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PHYSICS AND TECHNOLOGY I
INDTT 170
OLYMPIC COLLEGE
Name____________________
Date_____________________
PHYSICS AND TECHNOLOGY I
Jack Kinert
Bob Abel
Olympic College
Bremerton, WA
Copyright 2015
The Video version of the textbook is on-line. Go to
http://www.youtube.com/olympiccollege. Scroll down to where it says “Physics with Bob
Abel”. Click on the words (not the picture!) and you’ll find the entire list of Physics
videos in chronological order. Try to figure out which video was later made into a full
length movie with Jennifer Lawrence.
The textbook and the lab directions may be found on-line at
instructors.olympic.edu/psnsphysics/default.aspx. You’ll find the text under “PDF Files”
and the lab directions under “Links”.
Tutoring assistance is available Tuesdays and Thursdays after work in the Physics lab
(around 4:20). The main campus also has physics tutoring services.
.
Access Services coordinates accommodations for eligible students with disabilities and
works to ensure equal access to educational programs, services, and activities at Olympic
College.If you are in need of and/or eligible for such assistance, call them at 360-5757540 or visit their website at
www.olympic.edu/Students/StudentServices/AccessServices. Students may also take
advantage of the excellent Counseling Services on campus.
Acknowledgments
The authors gratefully acknowledge the assistance of Applied Physics Faculty and Staff
and many helpful Apprentices in the preparation of this text.
INDTT 170 - INTRODUCTION TO APPLIED PHYSICS
TABLE OF CONTENTS
I.
INTRODUCTION
1. SURVIVAL SKILLS
2. CLASS SYLLABUS
II. SOME FUNDAMENTALS
i
i
v
1
1. SIGNIFICANT FIGURES
2. POWERS OF TEN
3. SCIENTIFIC NOTATION
4. DIMENSIONAL ANALYSIS
5. FUNDAMENTAL MEASUREMENTS
6. SYSTEMS OF UNITS
7. SCALARS AND VECTORS
PROBLEM SET 1: FUNDAMENTALS
INTRODUCTORY EXPERIMENTS
1
1
2
3
4
5
6
7
8
III. FORCE AND FORCE-LIKE QUANTITIES
12
1. MECHANICAL TRANSLATIONAL SYSTEMS: FORCE
PROBLEM SET 2: FORCE
LAB #1: FORCES IN EQUILIBRIUM
12
25
26
2. MECHANICAL ROTATIONAL SYSTEMS: TORQUE
PROBLEM SET 3: TORQUE
LAB #2: TORQUE
31
35
36
3. DENSITY AND SPECIFIC GRAVITY
PROBLEM SET 4: DENSITY/SPECIFIC GRAVITY
LAB #3: MEASURING SPECIFIC GRAVITY AND DENSITY
40
45
46
4. FLUID ENERGY SYSTEMS: PRESSURE
a) Atmospheric Pressure
b) Absolute and Gauge Pressure
c) Pressure in a Liquid
d) Hydraulic Lift
e) Equilibrium in Fluid Systems
f) Measuring Pressures
PROBLEM SET 5: PRESSURE
LAB #4: MEASURING PRESSURE
50
50
51
53
54
55
56
58
60
5. ELECTRICAL ENERGY SYSTEMS
a) Introduction
b) Force in Electrical Systems
c) The Electric Field
d) Voltage
e) Electrical Circuits
PROBLEM SET 6: VOLTAGE/CIRCUITS
LAB #5: MEASURING VOLTAGES
LAB #6: ELECTRICAL CIRCUITS
f) Generating Magnetic Fields
g) Force Produced by a Magnetic Field
h) Solenoids
PROBLEM SET 7: SOLENOIDS
LAB #7: SOLENOID OPERATION
64
64
64
66
67
68
70
72
77
81
82
83
85
86
6.
THERMAL SYSTEMS
89
a) Temperature and Temperature Difference
89
b) Thermocouples
90
PROBLEM SET 8: TEMPERATURE AND THERMOCOUPLES 93
LAB #8: MEASURING TEMPERATURE WITH THERMOCOUPLES 95
7.
SUMMARY
99
IV. WORK
100
1. MECHANICAL TRANSLATIONAL SYSTEMS
a) Introduction
b) Work and Efficiency
c) The Lever
d) The Inclined Plane
PROBLEM SET 9: LINEAR WORK
LAB #9: WORK DONE IN MOVING LOADS
100
100
104
106
107
109
111
2. MECHANICAL ROTATIONAL SYSTEMS
PROBLEM SET 10: ROTATIONAL WORK
LAB #10: ROTATIONAL WORK OF A WINCH
115
120
121
3. WORK IN FLUID SYSTEMS
a) Introduction
b) Open and Closed Fluid Systems
c) Basic Hydraulic Power System
d) Work in a Closed System
e) Examples of Work in Fluid Systems
PROBLEM SET 11: FLUID WORK
LAB #11: WORK DONE BY A PUMP
126
126
128
128
129
132
133
135
4. WORK IN ELECTRICAL SYSTEMS
a) Introduction
b) Effects of Electrical Work
c) Electrical Efficiency
PROBLEM SET 12: ELECTRICAL WORK
d) Work due to Magnetic Fields
e) The Electric Motor
LAB #12: WORK DONE BY A MOTOR
138
138
141
141
143
144
144
146
5. THERMAL SYSTEMS
a) Introduction
b) Change of State
PROBLEM SET 13: THERMAL ENERGY
LAB #13: THERMAL ENERGY
149
149
152
154
156
6. SUMMARY
161
V. RATE
1. MECHANICAL SYSTEMS: TRANSLATIONAL RATES
PROBLEM SET 14: LINEAR MOTION
LAB #14: LINEAR MOTION
162
162
171
172
2. MECHANICAL ROTATIONAL RATES
176
PROBLEM SET 15: ROTATIONAL RATES
179
LAB #15: MEASURING ANGULAR RATE WITH A STROBOSCOPE 181
3. CIRCULAR MOTION
PROBLEM SET 16: CENTRIPETAL MOTION AND FORCE
LAB #16 CENTRIPETAL FORCE AND ACCELERATION
APPENDIX A: SOLUTIONS TO PROBLEM SETS
APPENDIX B: VECTORS AND SCALARS
APPENDIX C: CONVERSIONS, CONSTANTS
AND USEFUL EQUATIONS
APPENDIX D: SAMPLE UNITS FOR SOME COMMON VARIABLES
APPENDIX E: GLOSSARY OF TERMS
APPENDIX F: THE GREEK ALPHABET
184
187
188
192
197
201
213
215
217
I. INTRODUCTION
1. SURVIVAL SKILLS
Why is learning physics part of your apprenticeship?
Physics can help you predict the outcome of an act by improving your knowledge of the formation,
transmission, and transformation of energy, and how such phenomena can be applied (and misapplied) in your
working environment. The process of learning physics improves your reasoning skills by teaching you methods
of approaching problems. It improves your understanding of the fundamental processes governing your trade as
well as many other trades.
The Laws of Physics have no jurisdictional bounds, however they are useful outside of work as well. Studying
physics allows us the opportunity to peek beneath the veil that shrouds our mysterious universe and the world
about us.
This class is the first of a five-part course, emphasizing laboratory experience and hands-on interaction with
physics. Each lecture is accompanied by a lab of equal length, in which the students build and run experiments
that apply and measure the physics concepts described in the lecture. This quarter you will be introduced to
three concepts: force and force-like quantities, work and rate, in the context of four energy systems: mechanical
(translational and rotational), fluid, electromagnetic and thermal. This is a building block course where
concepts learned in preceding quarters are used in each of the succeeding quarters as you learn new concepts
and applications.
There are some simple rules for getting through this class which, if you follow them day-by-day, will make
understanding physics much easier and much more enjoyable for you.
1. Always read the assignment for the next lesson and lab prior to the class. Also, try to work ALL of the
assigned problems. If you are able to understand the reading and to work all the problems, you are well
prepared for the laboratory and the exams. On the other hand, if there are parts of the reading that confuse you
and/or problems that you cannot correctly solve, you can now ask pertinent questions in class.
2. Attend and listen to the pre-lab lecture and take notes in your student manual. Ensure that all questions that
you have about the reading or how to work any problems are answered during the pre-lab lecture or that you
ask specific questions during the lecture to resolve these misunderstandings for you.
3. Attend all laboratory periods. They are an integral part of the learning process for this course and a great aid
to understanding the physics concepts covered. Actively participate in the lab with your lab partners. Ensure
that you see the readings and all that happened during the lab experiment. Actually seeing the experiment will
be a significant learning aid for you in understanding and remembering the concepts covered by the lab.
Participate in the completion of the lab form including all calculations. Actively take part with your lab group
in discussions to answer ALL of the analysis questions. Fill in the same data, calculations and answers to the
questions on your copy of the lab report for use in review and for the tests.
4. Participate in post-lab discussions and lecture. Ensure that all questions you have concerning the lab, the
reading and problems are fully answered and are cleared up for you. If you still have questions or areas of
i
confusion talk to the instructor or go to after hours tutoring. Since this course is a building block course, what
you don't understand in the lab will seriously impede your learning in the succeeding lessons and throughout
the remainder of your physics courses.
5. Take a few minutes after class to write a summary of all the formulas, definitions and concepts covered in the
lab and lectures. Write this summary in your student manual where you can find it and use it for further
reference, review and during the test.
6. When the instructor reviews the graded lab report with your group make any corrections necessary on your
copy of the lab report. Following each of these steps for every lab will be a significant learning aid for you
and the key to your successful completion of the course.
8. Use the five-step method of problem solving. Learn this technique as fast as possible. It is one of the
better reasoning tools you will pick up in college, and your grade is based on how well you solve the
problem. We are more interested in how you attack a problem than your final answer. You must show
your work for credit. The process, rather than the solution accounts for the bulk of your grade on an
exam, although if your process is sound your chances of arriving at the correct solution are high.
THE FIVE STEP METHOD OF PROBLEM SOLVING
1. READ THE ENTIRE PROBLEM CAREFULLY AND MAKE A SKETCH
Read through the entire problem completely before you start to write anything down. A sketch of the problem
situation will help you to clarify the ideas of the problem. If you can't visualize the situation, you might be
missing some important concepts.
2. LIST THE GIVEN INFORMATION AND IDENTIFY THE UNKNOWN QUANTITY ASKED FOR
IN THE PROBLEM.
Write down each magnitude (number and units) that is given and identify it with the appropriate letter. For
example, " a time of six seconds" is listed as "t = 6 sec".
It is important to use the letter symbol that will appear in the equations. For example, if a problem asks you to
find " how long it takes" for an event to occur, you would write "t =?".
3. FROM YOUR LIST OF EQUATIONS, SELECT THE EQUATION THAT RELATES THE
UNKNOWN QUANTITY TO THE GIVEN INFORMATION. REWRITE THE EQUATION, IF
NECESSARY, TO SOLVE FOR THE UNKNOWN QUANTITY.
For example, if you know the velocity (v) of an object and the distance it has traveled (d), and you wish to find
the time required to travel that distance (t), you would choose the equation that uses all three of these variables:
d
.
t =
v
The unknown (in this case, t) should appear alone on the left of the equal sign in your working equation. In the
d
example above, if the unknown was the velocity, the equation would be rewritten to read: v =
.
t
ii
4. SUBSTITUTE THE KNOWN INFORMATION IN THE WORKING EQUATION, INCLUDING
ALL UNITS.
If you are solving for the velocity in step 3, and the distance is 12 meters (“d = 12 m”) and it took 6 seconds (“t
12 m
d
= 6 s”), then we would write (after “ v =
”), v =
.
6s
t
5. SOLVE THE EQUATION, INDICATING THE CANCELLATION OF UNITS, AND CIRCLE YOUR
ANSWER.
12 m
m
Take for example, the relation v =
= 2 . The units combine (and can cancel) just like numbers.
6s
s
Notice that “meters per second” is a correct unit for velocity. Check your answer to see that it has the correct
units. For example, if you find that the weight of an object is in units of "square feet", then an error has
occurred.
Also, check the magnitude of your answer; if it is obviously physically impossible, go back and look for an
error. For example, if you find the speed of a car to be 4000 mph, the answer is not reasonable. Make sure you
circle your answer to avoid confusion.
Each chapter is composed of the subject material, problems to work, and an experiment relating to the material.
Solutions to the work problems, units, conversions, constants, useful equations and techniques are provided in
the appendices. Space is provided in this book for your notes and work problem solutions. As you perform
each experiment, you will record your data in this book. Your work group will submit a separate, formal report
to the instructor for evaluation. Do not remove pages from this book, since you are allowed to use it as a
reference during examinations.
Lab Supervisor’s Responsibilities
Members of each lab group must assume the responsibilities of operator, data recorder and supervisor. The
supervisor’s responsibilities are listed below.
1. Take out the illustration pages and place them on the lab bench so that everyone can refer to them.
2. Read the overview out loud to your partners. Stop after each paragraph, and be sure that everyone
understood what was covered. Read the overview in advance so that you will be prepared to explain it.
3. Read the objectives to your partners. They can record them in their lab records at this time, or write them in
later.
4. Read the equipment list, allowing time for your partners to obtain each item from the lab bench. DO NOT
HANDLE THE EQUIPMENT. It is important for you to keep the instructions in front of you at all times.
5. Read the procedure and set-up instructions to the operators. DO NOT HANDLE THE EQUIPMENT. Read
the instructions slowly, and SUPERVISE to see that they are being followed correctly. The operators will
refer to the illustrations when necessary.
6. When the experiment is being performed, all partners, including the data recorder and supervisor, should
take at least one reading. Be sure that everyone has a chance to learn the proper use of the equipment.
7. Continue to read the instructions and supervise the experiment. None of your partners should be reading
from the instruction manual. Be sure that each step is followed.
8. All partners should join equally in discussion of the questions and problems at the end of each lab. Ask for
help from the instructor if you disagree on any of the answers. Fill in your individual lab report records as
iii
you proceed. These reports should not be removed from the binding; they will be your reference notes
during tests.
9. Be sure that the operators return all the equipment to its correct location before you leave the lab.
Supervisor’s Lab Report
You are responsible for turning in a final report for evaluation. If possible, turn it in by the end of class. All
partners should remain in the lab until their report is turned in or the period is over. Have them initial the report
after it is complete. All partners will receive the same grade based on the report you turn in. Your report is due
no later than the beginning of the next lab period.
iv
2. PHYSICS FOR TECHNICIANS (INDTT 170) SYLLABUS
TEXT: PHYSICS AND TECHNOLOGY I: INDTT 170
OBJECTIVES: This course is a multi-level exposure to basic physics concepts which incorporates
mathematical skills, laboratory techniques, team learning/cooperation, analytical thinking, problem solving
techniques, developing leadership skills, effective oral/written communications and required self-study with
assigned problems to solve. This course is designed to help the student develop self-confidence in his or her
own abilities.
This exposure includes:
1. Introductory subject lecture with experiment briefing.
2. Assigned self-study reading and problem set
3. Laboratory experiment demonstrating concept/law conducted by lab group with instructor supervision
and guidance.
4. Experiment analysis by lab group extending observed results to other applications of the physics
concepts.
5. Concluding lecture on concepts and applications.
6. Review of lab report by instructor with each lab group to reflect on and correct misunderstandings and
reinforce what has been learned.
7. Test on the concepts/laws and applications covered.
8. Review of test individually with instructor to reflect on and correct misunderstandings while
reinforcing what has been learned.
9. After hours instruction and individual tutoring available.
Specifically, this course covers the following topics:
1.0 The International System of units (metric system), the British system of units, dimensional analysis,
laboratory introduction and techniques.
1.1 Force and torque in mechanical systems
1.2 Pressure in fluid systems
1.3 Voltage in electrical systems
1.4 Temperature difference in thermal systems
2.0 Principles of work and energy
2.1 Translational and rotational mechanical work
2.2 Work in fluid systems
2.3 work in electromagnetic systems
2.4 Work in thermal systems
3.0 Principles of rate measurements
3.1 Translational and rotational speed/velocity/acceleration
There are 15-20 laboratory experiments related to the subjects above, including three introductory experiments
on laboratory procedures, techniques and various analysis calculations. Problem-solving laboratories are
included to teach and demonstrate the Five-step problem solving technique.
v
Assessment methods include:
1. Periodic examinations, testing physics concepts, extension of concepts and applications of these
concepts to other situations.
2. Daily lab experiments with assigned lab group demonstrating concepts and submitting written report for
grade.
3. Daily lab group analyses of experiment, extending observed results to wider applications. This analysis
submitted for grade with the lab report by group.
4. Periodic group leadership/supervisory experience by each group member, leading lab group through the
experiment and analysis of the experiment. This effort is graded. Each student will be supervisor for 4
or 5 lab experiments.
5. Instructor review of each lab report with lab group. This provides the students with immediate feedback
on what they did properly and correctly, what was incorrect or incomplete, what was overlooked, and
how to improve their performance in the following lab experiments. Students are encouraged to correct
their copies of the lab report in their text.
6. Instructor review supervisory grade with supervisor for the experiments.
7. Instructor review of each test with individual concerned as part of the learning experience.
OUTCOMES AND ASSESSMENTS
The official outcomes and assessment methods are provided below to give you a better idea of what we want
you to gain from this class and how we assess your development:
I.
LEARNING OUTCOMES
ASSESSMENT METHODS
1. Apply and explain the basic concepts of force and force-like
quantities (torque, pressure, voltage and pressure
difference) in the context of mechanical, fluid,
electromagnetic and thermal energy systems, respectively.
1. Tests, experiments, lab reports, results reviewed with
student.
2. Apply and explain the basic concepts of work in the context
of mechanical, fluid, electromagnetic and thermal energy
systems.
2. Tests, experiments, lab reports, results reviewed with
student.
3. Apply and explain the basic concepts of rate in t he context
of mechanical energy systems.
3. Tests, experiments, lab reports, results reviewed with
student.
4. Solve basic physics problems for force, work and mechanical
rate using the 5-step method for SI and British systems.
4. Classroom problem solving, tests and lab reports reviewed
with students.
5. Follow detailed laboratory instructions and report on the
results of laboratory experiments.
5. Instructor observation, written lab reports, tests, reviewed
with student.
6. Perform the roles of the supervisor, data recorder and
operator in conducting experiments and completing lab
reports.
6. Instructor group observation/critique during lab. Graded
supervisor's report reviewed with student.
7. Behave responsibly and ethically through class attendance,
participation, discussions and completion of class tasks and
projects.
7. Attendance, observations of individual effort, initiative,
participation in lab group and classroom discussions,
completion of assignments. Reviewed with student.
8. Develop self-assessment skills to modify learning strategies.
8. Student keeps track of all grades on labs and test, reviews
individual progress with instructor.
vi
ASSIGNMENTS: The Laboratory Assignments book lists the reading assignment and
problems for each lab experiment. Prior to each class, complete the assigned reading in the text
and supplementary text, and do the assigned problems. If you need to brush up on your math
skills you may want to attend after hours Monday and Wednesday in Building 466 or the walk-in
math/science lab opened Monday-Friday on campus.
The answers to all text problems will be provided. If you are unable to get the correct answer,
ask for clarification in the lecture class or during the lab period.
The lab records from the introductory labs will be checked individually by the instructor. At the
completion of each regular experiment, #1 through #16, a supervisor's report must be turned in
for evaluation. If possible, turn in this report at the end of the lab period. The report must be
turned in no later than the beginning of the next lab class.
TESTS: There will be three one-hour tests in addition to a final examination. The Student
Resource Book and your individual Laboratory Assignments and Record Book may be used
during test. No other texts or notes will be allowed.
MAKE-UP WORK: You may arrange to take a test early if you know that you must be absent
on test day. If you miss a test, the make-up test must be taken within 7 calendar days. Make-up
tests will cover the same material as the classroom test, but the difficulty will be greater.
THERE WILL BE NO MAKE-UP LABS.
LAB REPORT GRADES: The report forms in your Lab Record book should not be removed
from the bound book; these will be your reference notes for tests. Work with your lab partners to
complete the answers to all questions and problems. The supervisor will submit a report on a
separate form, which will be evaluated in the following manner:
Experiment performed and report turned in .. 4 points
Accuracy and completeness of experiment .... 2 "
Questions & problems completed correctly.... 3 "
Neatness........... 1 "
10 points possible
All students in the supervisor's group will receive the same grade, based on the supervisor's
report.
You will be the supervisor of 4 or more experiments. You will receive a supervisor grade based
on the criteria above and the overall group performance during the lab (15 points possible). The
four highest supervisor grades will be used for your supervisor grade (maximum 60 points).
vii
QUARTER GRADE:
We expect to have up to 16 experiments, in addition to the 3
introductory labs. The estimated point count is as follows:
3 introductory labs....... 30 points
16 labs x 10 ..............160 "
Supervisor grade........... 60 "
3 tests x 50.............. 150 "
final exam................ 150 "
550 points total
Your grade will be determined by dividing your total points by the total possible. The % value
will be translated into the official grade according to the schedule below.
Percentage
Grade
95.5 – 100
94.5 – 95.4
93.5 - 94.4
92.5 - 93.4
91.8 – 92.4
91.0 – 91.7
90.3 – 90.9
89.5 – 90.2
88.5 – 89.4
87.5 – 88.4
86.5 - 87.4
85.2 – 86.4
Decimal Point Grading Scale for INDTT Physics
Point
Percentage
Point
Percentage
Grade
Grade
Grade
Grade
4.0
83.8 – 85.1
2.8
71.8 – 72.4
3.9
82.5 – 83.7
2.7
71.1 – 71.7
3.8
81.8 – 82.4
2.6
70.3 – 71.0
3.7
81.0 – 81.7
2.5
69.5 – 70.2
3.6
80.3 – 80.9
2.4
68.5 – 69.4
3.5
79.5 – 80.2
2.3
67.5 – 68.4
3.4
78.1 – 79.4
2.2
66.5 – 67.4
3.3
76.6 – 78.0
2.1
65.5 – 66.4
3.2
75.0 - 76.5
2.0
64.5 – 65.4
3.1
74.2 – 74.9
1.9
63.5 – 64.4
3.0
73.4 – 74.1
1.8
59.5 – 63
2.9
72.5 – 73.3
1.7
viii
Point
Grade
1.6
1.5
1.4
1.3
1.2
1.1
1.0
0.9
0.8
0.7
II. SOME FUNDAMENTALS
Scientific calculations have their own set of “grammar” that, when obeyed, streamline the
calculations and increase your chances of getting the correct answer. In this chapter we present
some of the most useful of these "grammatical" tools.
1. SIGNIFICANT FIGURES
The number of significant figures, or "sig fig's", is a measure of a number's precision. They are
the number of digits that were actually measured. The following are a few examples of some
numbers and the number of significant figures they contain:
Number
0.1150
45,430
10.004
150,000,000,000
0.0000000150
No. of Significant Digits
4
4
5
2
3
Any zero between non-zero numbers is also a significant figure. Note that a zero at the end of a
set of significant figures (sig fig's) can also be a significant figure if it is a measured value. For
instance, if we measured a distance and found it to be 150.0 centimeters (cm) to the nearest tenth
of a centimeter (0.1 cm), then those last two zeros were truly measured and the number has four
significant figures. However, unless it is stated that the zeros at the end of a number were
measured, we will assume that they are not sig fig's.
Suppose you're making calculations with measurements that have different numbers of
significant figures. How many sig fig's should the answer have? The number of sig fig's in the
answer should be the same as the measurement in the calculation that had the least number of sig
fig's.
For example:
(3.85132)(6.815)(2.71) = 71.1
2.71 has the smallest number of sig fig's (3) of the inputs, so the answer can only be good to
three sig fig's.
2. POWERS OF TEN
Powers of ten are useful because 1), they save space and 2), they're easier to multiply and divide.
For example, 1,000,000,000,000,000,000,000,000,000,000,000 can be written as 1033. 33 is the
exponent, and we express the number as "ten to the thirty third power". Now that's a space
saver! Negative powers of ten indicate numbers smaller than one. Here are some examples of
numbers in terms of powers of ten:
1
Number
Powers of Ten Equivalent
0
1
10
100
1,000
1,000,000
0.1
0.01
0.001
0.000001
10
1
10
2
10
3
10
6
10
-1
10
-2
10
-3
10
-6
10
2
Exponent
0
1
2
3
6
-1
-2
-3
-6
-2
Note that 0.01 = 1/100 = 1/10 = 10 , so if you have a power of ten in the denominator, you can
move it to the top (into the numerator) by changing the sign of its exponent.
Multiplying is easier using powers of ten. When you multiply two numbers written as powers of
5
5
10
ten, you just add the exponents together. For example, 100,000 x 100,000 = (10 )(10 ) = 10 .
Numbers in the denominator should be moved up to the numerator before combining. You can
4
2
4
-2
2
do this by changing the sign of the exponent, so, for example, 10 /10 = (10 )(10 ) = 10 . Using
powers of ten can save a lot of space, eliminate copying errors (you don't have to write all the
zeros!) and make multiplication and division much simpler (you just add or subtract the
exponents!).
Some powers of ten occur so often that we abbreviate them with symbols. The following are
some common examples:
Factor of:
one millionth
Name
micro
one thousandth
one thousand
one million
milli
kilo
Mega
Abbreviation

Power of Ten
-6
10
-3
m
k
M
10
3
10
6
10
3. SCIENTIFIC NOTATION
Scientific notation is a common way to express very large or very small numbers. For example,
the mass of the Earth is 5,980,000,000,000,000,000,000,000 kilograms (kg). It's very easy to
write this number incorrectly when you copy it down, and including it in an equation takes up a
lot of space. This number is easier to remember if it is written as the product of two components:
the significant digits, and the powers of ten. The significant digits are the numbers in front of
all the zeros (598). They're usually listed as a decimal number between one and ten (5.98),
which has three significant digits. You need to multiply this number by some power of ten to get
24
the actual value. In this case 5.98 must be multiplied by 10 , so the mass of the Earth can be
2
24
written in scientific notation as 5.98 x 10 kg. Using scientific notation reduces your chances of
making errors when you copy numbers down and perform calculations. For example,
5,980,000,000,000,000,000,000,000 kg x 0.00593
m
s2
can be rewritten as
. x 10
598
24
m
kg  m

kg 5.93 x 10-3 2  = 3.55 x 1022

s 
s2
It's also much easier to estimate the answer this way. 5.98 and 5.93 are both about 6, and 6 x 6
24
-3
21
21
22
= 36. 10 x 10 = 10 , so the answer should be about 36 x 10 = 3.6 x 10 . Understanding
and using scientific notation will improve your ability to grasp physical processes more quickly
and reduce computational errors, which means better grades. Ask your instructor how to input
numbers into your calculator in terms of scientific notation.
Examples of Numbers Expressed in Terms of Scientific Notation:
Number
6024
0.000003450
750,000
2.150
Scientific Notation
3
6.024 x 10
-6
3.45 x 10
5
7.5 x 10
0
2.15 x 10
4. DIMENSIONAL ANALYSIS
The dimensions of an object are the smallest set of quantities we can use to describe it. Typical
dimensions are length, time, mass, force, pressure, charge and current. Units are the reference
scales we choose to use for a given dimension. For example, for the dimension of length we may
choose to use scales of inches, feet, miles, centimeters, meters, or kilometers. You must always
include the scale associated with a number. If you tell us the answer is 8, for example, we will
ALWAYS respond with, "8 WHAT?". Do you mean 8 feet, or 8 days, or 8 supervisors, or 8
pounds or 8 slugs or WHAT? Feet, days, supervisors, pounds and slugs are all units; they are
examples of scales of reference we use to describe different dimensions.
2
Dimensions can be combined and simplified just like numbers. Meters x meters = (m)(m) = m ,
meters
m
for example.
x kilograms x seconds = ( )( kg)(s) = kg  m . The seconds cancel,
second
s
because one was in the numerator and one was in the denominator, the same way identical
numbers would cancel if one was on top and one was on bottom. Combining dimensions in this
manner (algebraically) is known as dimensional analysis.
3
2
Example II-1: In the international (SI) system of units, the units for acceleration are m/s
(“meters per second squared”). It is the rate of change of velocity (meters per second per
second). Force, given in units of newtons (N), is the product of mass (in kilograms, or “kg”) and
acceleration. The units of force are therefore the product of mass units and acceleration units:
1N = 1
kg  m
s2
Joules (“J”, pronounced “jewels”), a measurement of energy, are the product of newtons (a unit
of force) and meters (a unit of length):
kg  m2
 kg  m 


1 J = 1 Nm = 1  2  m = 1
 s 
s2
A watt, a unit of power, is a measure of the rate at which energy is used. Energy is commonly
 J
given in units of joules and watts are joules per second   .
 s
1W = 1
 kg  m2   1
J
kg  m2
= 1
=
1



s
s3
 s2   s 
Do not use diagonal lines to separate numerators and denominators of units (for example,
“1 W = 1 J/s”); it makes dimensional analysis much more difficult. Do use horizontal lines
J
(“ 1 W = 1 ”).
s
Sometimes the units in a calculation will cancel completely and there will be no units left. For
d
8 ft
example, the ratio of the distance d1 = 8 ft to the distance d2 = 4 ft is: 1 =
= 2 . Ratios
d2
4 ft
of numbers with the same dimensions (feet, in this example) have no units. Such quantities are
said to be dimensionless.
5. FUNDAMENTAL MEASUREMENTS (those upon which all others are based):
i) Length (usually given by x or d): The magnitude of position from some reference point.
Common units are inches (in), feet (ft), meters (m), centimeters (cm) and kilometers (km).
ii) Mass (m): The amount of matter in an object or region. Mass is not equal to weight! Weight
is a force, the product of mass and acceleration. Common units of mass are grams (g),
kilograms (kg) and slugs (yes, slugs!). For example, if you are accelerating up or down in an
elevator your weight changes, but your mass does not. Your weight is different on the moon
than on the surface of the Earth, but your mass is not. A pound is a unit of force, not mass.
iii) Time (t): Means of reference for discerning sequential events. Common units are seconds
4
(s), minutes (min), hours (hr), days (d) or, longer still, an entire physics class.
iv) Charge (q): The characteristic of matter that produces electromagnetic attraction and
repulsion. Charges come in two flavors, which Benjamin Franklin named positive and
negative. Like charges repel and opposite charges attract. The unit for charge is coulombs
(C, or Coul).
v) Temperature (T): Temperature is a measurement of the average kinetic energy of a group of
particles. The SI units for temperature are degrees Celsius (oC) or Kelvins (K). The English
units are degrees Fahrenheit (oF) or degrees Rankine (oR).
All other measurements are based on fundamental units and are known as derived units.
Velocity, for example, is the rate distance changes with time. Common units are m/s and ft/s.
2
2
Acceleration is the rate velocity changes with time: m/s or ft/ s . When you push or pull an
object, you are exerting a force on it. Force is the product of mass and acceleration, F = ma.
Common units are kgm/s2, also called newtons, after Isaac Newton. Your weight is generally
given in the English unit of force, pounds (lbs), which is the force with which you are attracted
to the Earth. Dividing your weight by the gravitational acceleration, in ft/s2 (English units) or
m/s2 (SI units) will give you your mass.
Fundamental Units
Length
SI
English
m
ft
Time
Mass
s
s
kg
slug
Temperature
o
o
C, K
F, oR
Charge
C
C
6. SYSTEMS OF UNITS
In this course we will work exclusively with two sets of units: the International System (SI), also
known as the "mks" system (for "meters, kilograms, seconds"), and the English system. The
standard SI unit of length is the meter, while the Standard English unit is the foot. The standard
SI unit of mass is the kilogram, while the Standard English unit is the slug. Yes, that's right, the
slug. Not the pound. The pound is a unit of force, not mass. The second is the standard unit of
time for both measurement systems. Converting from one system to another is accomplished via
unit conversions. For example, there are 3.28 feet in one meter:
3.28 ft = 1 m
Dividing both sides by 1 m (one meter) we get
3.28
ft
= 1
m
5
1m
ft
is the conversion factor for converting from meters to feet. Conversely,
is the
3.28 ft
m
conversion factor for converting from feet to meters.
3.28
Example II-2: How many feet are in 33 meters?
ft 

