energy and work for a mechanical system.
3. State the First Law of Thermodynamics, "Conservation of Energy."
4. State the mathematical expression for power.
5. Calculate power in a mechanical system.
Rev 1
95
Physics Summary
The physics module covered the concepts of force, motion, momentum,
energy, work, and power. The rest of this course will use these concepts. It
is necessary for you to understand these concepts in order to understand
power plant equipment. These concepts are also key to your ability to
operate power-plant equipment safely and efficiently.
Now that you have completed this module, you should be able to
demonstrate mastery of this topic by passing a written exam with a grade of
80 percent or higher on the following TLOs:
1. Convert between units of measure associated with the English and
System Internationale (SI) measuring systems.
2. Describe vector quantities, and how they are represented.
3. Solve resultant vector problems.
4. Describe the measurement of force and its relationship to free body
diagrams.
5. Apply Newton's laws of motion to a body at rest.
6. Calculate the change in velocity when two objects collide, with
respect to the conservation of momentum.
7. Describe energy, work, and power in mechanical systems.
96
Rev 1