Solution:  33 m  3.28  = 108 ft . Dimensional analysis shows us that the meters cancel out

m
of the equation.
Example II-3: How many meters are in 33 feet?
Solution: We need to wind up with meters in the numerator this time, which means we have to
1m
invert the conversion (put it in the denominator), so we'll use
:
3.28 ft
 33 ft 
1m 
 = 10 m
 3.28 ft 
Dimensional analysis shows us that the feet cancel and the meters wind up on top, just like we
wanted. Proper use of dimensional analysis is one of the most important skills you can learn this
quarter.
7. SCALARS AND VECTORS
As we study physics we will learn the definition of a number of new quantities. A scalar
quantity is one that can be completely described by its magnitude and units. Examples of scalar
quantities include mass (10 g, 40 kg, 20 slugs), time (1 hr, 30 min, 10 s), distance or dimensions
(5 miles, 2 km, 2 in by 8 in, 2 cm by 7 cm), speed (30 m/s, 40 ft/s, 60 km/hr, 45 mph),
temperature (40 oC, 20 oF) and charge (20 C).
Other quantities, called vectors, must include magnitude, units and direction to be completely
described. Examples include force (10 lb at 10o, 20 N to the west), velocity (40 mph to the east,
25 m/s to the south), acceleration (2 ft/s2 to the right, 1 m/s2 to the north) and torque (3 ftlb
clockwise, 15 Nm counterclockwise). More information on vectors and scalars may be found in
Appendix B of the 170 text.
6
PROBLEM SET 1: FUNDAMENTALS
1. Multiply the following numbers and write the answer to the appropriate number of
significant figures: 2.16, 7.49, 4.262 and 3.34.
2. Multiply and divide the following numbers, writing the answer to the appropriate number
of significant figures:
(237.1)(19.7)(456.27)
(345.2)(78.8)
3. Change the numbers in the following problem into scientific notation and perform the
indicated operations, writing the answer to the appropriate number of significant digits:
(10,479.2)(436,900)(22.6)
(47,900)(77.8)(679.3)
4. Convert 147 cm into meters.
5. Convert 0.427 meters into feet, and then into inches.
6. Convert 6.21 meters into millimeters, writing the answer in scientific notation.
7
INTRODUCTORY EXPERIMENTS (3 PARTS)
OVERVIEW
In these experiments, you will learn to use a metric ruler and Vernier Calipers. You will
determine the volume of a cylinder in metric units. You will also practice converting metric
units of mass to weight units.
LEARNING OBJECTIVES
The learning objectives for the introductory laboratory exercises are:
1. Learn how to use a ruler, meter stick, triple beam balance scale and Vernier Calipers to
make measurements, calculate volume and convert measurements from the SI to English
unit systems.
2. Identify methods to avoid or minimize parallax error when making measurements.
3. Learn how to calculate weight and mass in both the SI and English unit systems.
8
INTRODUCTORY LAB #1
Date__________
METRIC LENGTH MEASUREMENTS
Measurement Unit:
cm
mm
inches
Measured Width of
Paper
Show your conversions:
Comparison of Measurements (write a sentence):
VERNIER CALIPERS
Sketch of slotted weight:
Recorded measurement
with Vernier Caliper:
Recorded measurement
with ruler:
Caliper reading rounded
to nearest mm:
Sketch of hollow cylinder:
Do the values agree?
Inside diameter of cylinder ____________
9
INTRODUCTORY LAB #2
Date______________
VOLUME
SKETCH:
DIAMETER:
HEIGHT:
VOLUME (show your work):
Rounding off (write a sentence):
Rounded off value of volume:
CONVERSIONS (show your work):
10
INTRODUCTORY LAB #3
Date______________
MASS AND WEIGHT
1) Weight of cylinder in newtons (N):
2) Mass of cylinder in grams (g)
3) Calculated weight using w = m∙g:
4) Write a sentence comparing the measured weight and the calculated weight:
5) Mass of calibrated weight in grams
6) Calibrated weight in lbs
7) Ratio of lb to kg:
How does this compare with the textbook value? ______________________________
8) Weight in newtons (N):
10) Mass from balance scales (g):
9) Computed mass in kg:
Do the values for computed mass and measured mass agree?______________________
11
III. FORCE AND FORCE-LIKE QUANTITIES
Any change in the motion of an object in a translational mechanical energy system is due to
forces acting on it. Force-like quantities are the result of forces acting within other types of
energy systems (mechanical, fluid, electric and thermal). They are physical quantities that act as
the source of motion in those systems. In any energy system, an unbalanced force-like quantity
produces a change in the motion (a displacement) of some quantity. Although force-like
quantities do not act in their respective systems in exactly the same way that force acts in a
translational mechanical system, all have certain characteristics in common with force.
1. MECHANICAL TRANSLATIONAL SYSTEMS: FORCE
Isaac Newton defined force in the classical sense. Newton's Laws of Motion are as follows:
Newton’s First Law:
1. An object in motion will remain in motion and an object at rest will remain at rest unless
acted upon by an external force.
Newton got this first law from Galileo, who died in 1642, about the time that Newton was born.
We don't see this effect very often because friction acts against motion, but if you keep reducing
friction, you can see its effects. Try the following thought experiment. In your head imagine
sliding a hockey puck on a concrete surface. Then imagine sliding it on a smooth piece of wood,
then slide it on linoleum, and finally on a sheet of ice. As your experiment proceeds you find
that the puck slides further and further because the force of friction gets smaller and smaller in
each case. If there were no frictional force working against the motion, the puck would slide
forever at the same velocity! Planets orbiting the sun have extremely small frictional forces
working against them; their orbital trajectories have changed very little in the last 4 billion years.
Newton’s Second Law:
2. The acceleration of an object is directly proportional to the resultant force acting on it, and
inversely proportional to its mass.
Using a for acceleration, F for force and m for mass, we can write the above sentence in
equation form:
a =
F
m
III-1
Acceleration is a change of velocity with time, like when you accelerate to get on the freeway.
To accelerate a mass (an object), you must apply a force. If you multiply both sides of the
equation by m (the mass), you get Newton's most famous equation:
F = ma
III-2
Force is the product of mass and acceleration. It’s the amount of "push" or "pull" exerted on an
object. Since force is a vector, it is described in terms of its magnitude, units and direction.
The hockey puck we just talked about requires a force to get it moving (you slide it) and a force
to stop it from moving (friction from the floor and air resistance). Force makes a moving body
move faster or slower; it can even change the shape of an object.
12
Force Units
The units of force are obtained by multiplying the units of mass and acceleration. In the metric
system, mass is typically expressed in kilograms (kg), while acceleration is commonly given in
units of meters per second squared (m/s2). Therefore the most common force unit in the metric
system is the kg·m/s2. We use this unit so often that it has its own name: the newton
(abbreviated as "N"). In the English system of measurement, the most common unit for mass is
the slug, and we generally use ft/s2 as acceleration units. The force unit is then the slug·ft/s2,
which we always refer to as the pound (lb).
Adding Vectors: The Head-to-Tail (or Graphical) Method
If more than one force is acting on an object, the net, or resultant, force is the sum of the
individual force vectors, which we can calculate using trigonometry or graphically. If the sum of
the individual forces is zero, the object is said to be in equilibrium.
The following two examples illustrate the head-to-tail method of adding vectors. We’ll find the
resultant force graphically by adding the force vectors together. We start with one vector, then
place the tail of the next vector at the head of the previous vector until all of the individual
vectors are connected. The resultant force (FR) vector extends from the tail of the first force
vector to the head of the last vector added. This method of adding vectors is known as the
graphical, or head-to-tail method of adding vectors.
The equilibrant force is the force required to produce a net force of zero. In other words, if
there is no net force, the system is in equilibrium. The equilibrant force is of the same
magnitude, but in the opposite direction of the resultant force. The resultant force and the
equilibrant force for the preceding example are shown below.
Example III-1: Adding Vectors
Adele, Sven, Kareem and Darth Vader are all pushing on a table at the same time. A sketch and
a free-body diagram of the forces being applied to the table are shown below. The free-body
diagram depicts the forces as vectors, all acting on one point (or line, as for the torque
experiment in Lab #2). We’ll call the force applied by Adele “FA”, the force applied by Sven
“FS”, and so on:
FDV = 3 lb
FA = 16 lb
•
FS
FK = 4 lb
FK
FA
•
FDV
FS = 8 lb
Sketch
Free-body Diagram
13
We’ve tried to make the length of the force vectors proportional to their magnitude. FA (= 16 lb),
for example, is twice as long as FS (= 8 lb), which is twice as long as FK (= 4 lb). Notice that
FDV and FS act along the same line, and so do FA and FK. We say that FDV and FS (and also FA
and FK) are collinear.
Find the net, or resultant, force (FR) acting on the table, and the equilibrant force.
Resultant Force Solution:
Let’s start with Adele. I need to pick a scale for the vectors, so I’ll say 1 cm = 2lb (one
centimeter is equivalent to two newtons). FA = 16 lb to the right, so that’s 8 cm to the right.
FA
Now let’s add Sven. FS = 8 lb up, that’s a force vector going up for 4 cm.
FS
FA
Let’s add Kareem next. He’s pushing with a force FK = 4 lb to the left, so that’s 2 cm to the left.
FK
FS
FA
And finally, Darth Vader, using the force. FDV = 3 lb down, or 1.5 cm down:
14
FK
FDV
FS
Now draw the
vector (FR),
force. It begins
starting point)
head of FDV (the
FA
resultant force
which is the net
at the tail of FA (the
and extends to the
end point).
FK
FDV
FS
FR
23°
FA
FR is the resultant force. We measure its magnitude as 6.5 cm, which would be 13 lb. Its angle
with respect to the horizontal is 23°, which we measure with a protractor. We commonly
assume that 0° is straight to the right, and the angle increases as you move
counterclockwise. The resultant force is given as:
FR = 13 lb @ 23°
Equilibrant Force Solution:
The equilibrant force is the force that would have to b added to make the net force be zero. The
equilibrant force (Feq) has the same magnitude as the resultant force (FR), but it’s in the opposite
direction, that is, it’s 180° from FR. The direction would be
 = 23° + 180° = 203°,
where the symbol  (“theta”) is used to denote the angle. The equilibrant force is given as:
Feq = 13 lb @ 203°
15
FDV
FK
203°
FS
Feq
FA
16
Example III-2: Adding Force Vectors (Again)
Two forces are acting on the same point. The first force is 45 N (newtons) at an angle of 40°.
The second force is 50 N at 253°.
a) Find the resultant force.
Solution: F1 = 45 N @ 40°
F2 = 50 N @253°
Scale: Let 1 N = 1 cm
Draw F1 = 45 cm @ 40°:
Add F2 ( = 50 N @ 253°) to F1:
Draw the resultant force from
the tail of F1 (the starting point)
to the head of F2 (the ending
point):
F1
40°
NEXT PAGE 
17
FR is 27.4 cm long at an angle of about 314°, so
FR = 27.4 N @ 314°
253°
F1
314°
40°
F2
FR
b) Find the equilibrant force (Feq)
Feq is the same magnitude as FR, but in the opposite direction. The angle would be
314° - 180° = 134°
18
so Feq = 27.4 N @ 134°
253°
F1
40°
F2
Feq
134°
Vectors and the graphical addition of vectors are discussed in even greater depth in Appendix B.
Newton’s Third Law:
3. For every action, there is an equal and opposite reaction.
This means that if you exert a force on an object, the object exerts an equal but oppositely
directed force on you. If your mass and the object's mass are different, the two accelerations will
be different as well.
Example III-3: A 50 kg fishmonger standing on frictionless ice throws a 10 kg fish at a
customer with a horizontal force of 40 N. What is the fish's net horizontal acceleration? What's
the fishmonger’s net horizontal acceleration?
19
The Fishmonger
Solution: The fish's net acceleration can be solved via
F
a =
m
2
Remember that newtons are kgm/s , so
kg  m
40
F
m
s2
a =
=
= 4 2
m fish
10 kg
s
The acceleration of the fishmonger is in the opposite direction and of magnitude
a=
F
m fishmonger
kg  m
s 2 = 0.8 m
50 kg
s2
40
=
The fishmonger is accelerated in the opposite direction by a smaller amount because he has more
mass.
Newton's Law of Gravitational Attraction
Imagine you're sitting under an apple tree on a warm autumn afternoon, contemplating the
universe. You look up into the sky just in time to see an apple drop from a low lying branch. As
it hits you smack between the eyes you pause to consider how much more it could have hurt, had
the apple fallen from a higher branch. You instinctively know that the farther the apple falls, the
faster it moves. It accelerates! An accelerating mass implies a force (F = ma) is at work. Isaac
Newton was the first to derive an expression for the mutual attraction between two masses. He
theorized a gravitational force between two masses (m1 and m2 in Figure III-1) that is directly
20
proportional to each of the masses and inversely proportional to the square of the distance (r)
between the centers of the two masses.
F
F
r
Figure III-1. Mutually attractive force between two masses.
Newton's Law of Gravitation can be expressed as
F
Gm 1m 2
r2
III-3
where G is the Universal Gravitational Constant, approximately equal to 6.67 x 10-11 N·m2/kg2.
From his Third Law, Newton determined that each of the two bodies exerts the same force on
each other.
Example III-3: A 2 kg rabbit sits by the seashore, 6,371 km (6.371 x 106 m) from the Earth's
Rabbit
center. The mass of the Earth is 5.97 x 1024 kg.
a) What force do the rabbit and the planet exert on each other?
RE = 6,371 km
EARTH
Solution: The distance from the rabbit's center of mass to the Earth's center of mass is
essentially the radius of the Earth, 6.371 x 106 m. Let m1 be the mass of the rabbit and m2 be the
mass of the Earth. The gravitational force is then:
𝐅=
𝐆𝐦𝟏 𝐦𝟐
=
𝐫𝟐
𝐍 ∙ 𝐦𝟐
) (𝟐 𝐤𝐠)(𝟓. 𝟗𝟕 𝐱 𝟏𝟎𝟐𝟒 𝐤𝐠)
𝐤𝐠 𝟐
(𝟔. 𝟑𝟕𝟏 𝐱 𝟏𝟎𝟔 𝐦)𝟐
(𝟔. 𝟔𝟕 𝐱 𝟏𝟎−𝟏𝟏
= 𝟏𝟗. 𝟔
𝐤𝐠 ∙ 𝐦
= 𝟏𝟗. 𝟔 𝐍
𝐬𝟐
b) If the rabbit fell into a hole, at what rate would it accelerate?
Solution: Let’s call the acceleration of the rabbit “a1”. From Newton's Second Law,
21
a1 
F

m1
kg  m
s 2  9.80 m
2 kg
s2
19.6
Any object near the Earth's surface will fall at this rate, in the absence of other forces.
This acceleration is so commonly used that it has its own variable, "g", the gravitational
acceleration at the Earth's surface (sea level), where g = 9.80 m/s2 (= 32 ft/s2 in English units).
Anything or anyone experiencing an acceleration of this magnitude is said to be accelerating at
"one g".
b) Since "for every action there is an equal and opposite reaction", the Earth must also
accelerate toward the rabbit! What is the Earth's acceleration?
Solution
𝐤𝐠 ∙ 𝐦
𝟏𝟗. 𝟔
𝐅
𝐦
𝐬𝟐
𝐚𝟐 =
=
= 𝟑. 𝟐𝟖 𝐱 𝟏𝟎−𝟐𝟒 𝟐
𝟐𝟒
𝐦𝟐
𝟓. 𝟗𝟕 𝐱 𝟏𝟎 𝐤𝐠
𝐬
Which is about a millionth of a billionth of a billionth of a meter per second squared! Since the
Earth's mass is so big, its acceleration toward the rabbit is correspondingly small.
Weight and Mass
The product of an object's mass (m) and its gravitational acceleration (g) is called its “weight”
(w):
w = m·g
III-4
where g is approximately 9.80 m/s2 (metric units) or 32 ft/s2 (English units). While mass is
given in terms of kilograms or slugs, weight is given in terms of newtons or pounds. Mass and
weight are different quantities. While your mass is the same whether you are here or on the
moon, your weight will differ, because the gravitational acceleration on the moon is much less
than it is here.
Example III-4: Griswolda the Wonder Dog hangs motionless from a frisbee clenched between
her teeth that is being held by her owner, Corey the Above Average. What is the net force on
Griswolda?
22
Corey and Griswolda
Solution: Griswolda experiences a downward force (Fdown) due to her weight (w), which is
equal to the product of her mass (m) and the gravitational acceleration (agravity = g):
Fdown = w = -mg
We can use a minus sign if we define "down" as being in the negative direction. Griswolda also
experiences an upward force (Fup), that of Corey holding up the frisbee, that is slowly pulling her
teeth out of her mouth. Since she is motionless, the upward force must be of equal magnitude,
but in the opposite direction of the downward force:
Fup = ma = mg
The resultant (or net) force is the sum of the individual forces, which in this case is zero:
Fup + Fdown = mg - mg = 0
Free-Body Diagrams
Notice that we have drawn two force vectors adjacent to Cory and Griswolda in the previous
example. Often it is easier to see which forces are at work by replacing the figures with the
vectors that they represent. As we explained in Example III-1, a “free-body diagram” is a
sketch that includes only the vectors involved in the problem, all acting on one point (or line, as
we shall see when we study torque). By drawing only the vectors (and labeling them!), we can
isolate the crucial components of the problem. Free-body diagrams can be used to analyze any
set of vectors such as forces, velocities, or accelerations.
23
Examples of Free-body Diagrams
Ceiling
F1
F2
F2
F1
Box
FBox = wBox
Free-body Diagram
(the force due to the box
is equal to its weight, w)
Sketch of Box Suspended
from Ceiling
F1
F2
F1
F2
Free-body Diagram
Sketch of Tug-of-War
F3
F2
F3
F2
F1
F1
Sketch of Two People on
a Teeter-Totter
Free-body Diagram
24
PROBLEM SET 2: FORCE
1. Three forces act upon an object: 40 lb to the right, 180 lb to the left and 200 lb to the right.
Determine the resultant force and its direction.
2. Three forces act upon an object: 50 lb to the left (1800), 45 lb at 450 and 60 lb at 2800. Using the
graphical method, find the resultant force with a metric ruler and a protractor.
3. Two horizontal forces act upon an object: 40 lb to the north and 60 lb to the south. A. Sketch a
"free body" diagram showing the two forces. B. Find the resultant force, both magnitude and
direction.
4. An object is acted on by three forces: Force A is 140 lb to the right, Force B is 180 lb to the left
and Force C is 200 lb to the right. A. Sketch a "free body" diagram indicating the forces. B. Find
the resultant force in both magnitude and direction.
5. In the laboratory, a force board is set up with three strings pulling on a center ring. The first pull
is 70 newtons at 0 o (to the right), the second pull is 80 newtons at 90o.
a) Draw a "free body" diagram of the ring.
b) Draw a scaled drawing setting 1 cm = 10 newtons showing the two forces using the head-to-tail
method. Find the third force by drawing an arrow from the end of the second vector to the
beginning of the first. This is the equilibrant or third vector, since it has both magnitude and
direction.
c) Add the third vector to the free body diagram.
6. The values below can be classified as either scalar (S) or vector (V) quantities. Examine each and
identify which it is.
a. 15 N at 30 o east of north____
b. 30oC____
c. 80 kg____
d. 105 mph southwest____
o
e. 20 lb at 60 f. 25 lb/in____
]
7. An object is acted upon by four forces: 10 N at 30 o, 8 N at 86 o, 15 N at 135 o and 4 N at 290 o.
a) Sketch the object and forces as a free-body diagram.
b) Using the graphical head-to-tail method, find the magnitude and direction of the resultant force.
8. Three forces act upon an object: 6 lb at 240 o, 12 lb at 90 o and 7 lb at 120 o.
a) Sketch the object and forces as a free-body diagram.
b) Using the graphical head-to-tail method, find the magnitude and direction of the resultant force.
25
LAB #1: FORCES IN EQUILIBRIUM
OVERVIEW
A force is the push or pull exerted on an object. Sometimes two or more forces act on an object at
the same time. The result of this action is called the resultant force. If the resultant force is zero,
then the body experiences no net force. When this happens, the body is said to be in equilibrium.
When several forces act on an object resulting in a net force, they may be balanced by a single
force called the equilibrant. The equilibrant exactly cancels the resultant of all the other forces,
so that the object will be in equilibrium.
Force is a vector; it has both magnitude and direction. When two forces act in the same direction,
their magnitudes are added. When they act in opposite directions, their magnitudes are subtracted
to obtain the resultant force.
When two or more vectors act on an object, and their directions are not along the same line, the
resultant force is not simply the sum or difference of the magnitudes. For example, two forces of
five pounds each may act on an object with their directions at 90o with respect to each other (see
Figure III-2). The resultant magnitude of (52 + 52)1/2 lb = 7.07 lb will not be as great as if the two
vectors were in the same direction (10 lbs) and it will not be as small as if they were in opposite
directions (zero). Also, the direction of the resultant force will be at a 45o angle to each original
vector. Vectors may be represented graphically in order to "add" them; both the correct magnitude
and the correct direction of the resultant can be determined by graphical methods.
=
FR
a
º
7.
F2 = 5 lb at 90º
07
lb
5
t4
F1 = 5 lb at 0º
Figure III-2. Force Vectors
In this experiment you will place a system of forces in equilibrium. Then you will measure the
forces, and graphically add them to find the resultant. The resultant of forces in equilibrium should
be zero. If the graphical method does not give a resultant of zero, there may be an unmeasured
force in the system, probably due to friction. Also, when drawings are done manually, the accuracy
is limited.
The process of making a drawing to determine the resultant of two or more vectors is called the
26
graphical method of adding vectors. It is illustrated in Appendix B.
LEARNING OBJECTIVES FOR LAB 1
The learning objectives for Labs 1-2 are:
1. Learn how to use the 5-step problem solving method.
2. Define the concept of force and force-like quantities in each energy system
(mechanical, fluid, electrical and thermal).
3. Identify devices used to measure force and force-like quantities and use these devices.
4. Learn to use the units of force and torque in both the SI and English systems.
5. Understand and be able to use the conditions for equilibrium for solving problems
with forces and torque involved.
6. Distinguish between the following: force, torque, mass, weight, scalar quantity
and vector quantity.
7. Learn how to use the protractor.
8. Solve for the resultant force using the graphical method when two or more forces
are acting on an object.
9. Solve torque problems using the conditions for equilibrium.
27
LAB #1 FORCES IN EQUILIBRIUM
Date_______
OBJECTIVES:
SKETCH OF
LAB-SETUP:
VECTOR SOLUTION DRAWING:
28
FREE-BODY
DIAGRAM:
Lab #1 ANALYSIS:
1. Two horizontal forces, of 8 N and 6 N respectively, each pull on an object due East.
The resultant force is ___________
2. The equilibrant (magnitude & direction) is _________________________________
3. If the two forces of Question #1 are pulling in opposite directions, the magnitude of the
resultant is ____________
4. Two forces of 15 lb and 20 lb are acting on an object in unspecified directions. What are the
maximum and minimum magnitudes of their resultant? Explain.
______________________________________________________________________________
______________________________________________________________________________
5. When an object is at rest, or moving in a straight line at constant speed, what is known about
the resultant of all the forces acting on the object?
_________________________________________
FOR QUESTIONS #6, #7, & #9, SKETCH A DRAWING, FREE-BODY DIAGRAM, & THE
VECTOR SOLUTION DRAWING.
6. A metal ring has a pull of 10 lb at 0o, 120o, and 240o with the x-axis. Find the resultant force.
7. A 10 lb weight is supported by two cords, both making an angle of 60o with the vertical. Find
the tension in each rope.
29
8. Questions # 6 & 7 refer to a horizontal plane and a vertical plane respectively. Compare the
two situations.
______________________________________________________________________________
______________________________________________________________________________
9. Show how a 10 lb weight supported by two cords could cause a much greater tension in the
cords than 10 lb. Explain.
_______________________________________________________________________
_____________________________________________________________________________
30
2. MECHANICAL ROTATIONAL SYSTEMS: TORQUE
Torque ("") is the force-like quantity in mechanical rotational energy systems. It causes a
change in the state of rotational motion about an axis, just as force produces a change in the state
of linear motion. Torque results when a force is applied at some distance from an axis of
rotation. The perpendicular distance from the line of action of the force to the axis is called the
lever arm, or the moment arm of the force. Torque is the product of the moment arm and the
force acting perpendicular to the moment arm, as shown below.

F
Force
perpendicular
to moment arm
moment
arm
Figure III-3. Torque is the product of the moment arm and the force acting perpendicular
to the moment arm:  = F x.
It can be expressed in equation form as
=Fx
III-5
where  is the torque (in units of Nm or ftlbs),
F is the force (in units of newtons or pounds),
and  is the length of the lever arm (generally in units of meters or feet).
Example III-5: Gustav attempts to turn a nut with a socket wrench. He applies a force of 600 N
perpendicular to the axis of the wrench at a distance of 0.30 meters from the nut. What is the
applied torque?
31
Wrenchmeister Gustav
Solution:
= F x = (600 N)(0.3 m) = 180 Nm
Note: newtonmeters (“Nm”) and foot pounds (“ftlb”) are units of energy, which we will
discuss later in the quarter. Newtonmeters occur so often that they are commonly called joules
("J"), but we'll stick with Nm for a while.
Conditions for Equilibrium
Two conditions must exist for a rotational system to be in equilibrium:
1. The sum of all forces acting on the system must be zero. Mathematically, we denote
a summation of values with the symbol “”, which is a capital “sigma” in the Greek
alphabet. Say there are n different forces acting on the system, where n is the total
number of forces. We can denote each force as Fi, where i = 1, 2, 3, …, n. We can
mathematically represent the sum of all the forces acting on the system, from F1 to
n
Fn, as
F
i 1
i
. Thus we can re-write our statement that the sum of all forces acting on
the system is zero as:
n
F
i 1
i
0
III-6
2. The sum of all the torques acting on the system must also be zero. We can express
32
this mathematically as:
n

i 1
i
0
III-7
In other words, if you add up all of the clockwise torques, they will be of the same
sign as, but in the opposite direction of, the sum of all counterclockwise torques.
Example III-6: Stella and Arsenio are playing on a teeter-totter with a total length of 8 meters,
and the fulcrum in the center (see the figure below). If Stella (m = 80 kg) is at the far end of her
side of the teeter-totter, where must Arsenio and his robot pal (total mass = 160 kg) be so that
they are perfectly balanced?
A typical day at the park.
Solution: Arsenio and Stella must be applying equal torque in opposite directions in order to
have the teeter-totter be perfectly balanced. The force they are each applying is their weight, a
vector that is always pointing downward, therefore Arsenio must be on the opposite side of the
fulcrum as Stella. For the calculations, the subscript “S” refers to Stella and the subscript “A”
refers to Arsenio and his robot pal.
First we calculate Stella's torque (“S”):
m
kg  m 2

S  FS x   m S g x   80 kg  9.80 2 4 m   3136
 3136 N  m
s 
s2

33
This must also be the torque produced by Arsenio and his robot pal:
A = FAx = mAg x
Now we can solve for Arsenio's distance from the fulcrum:
 A
kg  m 2
3136
A
s2
=
=
= 2m
m
mA  g

160 kg 9.80 s2 
34
PROBLEM SET 3: TORQUE
1. Circle the correct answer. Torque is defined as:
a. the product of the length, in pounds, and the force, in feet.
b. the product of the force applied and the length of the lever arm.
c. the product of the force, lb, and the length, N.
d. the speed at which a body rotates.
2. A force of 15 lb is used to produce a torque of 45.8 ftlb. Find the lever arm.
3. A force of 15 N is used to produce a torque of 130.5 Nm. Find the lever arm.
4. A force of 9 lb is applied at right angles to a torque wrench. The handle of the wrench is 18 in
long.
a) What is the lever arm in feet?
b) Find the magnitude of the torque in ftlb.
5. A uniform rod 6 ft long is balanced at its center. A weight of 5 lb is suspended 1.25 ft from the
center on the right. A second weight is placed on the left 6 in from the center.
a) Draw a free-body diagram of the rod showing the forces as arrows.
b) Compute the clockwise torque.
c) What is the value of the counterclockwise torque? Explain how you know.
d) How many pounds is the second weight?
6. Four weights hang from a uniform 1 m long rod, which is supported at its center (50 cm). A weight
of 30 N hangs at 10 cm, a 40 N weight hangs at 35 cm and a 25 N weight hangs at 70 cm. The
rod is balanced (in equilibrium). The fourth weight is 50 N.
a) Sketch and draw the free-body diagram.
b) Using torques and conditions for equilibrium, find the location of the fourth weight.
7. When we use the English system of units, mechanical force is measured in pounds. When we use
the metric (or SI) units, mechanical force is measured in
a. newtons
b. newtons per square meter
c. pounds
d. pounds per square foot
In questions 8 -12 use the following vocabulary: vector, scalar, mass, weight or torque.
8. A measure of the amount of matter contained in an object is ___________.
9. A physical quantity described by both magnitude and direction is ___________.
10. A physical quantity described only by magnitude is _____________.
11. A measure of gravitational pull is _____________________.
12. The product of the force
_____________________________.
LAB #2: TORQUE
applied
times
the
length
of
the
lever
arm
OVERVIEW
Torque is a vector, because it has both magnitude and direction. The direction can be indicated
35
is
by “clockwise” or “counterclockwise”. When two or more torques are applied to the same object
in the same direction their values add. When the applied torques are in the opposite direction,
their values subtract..
In this lab, we will balance a meter stick at its own center of gravity so that it is free to rotate.
Then a weight suspended from one side of the rotation point can be balanced by one or more
weights placed on the opposite side of the stick.
The formula for torque is  = F x , where F is the applied force and  is the lever arm. In the
metric system, the unit of torque is the newtonmeter (Nm). The slotted weights which we will
use are marked in “grams”, which is a mass unit. The mass in grams must be converted to
weight in newtons before the torque can be calculated.
In the English system, the unit of torque is the lbft (more commonly written as ftlb).
LEARNING OBJECTIVES
The learning objectives for lab 2 are:
1. Define and understand the concepts of torque and center of gravity.
2. Understand the relationship between mass and weight.
3. Understand the difference between a vector and a scalar.
36
LAB #2: BALANCED TORQUES
OBJECTIVES:
Date_______
SKETCH OF LAB SET-UP:
LOCATION OF METER STICK BALANCE POINT
(CENTER OF GRAVITY): ___________
TRIAL
MASS
grams
FORCE
newtons
METER
STICK
LOCATION
MOMENT
ARM
meters
TORQUE (MOMENT)
CLOCKW.
A: #1
CCW.
-------
#2
-------
B: #1
-------
#2
-------
#3
------SUM OF MOMENTS:
C:#1
#2
----UNKNOWN-
COMPUTED
COMPUTED
WEIGHT_________
MASS__________
MASS
(BALANCE
SCALES)_______
CALCULATIONS:
37
ERROR______%
PART D:
LOCATION
METER STICK:
OF 250 g MASS________
MASS_________WEIGHT_________
SKETCH
FREE-BODY DIAGRAM:
EFFECTIVE LEVER ARM__________ POSITION ON METER STICK_________ How
does this position compare to the meter stick center of gravity (the original balance point)?
__________________________
38
LAB #2 ANALYSIS
1. How can you describe the direction of torque without using positive and negative signs?
2. The unit of torque in the English system is ________________________.
3. The common unit of torque in the SI system is_______________________.
4. Explain why torque is a vector rather than a
scalar____________________________________
______________________________________________________________________________
USE THE 5-STEP METHOD TO SOLVE THE FOLLOWING PROBLEMS:
5. A meter stick is suspended from its center of gravity (50.0 cm mark), and a 440 N weight is
suspended from the 40.0 cm mark. Where must a 220 newton weight be suspended to balance the
system?
6. Two children are sitting on a see-saw. A 68 lb child sits at one end, 6.0 ft from the fulcrum.
Where must a 100 lb child sit in order to balance the system?
7. A wheel and axle system has a weight of 80.0 lb suspended from the axle. The radius of the
axle is 3.20 in., and the wheel has a radius of 5.50 inches. How much weight must be suspended
from the wheel in order to balance the system?
80 lb
?
39
3. DENSITY AND SPECIFIC GRAVITY
In modern industries, fluid systems frequently are used to drive robots and circulate cooling liquids.
They're also used to operate brakes and lubricate moving parts with oil. In an automobile engine,
rapidly burning gases in cylinders cause high pressures that move pistons. A technician must
understand fluid systems to be able to measure pressures in order to operate or maintain modern
equipment.
A fluid is a gas or liquid that conforms to the shape of its container. In a car, antifreeze mixtures,
brake fluid, lubricating oils and gases in cylinders are examples of fluids. A hydraulic system is a
fluid system that uses a liquid as the fluid. A pneumatic system is a fluid system that uses air or
gas as the fluid.
Air conditioning systems use fluids to control temperature and humidity in homes, offices and
factories. They operate by using fans to create pressure in duct work that circulates cool air.
a) Density
Have you ever wondered why motor oil floats on water? Or why a balloon filled with hot air rises?
The answer is found in a physical property of materials called "density". The mass density ()of a
substance is the mass of that substance (m) divided by its volume (V):
Mass
Volume
Mass Density =
or
 =
m
V
III-8
The weight density (w) is also defined in a similar manner:
Weight Density =
or
w =
Weight
Volume
w
V
III-9
Weight density is more commonly used in the English system of units.
Mass density is expressed in units of mass divided by units of volume. In SI, mass often is
measured in kilograms. A smaller unit of mass, called the "gram," is also used. A gram is 1/1000
of a kilogram. A weight of one pound at sea level has a mass of about 454 grams (remember that
weight (w) is the product of an object’s mass (m) and the gravitational acceleration (g)). Volume is
measured in cubic meters (m3) or cubic centimeters (cm3). Thus, in SI, density is expressed in
kg/m3 or in g/cm3. In the English system density is given as mass/volume (slugs/ft3) or, more
commonly, weight/volume (lb/ft3). If density is given as pounds/volume, it's called "weight
density." Water is used as a basic reference. In SI units, water has a mass density of 1 g/cm3 or
1000 kg/m3. In English units, water has a weight density of 62.4 lb/ft3.
40
b) Specific Gravity
The specific gravity (“SG”) of a substance is the density of that substance divided by the density of
water. The following equation explains this relationship:
S pecificGravity =
Densityof S ubstance Weight Densityof S ubstance

Densityof Water
Weight Densityof Water
III-10
or
SG
 (substance)  w (substance)

 (water)
 w (water)
Since specific gravity is “density divided by density”, the units cancel out. Therefore, specific
gravity is always a pure number. For example, the specific gravity of mercury is 13.6. The table
below lists the density of some solids and liquids. In the SI system of units, specific gravity has the
same numerical value as density for any given substance, because the density of water is 1.0 g/cm3,
but carries no units. For example, copper has a density of 8.9 g/cm3 , and a specific gravity of 8.9.
DENSITY AND SPECIFIC GRAVITY OF REPRESENTATIVE SUBSTANCES
Material
Solids
Gold
Lead
Silver
Copper
Steel
Aluminum
Tin
Balsa Wood
Oak Wood
Liquids
Mercury
Water
Oil
Alcohol
Ethylene Glycol
(Antifreeze)
Density
Specific Gravity
19.3 g/cm3
11.3 g/cm3
10.5 g/cm3
8.9 g/cm3
7.8 g/cm3
2.7 g/cm3
7.29 g/cm3
0.3 g/cm3
0.8 g/cm3
19.3
11.3
10.5
8.9
7.8
2.7
7.29
0.3
0.8
13.6 g/cm3
1.0 g/cm3
0.9 g/cm3
0.8 g/cm3
1.125 g/cm3 (at 32oF)
1.098 g/cm3(at 77oF)
13.6
1.0
0.9
0.8
1.125
1.098
Example III-7: A volume (V) of 500 cm3 of a certain fluid has a mass (m) of 550 g. Find the
density () and specific gravity (SG) of that fluid.
Solution:
m
550 g
g
= =
= 1.1
3
V 500 cm
cm 3
41
SG=

H O
2
g
cm 3 = 1.1
=
g
1.0
cm 3
1.1
The density of the fluid is 1.1 g/cm3. The specific gravity is 1.1
If you know the density and volume of a fluid, you can find the mass of the fluid by multiplying
volume times density:
mass = mass density x volume or m = V
This is shown in Example III-8.
Example III-8: The fluid in a container has a volume of 400 cm3 and a specific gravity of 0.9. Find
the mass of the fluid.
Solution:
Since Specific Gravity = 0.9, Density = 0.9 g/cm3
Mass = Density x Volume
m = V = (0.9
g
)(400 cm3) = 360 g
cm3
Example III-9: Weight Density of a Fluid
A container of fluid has a volume of 2.3 ft3 and weighs 89 lb. What is the weight density of the
fluid?
w =
Solution:
w
V
and since w = 89 lb and V = 2.3 ft3,
w =
89 lb
lb
3 = 38.7
2.3 ft
ft 3
c) Buoyant Force
When an object floats or sinks in a fluid, there is always an upward force on it equal to the weight
of the fluid displaced. This upward force is called the “buoyant force”. The fluid displaced is the
amount of fluid "pushed out of the way" by the object in the fluid. If the density of the object is
greater than the density of the fluid, its weight is greater than the buoyant force, so it sinks. For
example iron sinks in water. If the object is less dense than the fluid in which it is placed, its weight
is less than the buoyant force, and the object floats. It will float at a level such that its weight equals
42
the weight of the displaced water (the buoyant force). For example, balsa wood floats easily in
water. Use the density/specific gravity table to help you answer this question: "Will lead float or
sink in mercury?"
Figure III-4 shows an object floating in a liquid. The weight of the displaced liquid equals the
weight of the object. If the density of the object is increased (but not so much that it equals or
exceeds the density of the liquid), it floats lower in the liquid, just as a boat floats lower in the water
when it is loaded. The same object will float higher in liquids of higher density and will float lower
in liquids of lower density. This is the idea used in the hydrometer, which is an instrument that
measures density or specific gravity of liquids.
Figure III-4. A block floating in water.
The apparent weight (wapp) of an object in a fluid is the difference between its weight (w) and
the weight of the fluid it displaces (the buoyant force, FB):
wapp = w – FB
43
III-11
PROBLEM SET 4: DENSITY/SPECIFIC GRAVITY
1. All fluid systems can be classified as either:
a. solid or liquid
b. hydraulic or liquid
c. pneumatic or hydraulic
d. pneumatic or gas
2. What is the density of a substance with a mass of 105 g, and a volume of 10 cm3?
3. If the density of a substance is 0.9 g/cm3 and water has a density of 1 g/cm3, the specific gravity is
________________.
4. Which is the most complete definition of buoyant force?
a) the apparent weight of an object that floats in water.
b) the upward force on an object that is equal to the weight of the fluid displaced by the
object.
5. A cube has a volume of 23.45 in3 and a weight density of 9.62 lb/ft3. Using the five-step
method, find the weight of the cube.
6. A substance has a specific gravity of 4.5 and a volume of 2 m3.
a) Find the density of the substance
b) Find the mass of the substance
7. A cube of steel has a volume of 1.6 ft3.
a) Find its weight density
b) Find the weight of the steel
8. A canoe displaces 2.3 ft3 of water when it is floating empty. What is its weight?
9. The mass of a cabin cruiser is 256.7 kg fully equipped.
a) Find the volume of water it displaces when floating on a lake.
b) What is the mass of the water displaced?
44
LAB #3: MEASURING SPECIFIC GRAVITY AND DENSITY
OVERVIEW
The mass density of a substance is a measure of the mass that that substance contains in a given
m
volume. Mathematically this is written:
 =
V
Mass 

 Mass Density=

Volume 

In the SI system, the units of mass density are g/cm3 or kg/m3. In the English system, weight
density (w) is commonly used.
w
Weight 

w =
 Weight Density=

V
Volume 

In the English system, the unit of weight density is lb/ft3.
The density of a solid may be determined by measuring the mass with balance scales, and
dividing by the volume. The volume may be found by measurement, if the solid is regular. The
volume of an irregular solid may be determined by observing the volume of liquid displaced
when it is immersed in water.
A solid object immersed in water displaces an equal volume of water. The object is lighter in
water than in air because of the displaced water. The apparent weight (wapp) of the object in
water is the weight of the object in air (wactual) minus the weight of the displaced water (wwater):
wapp = wactual - wwater
The buoyant force (FB) on an object immersed in water is equal to the weight of the water it
displaces (wwater).
In this lab you’ll measure the mass and volume of an object and calculate its mass density. You
will also weigh another object in air and in water to see the difference between actual and
apparent weight.
The specific gravity (SG) of a substance is a number that states the ratio between the density of
a substance and the density of pure water. It is written:
ρ
SG 
ρ(H 2 O)
Either mass density or weight density may be used to find specific gravity, but the units must be
identical for the substance and water. If you’re calculating specific gravity using the weight
density, the formula would be:
SG 
w
 w (H 2 O)
In either case, specific gravity has no units; it is a pure number ratio.
45
A hydrometer is a device that measures the specific gravity of liquids. You will use a
hydrometer in this experiment to measure the specific gravity of some liquids.
The scaled hydrometer is an air-filled tube that is weighted at the bottom and sealed so that it
floats upright in liquid. It is placed in water, and the level at which it floats is marked 1.000; any
other liquid in which it floats at the same level must have a density equal to water, so its specific
gravity is 1.000. If the tube floats higher in an unknown liquid, that liquid is denser than water.
If the tube sinks to a lower level, then the liquid is less dense than water. The tube is calibrated
in very small increments so that it is very precise, and the specific gravity can be read directly.
The density of the liquid may be determined by measuring the specific gravity with a hydrometer
and computing with the equation:
  SG(H 2 O)
The learning objectives for Lab 3 and 4 are:
1. Distinguish between hydraulic and pneumatic systems.
2. Define and understand the concepts of mass and weight density, specific gravity, pressure and
buoyant force and be able to use these concepts in fluid problems.
3. Learn how to use a hydrometer, manometer, pressure and vacuum gauges to take readings.
4. Understand the differences between gauge, atmospheric and absolute (total) pressure, and how to
convert from one to the other.
5. Explain the effect of depth on the pressure in a fluid and be able to solve for the pressure at any
depth.
6. Be able to convert pressure from the SI unit system to the English system and vice versa.
46
LAB #3: SPECIFIC GRAVITY AND DENSITY
Date_______OBJECTIVES:
SKETCH OF LAB SET-UP:
TABLE 1: DENSITY OF A SOLID OBJECT
Sketch:
Dimensions
Volume
V (cm3)
Mass
m (grams)
Mass
Density
 (g/cm3)
Material:____________________Text Book Density Value__________
TABLE 2: ACTUAL AND APPARENT WEIGHT OF AN OBJECT
Actual Weight of Aluminum
Cube (N)
Apparent Weight of Cube (N)
Buoyant Force (weight of
water displaced) (N)
TABLE 3: SPECIFIC GRAVITY OF A LIQUID
Mass of Beaker:__________ Mass of Beaker and Sample: 1) _________ 2) _________
Specific
Gravity
(Float
Reading)
Mass
Density

(g/cm3)
Sample
Volume
V
(cm3)
Calculated
Sample
Mass
m (grams)
1
2
47
Calculated
Sample
Density
 (g/cm3)
% Difference
 |  calc -  meas | 


=  BestValue 
x 100%
Lab #3 ANALYSIS: 1. How did the measured mass density of Sample #1 (balance scales)
compare with the computed mass density using the specific gravity?
_________________________________________________________
2. Which value was the most accurate? Explain your reasoning.
_____________________________________________________________________________
3. What range of specific gravity can be measured by the scaled hydrometers?
________________________________________
4. How can you tell from the specific gravity of an object whether it will sink or float in water?
______________________________________________________________________________
Use the 5-step method to answer the following problems.
5. If the specific gravity of a statue of Norm Dicks is 3.2, find its mass density in g/cm3. Also
find its weight density in lb/ft3.
6. A cylinder of material has a length of 1.5 ft and a radius of 8.0 inches. It weighs 89.0 lbs. A)
What is its weight density? B) What is its specific gravity? C) Will it sink or float in water?
7.
Find the mass, in grams, of a 90 cm3 steel block.
8. The buoyant force on an immersed object is equal to the weight of the liquid displaced
(Archimedes' Principle). An irregular object weighs 83.4 lb. When placed in water, its
apparent weight is only 21.0 lb. A) What was the buoyant force on the object? B) What
volume of water was displaced? C) What is the weight density of the object?
48
4. FLUID ENERGY SYSTEMS: PRESSURE
The source of motion, or force-like quantity, in fluid systems is pressure. Imagine a molecule or
atom inside a balloon. This particle moves with some velocity, and the more energy it has, the
greater its velocity is. It pushes against the inner wall of the balloon as it repeatedly bounces
back and forth against it. If more molecules were inside the balloon, it would look like a bag of
microwave popcorn cooking as the energetic particles pushed against the inner wall. As we
filled the balloon with more and more particles, the motion of individual particles would become
blurred, and the balloon would appear to bulge uniformly outward in all directions. This is why
we can't detect individual collisions of particles against a balloon wall. An average balloon
23
contains about 10 particles! The force exerted by all of the molecules inside the balloon on its
inner wall is therefore extremely uniform in all directions. Rather than calculate the force of
each individual particle on the wall, we speak in terms of the average force per unit area, that is,
the pressure on the inside of the balloon. We can express the definition of pressure in equation
form as
P =
F
A
III-12
where P = pressure, F is the force and A is the area over which the force is applied. Pressure has
2
units of force divided by area, such as newtons per square meter (N/m ), also known as pascals
2
2
(Pa), pounds per square inch (lb/in or psi) or pounds per square foot (lb/ft or psf).
The energy and velocity of the particles within a substance increase with temperature, and so
does the pressure. Therefore the pressure in a gas-filled container rises if you heat it inside, and
it decreases if you cool it. Another way to increase the pressure in a container is to add more
22
particles. If, for example, you have 5 x 10 air molecules (about 2 liters at atmospheric
22
pressure) in a sealed metal tank and then force in another 5 x 10 air molecules, the pressure in
the tank will double.
Suppose you have a bag made of leak-proof material with a hole at the bottom. If you heat the
air inside, the pressure in the bag will increase. Motion occurs in a fluid as a result of a pressure
difference, which is why pressure is considered to be a force-like quantity. As the pressure
increases, air will leak out through the hole at the bottom, because a gas always moves from a
region of higher pressure to a region of lower pressure. The larger the pressure difference, the
faster the molecules move. Less molecules inside the bag means lower density, and if the
density inside the bag is lower than the density outside, the bag will be buoyant and will rise.
Congratulations! You've just made a hot air balloon!
Hot air rises because it expands outward until it exerts the same pressure as the cooler air around
it. To do this it must become less dense than the cooler air surrounding it. In meteorology,
rising pockets of warmer air are called "thermals", and larger birds circle inside them to increase
their altitude.
a) Atmospheric Pressure
On Earth, we live at the bottom of a thick blanket of air. Since air has weight, and the blanket of
air is mostly above us, we feel its weight as a pressure pushing on us from all directions. At sea
level, the atmosphere presses on us (and everything else) with a force of 14.7 pounds for every
square inch of body or object surface area. At higher altitudes - like the peak of a high mountain the pressure is a little less. We call the sea level value of 14.7 lb/in2 (or 1.013 x 105 N/m2) the “one
standard atmosphere”, or “one atmosphere”. It acts equally in all directions-upward, downward or
49
sideways. Since pressure does act equally in all directions at any point in a fluid - unlike a force that
acts in specific directions - pressure is not a vector. Pressure is completely described by specifying
its magnitude; no direction is required. The table below lists units of atmospheric pressure often
used in SI and the English system.
UNITS OF ATMOSPHERIC PRESSURE
1 atmosphere = 14.7 lb/in2 (often written as psi)
= 2117 lb/ft2
= 1.013 x 105 N/m2 or pascal (Pa)
= 33.92 ft of water
= 760 mm of mercury (mm Hg) (760 torr by international agreement)
= 29.92 in. of mercury (in Hg)
Note: The units "ft of water", "mm of mercury", and "inches of mercury" do not appear to be
pressure units because they do not indicate a force unit divided by area. However, they are
commonly used; they indicate the pressure at the bottom of a column of a given height of that
liquid.
b) Absolute and Gauge Pressure
When working with fluid systems, pressure measurements often are reported either as absolute
pressure or as gauge pressure. It is important to know the difference. For example, when filling a
tire with air, we use a pump which forces more and more air into the tire. As the tire fills - and the
pressure inside increases - it takes on a rigid shape. If we check the air pressure with a tire gauge it
might read 30 lb/in2. But what does this reading mean? Is it 30 lb/in2 in gauge pressure or 30 lb/in2
absolute pressure?
To answer these questions, we need several definitions. Absolute pressure is the total
pressure measured above a reference of zero pressure (a perfect vacuum). Gauge pressure is the
pressure measured above atmospheric pressure. Gauge pressure is a measure of how much greater
the air pressure inside the tire is over the air pressure outside the tire. Gauge pressure is generally
measured with a gauge, hence it's name "gauge pressure."
Total pressure, atmospheric pressure and gauge pressure are related in a simple equation, as follows:
Total Pressure = Gauge Pressure + Atmospheric Pressure
Ptot = Pabs = Pg + Patm
III-13
Example III-10: A tire gauge is used to measure air pressure in a tire. It shows 30 lb/in2 - the
gauge pressure. If the atmospheric pressure equals 14.7 lb/in2, the total pressure is equal to:
Ptot = 30
lb
lb
lb
(gauge) + 14.7 2 (atmospheric) = 44.7 2
2
in
in
in
50
The trapped air inside the tire pushes out on each square inch of wall surface with a pressure of 44.7
lb/in2. The atmospheric air (air on the outside) pushes in on each square inch of the tire wall with a
pressure of 14.7 lb/in2. The difference - 30 lb/in2 - is the gauge pressure. That's what the gauge
measures.
The tire pressure gauge is a useful pressure-measuring device. Its operation is quite simple (see
Figure III-5). The tire gauge is made of a moveable bar indicator and coiled spring housed in a
cylindrical tube. When the gauge is placed over the valve stem of a tire, the gauge chamber and tire
become sealed. The pressurized air from the tire flows into the gauge chamber. This forces the coil
spring to compress. As the spring compresses, it pushes the calibrated bar indicator out of the
cylinder housing. When the force of the compressed spring equals the force caused by the pressure
within the gauge chamber, the forces are balanced. The exposed calibrated bar indicates the correct
gauge pressure - the pressure within the tire.
Figure III-5. Tire pressure gauge.
Let's apply the equation F = PA, that is, Force = Pressure x Area, to calculate the total force
pushing outward on a window of a commercial airplane when the plane is flying at a high altitude.
To determine the strength of window material to use, the designers need to know the total force on
the window so it can be fastened in place and not "blown out" of the airplane cabin at high altitudes.
Example III-11: Force on an Airplane Window
An airplane window has a surface area of 144 square inches. Air pressure inside the cabin is 14.7
lb/in2. Air pressure outside the window is 6.7 lb/in2. Find a) the force pushing inward on the
window, b) the force pushing outward on the window, and c) the net force on the window.
51
Pout = 6.7 psi
Pin = 14.7 psi
Force on an airplane window




lb 

a) Fout = Pout  A =  6.7 2  144 in2 = 964.8 lb (outside pressure)
in 

lb 

b) Fin = Pin  A =  14.7 2  144 in2 = 2116.8 lb (inside pressure)
in 

c) Fnet = Fin – Fout = 2116.8 lb - 964.8 lb = 1152 lb
Hopefully the structure of the window provides a force of equal magnitude in the opposite
direction!
c) Pressure in a liquid
Liquids are different from gases in that they compress very little, so little that they are generally
regarded as being "incompressible". Since the density in an incompressible fluid remains
constant, the pressure at any depth in a liquid-filled container depends only on the height of the
liquid above that point (h), the density of the liquid (), and the acceleration of gravity (g):
P = gh
III-14
The equation can be written in terms of weight density (w)as
P  w  h
III-15
3
Example III-12: The density of water is 1000 kg/m , and the gravitational acceleration at sea
52
2
2
5
2
level is 9.80 m/s . Atmospheric pressure at sea level is 101,000 N/m (or 1.01 x 10 N/m ). At
what depth will the pressure be equal to twice the atmospheric sea level pressure (also known as
two atmospheres ) ?
Solution: At the surface the pressure is already equal to one atmosphere (1 atm), so we need to
find the depth that will give us one additional atmosphere of pressure. We can rearrange the
above equation to read
N
1.01 x 105 2
P
m
h =
=
= 10 m
kg
m
 g


3
 1.0 x 10
  9.80 2 

m3  
s 
The tricky part in this solution is the dimensional analysis, which becomes easier once you
2
replace newtons with kgm/s . The solution tells us that for every ten meters of depth the
pressure increases by one atmosphere, so at 0 m the pressure is 1 atm, at 10 m the pressure is 2
atm., at 20 m the pressure is 3 atm., etc. 10 m is equivalent to about 33 ft. Sea water is slightly
denser than fresh water; here 1 atm. corresponds to only 32 ft. of depth. The hull of a submarine
320 ft. below the surface experiences a pressure differential between the inside and outside of the
bulkhead of 10 atmospheres.
d) Hydraulic Lift
A hydraulic lift at the local service station is used to lift cars and trucks off the ground. It works
because liquids are incompressible (liquid can't be squeezed into a smaller volume), and liquids
transmit pressure equally in all directions over a long distance. The air compressor shown in Figure
III-6 increases the pressure above the hydraulic fluid. This pressure is then transmitted to the
bottom surface of the lifting "piston." Because the pressure in the fluid is high, and the area of the
lifting piston is large, a large pushing force is exerted on the lifting piston. This force is enough to
lift the truck.
Figure III-6. The hydraulic lift.
Example III-13: Fluid Pressure in a Hydraulic Jack
The hydraulic jack shown below is rated at 4000 lb lifting capacity and has a large lifting piston
with a diameter (D) of two inches. Find the fluid pressure in the jack at maximum load.
Solution: First, compute the area of the piston: A =  
53
D2
(2 in) 2
= 
= 3.14 in 2
4

Calculate the pressure: P =
F
4000 lb
lb
=
2 = 1274
A
3.14 in
in 2
The hydraulic jack.
e) Equilibrium in Fluid Systems
Liquids or gases move in a fluid system when pressure differences exist between different points in
the system. If there's no pressure difference, there's no movement. For this reason, it's useful to
think of pressure acting like a force in fluid systems. Let's make this point more clear. Figure III-7a
shows two tanks connected by a pipe that contains a gate valve. The pressure at the bottom of each
tank is different. That's because the water level in tank 2 is higher than in tank 1. Pressure P2, at the
bottom of tank 2, is higher than P1, the pressure at the bottom of tank 1. When the pipe between the
two tanks is connected at the bottom of each tank, the pressure on the left side of the valve is P1 and
the pressure on the right side is P2. Since P1 is less than P2, there is a pressure difference across the
valve. What happens if the valve is opened?
TANK 1
Valve
TANK 2
P1
TANK 1
P2
Valve
TANK 2
P1
Figure III-7a. P1 < P2
P2
Figure III-7b. P1 = P2
54
Since pressure P2 is greater than P1, there will be a force per unit area on the right side of the valve
greater than the force per unit area on the left side of the valve. Water then will be pushed through
the valve from Tank 2 to Tank 1. Water will flow until levels in the two tanks are equal. Then, the
system will be in equilibrium. This situation is shown in Figure III-7b.
Now consider Figure III-8. Here, there are two tanks filled to the same level. (h1 = h2). Tank 2 has
a larger diameter than tank 1, so it contains much more water. But since pressure on the bottom of
the tank depends only on height of water contained, the pressure on the bottom of tank 1 and tank 2
is the same. Pressure on the bottom doesn't depend on the shape of the tank or the amount of water
it contains, just the height of the water column above the bottom. What happens when the gate
valve in Figure III-15 is opened?
Figure III-8. Water Tanks with different diameters
f) Measuring Pressures
The same ideas discussed in the illustration of the two tanks connected by a pipe are used in an
instrument that measures fluid pressure. This useful instrument is called a "manometer."
Manometers are used throughout industry to measure gas pressure and pressure difference between
two points in a system. For example, measuring pressure difference across a filter in an airconditioning duct will show if the filter is clean or clogged.
Manometers come in various shapes and often use different liquids (like water or mercury) as
indicators. The simplest manometer has a U-tube shape (see Figure III-9). If pressure differences to
be measured with the manometer are large, mercury is used as the fluid. If pressure differences are
small, water is used as the fluid, since water is much less dense than mercury. Mercury has a
weight 13.6 times larger than water. The weight density of water is 62.4 lb/ft3; the weight density of
mercury is 848.6 lb/ft3. In Figure III-9a, the gas pressures in containers A and B are equal. Each
exerts the same force per unit area on the top surface of the two mercury columns in the U-tube.
Since PA = PB at the top of each mercury column, the mercury must be at the same level in each
arm, hA = hB. In Figure III-9b, the pressure in container C is higher than in container A. Gas C
pushes down harder on the right column of mercury than Gas A pushes down on the left column.
Therefore, the mercury column moves down in the right arm-and up in the left arm. By measuring
the difference in height between the two columns of mercury in the U-tube, and using the equation
P = wh (discussed earlier), we can find the difference in pressure between gas A and gas C. Then,
if we know the pressure of gas A, we can find the pressure of gas C.
55
Figure III-9a. Pa = Pb
Figure III-9b. Pa < Pc
56
PROBLEM SET 5: PRESSURE
1. What is the gauge pressure at a depth of 60 ft in fresh water? Is this pressure equal to, less than or
greater than the gauge pressure at the same depth in salt water (w=64.2 lb/ft3)?
2. A 130 lb woman, wearing spike-heel shoes with a heel 1/4 in by 1/4 in is standing on the floor.
Calculate the pressure she exerts on the floor if all of her weight is on one heel.
3. A column of air that extends from sea level to the edge of outer space and has a cross sectional
area of 1 in2 (circle correct ones)
a. weighs 14.7 lb
b. is the cause of atmosphere pressure
c. applies a pressure of 2117 lb/ft2
d. weighs the same as a 1 in2 column of water 33.92 ft
high
4. A tire pressure gauge measures air pressure in a 10-speed bicycle tire at 55 lb/in2. Atmospheric
pressure is 14.7 lb/in2. The absolute pressure in the tire is __________. The gauge pressure of air in
the tire is ____________.
5. Pressure in fluid systems is defined as
a. force times the area upon which the force acts
b. force divided by the area upon which the force acts
c. force times the volume of the fluid
d. force divided by the volume of the fluid
6. Explain why pressure is not a vector quantity.
7. What must exist between different points in a fluid system for liquids or gases to move in that
system
a. equal pressure
b. temperature stability
c. resistance
d. pressure difference
8. Calculate the absolute pressure in pounds per square inch at the bottom of a lake that is 250 ft
deep (the weight density of water is 62.4 lb/ft3).
9. Calculate the depth of water in an open tank if the gauge pressure at the bottom is 105 N/m2 (the
density of water is 1000 kg/m3).
10. What is the average pressure on the face of a dam if the water exerts a total force of 12 x 108 N
and the surface area of the dam is 600 m2?
11. What is the lifting capacity of a hydraulic jack if the fluid pressure at maximum load is 1600 lb/in2
and the piston that lifts the load has a radius of 2 inches?
12. How much would a man have to weigh, when standing on one foot, to exert the same pressure on
the floor as did the woman in problem 2? Use 18 in2 as the area of the bottom of the man’s foot.
Is this reasonable?
13. A tank is filled to a depth of 6 m with a certain liquid. What is the density of the liquid if the
pressure at the bottom of the tank is 105 N/m2?
57
14. The porthole of a small submarine will break when the average force distributed over its area
exceeds 106 N. If a submarine’s porthole has an area of 2 m2 how deep below the surface of the
ocean can this submarine go before the glass in the porthole will break?
15. What type of unit does 15 ft of water represent?
16. Show how P = gh equals P = wh.
17. Can you ever experience an absolute pressure less than zero?
18. What pressure, in lb/ft2, does 2.65 x 105 N/m2 represent?
58
LAB #4: MEASURING PRESSURE
OVERVIEW
In order for a gas or liquid to be controlled, it must be confined to a container. Pressure is the
force that a gas or liquid exerts on the walls of its container, divided by the area of the container.
P = F/A
(Pressure = Force/Area)
The basic unit of pressure in the SI system is the pascal (Pa). A pascal is equivalent to
1
Other units are also used in the SI system, such as the bar, millibar and mm of Hg
(also known as a torr). In the English system, the common unit is the psi (lb/in2). Other units
used include lb/ft2 and inches of Hg. The equivalencies for these units are given in a table in the
back of this lab manual.
N/m2.
All pressures that we measure are relative to atmospheric pressure. This is because our
atmosphere surrounds our measuring systems. Absolute Pressure is the pressure relative to a
perfect vacuum, or zero pressure. Absolute pressure is found by adding atmospheric pressure
to measured pressure (also known as gauge pressure). The formula for finding absolute
pressure is given below.
Pabsolute = Pgauge + Patmospheric
In this experiment you will be using pressure gauges and a water manometer to measure
both pressure and vacuum. A manometer is a U-shaped tube containing water or mercury
(connected to a hand pump in Figure III-10). When the liquid is at the same level on both sides,
the pressures on each side are equal. In the figure below, the pressure on the left is greater than
atmospheric, and so the liquid on the right side is pushed to a higher level. The distance between
the two levels indicates the pressure difference.
EXAMPLE OF UNIT CONVERSIONS:
Convert a pressure difference of 12900 pascals to units of psi.
Use the values given in the reference tables in the back of this manual.
12900 Pa x
1 psi
= 1.87 psi
6900 Pa
59
THE MANOMETER:
Figure III-10. Manometer connected to a hand pump.
LEARNING OBJECTIVES
The learning objectives for Lab 4 are:
1. Understand the relationship between gauge (or gage), absolute and atmospheric pressure.
2. Learn how to calculate pressure differences using a manometer.
3. Learn how to read a pressure gauge.
60
LAB #4 MEASURING PRESSURE
OBJECTIVES:
SKETCH OF LAB SET-UP:
DATA TABLE 1
TRIAL
Date:_______
GAUGE & MANOMETER READINGS
Mechanical
Gauge
Pressure
Manometer
Readings
PM (mm H2O)
Water Column
Height
h (mm)
Computed
Pressure
Pgauge
Pgauge
#1
#2
#3
-------
-------
---------
-------------
#4
-------
-------
--------
-------------
DATA TABLE 2 ABSOLUTE PRESSURE CALCULATIONS
TRIAL
Measured Pressure
Atmospheric Pressure
Absolute Pressure
Pgauge
Patm
Pabs = Pgauge + Patm
#1
#2
#3
#4
61
Lab #4 ANALYSIS:
1.
Using the pressure conversion chart, determine the pressure in psi represented by a height of 1.0
mm of water in the manometer. (Show your work.)
2. Using the pressure conversion chart, determine the pressure in psi represented by a height of
1.0 mm of Hg (mercury) in the manometer. (Show your work.)
3. The maximum height of liquid in the manometer is 1000 mm. Compute the maximum
pressure difference that the manometer can measure when filled with water and when filled with
mercury:
1000 mm water =__________ psi
1000 mm mercury = _________psi
4. Explain the difference between absolute pressure and gauge pressure.
5. The “ 30-0-30" gauge reads pressures from +30 psi to -30 inches of Hg. Would a gauge ever
be calibrated to read a vacuum of -30 psi? Explain your answer.
______________________________________________________________________________
_____________________________________________________________________________
6. A mercury manometer has a height of +6.0 inches on the right side, which is open to the
atmosphere. The left side reads +16.0 inches. Draw a sketch. A) is the measured pressure
greater or less than atmospheric? B) Find the gauge pressure in psi, using a pressure conversion
chart. C) Find the absolute pressure in psi.
SKETCH & CALCULATIONS:
ANSWERS:
A)_________________ B)__________________ C)___________________
62
5. ELECTRICAL ENERGY SYSTEMS
a) Introduction
Voltage, or potential difference, is the force-like quantity in electrical systems that causes the
displacement, or a change in the rate of displacement, of charged particles. The potential
difference in a system is a measure of charge separation.
Charge is a fundamental characteristic of matter, just like mass. Atoms are composed of
(positively charged) protons and neutrons (no charge) in the nucleus, which is surrounded by
negatively charged electrons (see Figure III-11). The terms "positive" and "negative" serve only
to differentiate between the two types; Benjamin Franklin could just as easily have defined the
electron charge as positive. The electron and proton charges are of exactly the same magnitude,
although the electron mass is nearly 2000 times less massive than either protons or neutrons,
which are of nearly equal size.
Figure III-11. Structure of an atom.
The electric force causes like charges to repel each other and opposite charges to be mutually
attractive. The positive nucleus attracts negative electrons, but is repelled by other (positive)
nuclei. Charges of opposite sign can cancel each other out; an atom with the same number of
protons as electrons carries no net charge.
b) Force in Electrical Systems
We have described how the gravitational force between masses is proportional to the product of
the two masses and inversely proportional to the square of the distance between their centers of
mass (see Figure III-2 and Equation III-3). Similarly, the electric force between two objects is
proportional to the product of the charge on each object and inversely proportional to the square
of the distance between them. Coulomb’s Law describes the magnitude of the electric force
between the two objects,
F =
kq1q 2
2
r
63
III-16
Where F is the electric force between the two objects, k is a constant, approximately equal to 9 x
109 N·m2/C2, q1 and q2 are the net charges on objects 1 and 2, respectively, and r is the distance
between the two particles. There is however, one important difference between the gravitational
force and the electric (or Coulomb) force. While the gravitational force between masses is
always mutually attractive, the electric force between two charged particles can be either
attractive or repulsive, depending on the signs of the charges. Like charges (that is, two positive
charges or two negative charges) will exert a repulsive force on each other, while charges of
opposite (positive and negative) sign will attract each other. In other words, if the product of q1
and q2 is negative, the force is attractive; if the product is positive, the force will be repulsive
(see Figure III-12).
a)
b)
+
+
c)
-
-
Figure III-12. The electric force between oppositely charged particles is mutually attractive
(a), while the electric force between like charges, as in (b) and (c), is repulsive. Arrows
denote the direction of the force.
Example III-14: The electric force between an electron and a proton.
A distance of 1 cm separates an electron and a proton. What is the electric force between the
charges?
-
1 cm
+
Solution:
Electrons and protons each carry a charge (q) in coulombs of 1.602 x 10-19 C,
although the electron's charge is negative and the proton's is positive. The distance between the
two particles is 0.01 m, so the electric force is
F =
kq1q 2
r2
2

9 x 109 N  m  1.602 x 10-19C -1.602 x 10-19 C
2

C 
=
0.01 m2
= -2.3 x 10-24 N, where the (-) sign indicates an attractive force.
This force may seem small, but consider that the gravitational force between these two particles
at this distance is about 10-63 N, or about 10 thousand billion billion billion billion times smaller
than the electric force!
Example III-15: The electric force between an oxygen ion and a proton.
64
A doubly charged magnesium ion (Mg++) ion has a positive charge q1 = (2)(1.602 x 10-19 C) =
3.204 x 10-19 C. It is 1 cm away from a proton. What is the electric force between the two
charged particles?
+
+
1 cm
+
Solution: q2 = 1.602 x 10-19 C and r = 0.01 m, so that the electric force is
F =
kq1q 2
r2
2

9 x 109 N  m  3.204 x 10-19 C 1.602 x 10-19 C
2

C 
=
0.01 m2
= 4.6 x 10-24 N, where the positive sign indicates a repulsive force.
c) The Electric Field
The magnitude of the electric field (E) at any point in space is defined as the electric force per
unit positive charge at that point. We can write this as
E = F/q
III-17
where F is the electric force in newtons and q is the charge at that point in coulombs. The
electric field strength is therefore given in units of N/C. Since 1 volt (= 1 V) equals 1 N·m/C,
volts/meter is an equivalent unit for the electric field strength. The electric field is a vector
quantity; it has both magnitude and direction. Electric field lines can be used to describe the
direction of the electric field vector at any point in space. An electric field line is the path a
single positive charge would move if it were only acted upon by the electric field. Figure III-13
shows the electric field lines for a proton, an electron, a proton-proton pair, and a proton-electron
pair.
a)
b)
c)
d)
65
Figure III-13. Electric field lines for an electron, a proton, a pair of protons and a protonelectron pair.
d) Voltage
Separating charges takes energy; the further apart they're separated, the greater the amount of
energy required. The potential difference between two points on an electric field line is given
by
V = E·d,
III-18
where V is the potential difference, or voltage difference, E is the electric field strength and d
is the distance along the electric field line between the two points. V is measured in units of
volts. Whenever positive and negative charges are separated, a potential difference (voltage)
results. A particle in this electric field feels a force proportional to its own charge, and is
accelerated in the direction of maximum opposite charge. The current is the rate at which charge
moves. It is a vector quantity, having both magnitude and direction. If there is no potential
difference, the system is in equilibrium, and no charge flows. Current is measured in units of
amperes (A). One ampere of current is equivalent to a charge flow rate of one coulomb per
second.
Voltage sources include batteries, generators, solar energy and rubbing a cat's fur on a dry day
(static friction!). The potential difference between any two points in an electrical system is
measured in volts (V). If a circuit contains more than one voltage source in series, the total
voltage is the sum of the individual sources.
In Figure III-14 below, the batteries in series at left have a net voltage of 10 V since they both
cause current to flow in the same direction. In the figure at right, however, the two batteries
produce the same amount of flow in opposite directions, and there is no current.
Current
-
+
Current
-
5V
Current
+
-
5V
Current
+
5V
LOAD
+
-
5V
LOAD
Figure III-14a. 10 V circuit.
Figure III-14b. 0 V circuit.
e) Electrical Circuits
An electrical circuit typically includes a voltage source, a conductor (wire, for example), a control
device and a load. Rather than use pictures of the components comprising a circuit we use a
shorthand system of symbols to identify each component in a circuit. The use of these symbols to
66
draw a circuit is called a schematic diagram of the actual circuit. In Figure III-15 the symbols for
some commonly used circuit components are given and a sample schematic diagram is drawn.
voltage source
DC
AC
resistor (or any other load)
or
67
switch
light bulb
Voltmeter
Ammeter
V
A
+
-
Figure III-15. Circuit components.
There are two common types of current sources: direct current (DC) and alternating current (AC). In
DC circuits the electric charge or current always moves in one direction. In AC circuits the electric
charge or current moves first in one direction, then reverses its direction. The number of times that
the AC current reverses direction per second is called the frequency of the AC voltage supply. Each
cycle includes two changes in current direction. The time for one cycle of the AC current to pass a
point in a circuit is called the period of the current. The period and frequency are inversely related,
that is the period is the reciprocal of frequency and vice versa. In the United States we use a
frequency of 60 cycles per second for our AC voltage sources while in Europe and Japan 50 cycles
per second is used. The units of cycles per second are typically referred to as “hertz” (Hz), in honor
of physicist Heinrich Hertz. Batteries are the most common source of DC current, while alternators
and generators are the most common source of AC current.
To measure the voltage in a circuit we use a voltmeter and install the voltmeter probes in parallel
with the component whose voltage we wish to measure. The voltmeter is designed to affect the
circuit as little as possible. To measure the flow of electric charge or current in a circuit the
ammeter is installed in series with the circuit.
Resistance restricts the passage of current in the system. Any load in an electric circuit provides
resistance, but those specifically designed for such a purpose are called, appropriately enough,
resistors. Under normal operating conditions, Ohm’s Law generally gives the relation between the
resistance (R) in a DC circuit and the current (I) and voltage (V):
OHM’S LAW:
V = IR
We will discuss electrical resistance in greater detail next quarter.
68
III-19
PROBLEM SET 6: VOLTAGE/CIRCUITS
1. When electrical charge moves continuously in one direction through a conductor, it is called
_____________________.
2. When electrical charge moves back and forth through a conductor, it is called
_________________________________.
3. When frequency and hertz are used to describe electrical current, they are describing _________
voltage.
4. Voltage is considered a force-like quantity because it (select all correct answers):
a. moves electrons through a circuit
c. acts as a prime mover of electrical charge
b. pushes conductors through a circuit
d. is found in electrical circuits
5. Determine the total voltage, VAB, in each circuit below
6V
6V
-
+
+
+
9V
-
-
+
9V
-
6. The term, 115 volt AC, 60 Hz, means that the voltage is 115 volts and the alternating current
a. alternates 60 times a second
c. has a frequency of 60 cps
b. changes direction 120 times/s
d. all of the above
7. “Schematic diagrams” are used to represent electrical circuits. Draw the symbol(s) for each circuit
element below:
a. A battery (show polarity)
b. A switch
c. A voltmeter
d. A light bulb
e. Two batteries in series
f. A complete circuit with a battery, bulb and switch
8. a) Draw a circuit with a 6 V battery and a 9 V battery in series with a closed switch and one light
bulb. Show the polarity of the batteries. With an arrow, show the direction of the electrons flowing in
the circuit.
69
b) Show where the voltmeter would be placed to measure the total voltage of both batteries. C.
What is the total voltage?
9. In an alternating current, the period is the inverse of
a. charge
b. frequency c. voltage
d. time per cycle
70
LAB #5: MEASURING VOLTAGES
OVERVIEW
Voltage measures potential difference, which causes electrons to move through a completed
circuit. Current is the rate of flow of these electrons. The power source in a circuit provides the
voltage. Polarity describes the orientation of the positive (+) and negative (-) terminals of a
power source or measuring device used in a circuit.
Electrons move through a circuit in two ways. Direct current (DC) moves in only one
direction, and is the result of a constant polarity power source such as a battery. Alternating
current (AC) moves in two opposite directions, in a regularly alternating pattern. AC is the
result of a power source with changing polarity, such as an AC generator.
Batteries are voltage sources that derive their energy from a chemical reaction. Batteries can be
connected together to increase or decrease the total voltage of the power source. To increase
voltage, the positive terminal (cathode) of one battery is connected to the negative terminal
(anode) of the next battery. (Fig. III-16a)
If you reverse the polarity of one of the batteries, that battery opposes the others, and lowers the
total voltage. (Fig. III-16b)
Current
-
+
Current
-
Current
+
-
LOAD
Current
+
-
+
LOAD
Figure III-16. a) Voltages add. b) Net Voltage is the difference between the individual
batteries.
Batteries may also be connected in parallel. They must be equal in voltage. The voltage is not
increased, but each battery supplies only part of the current, so they will last longer.
71
The learning objectives for Labs 5-7 are:
1. Be able to differentiate between AC and DC energy sources and between series and parallel
circuits.
2. Learn how to use an analog and digital multimeter as a voltmeter including placement in a
circuit.
3. Learn how to represent electrical circuits and components using schematic symbols and
diagrams.
4. Describe how batteries may be connected in a series circuit and/or a parallel circuit and the
advantages of each.
5. Learn how a solenoid works and give examples of how they can be used.
6. Differentiate between solenoid holding force and pulling force.
72
LAB #5 MEASURING VOLTAGES
OBJECTIVES:
Date__________
SKETCH OF LAB SET-UP:
TABLE 1
Analog Multimeter
Digital Multimeter
Source
Voltage
Reading
VA
Switch Settings
Function
Switch
Range
Switch
(volts)
Wall
Outlet
9-V
Battery
6-V
Battery
73
Voltage
Reading
VB
Switch Settings
Function
Switch
Range
Switch
(volts)
DATA TABLE 2
Battery
Comb.
Analog Multimeter
Voltage
Reading
VA
Digital Multimeter
Switch Settings
Function
Switch
Voltage
Reading
VB
Range
Switch
(volts)
Switch Settings
Function
Switch
Range
Switch
(volts)
1
2
3
4
LAB #5
ANALYSIS:
1. A voltage with constant polarity (positive or negative) is called
______________________________
2. A voltage that changes polarity in a regular fashion is called
______________________________
3. Explain what range setting should be chosen to avoid damaging the voltmeter when measuring
an unknown voltage:
____________________________________________________________
4. Show a schematic diagram indicating 1.5 V batteries in series that will provide 4.5 volts:
74
5. Show schematic diagrams of two different combinations of 1.5 V, 6 V, and/or 9 V batteries
in series which will produce a total of 18 volts. (Don't use reverse polarity for either
combination.)
6. Show a schematic diagram with three 1.5 volt batteries in parallel.
A) What is the total voltage?___________
B) What advantage is obtained from connecting batteries in parallel?
75
LAB #6
ELECTRICAL CIRCUITS
OVERVIEW
Electrical energy is useful when changed to heat, light, or mechanical energy. An electrical circuit
is required to convert electrical energy to a useful form. An electrical circuit has four essential
parts: a power source, a control element, conductors, and a load (see Figure III-17A). The power
source supplies electrical energy to the circuit. The control element is a switch, a dimmer, or a
variable speed control for the circuit. The conductor provides the path for the current flow through
the circuit. The load converts electrical energy to a useful form.
A series circuit contains only one conductive path for the current (see Figure III-17B). As the
current travels around the circuit, the voltage drops across each load when converting the electrical
energy to other energy forms. The total voltage across the circuit must equal the sum of the
voltages used up in each load. This can be expressed by the following equation:
Vsource = V1 + V2 + V3
A parallel circuit contains two or more current paths (see Figure III-17C). The current in a parallel
circuit divides, and each path conducts only a portion of the total current. The sum of these currents
equals the total current leaving the battery. In a parallel circuit, the voltage across each load is the
same as the voltage read directly across the battery.
Itotal = I1 + I2 + I3
Vsource = V1 = V2 = V3
In this experiment you will construct a series and a parallel circuit, using light bulbs as the load.
The voltage is measured "across the load", which means that the meter must be connected in
parallel with the load. This means that the voltmeter is connected with the + probe on the side of
the bulb closest to the positive (+) terminal of the battery, and the - probe on the other side of the
bulb. Notice that the meter is measuring potential difference (voltage), not current (amperage).
+
V
R
A
V3
I1
+
I2 I3
+
V2
+
V1
-
B
C
Figure III-17. Electric Circuits.
76
LAB #6
ELECTRICAL CIRCUITS
OBJECTIVES:
Date________
SKETCH OF LAB SET-UP:
DATA TABLE 1: SERIES CIRCUIT
Power Supply Voltage:
Vs =
Lamp #1 Voltage:
V1 =
Lamp #2 Voltage:
V2 =
SCHEMATIC DIAGRAM OF
SERIES CIRCUIT
LAMP BRIGHTNESS: OBSERVATIONS
DATA TABLE 2: PARALLEL CIRCUIT
Power Supply Voltage:
Vs =
Lamp #1 Voltage:
V1 =
Lamp #2 Voltage:
V2 =
SCHEMATIC DIAGRAM OF
PARALLEL CIRCUIT
77
Lab #6 ANALYSIS:
1. Using the voltage measurements from Data Table 1, verify that, for a series circuit that
Vs = V1 + V2.
2. Explain why the sum of the lamp voltages from question 1 might not equal the source voltage.
___________________________________________________________________________
3. Using the measurements from Table 2, verify that Vs = V1 = V2
4. In which circuit did the bulbs have the highest voltage?
______________________________________________________________________________
5. Explain why a difference in bulb brightness occurs for the two types of circuit.
______________________________________________________________________________
6. In the space below, draw a schematic diagram for a four load series circuit. Use lamps as the
loads, a battery for the source, and include an ON/OFF switch.
7. In the space below, show a schematic diagram a three load parallel circuit. Use two batteries
in series for the source and lamps as the loads. Include three switches in the circuit, each one
controlling a separate lamp.
78
f) Generating Magnetic Fields
Magnetic fields are generated by moving charges. The magnetic field lines generated by a
moving charge are perpendicular to the motion and can be depicted as concentric circles about
the charge, with the intensity diminishing with distance from the moving particle. The greater
the charge and the faster it moves relative to the observer, the greater the observed magnetic field
strength.
X
Figure III-18. The magnetic field lines of a positive charge with motion relative to an
observer. The "X" signifies the position of the charge and that it is moving into the sheet of
paper. Field lines further away from the charged particle are less intense.
A charge moving in a circle, such as an orbiting electron, would generate the following field line
geometry, which is called a dipole ("two pole") field. Field lines appear to exit from the North
Pole ("N" in the figure) and enter the South Pole ("S"), as in Fig. III-19. A bar magnet placed on
a field line would orient itself such that it was parallel to the field line with its North Pole
pointing in the direction of the field line (see arrows).
Figure III-19. Magnetic field lines associated with a charge (or charges) in circular motion,
typically referred to as a dipole magnetic field.
As is the case with electric fields, the magnetic field strength ("B") is proportional to the density
of the field lines. Like poles produce a repulsive force and opposite poles produce an attractive
force. In permanent magnets enough of the atoms are aligned in the same direction to produce a
large-scale dipole field; in non-magnetic substances the random orientation of the atomic
magnetic fields causes them to cancel each other out. The Earth also has a dipole field, probably
due to the rotation of its metallic core
Running a current through a coil of wire is another way to generate a dipole field. In this
configuration the wire is known as a solenoid or electromagnet. The intensity of the magnetic
field (B) is proportional to the product of the number of windings in the coil of wire (N), the
79
magnitude of the current (I), the magnetic permeability of the substance inside the coil () and
inversely proportional to the length of the solenoid ():
B =
NI
III-19
The magnetic permeability describes the ease with which the magnetic dipole fields produced by
orbiting electrons in a substance can reorient themselves in the direction of the applied field. A
typical solenoid configuration is shown in Figure III-20. It is essentially a series of dipole
magnetic fields (the windings), stacked together to form a much stronger dipole field (the coil).
current

core
Figure III-20. An electromagnet, or solenoid, consists of a coil of wire through which a
current flows. This configuration produces a dipole magnetic field. The core material also
affects the intensity of the magnetic field produced.
A stronger magnetic field can be produced by increasing the current, the number of windings, or
by inserting a core with a relatively large magnetic permeability. Soft iron cores are typically
inserted in solenoids for this very purpose.
The quantity N·I is often referred to as the magnetomotive force, or mmf:
mmf = N·I
III-20
The magnetic force exerted on a charged particle is directly proportional to this value and so,
although it is not actually a force, the mmf is considered to be a force-like quantity.
g) Force Produced by a Magnetic Field
The magnetic force experienced by a charge moving through an existing magnetic field,
perpendicular to the direction of the charge's motion, is given by
F = qvB
III-21
where q is the charge, v is the charge's velocity, B is the intensity of the field perpendicular to
the motion of the charge and F is the force on the charge, which is perpendicular to both the
charge's direction of motion and the direction of the magnetic field. The motion of an electron
through a uniform magnetic field is shown in Fig. III-21. In this case the magnetic field is
directed into the paper. Since the force is always perpendicular to the direction of motion, the
electron will follow a circular path in the presence of an unchanging, uniform magnetic field, as
shown.
80
X
X
X
X
X
X
X
X
X
X
X
X
X
X
Force to due magnetic field
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
proton
original
trajectory
Figure III-21. For a constant magnetic field directed into the paper (denoted by the "X"s),
a proton will experience a force to the left of its trajectory. Since the direction of the force
is always perpendicular to both the charge's motion and the magnetic field direction, the
electron will continue to move in a circular path.
Since v is length divided by time (v = /t) and current is charge divided by time (I = q/t), we can
re-write eq. I-4 for a steady current through a wire as
F = IB
III-22
where I is the current and in this case  is the length of the wire.
h) Solenoids
A solenoid is an electromechanical device consisting of a tightly wrapped coil of conducting
wire and a ferromagnetic plunger, which moves inside the hollow core of the coil. When a
current is passed through the coil, the movement of the charge through the coil establishes a
magnetic field. Moving charge is the source of all magnetic fields. This magnetic field
attracts the plunger, which is the pulling force of the plunger. The pulling force is a function of
both the magnetic field strength and the plunger depth inside the core of the field.
The magnetic field strength increases with increasing current through the coil and thus the
pulling force on the plunger increases. When the plunger is fully inserted in the core, it has the
greatest magnetic coupling with the magnetic field and thus its strongest force of attraction,
which is called the holding force.
Generally the holding force of the solenoid will increase as the square of the voltage increase.
As an example, if you were to double the voltage to the coil the holding force would increase by
a factor of four. In Lab #7 we will probably see somewhat less than a four-fold increase in
holding force by doubling the voltage.
The pulling force of the solenoid will increase as the plunger’s initial position gets closer to the
center of the core. The pulling force is not constant as the plunger moves toward the center of
the core.
In our experiment with the solenoid we will see the effect of increasing voltage on the holding
force and the effect of starting the plunger closer to the center of the core on the pulling force.
The instrumentation we will use to measure these effects will prevent us from getting precise
81
measurements, but the accuracy should be enough for our purposes. We will also verify that the
holding force of a solenoid is always greater than the pulling force.
There are numerous applications for the solenoid, including:
 engaging the starter motor to start an automobile engine
 electric staple guns where the plunger pushes the staple in
 activating the proper letter/number on the daisy wheel in an electric typewriter
 striking the chimes or bell on a door bell
82
PROBLEM SET 7: SOLENOIDS
1. A mass is hanging from a pulley by a string connected to an energized solenoid. If the mass is 537
g, what is the holding force of the solenoid?
2. If the voltage to the solenoid is increased will the mass that can be supported increase, decrease or
stay the same? Why is this?
3. If the plunger of a solenoid is decreased in size with everything else remaining the same, what is
the effect on the pulling force of the solenoid?
4. What are some uses for solenoids?
83
LAB #7: SOLENOID OPERATION
OVERVIEW
A solenoid is an electromechanical device. It consists of a wire wound coil (the electrical part)
and the steel plunger that moves inside the hollow core of the coil (the mechanical part). If the
turns of the coil are close together and the solenoid is long relative to its diameter, then the
magnetic field inside the hollow core is uniform and parallel to its axis except near the ends.
The magnetic field is established when current passes through the coil. Moving charge is the
source of all magnetic fields. This magnetic field attracts the steel plunger with a force called the
pulling force of the solenoid. The pulling force strength depends upon both the magnitude of the
magnetic field and the plunger depth inside the core of the coil. If the current to the coil is
increased, the magnetic field strength is increased and the plunger is attracted with a greater
force. As the plunger is inserted deeper inside the core, there is more magnetic coupling between
the field and the plunger, which increases the force of attraction. When the plunger is fully
inserted in the core, it has its greatest force of attraction, which is called the holding force.
Solenoids are used in a number of common, everyday devices. Solenoids are used to force
the gears of an automobile starter motor to momentarily mesh with the gear teeth of the engine
flywheel and allow the engine to be cranked over and thus started. They are also used to strike
the tone bars of a door chime, to force out staples in an electric stapling gun, and to strike the
spokes in a daisy-wheel printing elements in typewriters.
Figure III-22. Cross-sectional view of a solenoid.
84
LAB #7 SOLENOID OPERATION
OBJECTIVES:
SKETCH OF LAB SET-UP:
DATA TABLE 1
SOLENOID
VOLTAGE
V (volts)
Date________
SOLENOID HOLDING FORCE
SOLENOID
CURRENT
I (amps)
MASS NEEDED TO
RELEASE SOLENOID
m (grams)
SOLENOID
HOLDING FORCE
F (newtons)
DATA TABLE 2: SOLENOID PULLING FORCE
6 VOLTS
PLUNGER DEPTH
(mm)
PULLING
FORCE
(newtons)
CURRENT
(amperes)
85
10 VOLTS
PULLING
FORCE
(newtons)
CURRENT
(amperes)
Lab #7
ANALYSIS:
1. Convert the mass needed to release the plunger from the solenoid from grams to force in
newtons. Enter these results for each trial into Data Table 1. Show a sample calculation below:
2. Could you use an aluminum or plastic plunger in the solenoid to reduce weight? __________
Explain the reason for your answer.
_________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________
3. What causes the holding and pulling force in the solenoid?
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
4. Which is greater, pulling force or holding force? ______________
What is the cause of this difference?_______________________________________________
____________________________________________________________________________
____________________________________________________________________________
5. Explain how the magnitude of the holding force varies as the voltage increases.
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
6. Explain how the magnitude of the pulling force varies as the plunger is inserted further into
the core of the solenoid.
_____________________________________________________________
____________________________________________________________________________
86
THERMAL SYSTEMS
a) Temperature and Temperature Difference
Temperature difference (T) is the force-like quantity that transports heat energy in thermal
systems. Heat energy always flows from warmer regions to cooler regions. The heat flow rate
in any system depends on temperature difference, just as fluid flow depends on pressure
difference, and charge flow depends on the voltage difference.
All matter is composed of molecules (single atoms or groups of atoms) that are in constant
motion. Heat energy is the energy of motion (kinetic energy) of molecules. This energy is
transferred through a material by the action of the forces that exist between molecules of that
material. A glass of ice water sitting in a warm room gains heat energy from its surroundings
until all the ice melts and the water reaches the same temperature as the room. No further heat
energy flows in the final equilibrium condition, because a temperature difference no longer
exists.
Temperature is a measure of the average kinetic energy of the molecules of a substance. In
thermal systems, temperature difference causes heat (heat energy) flow. Therefore, temperature
difference acts as a force-like quantity.
The two most common temperature scales are the Fahrenheit scale and the Celsius scale. The
Fahrenheit scale is a part of the English system of units; the Celsius scale, sometimes called
"centigrade", is a part of the SI system. Both systems use the freezing and boiling points of
water as references, and divide the temperature difference between these points into an arbitrary
number of divisions called degrees. The formulas for conversion of temperature in one system to
the equivalent temperature in the other are shown below:
Conversion of T(°C) to T(°F):
9
T( o F) = 32 o F + T( o C)
5
Conversion of T(°F) to T(°C):
5
T( o C) =  T( o F) - 32 o F
9
III-23
III-24
Temperatures are specified in degrees Celsius (°C) or degrees Fahrenheit (°F), while
temperature difference is expressed in Celsius degrees (C°) or Fahrenheit degrees (F°). When
converting temperature differences, the value of 32oF is not necessary:
87
9
T (F o )  [T (Co )]
5
III-25
5
T (Co )  [T (F o )]
9
III-26
and
Example III-16: Temperature Conversion
A block of ice at 32°F is left in a room at 78°F.
a) Find the temperature difference in F°:
T = T2 -T1 = 78°F - 32°F = 46 F°
The temperature difference is 46 Fahrenheit degrees.
b) Find T in Celsius degrees:
T(Co ) =
5
5
T(Fo ) =  46Fo  = 25.6 Co

9
9
The temperature difference is 25.6 Celsius degrees. Note that when changing T
from Fahrenheit to Celsius degrees or vice versa, it is not necessary to add or
subtract 32° since it is a temperature difference, as opposed to a specific
temperature.
c) Find the room temperature in degrees Celsius:
5
5
T( o C) =  T( o F) - 32 o F =  78 o F - 32 o F = 25.6 o C
9
9
The temperature is 25.6 degrees Celsius.
b) Thermocouples
The thermocouple is a temperature measuring device that produces a small voltage output in
response to a temperature difference. It is a simple, rugged device, which can be used to
measure temperatures in environments unsuitable for thermometers, and the electrical signal can
be displayed remotely in either analog or digital form. The ability to read temperatures remotely
is a major advantage in the operation of nuclear power plants, aircraft engines and countless
other applications. It is widely used as a temperature monitoring device in industry.
88
The thermocouple consists of three wires of two dissimilar metals, which are exposed to the
same temperatures at each of two different sampling points. If there is a temperature difference
between the two points, a proportional voltage difference between the two metals results. If zero
voltage is read between the two points, the temperature difference is zero and both points are at
the same temperature.
We use the type-E thermocouple, which uses chromel and constantin alloys as the two dissimilar
metals. A calibration table for the type-E thermocouple is in Appendix C. It is important that
you learn how to use the thermocouple for determining temperature readings. While many
commercial thermocouples read out temperatures directly on the meter, the type-E thermocouple
reads out the voltage difference in millivolts (mV). The calibration table is then used to convert
the voltage readings to temperature readings. This will often require interpolating between
points on the calibration chart.
The two points or junctions of the thermocouple are the reference junction and the measurement
junction. The reference junction exposes both metals to the same known temperature. We will
normally use ice water or room temperature as the reference junction. The measurement
junction is then placed where the temperature is to be determined. The measured voltage
difference reading, Vmeas, must be added to the voltage corresponding to the reference
temperature, Vref (from the calibration table), to get the total voltage, Vtotal.
Vtotal = Vref + Vmeas
III-27
The actual temperature, Tact, is obtained from the thermocouple calibration table.
The calibration table for the type-E thermocouple is based on a reference temperature of 0oC.
Each row lists the thermocouple voltage in mV for even temperatures over a ten degree span. If
you look at the fifth row (10oC in the first column), the second column lists the voltage
difference in mV for 10oC, the third column for 12oC, and so forth. Notice that the fifth column
is the potential difference for 20oC, which is the same for the value in the second column of the
next row. If the voltage reading (Vtot) is between two listed numbers, you must interpolate to
determine the actual temperature value. In this case the equation to use is
 T -T
Tact  Tmin  Vtot - Vmin  max min
 Vmax - Vmin



III-28
where Vtot is the total voltage, Vmin is the voltage in the calibration table just below Vtot, Vmax
is the value just above Vtot, and Tmin and Tmax are their corresponding temperatures.
89
PROBLEM SET 8: TEMPERATURE AND THERMOCOUPLES
1. Heat always moves from regions of (higher, lower) temperature to regions of (higher, lower)
temperature.
2. When you are measuring the average energy of motion of the molecules that make up a
substance, you are measuring its _______.
3. If the temperature outside is 20 0C, what is the temperature in 0F?
a. 68 0F
b. 43 0F
c. 72 0F
d. -6.77 0F
4. Place the degree symbol (o) in the proper position for the following temperatures:
a. Room temperature of 24 F
b. A temperature change of 10 F
c. Boiling water at 212 F
d. A difference of 10 C
5. A thermocouple with the reference junction at 20oC and the measurement junction at 65oC will
read what voltage on the multimeter?
6. A cylinder of metal is heated up from 20oC to 150oC.
a) What is the temperature change, T, in Co?
b) What is the T in Fo?
7. A container of boiling water (212oF) is sitting in a room that has a temperature of 72oF.
a) Find the temperature difference, T, between the water and the room in Fahrenheit degrees.
b) Find the temperature difference in Celsius degrees.
8. A thermocouple is set up as in the Lab, with the meter readings given below. Find the value of T
to the nearest Co. Use the Type-E thermocouple chart.
A. The voltmeter reads 1.43 mV, T = ____________
B. The voltmeter reads 3.35 mV, T = ____________
C. The voltmeter reads -0.81 mV, T = ____________
D. What does the negative reading on the voltmeter indicate?
_________________________________________________________________
_________________________________________________________________
9. Convert the following 0F temperatures to 0C temperatures:
a. 413 0F
b. 26 0F
c. -13 0F
10. Convert the following 0C temperatures to 0F temperatures:
90
a. 25.5 0C
b. 5 0C
c. -40 0C
11. Convert the following  temperatures from F0 to C0 or C0 to F0:
a. 42 F0
b. 162 F0
c. 10 C0
d. -25 C0
91
LAB #8: MEASURING TEMPERATURE WITH THERMOCOUPLES
OVERVIEW
In 1826 Thomas Seebeck discovered that a temperature difference existing at a junction of two
dissimilar metals results in an electric voltage. This type of junction is known as a
"thermocouple" and is used widely today to measure temperatures.
The illustration below shows a typical thermocouple junction. Two sensing junctions are formed
by connecting together wires of different metals. The two different metals interact with each
other to create a voltage between them. That voltage increases as the temperature at the junction
increases. A voltmeter is installed between the two junctions, to measure the potential difference
(voltage) between them.
If both junctions are at the same temperature, then both will produce the same voltage. The
voltmeter then reads the voltage difference, which will be zero. If one junction is at a higher
temperature, it will produce a greater voltage. The voltmeter will read the difference in voltage,
and it will be greater than zero. When you know how these junctions respond to temperature,
you can use this voltage difference to find the temperature difference between them.
When you use a thermocouple to measure temperature, one of the junctions must be kept at a
known temperature. This is called the reference junction.
The other junction is called the measurement junction. You place this junction where you want
to take the temperature measurement. Then you take a voltage reading from both junctions,
using the thermocouple voltmeter.
You will use the calibration table in the back section of this manual to determine what
temperature difference the voltage reading represents.
chromel: a trade name for a nickel-chromium alloy
constantan: a copper-nickel alloy
Figure III-23. Thermocouples.
92
LEARNING OBJECTIVES
The learning objectives for Lab 8 are:
1. Learn how to use a thermometer and a thermocouple to obtain temperature readings
2. Convert temperature in Celsius to temperature in Fahrenheit and vice versa.
3. Understand the mechanism for transfer of heat energy through a material(s)
and the direction that heat flows in.
4. Describe how to calibrate a liquid-in-glass thermometer in oC or oF.
93
LAB #8 MEASURING TEMPERATURE WITH A THERMOCOUPLE
OBJECTIVES:
SKETCH OF LAB SET-UP:
Date______
TABLE 1: DATA
REFERENCE JUNCTION TEMPERATURE: Tref
Ice Water: Ti =______oC
Warmer Water: (Trial #5) Tw =______oC
TRIAL
LABORATORY
THERMOMETER
READING: TT
THERMOCOUPLE
VOLTAGE
#1 UNHEATED WATER
#2 WARMED TO ABOUT 40OC
#3 BETWEEN 60OC & 70OC
#4 Boiling Water
#5 Boiling Water
TABLE 2: CALCULATIONS
To find T: Use the chart of T vs. V
To find T: Add the T value to the temp. of the reference junction.
Trial
T from chart
Actual T (Tref + T)
#1
#2
#3
#4
#5
94
LAB #8 ANALYSIS
1. To work properly, a thermocouple requires a _________________________ junction
and a ________________________________junction.
2. The two junctions of a thermocouple are producing the same voltage. What do you know
about their temperatures?
_________________________________________________________
3. The reference junction of a thermocouple is at a temperature of 24oC. The voltmeter reading
is 8.03 mV. Find the actual temperature of the measurement junction to the nearest degree
Celsius. (Show your work:)
4. Convert the answer to #3 to a Fahrenheit temperature.
5. The reference junction of a thermocouple is at a temperature of 0o C. The voltmeter reading
is -0.99 mV ( a negative reading). Find the actual temperature of the measurement junction.
(Show your work:)
6. Convert the answer to #5 to a Fahrenheit temperature.
7. The reference junction of a thermocouple is at a temperature of 75oF, and the voltmeter
reading is -1.32 mV a negative reading). Find the actual temperature of the measurement
junction, in degrees Fahrenheit. (Show your work:)
8. There were two reference temperature sources used in this experiment, ice water and warmer
water (approximately room temperature.) Which provided a more stable reference junction
temperature? Explain.
_____________________________________________________________________________
95
7. SUMMARY
Force-like quantities are the prime movers in all physical systems. Although net force, net
torque, pressure differential, voltage and temperature difference affect their respective energy
systems in different ways, they play strikingly similar roles. The table below summarizes the
force-like quantities in the four energy systems.
FORCE-LIKE QUANTITIES
Energy System
MECHANICAL
Translational
Force-like
Quantity
Units
SI
English
Quantity Moved
Force
N
Mass moved through
lb
Rotational
Torque
Nm
ftlb
FLUID
ELECTRICAL
Pressure
Potential
Difference
Temperature
N/m2
V
lb/in2
V
THERMAL
C°
F°
96
a distance
Mass rotated through
an angle
Fluid volume
Charge
Heat Energy
IV. WORK
Work is the energy expended whenever a force-like quantity produces a change in a system. The
general expression for work performed on a system is
Work = Force-like Quantity x Displacement-like Quantity
1. MECHANICAL SYSTEMS
a) Introduction
In a mechanical translational system
W = Fd (units are Nm or joules (J), or ftlbs)
IV-1
where W is the work performed, F is the magnitude of the force acting in the direction of motion
and d is the distance through which the force acts. The object must be displaced in the direction
of the force, or work is not done. An applied force must act on the object, and the object must
move while the force is being applied.
A Home Experiment
Hold this book as far as possible from you as you read this example. I mean it. Make sure your
arms are fully outstretched and parallel to the floor. Hold the book in this position for 30
seconds. Count it out. Right now. Think about how much work you're doing while you count.
If you're reading this, it should be 30 seconds later. Feel the burn? Keep holding the book out as
you read this:
You are not doing any work at all right now.
Your arms feel like they're exercising (which is true), but no
mechanical work is being done because the book is not being
displaced. Now move the book in, and then back out. You did
work when you moved the book in, and then you did the same
amount of work (but of opposite sign) when you moved it back
out, although the net amount of work performed was zero. Now
hold the book above your head. The net work is the product of the
force required to move the book and the distance it was moved,
W=Fd. The moral of the story is, no matter how tired you get,
97
you're not doing any work unless your efforts produce a net
displacement of something.
When mechanical work is done on an object, the object must move. Therefore, work changes an
object's position. The work done may change the object's vertical position by raising it, or the
object's horizontal position may be changed. The applied force may move the object forward or
backward, to the right or to the left. For example, an elevator cable changes the vertical position
of an elevator as it lifts a load of people. An electric truck causes a change in horizontal position
when it pulls an industrial cart from one place to another (on a level factory floor).
In some cases, a force acting on an object does two things. Let's look again at the example of
throwing a baseball. First a person's hand, while holding the ball during the throwing motion,
changes the position of the ball. Second, during the throwing motion, the hand also changes the
speed of the ball. The work done by the hand in throwing the ball (a) changes the ball's position,
and (b) increases the ball's speed. Later when you study energy, you'll learn that when work is
done to move an object, its energy is increased. The increase may happen because (1) the object
has been moved to a higher position with respect to its starting point, or (2), its speed has been
increased, or (3), both. Friction and drag on an object tend to lower its energy.
Example IV-1: Work Required to Lift a Mass
A 15 kg mass is raised 5 meters. How much work (w) was done on the mass?
5m
Work required to lift a 15 kg mass.
Solution: The force is in opposition to the gravitational pull:
98
m

 kg  m
W = F  d = m  g  d = (15 kg) 9.8 2  (5 m) = 735  2   m = 735 N  m = 735 J


 s 
s
where J stands for joules (pronounced "Jewels"), a unit of energy. The most common unit of
energy in the English system is the footpound (ftlb).
Example IV-2: Work Performed on a Roller Skating Cat
A horizontal force of 65 N is applied continuously to pull Jasmine the Roller Skating Cat a
distance of 8 meters. How much work was performed on Jasmine?
F = 65 N
The famous roller skating cat.
Solution:
W = Fd = (65 N)(8 m) = 520 Nm = 520 J
Note that the cat will probably keep moving after we stop applying the force (after all, it’s on
roller skates), but no work is being done, since no force is being applied. Eventually the cat will
stop moving due to friction, so the work put into the cat using the 65 N of force will be removed
by the friction.
In determining the work done by a force, only the component of the force in the direction of
motion is considered. The woman in Figure IV-1 does not push the lawnmower in just the
direction of motion. Only the component of the force in the direction of motion (Fx) is
responsible for the work performed. Her vertical force component (Fy) does no work since the
lawnmower is not displaced in that direction.
99
Fx
For
ce
Fy
F
Motion
Figure IV-1. The lawn mower on a bad hair day.
b) Work and Efficiency
When machines do work, they're often rated by efficiency. Efficiency is a comparison of output
work to input work. The efficiency of a machine is nothing more than a comparison of the
output work delivered by the machine to the input work done on the machine to operate it. The
formula for % efficiency () is
% Efficiency =
Output Work
x 100%
Input Work
 Wout 
 100%
 = 
 Win 
IV-2
A little algebra shows that the following formula can be used as well:
 AMA 
 = 
 100%
 IMA 
IV-3
A block and tackle is shown in Figure IV-2. A worker pulls on the free end of the cord with force
Fi. This force moves the rope through a distance D. At the same time, a load of weight w is raised
through distance y by the action of the block and tackle. Note that the output force must equal the
weight of the load.
The input work is the work done by the force (F) that causes the block and tackle to operate. The
output work is the work done by the block and tackle to raise the load. Since work is equal to "force
times distance," the input work is equal to FD and the output work is equal to wy.
100
Figure IV-2. Block and Tackle.
Example IV-5: Calculation of Work and Efficiency for a Block and Tackle
The block and tackle in Figure IV-2 is used to lift an engine. The engine weighs 540 lb and is
raised 1 ft. The operator pulls with a force of 100 lb over a distance of 6 ft. Find a) the input work
(Win), b) the output work (Wout), and, c) the efficiency (), of the system.
Solution:
a. Input work = FD (where F = 100 lb and D = 6 ft)
Win = (100 lb)( 6 ft) = 600 ftlb
b. Output work = wy (where w = 540 lb and y = 1.0 ft)
Wout = (540 lb)(1.0 ft) = 540 ftlb
 Wout 
 100% 
c. Efficiency of block and tackle =  = 
 Win 
 540 ft  lb 
 = 
 100%  = 90%
 600 ft  lb 
When a machine is used to do work, there is always some loss of energy (no machine is 100%
efficient). The purpose of using the machine is often to increase the force in order to lift heavy
objects. In the previous example of the block and tackle, the person pulled with 100 lb (input force)
but the output force lifting the engine was 540 lb, equal to its weight. The ratio of the output force
to the input force is called the mechanical advantage.
In the block and tackle example, the input force moved a distance of 6 ft, and the output force
moved only one foot; the distance ratio was 6 to 1. If there were no friction loss, the output work
101
would equal the input work. An output of 600 ftlb would mean that the output force was 600 lb,
and the output force would have a 6 to 1 ratio compared to the input force.
In any machine, if there were no friction, the ratio of forces would be equal to the inverse of the
distance ratios. The distance ratio is used to compute the "Ideal Mechanical Advantage", the
mechanical advantage which would be achieved if the machine were 100% efficient.
c) The Lever
The lever is a rigid bar that can be pivoted at some point. Consider the crowbar (Figure IV-3)
used to move a heavy rock. A smaller stone serves as the pivot point. The input force Fin is
transformed to an output force w large enough to move the boulder. Figure IV-3 illustrates the
lever with dimensions before and after the boulder is raised. The dashed line indicates the final
position of the lever. Notice that the applied force Fin moves through a larger distance D than
the load, which is lifted a distance y. The force advantage is gained as with the pulley, at the
expense of displacement. The input work (Win) is
Win = FinD
IV-3
Wout = wy
IV-4
The output work (Wout) is
F
D
in
w
y
in
out
out
Figure IV-3. Mechanical advantage of a crowbar.
The input and output work can also be calculated as the product of the force (F2), the lever arm
( ) and the angle the lever arm is rotated through (, which goes by the name “theta”):
Win = Fininin
IV-5
Wout = Foutoutout
IV-6
102
d) The Inclined Plane
A crate is moved more easily into the back of a truck by use of an inclined plane or ramp than by
a direct vertical lift (Figure IV-4). The inclined plane serves as another coupling device that
enables a small force to do a big job at the expense of a larger displacement. In this case, the
crate is rolled along the incline farther than it is lifted, that is, Di > Do. This "sacrifice" in
displacement shows up as a gain in force, or a mechanical advantage.
Figure IV-4. Example of an inclined plane.
Input work (Win) and output work (Wout) are:
Win = FiDi
IV-7
Wout = FoDo
IV-8
103
Example IV-4: Mechanical Advantage of an Inclined Plane
A force of 200 lb (= Fi) is required to push a 300 lb safe (w = Fo = 300 lb) up an incline 15 ft
long (Di = 15 ft) and 5 ft high (Do = 5 ft).
D i=
ft
5
1
w = 300 lb
Do = 5 ft
Moving a safe up an inclined plane.
a) What is the input work (Win)?
Win = FiDi = (200 lb)(15 ft) = 3000 ftlb
b) What is the output work?
Wout = wDo = (300 lb)(5 ft) = 1500 ftlb
c) What is the efficiency of the system?
η
Wout
1500 ft  lb
x 100% 
x 100%  50%
Win
3000 ft  lb
104
PROBLEM SET 9: LINEAR WORK
1. Mechanical work is done when force or torque causes an object
_____________________________________________(complete )
2. The correct units for work (select all correct answers)
a. ftlb b. ft/lb c. Nm d. N/m
3. The work unit of 1 joule is also equal to 1 ________. (fill in)
4. In questions a-f below indicate whether work is being done (“yes”) or not (“no”):
a. ____ engine moving car uphill
b. ____ parking brake holding car on hill
c. ____ elevator stopped at 5th floor
d. ____ elevator moving from 5th to 6th floor
e. ____ box sitting on edge of table
f. ____ box being lifted from floor to table
5. The ratio of output work to input work for a system is called
_________________________________.
6. A block and tackle is used to lift a 600 lb load a distance of 3.25 ft. It takes a force of 200 lb
pulling over 12 ft.
a) Find the input work
b) Find the output work
c) Find the efficiency.
7. A horizontal force of 45 N is applied to push a filing cabinet 40 meters down the hall.
a) How much work is being done?
b) What type of work is done?
8. A test reveals that 175 ftlb of work is required to lift an object 3 ft at constant speed. What is the
weight of the object?
9. A block and tackle requires 1400 ftlb of input work to accomplish 1200 ftlb of output work
while lifting a 600 lb load.
Find the efficiency of the system.
10. To lift a load straight up from the floor 140 cm requires 240 joules of work. What is the weight
of the load?
11. A 40 kg cart is pulled up an inclined plane by a force of 200 N. The length of the plane is 12.8
meters, and the height of the plane is 4.0 meters. Find A. the output work. B. the input work and C.
the efficiency of the system.
12. When a 30 lb weight is lifted from the floor to a shelf, 126 ftlb of work is done. What is the
height of the shelf above the floor?
105
13. The work done is 357 ftlb when a box is lifted to a height of 18 ft. How much did the box
weigh?
14. A man applies a constant force in a horizontal direction to a 12 kg mass to accelerate the mass
across a frictionless surface. If 32 joules of work was done to move the mass 6 meters, what is
the acceleration of the block using Newton’s Second law (Equation III – 1)?
106
LAB #9: WORK DONE IN MOVING LOADS
OVERVIEW: The equation for work is W = Fd
Work = Applied Force x Distance Moved in the Direction of the Force
The SI unit of work is the joule (J). 1 J = 1 Nm
The English unit of work is the ftlb.
Elevators perform work by lifting loads vertically. Sometimes, loads are raised to higher
elevations by using an inclined plane. That is because movement along an incline requires less
force than lifting the load directly upward. In the absence of friction, moving a load along an
incline would require the same work input as direct lifting. The elevator would apply a larger
force over a shorter (vertical) distance. With the inclined plane, a lesser force operates along the
incline over a longer distance.
When friction is involved, the force needed to move the load along the incline must be large
enough to overcome friction and sustain motion. Therefore, the input work will be greater than
if there were no friction.
In this experiment, you will determine the output work by lifting a load directly. In this case,
friction losses are negligible, and the output work will be equal to the applied force times the
vertical distance lifted. The applied force is equal to the weight of the load.
Wout = mgh = wh
(units of joules or ftlb)
The input work will be found when the load is moved up an inclined plane by measuring the
applied force and multiplying by the distance traveled along the plane.
Win = Fin  D
The efficiency of the inclined plane system is found by dividing the output work by the input
work. The units of work must be the same, so efficiency will be a dimensionless number.
Efficiency = Work Output  Work Input
Efficiency is expressed as a per cent with the symbol  (“eta”).

Wout
 100%
Win
Note: Ideal Mechanical Advantage may be indicated as Mi or IMA. Actual Mechanical
Advantage is Ma or AMA.
LEARNING OBJECTIVES
107
The learning objectives for Labs 9-10 are:
1. Describe what work means in general and specifically in mechanical systems
for both linear and rotational work.
2. Understand and be able to use both SI and English units in solving mechanical work problems
(both linear and rotational).
3. Identify the effects of work in a mechanical system.
4. Define the concepts of efficiency, ideal mechanical advantage and actual mechanical
advantage and how they relate to input and output work.
5. Define the radian measure for defining rotational angle in non-dimensional terms.
Be able to convert from degrees to radians and vice versa.
6. Learn that work requires both force (or force-like quantity) and movement in the direction
of the force (or force-like quantity).
7. Identify the workplace applications of mechanical work.
108
LAB #9 WORK DONE IN MOVING LOADS
OBJECTIVES:
SKETCH OF LAB SET-UP:
Date__________
TABLE 1 WORK TO MOVE A LOAD VERTICALLY
Weight of Load
w (newtons)
Distance Lifted
h (meters)
Output Work
Wout (joules)
TABLE 2 WORK WITH FRICTION PRESENT
TRIAL
INCLINED PLANE
Total Mass
(Weights + Hanger)
m (grams)
HORIZONTAL
PLANE
Distance Along
Plane
din (meters)
Total Mass
(Weights + Hanger)
m (grams)
Side 1
(Smooth)
Side 2
(Rough)
TABLE 3 CALCULATIONS
INCLINED PLANE
HORIZONTAL PLANE
Win-Wf
Side
Input Force
Fi (N)
Input Work
Win (J)
Eff.
 (%)
Force of
Friction
Ff
#1
#2
How does (Win - Wf)
compare with Wout?
109
Friction Work
Wf
Lab #9 ANALYSIS:
1. How does the work done against friction affect the efficiency of an inclined plane
system?
___________________________________________________________________________
2. How does the output work of an inclined plane compare with the input work?
______________________________________________________________________________
3. What is the benefit of using an inclined plane?
______________________________________________________________________________
Use the 5-step method to solve the following problems:
4. A trunk is pulled at constant speed along a level floor for a distance of 18.5 ft. If 1200 ftlb of
work is expended against friction, calculate the horizontal pulling force.
All of the work against friction is turned into heat. What would happen if a greater force were
applied?
_______________________________________________________________________
5. A 500 lb load is moved along an inclined plane by a force of 125 lb. The cart moves 40.0 ft
along the plane as it rises by a height of 6.0 ft. Find A) the input work B) the output work and
C) the efficiency of the inclined plane system.
6. A 60 kg cart is pulled along an inclined plane by a steady force of 350 newtons. The load rises
to a height of 3.5 m as it travels a distance along the plane of 8.0 m.
Find a) the input work, b) the output work and c) the efficiency of the inclined plane system.
110
2. MECHANICAL ROTATIONAL SYSTEMS
In mechanical rotational systems, the general equation for work takes the form
W = 
IV-9
where W is the mechanical work done,  is the magnitude of the torque and (“theta”) is the
angle of rotation through which the torque is applied. Angles can be measured in degrees or
number of revolutions, but in Equation IV-9, the angle must be expressed in radians, the most
natural of angular units. Figure IV-5 shows a picture of an angle  in a circle with radius r. The
angle traces out a distance along the perimeter that we’ll call s. The angle is the ratio of the
distance along the perimeter to the radius:
s
r
An angle of one radian is an angle that, if placed at the center of a circle, would intercept an arc
length equal to the radius of the circle.
The size of an angle of one radian is shown in Figure IV-5. This angle of one radian is
approximately 57.3°, but an identity derived from the formula for the circumference of a circle is
easier to remember. The circumference of a circle encloses an angle of 360°. Since
circumference is found by multiplying 2 times the radius, and an arc length of one radius
corresponds to an angle of one radian:
2 radians = 360°
IV-10a
 radians = 180°
IV-10b
or
1 radian
Figure IV-5. The angle is equal to 1 radian (57.3o).
111
The radian is not actually a unit in the same sense as a meter or a pound. It is the ratio of the
length of an arc to the length of the radius. The lengths are expressed in the same units, which
cancel to leave a dimensionless quantity. When angles expressed in radians are used in
calculations, we usually drop the term "radian", as shown in Examples 5 and 6.
Example IV-5: Work Done to Turn a Pulley
A force of 70 lb is required to turn a pulley with a radius of 1.5 feet at a constant speed. The
pulley turns through 5 revolutions. What is the work done on the pulley?
A pulley system.
Solution: Recall from the definition of torque,
= F (Torque = Force perpendicular to the Lever arm x Lever arm)
 = (70 lb)(1.5 ft) = 105 ftlb
The pulley is turned through 5 revolutions, and there are 2 radians (or 360°) per revolution:
 = (5 rev)(2 rad/1 rev) = 10 rad
where "rad" is an abbreviation for radians. The work done is then
W =  = (105 ftlb)(10)
= 3.30 x 103 ftlb
112
("rad" is dropped)
The applied torque does 3.30 x 103 ftlb of work.
Example IV-6: Energy Stored in a Rotating Flywheel
The spinning flywheel shown below is 1.20 meters in diameter. A braking force of 80 newtons
is applied at its outer edge, causing it to stop after rotating 484 degrees. How much energy was
consumed in stopping the rotation?
A flywheel.
The radius of the flywheel is half of the diameter, or 0.60 m. The angle in radians through which
the flywheel turns while the braking force is applied is
  rad 
 = (484 o )
 = 8.45 rad
 180 o 
The amount of energy consumed is equal to the work done:
W =  = F = (80 N)(0.60 m) (8.45 rad) = 405 Nm
113
The energy required to stop the flywheel is 405 Nm or 405 J.
Notice that the units of torque and work are the same - Nm in SI units and ftlb in English units.
This does not mean that torque and work are equivalent. You must take care to identify a
quantity expressed in these units as either torque or work. The multiplication of torque (Nm or
ftlb) by an angle of rotation in radians (no units) yields work (Nm or ftlb). If you were to exert
a torque of 20 Nm, but no rotation resulted from your efforts, you would have done no work.
114
PROBLEM SET 10: ROTATIONAL WORK
1. Change each angle in degrees to an angle in radians.
a. 30o = _____ b. 57.3o = ______ c. 270o = _____ d. 360o = ____
2. Change revolutions to angle in radians.
a. 0.5 rev = ______ b. 5 rev = _______ c. 3.2 rev = ______
3. When a torque causes a rotation through an angle, the size of the angle can be expressed in
degrees or in __________.
4. Work done by a torque is equal to the torque applied times
_____________________________________________________________.
5. To tighten a nut, a force of 50 lb is applied at the end of an 18 in torque wrench. The wrench
moves the nut through one revolution. Find a) the torque applied, and b) the work done.
6. A torque of 300 ftlb is applied to a lever. The lever does 400 ftlb of work in lifting a load. A.
Find the angle (radians) the lever moves through.
7. What torque is required to cause a wheel to rotate through 4 rev if the work done is 4500 J?
8. A flywheel requires 75 ftlb of work to stop its motion when a braking force of 3 lb is applied at
the outer edge. The diameter of the flywheel is 22 in.
a) Find the applied torque in ftlb.
b) Find how many revolutions the flywheel makes as it stops.
9. A 196 N force is applied to the end of a torque wrench. The arm of the torque wrench is 46 cm.
long. It causes the wrench to move through an angle of /4 radians. Find the work done in
joules.
10. When loosening a nut from a bolt, an average torque of 70 Nm must be supplied for 4.5
revolutions. How much work is done in Nm and ftlb?
11. What braking force applied at the edge of a heavy flywheel is required to bring the flywheel to
a stop in one revolution? The flywheel has a radius of 1 m and 600 J of energy are needed to
stop the flywheel.
115
LAB #10: ROTATIONAL WORK OF A WINCH
OVERVIEW
A winch is a machine consisting of two or more rotating cylinders. Rope or cable is attached to
the winch in a way that allows us to increase the force applied to the load. In doing so, the winch
decreases the output distance. When a machine increases force, at the expense of output
distance, it gives us a mechanical advantage.
The winch used in this lab is designed to give a mechanical advantage . This is accomplished via
a two-stage process.
In the first stage, the pulley and input shaft provide an increase in force. This is due to the
difference in their diameters. The force applied by the string to the larger diameter pulley is
increased at the smaller diameter input shaft. This force is applied to the second stage.
The second stage is a gear combination made up of the input and output shaft gears. The force is
increased further, due to the difference in the size of these two gears.
The input work to the winch is the product of the input force (the force applied to the pulley)
and the input distance (the distance the pulley string is extended). The common SI unit of work is
the joule, and the English unit is the ftlb.
Linear work: Win = FinDin
The input work may also be computed by the rotational work equation: input work is the product
of the applied torque and the angle ( in radians) which the pulley turns through.
Rotational work: Win = 
The input torque is the product of the applied force and its lever arm (radius of the pulley)
Torque:  = Fin
The input work should be the same, whether computed using the linear work equation or the
rotational work equation.
The output work of lifting a load is the product of the weight and the vertical distance through
which it is lifted.
Output Work = weight x height lifted
In all mechanical systems, some of the input work is changed to heat as a result of friction.
Therefore, the output work is never as great as the input work. The ratio of output work to input
work is called the efficiency of the winch.
Efficiency = Output Work  Input Work
Efficiency is expressed as a percent with the symbol  (eta):
 = (Wout/Win) x 100%
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LAB #10
WORK DONE BY A WINCH
OBJECTIVE:
Date___________
SKETCH OF LAB SET-UP:
TABLE 1 : DATA
HANGER
MASS
m
(grams)
INPUT
FORCE
Fin (newtons)
INPUT
DISTANCE
Din
WINCH
PULLEY
TURNS
OUTPUT
DISTANCE
Dout
WINCH
PULLEY
DIAMETER
(mm)
TABLE 2: CALCULATED VALUES (From worksheet)
LINEAR EQUATIONS
Output
Force
Fout
(N)
Output
Work
Wout
(J)
Linear
Input Work
Win
(J)
ROTATIONAL EQUATIONS
Input
Lever Arm

(m)
Torque

(Nm)
Angle

(rad)
Rotational
Input Work
Win
(J)
Eff.

(%)
Compare the values for Win that you obtained by the linear equation to the values from the
rotational equation:
117
Lab #10 WORKSHEET: LINEAR MECHANICAL EQUATIONS
Show your calculations below; enter the answers in Table 2.
1. The output force is equal to the weight of the hanging mass in newtons. Find the weight, using
the (sea level) gravitational acceleration factor of 9.8 m/s2.
2. Calculate the output work: Wout = Fout  Dout
3. Find the input work, using the linear mechanical work equation,
Work = Input Force x Input Distance. Win = Fin  Din
ROTATIONAL MECHANICAL EQUATIONS
4. Sketch the input pulley, showing the lever arm of the input force. Calculate  (radius of the
pulley) in meters.  = pulley diameter  2
5. Find the input torque: Torque = Input Force x Lever Arm
= Fin
6. Calculate the number of radians through which the input pulley rotated.
118
7. Calculate the input work, using the rotational mechanical equation: Work = Torque x Angle
(in radians) (W =  This answer should equal the linear input work, within experimental
error.
8. Compute the efficiency using the linear mechanical work value:  = (WoutWin)100%
ANALYSIS:
1. Why should the efficiency of this winch system, or any machine, be less than 100 % ?
______________________________________________________________________________
______________________________________________________________________________
Use the 5-step method to solve the following problem:
2. A force of 38.5 lb is used to turn a pulley with a radius of 29.0 inches at a constant speed. The
pulley turns through 7 revolutions. How many ftlb of work is done on the pulley?
119
3. WORK IN FLUID SYSTEMS
a) Introduction
In a fluid system, work is accomplished when a pressure difference causes fluids (liquids or
gases) to move. The general formula,
Work = Force-like Quantity x Displacement-like Quantity
takes the form shown by Equation IV-11 for the fluid system:
W = (P)V
IV-11
where W is the work done moving the fluid, P is the pressure difference through which the
fluid moves, and V is the volume of fluid transferred.
In fluid systems, the force-like quantity (that causes movement of the fluid) is pressure, and the
volume of fluid moved represents the displacement-like quantity. Winds blow because air
moves from high pressure regions to low pressure regions. Work is done by pressure differences
that move air as we breathe. When we inhale, we expand our chest cavity and create a region of
low pressure in our lungs. The higher pressure outside acts to force air into our lungs. When we
exhale, body muscles decrease our lung volume. This increases the air pressure in the lungs and
forces the air out. Pressure differences cause movement within a fluid system.
In a system involving liquids, which are practically incompressible, the volume of a given mass
of fluid remains the same, this is, the volume of fluid entering a pipe is equal to the volume
leaving the pipe during the same interval. Only fluid pressure changes in the system.
Example IV-7: Work Done to Fill a Water Tank
How much energy is required to fill a 500 cubic foot capacity water tank with water from a lake
150 feet below? The (weight) density of water is 62.4 lb/ft3.
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Filling a water tank.
Solution: We have shown earlier that
Pressure = weight density x height
lb 
lb

P = w  h =  62.4 3  150 ft  = 9360 2

ft 
ft
The pressure difference P between the top of the tank and the surface of the lake is 9360 lb/ft2.
The work performed on the water is
lb 

W = P  V =  9360 2  500ft 3  = 4,680,000 ft  lb

ft 
The work required to fill the tank is 4.68 x 106 ftlb.
In cases where the fluid is a compressible gas, the volume change of the gas must also be taken
into account when you calculate work.
As previously stated, there must be some movement in the system, or work has not been done.
For example, an air pressure of forty pounds per square inch in a tire is not work in itself, but
work is done in putting the air into the tire. The change in air volume involved in the work is the
difference between the volume of air under pressure in the tire and the volume that the same
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quantity of air would occupy if released to the atmosphere. The pressure involved is the pressure
of the air in the tire at any time, which increases gradually from one atmosphere (14.7 lb/in2) to
the final pressure in the tire.
Work done in filling a tire with air is partly stored as increased thermal energy of the air
molecules within the tire (as indicated by a rise in temperature), and partly in the stretched tire
itself. A mechanic must use caution in filling an automobile tire with air after repairing it,
because this stored energy can be released quickly. If the tire and wheel are not assembled
properly, the tire can explode and cause serious injury. Businesses involved in changing and
repairing large tires provide a heavy safety frame that holds the tire while it is being filled with
air.
Gas bottles used for many applications, such as welding, contain gases at thousands of psi (or
hundreds of atmospheres) of pressure. If an unsecured, full bottle falls and the valve breaks off,
it becomes a potentially lethal rocket, as the stored pressure energy is converted to thrust. Of
course the flammability of the stored gas could provide an additional cause for concern.
b) Open and Closed Fluid Systems
There are two types of fluid systems; open and closed. A closed fluid system is designed to retain
and re-circulate the fluid. Examples of a closed fluid system include a hydraulic jack, a hydraulic
brake system and the body's circulatory system. An open fluid system moves fluids, but does not
retain and re-circulate the fluids throughout the system. Open fluid systems include city water
systems, an irrigation system, and a fire truck water system.
c) Basic Hydraulic Power System
Hydraulic power systems are used widely in industry and transportation. The basic system, shown
in Figure IV-6, is a closed system. It's used to do mechanical work on a load attached to the
working piston. We will examine the different parts and see how they operate as a system.
The pump draws water from the reservoir, creating a pressure difference that forces liquid through
pipes under pressure. The pump is driven by a rotating shaft that is connected to a motor. The
control valves are used to direct the liquid to the cylinder to either push or pull on the load. When
valve #1 is closed and Valve #2 is open, high pressure liquid flows directly to the left side of the
piston. This creates a force on the piston that causes the rod to extend or push on the load.
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Figure IV-6. Closed hydraulic system.
When valve #1 is open and valve #2 is closed, high pressure liquid flows directly to the right side of
the piston. This causes the rod to retract or pull on the load. With control valves in either setting, as
the piston in the cylinder moves, liquid is forced out of the cylinder and directed back to the
reservoir to be recycled by the system.
d) Work in a Closed System
The equation for fluid work is used in calculating the work done by a piston in a hydraulic cylinder
(see Figure IV-7). Here the gauge pressure (P) acting on the piston, moves the piston. This
movement displaces a volume (V) of fluid.
Figure IV-7. A Piston in a hydraulic cylinder.
Heavy duty robots are used in industry to move large objects or exert large forces. These robots
usually use hydraulic cylinders to get the job done. Most hydraulic cylinders (such as the one in
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Figure IV-7) are made with a hollow cylinder, a piston and a fluid. Often the fluid is oil, delivered
under pressure to the chamber on one side of the piston. The oil pushes against the piston. This
causes the piston to move. The other end of the piston is connected to a load. As the piston moves,
it moves the load. This is useful mechanical work. The following example shows how work done
by a hydraulic cylinder is calculated.
Example IV-8: Work Done by a Hydraulic Cylinder
A hydraulic cylinder moves a piston 10 centimeters horizontally while pushing on a mechanical
load. The fluid pressure on the piston face is 80,000 pascals (1 Pa = 1 N/m2). The area of the piston
face (cross-sectional area of the cylinder) is 12 cm2. Find the work done by the hydraulic cylinder
while moving the piston 10 cm.
Work done by a hydraulic cylinder.
Solution: Work Done = (Pressure) x (Volume Displaced)
W = (P)V
Step 1: Find P (the gauge pressure):
The pressure (P) is given as 80,000 Pa = 80,000
N
.
m2
Step 2: Find V: The volume of fluid displaced as the piston moves 10 cm
is the same as the volume of a cylinder of cross-sectional area 12 cm2
and length 10 cm.
V = Area x Length = (12 cm2)(10 cm) = 120 cm3
Convert the units of volume from cm3 to m3:
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 1 m3 
V = (120 cm3)  6
 = 0.00012 m3
 10 cm 3 
Step 3: Now substitute known values for P and V in the equation W = PV
to find the work done:
W = (80,000
N
)(0.00012 m3) = 9.6 Nm
m2
The hydraulic cylinder does 9.6 Nm of work while moving the load attached to the piston at
a distance of 10 cm. Since 1 Nm is equal to one joule, the work done is also equal to 9.6 J.
e) Examples of Work in Fluid Systems
You have studied work done in an open fluid system (filling a city water tank) and a closed fluid
system (hydraulic cylinder). You've studied fluid work done when the intended outcome of the
work is (a) to move fluid from one location to another (filling the water tank), and (b) to do some
type of mechanical work (hydraulic cylinder).
Useful fluid work is often done simply to move fluids from one location to another. Examples of
open fluid systems that do work include:
- Movement of natural gas through pipes from supplier to user.
- Movement of oil through pipes from wells to refineries (Alaskan pipeline)
- Movement of air through ducts in air conditioning systems.
- Movement of water in fire truck pumpers from water hydrants to fire hoses.
- Movement of paint in paint spraying systems.
- Movement of water and steam in a coal-fired power plant.
- Movement of water from lakes or wells to storage tanks.
- Movement of air through compressors in jet engines.
. Examples of closed fluid systems that do work include:
- Operation of brakes in an automobile braking system.
- Movement of blood by the heart (pump) through the body's circulatory system.
- Use of hydraulic jacks to lift heavy loads.
- Use of hydraulic cylinders to push or pull on mechanical loads.
- Use of hydraulic motors to power (run) equipment.
If you look back at the basic fluid power system shown in Figure IV-16, you can see that three
different forms of work are involved in the operation of the entire system. Electrical work is done
on the motor (not shown) to operate the pump. The pump, in turn, provides the pressure to force the
fluid through the system. The moving liquid, under pressure, does mechanical work on the piston
and load. It's often true, as shown in this system, that several types of work are being done at the
same time. That is why understanding work in mechanical, fluid and electrical systems is
important. That's also why the general idea of WORK, as a UNIFYING CONCEPT, is helpful to
your understanding of how systems operate and interact. That general statement is repeated here:
Work is the product of a force (or force-like quantity)
and a displacement (or displacement-like quantity).
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PROBLEM SET 11: FLUID WORK
1. When a pump is used to raise water through a pressure difference, the work done by the pump is
equal to the ________________________________________ multiplied by the pressure difference.
(circle the correct answer)
a. fluid volume moved
b. fluid weight moved
c. distance fluid moves
d. weight density of fluid
2. The work done by a moving piston in a hydraulic cylinder is equal to the pressure exerted by the
piston multiplied by the (Circle the correct answer):
a. area of the piston.
b. fluid volume displaced.
c. weight density of the fluid.
3. Each system below is either an open (O) system or a closed (C) system. In each of the blanks
insert the appropriate letter.
a. ___ city water system
c. ___ auto brake system
e. ___ fire department pumper truck
b. ___ auto fuel system
d. ___ hydraulic jack
4. A hydraulic cylinder displaces 0.25 ft3 of water while the piston moves 1 ft. The pressure of the
face of the piston is 75 lb/ft2 greater than on the other side of the piston. Find the work done by the
hydraulic cylinder while moving the piston.
5. A pump under constant pressure uses 25,000 joules of energy (work) to move 30.0 m3 of water.
What pressure difference was required?
6. A pump doing work at a constant pressure difference of 85,000 N/m2 moves 420 cubic meters of
liquid. Find the work done in joules. Write the answer in scientific notation.
7. A constant pressure difference of 1500 kPa displaces a piston in a hydraulic cylinder. The
hydraulic cylinder volume through which the piston moves is 0.0060 m3. How much work is
required?
8. A pump lifts water from a container on the floor to fill a 6.0 cubic foot storage container. If 1240
ftlb of work is done, what was the pump pressure required (disregarding friction)? Change the
pressure units to psi.
9. An air conditioner fan provides a gauge air pressure of 0.3 psi. The air conditioner is in a room
that is 20 ft long, 15 ft wide and 8 ft high. How much work does the fan do each time it fills the
room?
10. A pressure difference of 100 psi is applied to the face of a hydraulic cylinder piston rod. This
causes a force of 70.6 lb on the piston face.
A. Find the diameter of the piston face.
B. Find the volume of fluid moved when the piston moves 10 in.
C. Find the work done when the piston moves 10 in.
126
11. It requires 5 x 107 joules of work to fill a storage tank from a reservoir some distance below the
tank. What must the pressure difference be between the top of the tank and the surface of the
reservoir if the tank has a volume of 300 m3?
127
LAB #11: WORK DONE BY A PUMP
OVERVIEW
A water pump is a machine used to move water from one place to another. The pump does
work on the water, when it lifts the water from one level to another. Some of this work shows up
as energy given to the water in the form of output work.
Some of the work is "wasted" in the form of heat, as the resistance of the system must be
overcome to move the water through the pipes.
Electrical energy is input into the pump motor, and converted to mechanical work. This energy
is expressed in the equation below.
Electrical Energy = V  I  t
V represents the voltage in units of volts (V)
I represents the current in units of amperes (A)
t represents the time in units of seconds (s)
This electrical energy input to the pump is the work input. Therefore,
Workin = Electrical Energy = V  I  t.
The energy unit of volt  amp  second is equal to one joule (J)
In this experiment, the output work is defined as in the previous experiments. The mechanical
work output of the pump is equal to the weight of the water being lifted (force) times the height
it is lifted (distance).
W = FD
Workout = Weightwater  Height
In this experiment you will be measuring the electrical energy input to the water pump. Then
you will compare it to the mechanical work done by the pump to lift the water. A low efficiency
indicates that much of the input electrical work is lost due to heat production in the motor. Some
is also lost overcoming fluid friction.
LEARNING OBJECTIVES
Learning objectives for lab 11 are:
1. Differentiate between open and closed fluid systems and give examples of each.
2. Describe what is meant by work in a fluid system.
3. Understand and be able to use both SI and English units in solving fluid work problems.
4. Identify the effects of work done on a fluid.
5. Identify the workplace applications of Fluid Work.
6. Use efficiency to determine the relationship between input and output work.
7. Understand the concepts behind the basic hydraulic system.
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LAB #11 WORK DONE BY A PUMP
OBJECTIVES:
SKETCH OF LAB SET-UP:
TABLE 1 : DATA
Date_______
VOLUME OF WATER PUMPED:_______________
#1 (12V)
#2(18 V)
#3(12V)
#4(12 V)
HEIGHT: D (meters)
D1=
D2=
D3=
D4=
TIME: t (sec)
t1 =
t2 =
t3 =
t4 =
VOLTAGE: V (volts)
V1=
V2=
V3=
V4=
CURRENT: I (amps)
I1=
I2=
I3=
I4=
TRIAL
TABLE 2: CALCULATIONS
ELECTRICAL WORK
INPUT Win (Nm)
WORK OUTPUT:
Wout (Nm)
______
EFFICIENCY:
 (%)
______
TABLE 3 : FLUID WORK CALCULATIONS
Weight Density of Water (N/m3): w =
Pressure Required to Lift
Water for Trial #1:
P =
Work Output of Trial #1
using Fluid Work Equation:
Wout =
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Lab #11 ANALYSIS
1. Find the difference, in joules, between the work input and work output for trial #1. _________
How does this value compare with the work needed to pump the water to a bucket at the same
level (trial #3)?
_________________________________________________________________
2. Using the results of question #1, explain why the efficiency of this pumping system is very
low.
______________________________________________________________________________
______________________________________________________________________________
3. The pump is rated at 12 VDC. What effect did operating the pump at a higher voltage have on
the efficiency? _________________________________________________________________
4. What effect did raising the water to a higher level have on the efficiency?
______________________________________________________________________________
5. How does the calculated work output from Table 2 compare with the calculated work output
from Table 3? Discuss this answer.
______________________________________________________
______________________________________________________________________________
USE THE 5-STEP METHOD TO SOLVE THE FOLLOWING PROBLEMS:
6. A pump lifts water from a lake to fill a 700-cubic ft tank. If the tank is 52.0 ft above the lake,
find the amount of work needed to fill it.
7. Calculate the work required to lift 3.5 liters of water to a height of 1.4 meters.
8. Suppose a pump with an efficiency of 25% is used for prob.#7. Calculate the electrical energy
required to operate the pump.
9. Given that the density of water is 1.0 g/cm3, show that this is equivalent to a density of
1.0 kg/liter.
130
4. WORK IN ELECTRICAL SYSTEMS
a) Introduction
Work in mechanical systems means that a force must be moving something. Likewise, work occurs
in an electrical system when a voltage difference causes a charge to move. This happens when there
is a conducting path (part of a complete circuit) between two regions of different voltage. Just as
pressure causes fluids to move and do work in a fluid system, voltage difference causes charge to
move through a conductor and do useful work.
Examples of devices that do useful work when voltage difference moves charge are found in
rotating motors, fluorescent lamps, radios and coffeepots. Note that in each instance, you can't see
the electrical work being done. You only see the results of electrical work. Electrical work
produces heat, rotates shafts, and creates light or sound. For example, when you look at the monitor
on a computer or television, you see light emitted from the cathode-ray tube. The light is produced
when electrically charged particles strike a phosphor surface inside the tube.
In an electrical system, work and energy are measured in joules (J). The general formula for
work takes the form
Work = Potential Difference x Charge Moved
W = Vq
IV-12
where V is the potential difference through which the charge is moved and q is the quantity of
charged moved. Potential difference, measured in volts (V), is the force-like quantity. Charge,
measured in coulombs (C), is that "substance" in electrical circuits that is moved by a potential
difference. Since charge is always associated with particles, the "substance" is charged particles.
Therefore charge is the displacement-like quantity. In almost all electrical circuits, the charge
carriers are electrons.
A net charge of one coulomb in an electrical dc circuit represents a net transfer of
6,250,000,000,000,000,000 electrons from the negative pole to the positive pole! Since this
number is so large, it almost always written in scientific notation as 6.25 x 1018. One joule of
work or energy in the electrical system is the amount of energy required to move one coulomb of
charge through a potential difference of one volt.
1 joule = 1 volt x 1 coulomb
1 coulomb = charge of 6.25 x 1018 electrons
When a battery is recharged, the actual number of electrons in the battery is not increased. By
131
"charging" the battery, we mean that as a result of a chemical reaction the electrons gain energy.
The work done stores up energy in the battery, which can be used to perform useful work at
some later time. When the battery discharges, the work it performs is again the voltage times the
charge moved.
Example IV-9: Energy Required to Charge a Battery
How much energy is required to charge a 6 volt automobile battery that stores 8,000 coulombs of
charge?
Charging a battery.
Solution:
W = Vq = (6 V)(8,000 C) = 48,000 J
4.8 x 104 J, or 48 kJ of energy are required to charge the battery.
In most electrical systems, the measurement of electrical charge is unnecessary and inconvenient.
Instead, electrical current is measured. Current (I) is the rate at which charge flows, and is
defined as
I =
q
t
132
IV-13
where I is the current in amperes (A), q is the charge transferred in coulombs (C), and t is the
time in seconds (s) required for the transfer. Note that one ampere equals one coulomb per
second. This equation will be examined in more detail in the next section, where we discuss
"rate". It is presented here only because of its importance in determining work done in an
electrical system.
Example IV-10: Work Done by an Electric Motor
How much electrical energy is consumed in 2 minutes by a 12 V dc electric motor drawing a
current of 4 A?
A DC electric motor.
Solution:
I =
q
t
so
q = It = (4 A)(120 s) = 480 C
The charge transferred is 480 coulombs.
W = Vq = (12 V)(480 C) = 5760 J
5760 J of work, or energy are consumed during the two minute period.
b) Effects of Electrical Work
Electrical work is done when a voltage causes charge to move. Electrical work produces
movement, heat, light and/or sound. In your daily life, you are exposed to the effects of electrical
work whether you are at home or at work.
Electric motors are behind the movement of air in ventilating systems. Electric relays and solenoids
open or close switches and valves. Electric motors help move fluids by operating pumps or pushing
133
pistons with the aid of mechanical drives. These are examples of electrical work causing mechanical
movement.
Another result of charge movement by voltage is the production of heat. Heat may also be
welcomed but a natural byproduct, such as in electric heaters , clothes dryers, hair dryers,
dehydrators, or electric ovens. Heat may also be an unwelcome byproduct, such as the heat
produced in electric motors or incandescent light bulbs.
Light may be produced when a voltage difference causes charge to move. The light source may be
an incandescent bulb, fluorescent tube, mercury-discharge lamp, or laser.
c) Electrical Efficiency
Motors convert electrical work into mechanical work. The work done by electricity shows up as
mechanical work. Some of the electrical work is lost; this lost work shows up as heat, thereby
raising the temperature of the motor. Unless the electrical device is designed to produce heat
(motors are not), then the undesired heat represents a loss of useful electrical work. The loss lowers
the electrical efficiency of the system.
The efficiency of an electrical device is defined by the following equation:
Efficiency = Useful Work Done by the Device x 100%
Electrical Energy (Work) Input
or

Wout
x 100%
Win
134
IV-14
Example IV-11: Efficiency of an Electrical Motor
The input energy (Win) to the electric motor in Example IV-10 is 5760 joules. That motor is
connected to an overhead crane that lifts a load of 2000 newtons through a height of 2.5 meters.
Find: a) The work done by the motor (Wout) and b) the efficiency of the motor.
Solution:
a) Wout = Fout x Distance = (2000 N)(2.5 meters) = 5000 J
W 
 5000 J 
b) Efficiency =  =  out  100%  = 
 100%  = 86.8%
 5760 J 
 Win 
135
c) PROBLEM SET 12: ELECTRICAL WORK
1. In electrical work, the displacement-like quantity is the charge that is moved. The force-like
quantity is:
a. measured in newtons
b. measured in joules
c. the voltage difference
d. the product of current and time
2. The coulomb is a unit of electrical charge that is equal to:
a. one ampere
b. the charge on 6.25 x 1018 electrons
c. 60 ampereseconds
d. one amperesecond
3. A 1.5 Volt battery delivers 6 coulombs of charge. What is the electrical work done?
a. 4 amperehours
b. 9 J
c. 4 J
d. 9 N/m
4. A 12 volt battery does 180 J of work. How much charge has it moved? How many electrons
were moved?
5. A 220 V electric motor draws a current of 40 amp for 5 seconds. How many coulombs of charge
are moved?
6. An electric motor ran for 2 minutes at a setting of 18 volts, drawing a current of 4.5 amps. A.
How many coulombs of charge were moved? B. How much work was done in joules?
7. An electric motor lifts a 6.20 kg object through a height of 2.0 meters in 4 seconds. The motor
draws 6.00 amps while operating at 12 volts. A. Find the work output. B. Find the electrical
input work. C. Find the efficiency of the system.
8. What is the operating voltage of a DC motor that uses 8000 joules of energy in 3 minutes,
drawing an average of 3.7 amps.?
9. It takes 10,000 J of energy to charge a battery from 0 to 12 volts. Using a one amp charging
current, how long does it take to charge the battery?
10. If an electric motor does 1680 joules of useful work and is rated at 70% efficiency, how much
input energy does the electric motor require?
136
d) Work due to Magnetic Fields
We have shown that a charge moving through a uniform magnetic field perpendicular to its
direction of motion will be confined to a circular trajectory. Thus, even though a magnetic force
is acting upon the particle, it will constantly retrace its path so that its net displacement is zero.
Hence no work can be done by a uniform magnetic field.
Work can be done by a changing magnetic field, or rather, changing magnetic flux. The
magnetic flux is the product of the magnetic field intensity (B) and the cross sectional area (A)
through which it passes:
 = B·A
IV -15
The work required to establish a magnetic field in a solenoid for example, is the product of the
magnetomotive force (mmf = N·I) and the change in the magnetic flux (). The work done by a
changing magnetic field can then be written as
W = mmf·
IV -16
where the mmf is the force-like quantity and , the change in the magnetic flux, is the
displacement-like quantity. Changes in the magnetic flux will cause the path of a charged
particle to be non-circular, producing a net placement and work performed on the particle (see
Fig. IV-8).
electron
trajectory
X
magnetic
field
directed
into paper
Figure IV-8. A locally-varying magnetic field produces a net displacement of a charged
particle, thus work is performed on the particle.
e) The Electric Motor
A common application of magnetic fields is the electric motor. It can be found in most small
fans, blenders, can openers, washers, dryers, vacuum cleaners, hair dryers, windshield wipers,
VCR's, computers, battery operated toys that move, garage door openers… - suffice to say that
the electric motor is a crucial component of technology. A simple DC motor is shown in Figure
IV-9. Wire is coiled around a cylinder that is free to rotate about an axle at its center. This
component is known as the armature. The commutator is a pair of plates that are connected to
the ends of the wire and form a collar about the axle. The entire assembly resides in the midst of
a magnetic field, which can be provided by either an electromagnet or a permanent magnet, as
shown. A DC power supply, such as a battery, can connect to the wires whenever the brushes
shown come into contact with the commutator. When this occurs, the armature becomes a
137
solenoid, and its North Pole is drawn to the South Pole of the external magnet. As the poles
come into alignment, the brushes lose contact with the commutator and the armature rotates
freely. When the brushes come into contact again the current, and thus the magnetic poles of the
armature, are reversed, causing it to continue its rotation. In this manner, by continuously
changing the polarity of the armature/solenoid current, the armature maintains a continuous
rotation. The rotational energy of the rotating axle can be used to perform any number of useful
functions.
Armature
Windings
Commutator
Brushes
North
South
Battery
+
Field Magnet
Figure IV-9. A Simple DC Motor.
138
LAB #12: WORK DONE BY A MOTOR
OVERVIEW
An electric motor is an electromechanical device. It converts electrical energy into mechanical
kinetic energy, or motion.
Assume you are using an electric motor to lift a load. The electrical energy used by the motor is
converted to mechanical energy, or work, when the load is lifted. The electrical energy input is
the product of the motor voltage, motor current, and the time required to lift the load. This
electrical energy is the electrical work input to the system. This can be expressed as follows:
Workin = Electrical Energy = VIt
Recall that the unit of voltampsecond is equal to one joule.
The mechanical work done by the motor is the product of the lifting force and the distance the
load was lifted. This is expressed in the equations below, where lifting force (output) equals the
weight of the load, and the output distance is the height through which the weight is lifted.
Mechanical Linear Work:
W = FD
Workout = Weight x Height
In this experiment you will use an electric motor to lift a load. You will measure the electrical
energy input to the motor. You will also measure the mechanical work output of the motor.
Then you will calculate the efficiency of the motor lifting system.
As in Lab #11, much of the input electrical work is lost due to heat production in the motor.
LEARNING OBJECTIVES
The learning objectives of lab 12 are:
1. Identify the basic unit of electrical charge as the coulomb which is the charge that
6.25 x 1018 electrons have.
2. Describe what is meant by work in electrical systems.
3. Understand and be able to use SI and English units in solving work problems including
electrical systems.
4. Identify the work place applications of work in electrical systems.
5. Identify the effects of doing electrical work.
6. Use efficiency to determine the relationship between input and output work.
7. Define current in terms of electrical charge moved per unit time.
139
LAB #12 WORK DONE BY A MOTOR
Date___________
OBJECTIVES:
SKETCH OF LAB SET-UP:
DATA TABLE 1
LIFTING HEIGHT: h =___________cm = ____________m
TRIAL
#
1
TOTAL MASS
OF LOAD
m (grams)
TIME TO
RAISE LOAD
t (seconds)
MOTOR
VOLTAGE
V (Volts)
MOTOR
CURRENT
I (amps)
150 grams
2
3
DATA TABLE 2 (COMPUTED VALUES)
TRIAL
#
WEIGHT OF
LOAD
w (newtons)
MECHANICAL
WORK
OUTPUT
Wout (joules)
1
2
3
140
ELECTRICAL
ENERGY
INPUT
Ein (joules)
EFFICIENCY
 (%)
Lab #12 COMPUTATIONS: Show your calculations for Trial #1 below.
WEIGHT:
COMPUTING
COMPUTING OUTPUT WORK:
COMPUTING INPUT WORK (ELECTRICAL ENERGY):
ANALYSIS:
1. What does the voltage rating of a motor indicate?
______________________________________________________________________________
2. What happened to the electrical energy that was not converted to mechanical work?
______________________________________________________________________________
______________________________________________________________________________
3. Why are the British units of work (ftlb) not used for electrical calculations?
______________________________________________________________________________
______________________________________________________________________________
USE THE 5-STEP METHOD TO SOLVE THE PROBLEMS BELOW:
3. A DC electric motor operates on a voltage of 24 V and draws a current of 3.8 amperes. It runs
for 7 minutes. A) Find the total charge, in coulombs, that was energized. B) How much work
(electrical energy) was used to recharge the battery?
4. A 12 volt automobile battery is recharged by a current of 1.8 A in a time of 95 minutes. A)
Find the total charge, in coulombs, that was energized. B) How much work (electrical energy)
was used to recharge the battery?
141
5. HEAT TRANSFER IN THERMAL SYSTEMS
a) Introduction
The general work equation
Work = Force-like Quantity x Displacement-like Quantity
is not suitable for thermal systems. In thermal systems, the force-like quantity is identified as a
temperature difference, and the displacement-like quantity as heat energy. To this point, thermal
systems are like the other energy systems. In the application of the work equation, however, the
analogy fails, because energy, rather than mass, is displaced, thus no actual work is performed.
Yet in thermal systems, it is quite evident that a temperature difference will cause heat energy to
flow. So, even here, it is useful to identify a force-like quantity (T) and a displacement-like
quantity (H, or H). The analogy is helpful, even though the general work equation (Work =
Force-like Quantity x Displacement-like Quantity) does not hold. For example, when a pot of
water is heated on a stove, the temperature of the water changes as heat energy moves from the
hot stove into the water. In this instance, it is the temperature difference T between the hot
stove and cool water that “moves” heat energy into the water and, as a result, the cool water rises
in temperature. You can determine the temperature change of the water from a knowledge of the
heat energy flowing into the water and properties of the water alone. The required property of
the water is called its “specific heat” - the amount of energy required to change the temperature
of a unit mass of water by one degree. The specific heats for various substances are given in
Appendix C.
For example, the quantity of heat energy required to change the temperature of a body (such as
the pot of water) is given by
H = mcT
IV-17
where
H = thermal energy added to the body to cause a temperature change T
T = temperature change of the body
m = mass of body (or weight, in the English system) that undergoes change T
c = the specific heat of the substance.
In this equation the quantity mc is the product of the mass and the specific heat of the body that
undergoes the temperature change. The quantity mc is called the “heat capacity” of the object.
It represents the amount of heat energy required to raise the temperature of that body by one
degree (a Fahrenheit degree in English units and a Celsius degree in SI units).
142
The heat energy transferred to a substance actually shows up as increased motion of the
molecules in the substance. Thermal properties of different substances vary widely and are
dependent upon the molecular structure of the substance.
In SI units, the quantity of heat is measured in calories (cal) or kilocalories (kcal). One calorie is
the amount of heat required to raise the temperature of one gram of water one Celsius degree.
One kilocalorie will raise the temperature of 1 kg of water one Celsius degree. Since the calorie
measures small amounts of heat, kilocalories (kcal) are often used to measure larger quantities,
so often that one food Calorie (C, with a capital “C”), is equal to one kilocalorie (kcal). For
instance, a 60 Calorie slice of bread actually contains 60,000 calories per slice. One calorie is
equal to approximately 4.18 joules.
In English units, the quantity of heat is measured in British Thermal Units (BTU). One BTU is
the amount of heat required to raise the temperature of one pound of water one Fahrenheit
degree. This unit is used in the United States to measure the cooling capacity of air conditioners
or the heating capacity of furnaces and stoves. One BTU is equal to approximately 252 calories.
Conversion factors for the various energy units may be found in the back of the book.
Specific heat indicates how many units of heat energy are required to raise a unit mass of some
substance a unit temperature difference. Water, for example, has a specific heat of 1 cal/gC°, or
1 BTU/lbF°. Since water has a large specific heat (that is, an ability to soak up large amounts of
heat energy at the expense of only a small increase in temperature), and it’s safe and cheap, it
makes an ideal coolant. That's why water is used as the coolant in most systems in which
considerable thermal energy is generated.
Specific heats of numerous substances have been determined experimentally, some of which
may be found in the appendix.
Example IV-12: Energy Required to Heat Water
How much heat energy is required to heat 417 pounds (about 50 gallons) of water from 70°F to
130°F?
143
50 gallons
Heating a container of water.
H=
wcT = (417 lb)(1
= (417 lb)(1
BTU
)(130°F - 70°F)
lb  F o
BTU
)(60 F°) = 25,020 BTU
lb  F o
Example IV-13: Heat Energy Removed to Cool an Object
A 25 kg brass ball at 100°C cools to 30°C when dropped into an insulated container of water,
increasing the water temperature from 10°C to 30°C. The specific heat of brass is 0.091
cal/gC°.
Cooling a brass ball.
a) How much heat energy is removed from the ball and added to the water?
144
H = (25 kg)(1,000
g
cal
)(0.091
)(100°C - 30°C) = 159,300 cal
kg
g  Co
b) Assuming no heat is lost to the container or the atmosphere, what is the mass of water in the
container?
H
1.593 x 105 cal
m =
=
= 7,965 g
cT
 cal 
o
o
1
 (30 C - 10 C)
 g  Co 
b) Change of State
If heat is supplied to matter, two possible changes can occur. The first possible change,
described above, is that the temperature of the material will be raised while its state remains
fixed. The second possible change is that the material will change from a solid state to a liquid
state, or from a liquid state to a gas state. Adding or removing heat energy changes the
temperature or the state, but not both at the same time.
If the temperature of a material is changing, heat supplied to the material is classified as
“sensible heat”, since the change can be “sensed” by means of a thermometer. If the state of the
substance is changing, the heat is termed “latent heat”, since the change cannot be sensed by
means of a thermometer. The word "latent" means "hidden".
Figure IV-9 shows a container of ice on the left, to which heat is applied. The resulting states
and temperatures at various times in the heating process are shown. Assume that the heating
process is very slow, so that the entire container and its contents are always in thermal
equilibrium (the same temperature throughout). The ice experiences first a rise in temperature to
its melting point (0oC at sea level pressure). Once the melting point is reached, no further
temperature increase can occur until all of the ice is melted. The added heat energy can not be
observed as a temperature increase at this stage; the heat added is “latent heat” (“hidden heat”).
Once all of the water is in the liquid state, the temperature will rise from 0oC to 100oC, the
boiling point. Once again, as the water undergoes a change of state, no change in temperature is
observed until all of the water is converted to steam. As soon as it is all converted to steam, the
temperature of the water vapor can continue to rise. If the steam were trapped, it could be heated
further, above 100°C, or “superheated”.
145
Figure IV-9. Heating a container of ice.
Latent heat is energy used in changing matter from one state to another. Change of state is a
very important principle used in air conditioners, refrigerators, and many other cooling and
heating systems.
146
PROBLEM SET 13: THERMAL ENERGY
1. A 200 g piece of material heats from 20oC to 100oC when 2400 calories of heat energy are added
to it.
cal
a. Find the specific heat in units of
.
g  Co
b. What is the value of its specific heat in units of
c. What is its specific heat in units of
kcal
?
kg  C o
BTU
?
lb  F o
2. Which requires more heat to increase in temperature by one degree Fahrenheit, one pound of
copper or one pound of brass?
3. How many BTU's are needed to heat 2 lb of tin by one degree Fahrenheit?
4. How much heat is required to heat 800 grams of aluminum from 20 oC to 1500C? Use the Five
Step Method.
5. Use the Five Step Method. If 200 calories are added to a 250 gram piece of brass, how much will
the temperature change?
6. Use the Five Step Method. How many BTU are given off when a 2 lb steel object cools from
300oF to room temperature (70o)?
7. Which is the larger unit, the BTU or the kilocalorie?
8. Which of the following is not a property of a material?
a. specific heat
b. heat capacity
c. melting point
d. latent heat of vaporization
9. The heat equation, H = mcT, is a description of _____________
_________________________________________________________________
_________________________________________________________________
10. If a beaker of water is heated from 35 0F to 85 0F by the addition of 90 BTU of heat energy,
what is the weight of the water in the beaker?
11. Define and give examples of latent heat and sensible heat.
12. What is the specific heat of the substance if it takes 1500 calories to raise the temperature of
147
400 grams of the substance 10 C0?
13. How do you explain the fact that it takes 1150 calories to change 10 g of ice at -7 0C to water
at 32 0C? Use 0.5 cal/gC0 as the specific heat of ice. Show that just using H = mcT yields
a value of thermal energy less than 1150 calories. How is the rest of the energy used?
Assume no heat energy is lost to the environment.
148
LAB #13: THERMAL ENERGY
OVERVIEW
Mechanical work on a system may be partially or completely turned into thermal energy. If
there is no change of state (solid to liquid, or liquid to gas), then increased thermal energy causes
the temperature of the system to increase. When energy is in the form of thermal energy, it is
usually measured in calories or BTU's. The factors for converting from calories or BTU's to
joules or ftlb may be found in Appendix C.
When two substances at different temperatures are placed in contact, heat energy will flow from
the hotter object to the colder one. However, each substance does not necessarily have the same
change in temperature for an equal increase in thermal energy.
Each pure substance has a property called its specific heat. Specific heat is a measure of the
heat required to change the temperature of one gram of the substance by one degree Celsius. The
specific heat of pure water is 1.00 calorie per gram per Celsius degree. The letter c indicates
specific heat, and the units of specific heat are cal/(gCo) or BTU/(lbFo) In this experiment, the
SI units will be used.
c = 1.00 cal/(gCo) for water.
When a solid or liquid of known specific heat increases in temperature, the amount of heat
absorbed may be calculated by the following equation:
H = mcT
The specific heat of a solid or liquid may be determined experimentally by using the equation
c=H/(mT), where m is the mass of the substance, H is the heat gained or lost by the substance,
and T is the temperature change of the substance. The mass and temperature of a sample are
easily measured in the laboratory, but the amount of heat, in calories, is very difficult to measure.
The specific heat of a substance is different for each state; gaseous, liquid, or solid. As the
substance changes state (melting or boiling), there is no specific heat value, since there is no
change of temperature. Instead, the quantity of heat required to change the state of a gram of
substance is given as the “latent heat of fusion” or the “latent heat of vaporization”.
In the following experiment, you will determine the specific heat of several substances and
compare your values with the textbook values.
149
LEARNING OBJECTIVES
The learning objectives of lab 13 are:
1. Understand the difference between mechanical, fluid and electrical work and the
movement of thermal energy.
2. Describe the movement of heat energy through a material due to a T across the material
and how it may be determined.
3. Understand the concepts of specific heat of a substance and heat capacity of a sample.
4. Understand and be able to determine the specific heat of a substance.
5. Solve for heat energy required to change the temperature of a substance a given number of
degrees.
6. Learn to convert one unit of energy into any other unit of energy in both the SI
and English unit systems.
7. Understand what is meant by charge-of-state of a material, latent heat of vaporization,
latent heat of fusion, thermal conductivity and the difference between latent heat and sensible
heat.
8. Be able to calculate heat energy problems for changes of temperature of a substance.
150
LAB 13: SPECIFIC HEAT
Date__________
OBJECTIVES:
SKETCH OF LAB SET-UP:
TABLE 1: Masses for water-only experiment
ITEM
MASS
(grams)
cup only
cold water + cup
hot water + cold water + cup
cold water only
hot water only
TABLE 2: Water-only Experiment
Material
Tinitial
(oC)
cold water
Tfinal
(oC)
Heat
(cal)
(heat gained)
hot water
Heat Transfer Efficiency =  
(heat lost)
H gained
H lost
 100%  _____________
TABLE 3: Metal Samples Experiment
Water
Type of
m Tinitial Tfinal T
Heat
Metal
(g) (oC) (oC) (Co) Gained
(cal)
Aluminum
Brass
Steel
Tin
m
(g)
151
Metal Sample
Tinitial Tfinal T
(oC) (oC) (Co)
Heat
Lost
(cal)
Heat
Transfer
Efficiency
()
LAB 13: ANALYSIS
TABLE 4: Calculation of experimental specific heat
Type of Metal
cexp
ctextbook
(cal/g∙Co)
(cal/g∙Co)
Aluminum
Brass
Steel
Tin
%
1. How did the heat gained by the cold water compare to the heat lost by the hot material? If
they were not equal, explain why.
_______________________________________________________________________
_______________________________________________________________________
2. How do the specific heats of the metals compare with the specific heat of water?
________________________________________________________________________
3. Were the specific heats of the experiment ranked in the same order, lowest to highest, as the
text book values?
___________
4. What happened to the heat from the hot substance that was not absorbed by the water?
________________________________________________________________________
________________________________________________________________________
Use the 5-Step Method to Solve the Following Problems
5. A sample of aluminum at 150 oC is immersed in 800 grams of water at 25 oC. The final
temperature of the mixture is 32 oC.
A) Find the heat gained by the water.
B) Assuming no loss of heat to the surroundings, find the mass of the aluminum.
152
LAB 13: ANALYSIS (cont’d)
7. How many calories does a 4.5 kg block of copper give off as it cools from 200 oC to
23 oC (room temperature) ?
8. A 60 lb block of brass was heated from 75oF to 350oF. How many BTU's were absorbed by
the metal?
153
6. SUMMARY
Work is done in any physical system when a force-like quantity causes a change (displacementlike quantity) within the system. Energy is the ability to do work, and is conserved within a
closed system.
When a system is in equilibrium, it contains energy, but no work is done, because no
displacement occurs overall as the result of force-like quantities within the system. When a
system is not in equilibrium, displacement of some kind occurs as the result of force-like
quantities, and energy is converted to work within the system.
Work and energy are equivalent and are measured in the same units. The table below shows the
general work equation and the form it takes in mechanical, fluid and electrical systems. The
equation for heat energy required for a temperature increase also is included, although it is not
derived from the general work equation.
Work as a Unifying Concept
ENERGY SYSTEM
Mechanical
Translational
Rotational
FLUID
ELECTRICAL
THERMAL
Work = Force-like Quantity x Displacement-like Quantity
Work = Force x Displacement (in direction of force)
Work = Torque x Angular displacement
Work = Pressure difference x Volume change
Work = Potential difference x Charge transferred
Heat Energy = Temperature change x Heat capacity
154
V. RATE
Rate is a quantity that describes how rapidly an occurrence takes place. A car that travels 120
miles in two hours travels at a rate of 60 miles per hour (mph). Rate is defined as a displacement
like quantity divided by time:
Rate =
Displacement - like Quantity
Time Elapsed
In a linear mechanical system, velocity is the ratio of distance traveled to elapsed time. In a
rotational mechanical system, angular velocity is the ratio of angle of rotation to elapsed time. In
a fluid system, volume flow rate is the ratio of volume of fluid moved to the elapsed time. In an
electrical system, current is the ratio of charge transferred to elapsed time. In a thermal system,
heat flow rate is the ratio of thermal energy transferred to (surprise!) elapsed time. This chapter
discusses mechanical rates and their formulation.
1. MECHANICAL SYSTEMS: TRANSLATIONAL RATES
The most fundamental rate in a mechanical system is speed. Speed indicates only the rate at
which the object moves through some distance; it contains no information concerning the
direction of motion. Average speed can be calculated by dividing the total length of the path of
an object by the time required for the movement along that path. Velocity, rather than speed, is
used whenever the direction of the moving body is important. Velocity specifies both the speed
and direction of the motion, and thus is a vector quantity. For example, "100 kilometers per
hour" denotes a speed, while "100 kilometers per hour due northeast denotes a velocity.
When an object moves in a straight line, it is said to be in linear or translational motion. The
object may accelerate, slow, stop or reverse direction, but its motion always lies along the same
straight line. This discussion of mechanical translational rates will deal only with cases of
straight-line motion. The term "speed" will be used when the direction of motion is not
important; velocity will be used when both speed and direction are important. The formula for
calculating speed (the magnitude of velocity) in a linear mechanical system is given by
v =
d
t
V-1
where v is the average speed (the magnitude of the velocity), d is the displacement, and t is the
elapsed time. The symbol v is commonly used for both speed and velocity. If the quantity
155
identified as v includes a direction, it is the vector quantity velocity. If it does not include a
direction, it is the scalar quantity speed.
Example V-1: Speed and Velocity of an Automobile
What is the average speed and velocity of an automobile that travels to the east 250 feet in 10
seconds?
Figure V-1. Speed and velocity of an automobile.
Solution:
v =
d
250 ft
ft
=
= 25
t
10 s
s
The speed of the automobile is 25 ft/s. Its velocity is 25 ft/s, due east.
Distance (displacement) often is measured in units such as meters, kilometers, feet, or miles, and
time in units such as seconds, minutes, or hours. Common units for speed are m/s, km/hr, ft/s or
mph.
Example V-2: Time to Run a Mile
A runner can average 13.34 mph. How long does it take her to run one mile?
156
Figure V-2. The jogger.
Solution:
v =
d
t
so
t =
d
=
v
1 mi
= 0.0750 hr
mi
13.34
hr
min 

=  0.0750 hr   60
 = 4.5 min

hr 
Both Examples V-1 and V-2 involved the average speed of an object. When speed is constant
throughout the entire measured time, it is identical with the average speed. In many cases the
speed of a moving object fluctuates as it moves in a fixed direction. The rate of change of linear
speed is called "linear acceleration". A constant or an average linear acceleration is calculated
by using the relationship
a =
vf - vi
t
V-2
where a is the linear acceleration, vi is the initial speed before the change, vf is the final speed
after the change, and t is the time in which the change in speed takes place.
157
The units of acceleration are velocity units divided by time units. The most commonly used
acceleration units are m/s2 and ft/s2. An acceleration of 1 m/s2 means that the velocity increases
by 1 m/s each second.
Example V-3: Acceleration of a Car
A driver increases the speed of a car uniformly from 35 miles per hour to 55 miles per hour in 15
seconds. What is the acceleration of the car?
Solution:
Figure V-3. Acceleration of a car.
v f - vi
55 mph - 35 mph
20 mp
mph
a =
=
=
h = 1.33
t
15 s
15 s
s
The car's velocity increased by 1.33 mph every second.
The units shown may look strange because two different time units appear (hours and seconds).
The units often are reduced in order that only one time unit is used. To reduce the units, the unit
for speed may be converted from mph to ft/s:
miles
ft  1 hr 
ft

a =  1.33
 5,280

 = 1.96 2
hr  s 
mile  3600 s 
s

Acceleration is illustrated by the graphs in Figure V-4, where speed is recorded on the vertical
axis, and where time is recorded on the horizontal axis. A uniform change in speed, indicated on
the graph by a straight line, means that linear acceleration is constant. A non-uniform change in
speed, indicated on the graph by a curving line, means that linear acceleration is varying.
158
speed
speed
time
time
Figure V-4. a) Uniform acceleration and b), non-uniform acceleration.
The concepts of speed and acceleration are illustrated further in Figure V-5, which represents the
changing speed of a car during a short trip. Section A of the graph in Figure V-5 shows
acceleration of the car as it starts from rest and increases its speed. Since the graph rises very
rapidly, the rate of increase in speed (acceleration) is large. The car is in low gear. Section B
represents a slowing down (deceleration) of the car as its gears are shifted from low to second.
Section C represents acceleration as the car increases its speed while in second gear. Section D
again represents deceleration as gears are shifted from second to third. Section E at first
represents an increasing speed and thus an acceleration, but notice that acceleration during
Section E does not remain constant. The graph shows that, after the initial high acceleration, the
car accelerates less rapidly. Section F shows a rapid deceleration, which must result from a
braking action on the part of the driver. Section G shows no acceleration, because the speed
does not change during this time. Section H represents a final deceleration, which brings the car
to a stop.
Figure V-5. Automobile acceleration.
Figure V-5 illustrates both accelerations and deceleration. "Acceleration" is the term generally
applied to a rate of increase of speed, and "deceleration" is the term applied to a rate of decrease
of speed. If a moving object is decelerating, the value of a change of the change in speed (the
159
acceleration) will be negative; thus deceleration is simply negative acceleration.
Acceleration is produced only when an unbalanced or net force is applied to an object. The
amount of acceleration produced depends upon both the magnitude of the unbalanced force and
the mass of the object. The greater the unbalanced force on a given object, the greater is the
acceleration. Also, the more massive an object, the harder it is to accelerate. Furthermore, the
object always accelerates in the direction of the unbalanced force. These physical relationships,
referred to as Newton's second law of motion, are expressed as
a =
F
m
V-3
where a is the acceleration, F is the unbalanced force, and m is the mass of the object. This
equation often is stated in the form
F = ma
V-4
In applying this equation, force and mass units should be in the same system. For example,
when mass is in kilograms and acceleration is in m/s2, force is in newtons (all in the SI system).
When mass is in slugs and acceleration is in ft/s2, force is in pounds (all in the British system).
Example V-4: Acceleration of a Mass
How much acceleration is produced when a force of 35 N is applied to a 7 kg object? Neglect
friction effects.
7 kg
F = 35 N
Figure V-6. Applying a force to a block on a frictionless surface.
Solution:
kg  m 

 35


F
35 N
m
s2 
a =
=
=
= 5 2
m
7 kg
s
7 kg
The acceleration is 5 m/s2 in the direction of the push.
160
Example V-5: Truck Deceleration
What is the constant braking force required to stop a 5,000 kilogram truck in 6 seconds if it is
initially moving at 20 meters per second?
Figure V-7. Force required to stop a truck.
The acceleration is
a =
vf - vi
t
m
 m
m
- 20 
0
- 20
 s
m
s
s
=
=
= - 3.33 2
6s
6s
s
Since it's a negative acceleration, it must be decelerating.
Now, find the force required:
m
kg  m

F = m  a = 5,000 kg - 3.33 2  = -16,650
= -16,650 N

s 
s2
A force of 1.665 x 104 N is required to stop the truck in the time indicated. The negative sign
indicates that the braking force must be applied in a direction opposite to the motion of the truck.
The acceleration of a free falling body due to Earth's gravity (symbol g) is 9.8 m/s2 or 32 ft/s2.
This means that a freely falling body is increasing its velocity an additional 9.8 m/s for every
second that it falls (assuming that there is not air resistance). After two seconds, a falling body
that started from rest will be falling with a velocity of 19.6 m/s and, after five seconds, with a
velocity of 49 m/s.
161
The weight of a body is the force or the pull due to gravity on the mass of that body. This
relationship is given as
w = mg
V-5
where w is the weight (N or lb), m is the mass (kg or slugs), and g is the gravitational
acceleration. Notice that this equation has the form
Force = mass x acceleration
Example V-6: Towing an Automobile
A 3700 lb automobile is being towed on a level road by a tow truck accelerating at 3.8 ft/s2.
What is the tension on the towline? Assume that no frictional losses occur in turning the
automobile wheels.
Figure V-8. Towing an automobile.
Solution:
F = ma
or
w = mg
so
m =
w
g
and


 3700 lb  
 w
ft 
  3.8 2  = 439 lb
F =     a = 
 g
s 
 32 ft  
2

s 
162
The tension in the towline is 439 lb
163
PROBLEM SET 14: LINEAR MOTION
1. Define the following rates and express their units in both the SI and the English system of units:
Linear speed and linear acceleration.
2. Calculate an airplane’s average speed if it travels 3200 kilometers in 3.5 hours.
3. A car traveling at 55 miles per hour is slowed to a stop in 16.3 seconds. Calculate the car’s
deceleration in ft/s2.
4. A net force of 200 N accelerates an object with a mass of 100 kg. If the object started at rest,
determine its speed after 3 seconds.
5. How long does it take an object to accelerate from 3 m/s to 15 m/s if the object has a mass of 12
kg and is accelerated by a force of 20 N?
6. When the velocity is not constant, the “v” in Equation V-1 represents the average velocity.
When the velocity changes at a constant rate, the average velocity is equal to half of the sum of
the starting and ending velocities. A 1 kg mass undergoes constant acceleration and, starting
from rest, travels 10 meters in 1 second.
a) Find the speed of the mass at the end of the first second.
b) Find the acceleration of the mass.
c) Find the force necessary to cause this acceleration.
7. How long will it take a car traveling 50 m/s to overtake a car traveling 20 m/s if the first car is
initially 2000 meters behind the second car?
164
LAB #14: LINEAR MOTION
OVERVIEW
Mechanics is that part of physics that is concerned with motion. The study of motion is very
important in almost every area of modern science and technology.
Motion is defined as a change of position. To describe motion or change of position of an object
we need to know the distance the object traveled as a result of its motion. We call this distance
displacement. Displacement is a vector as it has both magnitude (distance traveled) and
direction (the direction in which the motion took place).
The distance traveled divided by the time of travel is the speed of travel. Speed is a scalar
quantity showing magnitude but not direction. Average speed equals the total distance traveled
d
divided by the total time of travel. v av =
t
Velocity is the time rate of change of a body’s displacement or its rate of motion in a particular
direction. Velocity is a vector that gives both the direction of travel and the rate of travel.
v =
d
t
Acceleration is the change of velocity per unit time and it is a vector quantity. Average
vf - vi
v
acceleration is the change in velocity divided by elapsed time.
a =
=
t
t
A constant acceleration is produced whenever a constant net force acts on an object.
The magnitude of the acceleration depends on the magnitude of the net force and also on the
mass of the object. a = Fnet/m
This is known as “Newton’s Second Law”, and it is
commonly written in the following form: Fnet = m  a
For constant acceleration only,
vav = ½ ( vf + vi ),
and
d = vav t
165
LEARNING OBJECTIVES
Learning objectives for Lab 14-15 are:
1. Distinguish between linear and rotational motion
2. Describe what rate means in mechanical systems for both linear and rotational motion.
3. Understand the difference between speed, velocity and acceleration for both linear and rotational
motion.
4. Calculate speed, velocity and acceleration for problems in linear and rotational motion in both
the SI and English systems of units.
5. Measure linear and mechanical rates
6. Identify workplace applications where linear and rotation rates are important.
166
LAB #14 LINEAR MOTION
OBJECTIVES:
Date_________
SKETCH OF LAB SET-UP:
TABLE 1: SETUP DATA
Angle of Inclined Track
=
Distance Traveled on Incline
d1 =
Distance traveled on Level
d2 =
TABLE 2: TRIAL RUN DATA
TRIAL
Time on Incline
t1 (sec)
Time on level
t2 (sec)
#1
#2
#3
#4
#5
#6
Average
TABLE 3: COMPUTATIONS
Constant Velocity
on Level
Average Velocity
on Incline
Final Velocity
on Incline
v on
Incline
v = d2/t2
vav = d1/t1
vf = 2vav-vi
v = vf-vi
Acceleration
on Incline
a = v/t1
Compare constant velocity on the level (v of column 1) with vf: _________________________
167
Lab #14 ANALYSIS:1. The distance traveled under constant acceleration is given by the
equation d = 0.5 at2 + vit. Choose a time interval 0.2 seconds longer than your average time on
the incline. Using this new time interval and your calculated value of acceleration from Table 3,
predict the distance the ball would travel down your inclined plane, starting from rest. Show
your work:
Check your answer experimentally. Explain what you did and the results:
______________________________________________________________________________
______________________________________________________________________________
2. If the ball on the level plane traveled a longer distance, how would the velocity compare with
the one meter trial (assume negligible friction)?
______________________________________________________________________________
3. If the ball on the inclined plane traveled a longer distance, how would the velocity compare
with the 60 cm trial?
__________________________________________________________________
USE THE 5-STEP METHOD TO SOLVE THE FOLLOWING PROBLEMS.
4. A car travels at a steady speed of 47.5 mph. How far will it travel in 13.5 minutes?
5. A cart rolls along a level track at a constant speed of 12 m/s. How long will it take to travel
83.4 meters?
6. A car accelerates from 10 ft/s to 30 ft/s in 3.2 seconds. Find A) the acceleration in ft/s2 B) the
average speed during the acceleration. C) the distance traveled while accelerating.
168
2. MECHANICAL ROTATIONAL RATES
Rate of rotational motion is called angular velocity (""). Angular velocity is defined as the rate
of change of angle with time. Common units for measuring angular velocity are revolutions per
minute (rpm) and radians per second (rad/s). Recall that one revolution equals 2 radians.
Example V-7: Units for Angular Velocity
What is the angular velocity, in radians per second, of a phonograph turntable with an angular
velocity of 45 rpm?
Figure V-9. An ancient artifact known as a turntable.
Solution:
Angular velocity =  = (45
rad  1 min 
rad
rev
)(2
) 
 = 4.7
rev  60 s 
s
min
The angular velocity is 4.7 radians per second.
In the rotational mechanical system, the basic rate equation is
 =

t
V-6
where  is the angular velocity,  is the angular displacement, and t is the elapsed time. Note
that this formula is identical to the speed or velocity formula, but with angular terms substituted
in for linear terms:  for v,  for d and time remains the same.
Example V-8: Angular Velocity
169
A grindstone rotates 600 revolutions in 5 minutes. What is its average angular velocity?
Figure V-10. A grindstone.
Solution:
 =

600 rev
=
= 120 rpm
t
5 min
rad
rad
 120 rev   2 rad   1 min 
= 
= 12.56


 = 4
 1 min   rev   60 s 
s
s
When an unbalanced or net torque is applied to a rotary system, the angular velocity of that
system is changed. The rate at which this change takes place is called the "angular acceleration".
Angular acceleration can be determined by dividing change in angular velocity by the time
required for this change. The formula for angular acceleration is
 =
f - i
t
V-7
where  is the angular acceleration, i is the initial angular velocity, before the change in angular
velocity, f is the final angular velocity, and t is the time. This formula is identical to the
formula for linear acceleration, but with angular terms substituted for linear terms: f and i for
vf and vi, and  for a.
Example V-9: Angular Acceleration
A brake is applied to a flywheel for 5 seconds, reducing the angular velocity of the flywheel
from 2.2 rad/s to 1.8 rad/s. What is the angular acceleration produced?
170
Figure V-11. Angular acceleration of a flywheel.
Solution:
 =
f - i
t
rad
rad 

rad
- 2.2
1.8

- 0.4


rad
s
s
s
=
=
= - 0.08 2
5s
5s
s
Angular acceleration produced is -0.08 rad/s2. The negative sign indicates this is a deceleration
of 0.08 rad/s2.
171
PROBLEM SET 15: ROTATIONAL RATES
1. A wheel is 2 m in diameter. When it turns at 200 RPM, find the linear speed of a point on its
circumference in m/s.
2. A shaft rotates through 500 revolutions in 25 seconds. Find its rotational speed in A. RPM and B.
radians per second.
3. A wheel that starts from rest has an angular speed of 20 rad/s after being uniformly accelerated
for 10 seconds. Find A. the average rotational speed and B. the total angle through which it turned
in those 10 seconds.
4. A flywheel spinning clockwise completes 180 revolutions in 10 minutes. Find
a) its angular speed in RPM and rad/s and
b) its angular velocity.
5. On a graph of speed vs. time, constant positive acceleration will appear as
a. a horizontal line on the axis
b. a straight line sloping upwards
c. a horizontal line above the axis
d. a curved line
6. The scanner dish of an antenna completes 110 revolutions in one hour. Find the angular speed in
a) rpm, and b) rad/s.
7. An electric motor shaft starts from rest and reaches its design speed of 960 rpm 2.7 seconds after
the motor is started. Find
a) the angular acceleration in rad/s2
b) the average angular speed during acceleration
c) the total number of revolutions the motor turned through during acceleration.
8. Define the following rates with their units: angular velocity and angular acceleration.
9. A flywheel of radius 60 cm, initially at rest, requires 25 seconds to reach an angular velocity of
1750 rpm. Find the angular acceleration of the flywheel.
10. A motor rotates at 5000 rpm. How much time is required for one revolution?
11. What is the initial angular velocity of a flywheel if, after two minutes, the flywheel is rotating
clockwise at 250 rad/s and the acceleration is a constant 4 rad/s2 clockwise?
12. Through what angular distance does a flywheel turn while accelerating from 10 rad/s to 100
rad/s if it undergoes a constant acceleration of 14 rad/s2?
172
13. If the linear velocity of a point on a flywheel is v, then the angular velocity of that point is v/r,
where r is the radius out to that point. A car accelerates at 100 mi/hr2 around a curve that has a
radius of 50 ft. Just prior to accelerating the car speed is 30 mph. What is the angular speed of the
car as it leaves the curve if it takes 10 seconds to complete the curve?
173
LAB #15: MEASURING ANGULAR RATE WITH A STROBOSCOPE
OVERVIEW
Angular velocity is the rate at which a rotating body is turning. This velocity is expressed in
units of rotational displacement per unit of time. A revolution is a unit of rotational
displacement that is equal to one complete rotation of an object. A common unit of angular
velocity is revolutions per minute, or RPM.
Another unit of rotational displacement is the radian. When a point on the circumference of a
circle is rotated so that it moves through a distance equal to the radius of the circle, the rotation
angle is one radian. The distance around the circle (circumference) is equal to 2 times the
radius, so there are 2 radians in one complete revolution.
1 revolution = 2 radians = 360o
In this experiment you will measure the angular velocity of a rotating wheel using a
stroboscope. The stroboscope emits short bursts of high intensity light that appear to "freeze"
the position of rotating objects when the rates are synchronized. This can be tricky, since the
rotating wheel appears frozen in the same way whether you’re flashing on every rotation or
every two or three rotations. How do you tell the difference? The strobe gives a unique
signature every time it flashes twice per rotation. Your rotating wheel has two marks like this on
it:
but when it flashes twice per
rotation, the wheel looks like this 
Start the strobe at the highest setting and slow it down until you see the figure on the right, then
you’ll know that the actual rotation rate is half that value, so cut your strobe reading in half and
find the figure on the left.
A rotating motor shaft requires constant applied torque to maintain a steady speed. If a torque is
not applied, friction will cause the motor shaft to gradually decrease. You will measure the time
it takes a motor shaft to come to a full stop after the power is turned off.
The angular acceleration, , is defined as the change in angular speed () divided by the
corresponding time interval. You will calculate the angular acceleration, which will be negative,
in units of radians/sec2.
 = f - i
Units for  are rad/sec
or rev/sec or rev/min
 =   t
Units for  are rad/s2 or rev/s2
For constant acceleration only:
av = ½(f + i)
 = av  t
174
LAB #15 ANGULAR RATE
OBJECTIVES:
Date__________
SKETCH OF LAB SET-UP:
DATA TABLE
MOTOR
SPEED
SETTING
VOLTAGE
SETTING
SLOW
2.5 V
MEDIUM
3.5 V
FAST
4.0 V
 (ANGULAR
SPEED in RPM)
FLASH
RATE
TIME TO
STOP
CALCULATED VALUES
MOTOR SPEED
 in rev/sec
 in rad/sec
SLOW
MEDIUM
FAST
SAMPLE CALCULATIONS (For the fast speed only):
175
 in rad/sec2
Lab #15 ANALYSIS:
1. Find the ratio between the angular rate
in rpm and the rate in rad/sec: (round off to one digit)
This ratio allows a quick approximation for unit conversions._________________
2. What is the relationship between the voltage applied to the dc motor and the angular speed of
the motor shaft?
_______________________________________________________________________
______________________________________________________________________________
3. Suppose a wheel is turning at 100 rpm. A strobe light set at 50 flashes per minute is used to
measure the wheel's angular velocity. Why does the wheel appear to stand still even though the
two rates are not the same?
____________________________________________________________
______________________________________________________________________________
4. For the situation in question 3, what would happen with a flash rate of 200 flashes per minute?
______________________________________________________________________________
_____________________________________________________________________________
5. A wheel makes 2.50 revolutions in 7.20 seconds. Find its angular velocity in rad/s. Show
your work, including conversion factors.
Use the 5-step method to solve the following problems:
6. A motor shaft is rotating at 750 rpm. Find the time required for it to rotate through 2.00
radians.
7. A motor shaft is turning at 200 rpm. It accelerates for 3.50 seconds reaching a speed of 800
rpm. Find the angular acceleration, , of the motor in units of rev/s2.
176
3. CIRCULAR MOTION
An object traveling in a circle at a constant speed is said to be undergoing uniform circular
motion (see Figure V-12).
velocity (v)
Figure V-12. A body traveling at constant speed along a circular path.
The speed of the object never varies, but its velocity is constantly changing since its direction,
which is always tangent to the circle, is continuously being altered. A changing velocity means
that an acceleration is present, therefore the object must be acting under the influence of a force.
Since it is always directed toward the circle’s center, it is called a centripetal (from the Latin,
“center seeking”) force. The associated acceleration, known as centripetal acceleration, is also
directed toward the center of the circle due to Newton’s second law (F = ma).
If we whirl a ball on the end of a string in a horizontal circle above our head, and we keep the
ball moving at a constant speed, we have uniform circular motion. We see that the speed is
constant but the ball’s velocity constantly changes. The centripetal force is provided by the
tension your hand is applying to the string. This force produces the constantly varying
centripetal acceleration, which in turn produces the continuous change in the ball’s direction (and
velocity). If the string breaks, there is no longer a centripetal force acting on the ball, and it will
fly off with a constant horizontal velocity, the same velocity (in direction and magnitude) it had
at the instant the string broke. It will also fall to the ground, since its vertical velocity is
governed by gravitational acceleration.
Gravity provides the centripetal force that keeps the planets moving around the sun and the moon
artificial satellites around the Earth.
Centripetal acceleration has a magnitude given by
ac
v2
=
r
V-8
where ac is the magnitude of the centripetal acceleration, v is the speed of the object and r is the
radius of the circular motion.
The velocity of the object is given by
v =  r
V-9
where  is the angular speed of the object (in rad/s). We can thus rewrite Equation V-8 as
177
ac = r   2
V-10
The magnitude of the centripetal force can be determined using Newton’s second law, F = ma:
Fc = ma c
mv 2
=
= mr 2
r
V-11
The time required for an object in circular motion to complete one revolution (or orbit) is called
the period (T):
T =
distance traveled
2r
=
speed
v
V-12
When a car goes around a curve, the friction between the tires and the road provides a centripetal
force directed toward the center of the curve. Your speed may be constant, but your direction,
and hence your velocity, is constantly changing. If there is ice on the road or you are going too
fast, your tires may not provide enough centripetal force to navigate the curve and you could end
up in a ditch. When you see a caution sign signifying a curve, is it most appropriate to slow
down before or after entering the curve? Braking during a curve causes an additional horizontal
force on the tires perpendicular to the centripetal force they are already experiencing. If the
resultant of these two forces exceeds the maximum static friction force for the tires, the car could
go into a skid and leave the road. Definitely slow down before entering the curve!
Highways built for higher speeds bank the roads in curves so that a portion of the vehicle’s
weight acts along the radius of curvature (in the direction of the centripetal force). The banking
angle of the curve is based on the speed the road is designed for, as well as the radius of the
curve. If you are driving through the curve at the posted speed, the road pushing directly upward
on the vehicle provides most of the centripetal force, and tire friction provides an additional
safety factor.
Example V-10
Find the centripetal force acting on a 1,200 kg car negotiating a curve of radius 40 meters at a
constant speed of 25 km/hr.
v = 25 km/hr
m = 1200 kg
0m
4
r=
Solution:
Fc = ?
m
 km   1000 m   1 hr 
v =  25


 = 6.95
hr   1 km   3600 s 
s

178
Fc =
mv 2
r
1200 kg  6.95 ms
=
40 m
The centripetal force is 1449 N.
179
2
= 1449 N
PROBLEM SET 16: CENTRIPETAL MOTION
1. Centripetal force is the
a. outward force on an object moving in a curved path
b. inward force on an object moving in a curved path
c. tangential force acting on an object moving in a curved path.
2. Why is it dangerous to apply your brakes when making a turn in your car?
________________________________________________________________
________________________________________________________________
________________________________________________________________
3. Why do they bank highways? __________________________________
________________________________________________________________
________________________________________________________________
4. A cork is attached to a string and rotated in a horizontal circle. The cork has a mass of 350 grams
and makes 25 revolutions in 15.2 seconds. The string is 20 cm long. Find
a. the rotational speed in rev/sec
b. the rotational speed in rads/sec
c. the centripetal acceleration
d. the centripetal force
e. what causes the force and its direction
5. A cork is attached to a string and rotated in a horizontal circle. The string has a tension of 12 N
and it is 27 cm long. The cork has a mass of 280 grams. Find
a. the centripetal acceleration on the cork
b. the rotational speed in rad/s
c. the rotational speed in rev/s
d. the time to complete 10 revolutions
6. A 2500 lb car on a level road rounds a curve with a radius of 520 ft. The car is traveling at a
speed of 44 ft/s ( 30 mph).
a. Find the centripetal acceleration of the car
b. Find the centripetal force on the car
c. What direction is the centripetal force and what provides it?
180
LAB #16 CENTRIPETAL FORCE AND ACCELERATION
OVERVIEW
A body that is traveling in a circle at constant speed is undergoing uniform circular motion. In
Figure V-12, v = vector velocity where the direction of the velocity is continually changing as
the body moves around the circle, but the magnitude of the velocity remains constant.
velocity (v)
Figure V-13. A body traveling at constant speed along a circular path.
Since the velocity is changing direction there must be an acceleration, which requires that a force
act on the body. Since the path that the body is following is a circle, the force acting is directed
toward the center of the circle.
The acceleration on the body is called centripetal acceleration and the force acting on the body
is called centripetal force. Centripetal force is the net inward force on a body moving in a
curved or circular path. If a body is swung at the end of a string and the string breaks, the ball
will travel in a straight line tangent to the circle from the point the ball was located at the time
the string broke.
Planets are kept in their orbits by a gravitational force, which provides the centripetal force
required for circular motion. In a similar manner the gravitational force between the earth and
orbiting satellites provides the centripetal force to keep them in orbit. Cars derive their
centripetal force for their turns on the highway from the frictional force between the tires and
the pavement and/or the banking angle of the highway.
centripetal acceleration = ac = v/t
v = vB - vA. v is a vector, and v is determined by the change in direction, even though the
magnitude (speed) may remain constant. Mathematically, the magnitude of the acceleration is
given by the equation ac = v2/r where v is the speed, and r is the radius of the circular path.
The equation ac = 2r may also be used to find the centripetal acceleration.
Newton’s second law (Fnet = ma) can be applied to the case of uniform circular motion. The net
force causing an object to move in a circle in the centripetal force. Substituting Fc for Fnet,, and
using the formula for acceleration given above, we have Fc = mv2/r = m2r
181
v
r
ac
Figure V-14. Components of centripetal acceleration.
LEARNING OBJECTIVES
Learning objectives for lab 16 are:
1. Understand the concept of circular motion of a body including the centripetal force,
centripetal acceleration, orbital period and orbital velocity and their interrelationship.
2. Practice solving problems of circular motion in both the SI and English unit systems.
3. Identify workplace applications of centripetal motion.
4. Measure orbital period and centripetal force to determine orbital speed and acceleration.
182
LAB #16: CENTRIPETAL ACCELERATION
DATE:_______
OBJECTIVES:
SKETCH OF LAB SET-UP:
DATA TABLE 1
Mass of Rubber Stopper: __________ kg
Trial
Mass of
Washers
m (kg)
Tube-toTape
Distance
(cm)
Time
for 40
revs
t (sec)
Period of
one rev
T (sec)
Orbital
Radius
r (cm)
#1
#2
#3
DATA TABLE 2
Trial
Actual
Weight of
Washers
Fc (newtons)
Orbital
Speed
v
(m/s)
Experimental
Centripetal
Force
Fc = mv2/r
(newtons)
#1
#2
#3
183
Percentage
Difference
in Fc
Values
LAB #16 ANALYSIS
CENTRIPETAL ACCELERATION
1. Discuss the possible errors in the experimental values of FC that cause them to be different
from the actual values.
________________________________________________________________
______________________________________________________________________________
2. Show that the expressions mv2/r and mr2 both have the units of force in both English and SI
systems.
USE THE 5-STEP METHOD TO SOLVE THE FOLLOWING PROBLEMS:
3. A 0.5 kg ball moves in a circle 0.4 m in radius at a speed of 4.5 m/s. What is its centripetal
acceleration? What is the centripetal force on the ball?
4. A 4000 lb car travels at constant speed on a circular track with a 300 ft radius. The car
completes a single lap in 32 sec. Find the velocity of the car, its centripetal acceleration and the
centripetal force acting on the car. Lastly, describe the source of the centripetal force since no
cord is attached to the car.
184
APPENDIX A: SOLUTIONS TO WORK PROBLEMS
PROBLEM SET 1
1. 230
2. 78.3
(1.04792 x 10 4 )(4.369 x 105 )(2.26 x 101 )
= 4.09 x 101
3.
(4.79 x 104 )(7.78 x 101 )(6.793 x 10 2 )
4. 1.47 m
5. 1.40 ft, 16.8 in
6. 6.21 x 103 mm
PROBLEM SET 2
1. 60 lb to the right
2. 28.3 lb at 2540
3. 20 lb south
4. 160 lb right
5. 106 N at 2270
6. a. V b. S c. S d. V e. V f. S
7. 19.8 N at 90o
8. 14.4 lb at 116o
PROBLEM SET 3
1. b
2. l = 3.05 ft
3. l = 8.7 m
4. A. l = 1.5 ft B.  = 13.5 ftlb
5. B. 6.25 ftlb CW C. 6.25 ftlb CCW D. F = 12.5 lb
6. 50 N at 76 cm
7. a
8. mass
9. Vector
10. Scalar
11. weight
12. torque
PROBLEM SET 4
1. c
2. D = 10.5 g/cm3
3. SG = 0.9
4. b
5. W = 0.131 lb
6. a) 4.5 g/cm3 b) 9 x 106 g = 9 x 103 kg
7. a) 487 lb/ft3 b) 779 lb
8. 144 lb
9. a) V = 2.57 x 105 cm3 = 2.57 x 10-1 m3 b) 257 kg
192
PROBLEM SET 5
1. P = 3744 lb/ft2
This is less than pressure from sea water
2
2. P = 2080 lb/in
3. a, b, c, and d are all correct
4. Pabs = 69.7 lb/in2 Pg = 55 lb/in2
5. b
6. pressure act equally in all directions
7. d
8. 123 psia (absolute pressure)
9. 10.2 m
10. 2 x 106 N/m2 = 2 x 106 Pa
11. 2 x 104 lb
12. 37,440 lb; NO!
13. 1.7 x 103 kg/m3
14. 51 m
15. pressure (or depth or height)
16.  w  g   w h  gh  P
17. no
18. P = 5.53 x 103 lb/ft2
PROBLEM SET 6
1. direct current (DC)
2. alternating current (AC)
3. alternating
4. a and c
5. A. 15 V B. 3 V
6. d
7. Given in class
8. b) V = 15 V
9. b
PROBLEM SET 7
1. F = 5.26 N
2. mass increases
3. pulling force decreases
4. car starter, valve operator, switch operator
PROBLEM SET 8
1. higher; lower
2. temperature
3. a
4. a. 24 0F b. 10 F0 c. 212 0F d. 10 C0
5. V = 2.73 mV
6. A. 130 C0 B. 234 F0
7. A. 140 F0 B. 77.8 C0
8. A. 24 C0 B. 54.8 C0 C. -14 C0 D. Tref is warmer than Tmeas
9. a) 212 oC b) -3.3 oC c) -25 oC
10. a) 77.9 oF b) 41 oF c) -40 oF
11. a) 23.3 Co b) 90 Co c) 18 Fo d) -45 Fo
193
PROBLEM SET 9
1. to move or rotate
2. a and c
3. newtonmeter or Nm
4. Yes: a,d,f No: b,c,e
5. efficiency
6. A. 2400 ftlb B. 1950 ftlb C. 81.255 %
7. A. 1800 J B. linear or translational
8. 58.3 lb
9. 85.77 %
10. 171 N
11. A. 1570 J B. 2560 J C. 61.3 %
12. 4.2 ft
13. w = 19.8 lb
14. a = 0.44 m/s2
PROBLEM SET 10
1. A. 0.523 rad
B. 1 rad
C. 4.71 rad
D. 6.28 rad
2. A. 3.14 rad B. 31.4 rad C. 20.1 rad
3. radians
4. the angle turned through in radians
5. A. 75 ftlb B. 471 ftlb
6. 1.33 rad
7. 179 Nm
8. A. 2.75 ftlb B. 4.34 rev
9. 70.8 J
10. W = 1979 Nm = 1455 ftlb
11. F = 95.5 N
PROBLEM SET 11
1. a
2. b
3. A. open B. open C. closed D. closed E. open
4. 18.75 ftlb
5. 833.3 N/m2
6. 3.57 x 107 J
7. 9 x 103 J
8. 1.44 lb/in2
9. 1.04 x 105 ftlb
10. A. 0.948 in B. 7.06 in3 C. 706. inlb or 58.8 ftlb
11. 1.67 x 105 Pa = 1.67 x 105 N/m2
194
PROBLEM SET 12
1. c
2. b and d.
3. b
4. 15 C; 9.38 x 1019 electrons
5. 200 C
6. A. 540 C B. 9720 J
7. A. 121.5 J B. 288 J C. 42.2%
8. 12.0 V
9. 833 s (= 13 min. 53 s)
10. 2400 J
PROBLEM SET 13
1. A. 0.15 cal/gC0
B. 0.15 kcal/ kgC0
C. 0.15 BTU/lbF0
2. Copper
3. 0.110 BTU
4. 22.8 kcal
5. 8.79 C0
6. 52.9 BTU
7. kcal is larger
8. b
9. H = mcT is the amount of heat energy required to increase a mass of m, T degrees
10. w = 1.8 lb
11. Latent Heat: heat used to melt ice or vaporize water
Sensible Heat: heat used to raise the temperature of ice, water or steam.
12. c = 0.375 cal/gCo
13. Hint: calculate the latent heat of fusion needed to convert 10g of ice at 0C to 10 g of liquid
water at 0C.
PROBLEM SET 14
1. v = d/t, m/s, ft/s; a = v/t, m/s2, ft/s2
2. 914 km/hr
3. -4.95 ft/s2
4. 6 m/s
5. 7.20 s
6. a) 20 m/s b) 20 m/s2 c) 20 N
7. 66.7 s
195
PROBLEM SET 15
1. 20.9 m/s
2. A. 1200 rpm B. 125.7 rad/s
3. A. 10 rad/s B. 100 radians
4. A. 18 rpm B. 1.88 rad/s
5. b
6. A. 1.83 rpm B. 0.192 rad/s
7. A. 37.2 rad/s2 B. 480 rev/min or 8 rev/s C. 21.6 rev
8.  = /t, rad/s;  = /t, rad/s2
9. 7.33 rad/s2
10. 0.012 s
11. 230 rad/s CCW
12. 353 radians
13. 0.89 rad/s
PROBLEM SET 16
1. b
2. The brakes apply an additional force at right angle to the centripetal force. These two forces
exceed the static friction between the tires and the road so tires slide and car skids to outside of the
curve.
3. weight of car provides the centripetal force for turning at designed highway speed.
4. A. 1.64 rev/s B. 10.3 rad/s C. 21.36 m/s2 D. 7.48 N
5. A. 42.86 m/s2 B. 12.60 rad/s C. 2.01 rev/s D. 4.98 sec
6. A. 3.723 ft/s2 B. 291 lb C. to the center of curve, friction between tires and the road.
196
APPENDIX B: VECTORS AND SCALARS
Basic physical quantities such as speed, temperature, mass and pressure require only a
magnitude to specify their values. They are called “scalar quantities”, or simply
“scalars”. Other basic physical quantities such as force, velocity, acceleration and torque
require both a magnitude and a direction to specify their values. They are called “vector
quantities”, or simply “vectors”.
A scalar quantity is represented by its magnitude and its dimensions (in units).
Combining scalar quantities is easy. The hardest part is remembering to perform the
same operations on the units that you perform on the numbers.
Example B-1: An 800 kg horse accelerates at 5 m/s2 over a distance of 12 meters.
How much work does the horse do?
Solution: In this case we are working with a true scalar (mass) and the scalar components
of two vectors (acceleration and distance). This quarter you will learn the physics
definition of work. Work equals force times distance. You will also learn that force
equals mass times acceleration. We represent each quantity by a symbol. Work is
usually given by W, force is F, distance, or length, is often represented by d, mass is
denoted by m, and acceleration is given by a. In equation form:
Work equals force times distance  W = Fd
Force equals mass times acceleration  F = ma
Therefore
W = Fd = mad = (800 kg)(5 m/s2)(12 m) = 4,800 kgm2/s2
Note that we perform the same operations on the units that we performed on the numbers.
A vector quantity is represented by its magnitude, its dimensions (in units) and its
direction. For example, if I need directions to Gig Harbor and you tell me it’s 25 miles
away, you’re not giving me enough information. You’ve given me a scalar when I need a
vector. You’ve given me the magnitude and the dimensions (25 miles), but no direction.
I need to know that Gig Harbor is south of here. If the top of the page is north, then the
Gig Harbor vector can be drawn as shown below:
197
BREMERTON
N
Direction to
Gig Harbor
(magnitude
is 25 miles)
W
E
S
GIG HARBOR
Vectors are drawn as lines with arrows. The length of the line is proportional to the
magnitude of the vector. Vectors are drawn relative to a set of axes such as an x-y
coordinate system or a North-South coordinate system.
Example B-2: A car is driving at a constant speed of 40 mph in a circle. What is its
velocity in the y direction at points a, b, c and d?
b
a
c
y
x
d
Solution: At point A its velocity is 40 mph. At point b it’s y-velocity is 0 (velocity is all
in the negative x direction). At point c its y-velocity is -40 mph and at point d its yvelocity is again 0.
Vectors are also subject to the laws of addition, subtraction, multiplication and division.
Multiplication or division of a vector results in a change of magnitude only. Multiplying
a vector whose magnitude is 8 by a factor of 4 results in a vector in the same direction,
but with magnitude 32. Addition and subtraction of vectors is trickier. When one vector
is added to another, the resulting vector is obtained by placing the tail of vector B (“B”)
at the head of vector A (“A”) and drawing the resulting vector C (“C”) from the tail of A
to the head of B, as shown below:
198
y
B
C
A
x
What’s the magnitude of C? To get the magnitude of the resultant vector we need to
break each of the initial vectors into its x and y components and add them separately.
Let’s say A has a magnitude of 8 in the x direction and a magnitude of 6 in the y
direction, and B has a magnitude of -3 in the x direction and 8 in the y direction. The
magnitude of C in the x direction is 8-3 = 5. In the y direction C has a magnitude of 6+8
= 14. The total magnitude of C, which can be obtained by using the Pythagorean
Theorem, is the square root of (52 + 142) = 15 (approximately). A graphical
representation of C and its components is shown below:
Ctotal=15
Cy =14
Cx=5
Adding Vectors by the Graphic Method
Vectors can be added head-to-tail graphically using a ruler to measure lengths, and a
protractor to measure angles. The procedure is as follows:
1.
Choose a suitable scale and determine the length of the arrow that will represent the
vector according to the specified scale.
2.
Choose an origin or starting point on the paper and draw the arrow representing the
first vector. This arrow must be of proper length and must point in the appropriate
direction, indicated by its angle. The tail of the arrow is positioned at the origin, or
starting point.
3.
The arrow of the second vector is drawn so that its tail is joined to the head of the first
vector. The arrow of the second vector must also be drawn to scale and must point in
the correct direction indicated by its angle.
199
4.
If there is a third vector to be added, its tail is joined to the head of the second vector
and correctly oriented. This process of joining tail to head is continued until the
magnitude and direction of all vectors have been represented.
5.
The arrow representing the resultant vector is drawn with its tail at the origin (starting
point) and its head joined to the head of the last vector.
6.
The resultant arrow indicates the magnitude and direction of the resultant.
Measurement of the resultant arrow with the ruler and the protractor gives the
magnitude and direction of the resultant vector.
7. The length of the resultant vector is converted back to the units of the original vectors
by multiplying by the appropriate scale factor.
200
APPENDIX C: CONVERSIONS, CONSTANTS AND USEFUL EQUATIONS
WEIGHT AND MASS EQUIVALENTS
ALWAYS TRUE
1 lb (force) = 4.45 N (force)
1 slug (mass) = 14.6 kg (mass)
AT THE EARTH'S SURFACE ONLY
1 kg weighs 9.8 N or 2.2 lb
1 slug weighs 32.2 lb or 143 N
g = 9.8 m/sec2 = 9.8 N/kg
LINEAR MECHANICAL EQUATIONS
UNITS
SI
ENGLISH
W = FD
W = work
F = applied force
D = distance moved in the direction
of the force
J (joule)
N (newton)
m (meter)
ftlb
lb
ft
W = wh
W = mgh
W = work to lift a weight
w = weight
h = height lifted
m = mass
g = gravitational acceleration (at sea
level only)
J
N
m
kg
g = 9.8 m/s2
ftlb
lb
ft
slug
g = 32.2 ft/s2
v = d/t
v = speed or velocity
d = distance traveled
t = time of travel
a = acceleration
m/s
m
s
m/s2
ft/s
ft
s
ft/s2
F = net force
m = mass
a = acceleration caused by net force
N
kg
m/s2
lb
slug
ft/s2
 = efficiency
W = work
 is often expressed as %
a
vf - vi
t
F = ma
(Newton's
second law)

Wout
x 100%
Win
work units must agree
201
ROTATIONAL MECHANICAL EQUATIONS
UNITS
ENGLISH
SI
 = Fx
 = torque
F = applied force
 = lever arm
Nm
N
m
lbft or ftlb
lb
ft
W = 
W = work
 = applied torque
 = angle of rotation in
radians
J (joule)
Nm
ftlb
lbft
 : theta
radian (1 rev = 2 rad)
 = /t
 : omega
 = angular speed
rev/min (revolutions per minute or RPM)
rad/s (radians per second)
 =(f - i)/ t
 = angular or rotational
acceleration
rad/s2 (radians per second squared)
 = Wout/Win
 = efficiency
W = work
 is expressed as %
 : alpha
work units must agree
MECHANICAL ADVANTAGE
Mechanical Advantage for all machines:
IMA = Di/Do
AMA = Fo/Fi
Ideal Mechanical Advantage = input distance  output distance
Actual Mechanical Advantage = output force  input force
Efficiency = AMA/IMA
202
FLUID EQUATIONS
UNITS
SI
ENGLISH
 = m/V
 : rho
 = mass density
m = mass
V = volume
g/cm3, kg/m3
g
cm3, m3
slugs/ft3
(seldom used)
w = w/V
w = weight density
w = weight
N/m3
N
lb/ft3
lb
P = F/A
P = pressure
F = applied force
A = area
Pa (pascal)
1 Pa = 1 N/m2
lb/in2 or lb/ft2
P = gh
P = wh
or
P = gauge pressure at a
depth h
W = PV
W = work
P = pressure difference
V = volume of fluid
J (joule)
N/m2
m3
ftlb
lb/ft2
ft3
QV = V/t = vA
QV = volume flow rate
m3/s, m3/hr,
l/s
ft3/hr
gal/min (GPM)
Qm = m/t
Qm = mass flow rate
grams/sec
kg/s
slug/hr
(not commonly
used)
203
ELECTRICAL EQUATIONS
UNITS
SI
ENGLISH
VT = V1 + V2 + 
VT = total potential difference for
batteries in series
VT = V1 = V2 = 
VT = total potential difference for
batteries in parallel
I = q/t
I = electric current
q = charge
A (ampere)
C (coulomb)
W = Vq
W = electrical work
V = potential difference
q = quantity of charge moved
I = current
t = time in seconds
J (joule)
V
C (coulomb)
W = VIt
V (volt)
A (ampere)
s (second)
THERMAL EQUATION
SI
TC = 5/9(TF - 32o)
none
UNITS
ENGLISH
TC = Celsius temperature
TF = Fahrenheit temperature
oC
oF
(example:
TC = 20oC)
(example:
TF = 70oF)
T = temperature change or
temperature difference
Co
Fo
TF = 9/5(Tc) + 32o
TC = 5/9(TF)
TF = 9/5(TC)
examples:
T = 5 Co
T = 9 Fo
QH = H/t
QH = heat flow rate
H = heat energy
cal/s
cal (calorie)
Btu/hr
Btu (British
thermal unit)
H = mcT
(thermal energy
needed to heat or cool
an object)
AT
QH =

H = heat energy
m = mass of object
T = change in temperature
c = specific heat of the material
QH = heat flow rate
 = thermal conductivity
A = cross sectional area
T = change in temperature
cal
g (gram)
Co
cal/(gCo)
cal/s
cal/(mCo)
m2
Co
m
Btu
slug
Fo
Btu/(lbFo)
BTU/hr
BTUin/(ft2 Fo)
ft2
Fo
in
 = thickness of material
204
CONVERSION FACTORS & REFERENCE TABLES
CONVERSION FACTORS
LENGTH
1 inch (in) = 2.54 centimeters (cm) = 25.4 millimeters (mm)
1 meter (m) = 100 cm = 1000 mm = 39.4 in = 3.28 ft
1 kilometer (km) = 1000 m = 0.621 mi
1 mile (mi) = 5280 ft = 1.61 km
AREA
1 m2 = 10,000 cm 2 = 1550 in2 = 10.76 ft2
1 ft2 = 144 in2 = 929 cm2
VOLUME
1 m3 = 1000 liters () = 106 cm3 = 35.3 ft3
1 cm3 = 1 milliliter (ml) = 0.001 
1 ft3 = 1728 in3 = 28.3 liters () = 7.48 gal
1 gal = 4 qts = 3.785 liters
DENSITY
mass density of water = 1 g/cm3 = 1 g/ml = 1000 kg/m3 = 1 kg/ = 1.94 slug/ft3
weight density of water = 62.4 lb/ft3 = 9800 N/m3
SPEED
1 m/s = 3.28 ft/s = 2.24 mi/hr = 3.60 km/hr
1 ft/s = 0.305 m/s = 0.682 mi/hr = 1.10 km/hr
1 km/hr = 0.278 m/s = 0.913 ft/s = 0.621 mi/hr
1 mi/hr = 1.47 ft/s = 1.61 km/hr
MASS
1 kilogram (kg) = 1000 grams (g) = 0.0685 slug
1 slug = 14.6 kg
FORCE
1 newton (N) = 0.225 lb = 3.60 oz
1 pound (lb) = 16 oz = 4.45 N
PRESSURE 1 pascal (Pa) = 1 N/m2
1 bar = 105 Pa
1 lb/in2 (psi) = 144 lb/ft2 (psf) = 6900 Pa
1 atmosphere (atm) = 14.7 lb/in2 (psi) = 2117 lb/ft2 (psf) = 101,325 Pa
ENERGY
1 joule (J) = 0.738 ftlb
1 ftlb = 1.36 J
1 kilocalorie (kcal or Cal) = 1000 calories (cal)
1 cal = 4.184 J = 3.97 x 10-3 BTU = 3.077 ftlb
1 BTU = 252 cal = 778 ftlb = 1054 J
POWER
1 watt = 0.73760 ftlb/s = 3.41 BTU/hr = 0.239 cal/s
1 hp = 550 ftlb/s = 746 W
1 kilowatt (kW) = 1000 W
205
TEMPERATURE DIFFERENCE (T) 9 Fo = 5 Co
NOTE: FOR TEMPERATURE CONVERSIONS USE
𝑇(℃) =
or
𝑇(℉) =
5𝐶°
[𝑇(℉) − 32℉]
9𝐹°
9𝐹°
𝑇(℃) + 32℉
5𝐶°
THERMOCOUPLE EQUATIONS
Vtotal  Vref  Vreading
 T -T
T  Vtotal - Vmin  max min
 Vmax  Vmin



In our lab exercises, Vref is the voltage corresponding to the reference temperature (obtained
from the thermocouple table), Vreading is the measured voltage difference (the voltmeter reading)
and Vtotal is the total voltage difference, the value that is converted into a temperature using the
thermocouple table. Vmin and Vmax are the voltage readings in the thermocouple table just below
and just above Vtotal, respectively. Tmin and Tmax are the temperatures in the table corresponding
to Vmin and Vmax, respectively.
ANGLE MEASUREMENTS 1 revolution = 2 radians = 360o
206
TYPE-E THERMOCOUPLE
Chromel-Constantan Reference Junction at 0 oC
0 Co
2 Co
4 Co
6 Co
8 Co
10 Co
-20 Co
-1.15 mV
-1.26 mV
-1.38 mV
-1.49 mV
-1.60 mV
-1.71 mV
-10 Co
-0.58 mV
-0.70
-0.81
-0.93
-1.04
-1.15
0 Co
0.00
-0.12
-0.23
-0.35
-0.47
-0.58
0 Co
0.00
0.12
0.23
0.35
0.47
0.59
10 Co
0.59 mV
0.71
0.83
0.95
1.07
1.19
20 Co
1.19
1.31
1.43
1.56
1.68
1.80
30 Co
1.80
1.92
2.05
2.17
2.30
2.42
40 Co
2.42
2.54
2.67
2.80
2.92
3.05
50 Co
3.05
3.17
3.30
3.43
3.56
3.68
60 Co
3.68
3.81
3.94
4.07
4.20
4.33
70 Co
4.33
4.46
4.59
4.72
4.85
4.98
80 Co
4.98
5.12
5.25
5.38
5.51
5.65
90 Co
5.65
5.78
5.91
6.05
6.18
6.32
100 Co
6.32
6.45
6.59
6.72
6.86
7.00
110 Co
7.00
7.13
7.27
7.41
7.54
7.68
120 Co
7.68
7.82
7.96
8.10
8.24
8.38
130 Co
8.38
8.52
8.66
8.80
8.94
9.08
140 Co
9.08
9.22
9.36
9.50
9.64
9.79
150 Co
9.79
9.93
10.17
10.22
10.36
10.50
160 Co
10.50
10.65
10.79
10.93
11.08
11.22
170 Co
11.22
11.37
11.51
11.66
11.80
11.95
180 Co
11.95
12.10
12.24
12.39
12.53
12.68
The top three rows of the table provide voltages for temperatures below 0oC. In
that case, the column headings correspond to 0 oC, -2 oC, -4 oC, and so on.
NOTE: If the reference junction is not at 0oC, the table above gives the T value. To find the
measurement temperature, add T from the table to the reference temperature.
207
MASS DENSITY () & SPECIFIC GRAVITY
SOLIDS

g/cm3
Sp. G.
no units
LIQUIDS &
GASES

g/cm3
Sp.G.
no units
Gold
19.3
19.3
Mercury
13.6
13.6
Lead
11.3
11.3
Water
1.0
1.0
Silver
10.5
10.5
Oil
0.9
0.9
Copper
8.9
8.9
Alcohol
0.8
0.8
Steel
7.8
7.8
Antifreeze
1.125 (32OF)
1.098 (77oF)
1.125 (32OF)
1.098 (77oF)
Tin
7.29
7.29
Air
1.29 x 10-3
1.29 x 10-3
Aluminum
2.7
2.7
Hydrogen
9.0 x 10-5
9.0 x 10-5
Balsa Wood
0.3
0.3
Oxygen
1.43 x 10-3
1.43 x 10-3
Oak Wood
0.8
0.8
WEIGHT DENSITY (w)
To find the weight density of a substance, multiply the specific gravity of the substance
by the weight density of water. The weight density (w) of water is 62.4 lb/ft3 = 9800 N/m3
PRESSURE CONVERSION CHART
psi
in. of H2O
in of Hg
mm of H2O
mm of
Hg
bar
mbar
1.0
27.68
2.036
703.1
51.71
0.0689
68.95
APPROXIMATE COEFFICIENTS OF FRICTION ()
Material
static (s)
kinetic (k)
rubber on dry concrete
rubber on wet concrete
leather on wood
wood on wood
0.9
0.7
0.5
0.7
0.7
0.5
0.4
0.3
208
SPECIFIC HEATS (c) OF COMMON SUBSTANCES
SUBSTANCE
SPECIFIC
HEAT
SUBSTANCE
cal/(gCo) or
Btu/(lbFo)
SPECIFIC
HEAT
CONVERSION
FACTORS
cal/(gCo) or
Btu/(lbFo)
Air
0.24
Stone (average)
0.192
1 Btu = 252 cal
Aluminum
0.22
Tin
0.055
1 kcal = 1000 cal
Brass
0.091
Water
1.00
1 cal = 3.09 ftlb
Copper
0.093
Ice
0.50
1 Btu = 778 ftlb
Glass
0.21
Steam
0.48
1 cal = 4.18 joules
Iron (steel)
0.115
Wood (average)
0.42
HEAT OF FUSION AND VAPORIZATION
SUBSTANCE
MELTING
POINT (oC)
BOILING
POINT (oC)
LATENT HEATS
Solid  Liquid
Liquid  Vapor
OXYGEN
-218
-183
3.3 kcal/kg
51 kcal/kg
NITROGEN
-210
-196
6.1 kcal/kg
48 kcal/kg
ALCOHOL
-114
78
26.0 kcal/kg
204 kcal/kg
0
100
80.0 kcal/kg
540 kcal/kg
LEAD
327
1750
5.5 kcal/kg
205 kcal/kg
SILVER
961
2212
26.5 kcal/kg
563 kcal/kg
WATER
209
APPROXIMATE VALUES OF THERMAL CONDUCTIVITY ()
MATERIAL
SI
calcm
seccm2Co
ENGLISH
BTUin
hrft2 Fo
INSULATING
MATERIALS
ENGLISH
BTUin
hrft2 Fo
Air
1.96 x 10-8
5.7 x 10-4
Hair Felt
0.26
Water
1.14 x 10-4
0.33
Rock Wool
0.26
Corkboard
1.03 x 10-4
0.30
Glass Wool
0.29
Celotex
1.17 x 10-4
0.34
White Pine
0.78
White Pine
2.69 x 10-4
0.78
Oak
1.02
Marble
3.79 x 10-4
1.10
Cinder Block
2 to 3
Concrete
2.07 x 10-3
6.0
Building Block
3 to 6
Glass
1.99 x 10-3
5.8
Glass
5 to 6
Steel
0.12
350
Concrete
6 to 9
Aluminum
0.48
1400
Granite
13 to 18
Copper
0.93
2700
THERMAL CONDUCTIVITY OF METALS AT 0 OC
METAL
Thermal Conductivity
calcm
hrcm2Co
Btuin
hrft2 Fo
Aluminum
2031
1632
Copper
3450
Gold
2736
METAL
Thermal Conductivity
calcm
hrcm2Co
Btuin
hrft2 Fo
Iron
718
576
2784
Silver
3683
2964
2208
Tungsten
1566
1260
210
R-VALUES FOR TYPICAL INSULATION THICKNESSES
Batts or Blankets
Loose Fill (Poured-in)
R-Value
Glass Fiber
Rock Wool
Glass Fiber
Rock Wool
Cellulosic
Fiber
R-11
3½” - 4"
3"
5“
4"
3"
R-19
6" - 6 ½”
5¼”
8" - 9"
6" - 7"
5"
R-22
6 ½”
6"
10"
7" - 8"
6"
R-30
9 ½”-10 ½” *
9" *
13" - 14"
10" - 11"
8"
17" - 18"
13" - 14"
10" - 11"
R-38
12" - 13" *
10 ½” *
* Two batts or blankets required
211
RESISTOR COLOR CODES

COLOR
BLACK


FIRST
SIGNIFICA
NT FIGURE
SECOND
SIGNIFICA
NT FIGURE
MULTIPLIER
0
0
1

TOLERANCE
GOLD = ± 5%
SILVER = ± 10%
NONE = ± 20%
BROWN
1
1
101
RED
2
2
102
ORANGE
3
3
103
YELLOW
4
4
104
GREEN
5
5
105
BLUE
6
6
106
VIOLET
7
7
107
GRAY
8
8
108
WHITE
9
9
109
RESISTIVITY OF COMMON CONDUCTORS
MATERIAL
RESISTIVITY
cm
MATERIAL
RESISTIVITY
cm
Glass
1020
Lead
Silicon *
106
Aluminum
2.824
Germanium *
106
Gold
2.44
Carbon
105
Copper
1.724
Nichrome Wire
112
Silver
1.59
22
*The resistivity of semiconductors - like silicon and germanium - is very sensitive to
temperature.
212
APPENDIX D: SAMPLE UNITS FOR SOME COMMON VARIABLES
This certainly doesn’t cover all the possible ways each of these variables can be described, but
we’ve tried to include the units you will most commonly see.
ACCELERATION (a):
m N

,
s 2 kg
FORCE (F): newtons  N 
ft
lb

2
slug
s
kg  m
slug  ft
, pounds  lb 
2
s
s2
kg  m 2
slug  ft 2
TORQUE (): N  m 
, ft  lb 
, in  lb
s2
s2
N
J
lb
lb
PRESSURE (P): pascal  Pa  2  3 , psi  2 , psf  2
m
m
in
ft
kg  m
slug  ft 2
WORK (W) = ENERGY (E): joules  J  N  m 
,
, ft  lb 
s2
s2
calories =cal, British Thermal Units = BTU
2
CURRENT (I): amperes  amps  A 
FREQUENCY (f):
C
s
cycles
 hertz  Hz
s
m 3 cm 3  ft 3 gal
,
, ,
,
VOLUME FLOW RATE (QV):
s
s hr hr min
THERMAL RATE (RT):
cal BTU
,
s
hr
DRAG RESISTANCE (RD):
N
lb
,
m / s ft / s
FLUID RESISTANCE (RF):
Pa
N / m 2 Pa psi psf psi

,
,
,
m 3 / s m 3 / s  / hr gpm gpm ft 3 / hr
ELECTRICAL RESISTANCE (R, or RE):
THERMAL RESISTANCE (RT):
ohms   
C
F
,
cal / sec BTU / hr
213
V Js

A C2
ANGULAR VELOCITY ():
radians rad rev
rev

,
 rpm ,
s
s min
s
CAPACITANCE (C): farad  F 
INDUCTANCE (L):
J
C2
C2  s 2


J
V2
kg  m 2
henries H 
LINEAR MOMENTUM (p):
J
J  s 2 V  s kg  m 2



A
A2
C2
C2
kg  m
 N  s,
s
slug  ft
 lb  s
s
kg  m 2
ANGULAR MOMENTUM (L):
 N  m  s  J  s,
s
slug  ft 2
 ft  lb  s
s
POWER (P):
J N  m kg  m 2
watts  W  

,
s
s
s3
cal
BTU
horsepower  hp,
,
s
hr
IRRADIANCE ():
W mW hp
,
,
m 2 cm 2 ft 2
214
ft  lb slug  ft 2

,
s
s3
APPENDIX E: GLOSSARY OF TERMS
Acceleration (a) is the rate at which velocity changes with time. It’s typically measured in units
of m/s2 (or sometimes N/kg, which is the same thing) or ft/s2. The gravitational
acceleration (g) at sea level is 9.8 m/s2 or 32 ft/s2.
Angular acceleration () is the rate at which an object changes its angular velocity ().
Typical units are rad/sec2 or rev/sec2.
Angular Impulse is the product of torque and the time over which it is applied. It has units of
kg·m2/s = N·m·s.
Angular momentum (L) describes the momentum of a rotating object. It is the product of the
moment of intertia and the angular velocity and has units of kg·m2/s = N·m·s.
Angular velocity ()is the rate at which an object changes its angular position. Typical units
are rad/sec, rev/sec and rev/min (rpm).
British Thermal Unit (BTU) is an English unit of thermal energy. More specifically, it is the
amount of thermal energy required to raise the temperature of one pound of liquid water
(at one atmosphere pressure) by 1 Fo.
Buoyant Force (FB) is the upward force on an object due to the weight of the fluid it displaces.
calorie (cal) is a metric unit of thermal energy. More specifically, one calorie is the amount of
thermal energy required to raise the temperature of one gram of liquid water (at one
atmosphere pressure) by 1 Co.
Capacitance (C) describes a capacitor’s ability to hold charge for a given voltage difference. It
is in units of charge per potential difference: 1 farad = 1 F = 1coulomb/volt = 1 C/V = 1
C2/J = 1 C2·s2/kg·m2.
Charge (q) is that mysterious carrier of the electromagnetic force. The units of charge are
coulombs (1 coulomb = 1 C).
Coefficient of friction is the ratio of the frictional force to the normal force. It has no units. The
frictional force for the static coefficient of friction (s) is the maximum force that can be
applied (horizontal to the surface) without having the object slide. The frictional force
for the kinetic coefficient of friction (k) is the surface frictional force resisting an object
in motion. Both s and k are dimensionless.
Conductivity () is a measure of a substance’s ability to conduct charge or thermal energy.
Conductors are materials that readily transmit charge or thermal energy.
Cooling Rate (R) is the rate at which a temperature change occurs. Typical units are C°/min,
C°/sec, F°/min, F°/sec, etc.
Current (I) in an electrical system is the rate at which charge moves. 1 ampere = 1 amp = 1 A =
1 coulomb/second = 1 C/s.
Dynamic pressure (1/2v2) is the pressure due to the motion of a fluid.
Energy (E) is a property of the universe. Doing work means using energy. Units are typically
joules (1 J = 1 N·m = 1 kg·m2/s2) or foot-pounds (1 ft·lb = 1 slug·ft2/s2).
Frequency (f) is the rate at which a periodic (regularly repeating) event occurs. Its units are
generally cycles per second, also known as hertz (Hz). It is the inverse of period.
Force (F) is the product of mass (m) and acceleration (a), and so are its units: 1 newton = 1 N =
1 kg·m/s2, and 1 pound = 1 lb = 1 slug·ft/s2. If you want to change the motion of an
object, you have to apply a force (F) on it.
Gravitational acceleration (g) is the acceleration due to the presence of a gravitational field. At
sea level, g = 9.8 m/s2 or 32 ft/s2.
Heat (H) is more properly described as thermal energy. It comes in units of calories and British
thermal units (Btu’s).
215
Heat Flow Rate (QH) or heat transfer rate is the rate at which thermal energy is transferred.
Typical units are cal/sec or Btu/sec or even watts (1 watt = 1 W = 1 J/s = 1 kg·m2/s3).
Impulse is the change in momentum (p) of an object that is produced by an applied force
over a given time interval (p=F·t).
Inductance (L) is a measure of an inductor’s ability to store energy in its magnetic field, as a
function of the current flowing through it. It has units of henries (1 henry = 1 H = 1
V·s/A = 1 J/A2 = 1 kg·m2/C2).
Insulators are materials that inhibit the flow of charge or thermal energy.
Kinetic Energy (Ek) is energy of motion and has the units of energy (work).
Mass (m) is a measure of the quantity of matter in an object. It’s measured in units of kilograms
(kg) or slugs.
Mass density (), or just density, is mass per unit volume. Typical units are kg/m3 and g/cm3.
Moment of Inertia (I) describes an object’s resistance to rotational motion based on its mass,
the distribution of its mass from the axis of rotation and the axis of rotation. Its units are
kgm2 or slugft2.
Momentum (p) is the product of mass (m) and velocity (v). Its units are kg·m/s = N·s or
slug·ft/s = lb·s.
Period (T) is the time for one complete cycle of periodic (regularly repeating) motion. Its units
are those of time (or time per cycle). It is the inverse of frequency.
Potential Energy (Ep) is stored energy and has the units of energy (work).
Power (P) is the rate at which energy is used, or the rate at which work is done. Power is
measured in watts (1 watt = 1 W = 1 J/s = 1 kg·m2/s3), in foot-pounds per second (ft·lb/s),
in horsepower (hp), in calories per second (cal/s), or in Btu/s.
Pressure (P) is force per unit area (F/A). Units are pascals (1 Pa = 1 N/m2), lb/ft2, lb/in2, in.
Hg, in. H2O, mm Hg., mm H2O, bars and millibars, to name a few.
Resistance (R) in electrical system is the resistance to current flow in a circuit. The units are
ohms: 1 ohm = 1  = 1 joule-second/(coulomb squared) = 1 J·s/C2.
Resistivity () is a measure of a substance’s ability to restrict charge or thermal energy. Its units
are typically ·cm.
Reynolds Number (Re) is used to determine the level of turbulence in a fluid system. It is
dimensionless.
Scalar is a quantity with magnitude but no direction, like “24 oC”, or “32 years old”.
Specific Heat (c) is a measure of the amount of energy required to change the temperature of a
given amount of a certain substance. It is given in units of cal/g·C° or Btu/lb·F°.
Temperature (T) is a statistical measure of a substance’s kinetic energy.
Thermal Resistance (RT) is a substance’s (or an object’s) resistance to heat conduction.
Typical units are F°/(Btu/hr) and C°/(cal/hr).
Torque () is the result of a force (F) applied perpendicular to a moment arm (). Typical units
are N·m, ft·lb and in·lb.
Vector is a quantity with magnitude and direction like “4 newtons at 60o” or “6 miles east”.
Velocity (v) is the change in distance over time. Typical units are m/s, ft/s, km/hr and mph.
Voltage (V) or potential difference is the energy per charge in an electrical system. 1 volt = 1
V = 1 joule/coulomb = 1 J/C.
Volume Flow Rate (QV) is the rate at which volume flows. It typically has units of ft3/sec,
ft3/min, gal/min (=gpm), gal/hr, liters/min, m3/min, m3/sec, etc.
Weight (w) is the force due to an object¹s mass and the local gravitational acceleration (g).
Units are the units of force.
Weight density (w) is weight per unit volume. Typical units are lb/ft3 and lb/gal.
216
Work (W) is the product of an applied force and the distance over which it is applied. Units are
those of energy.
Appendix F: The Greek Alphabet
Capital Low-case Greek Name English
Alpha
a
Beta
b
Gamma
g
Delta
d
Epsilon
e
Zeta
z
Eta
h
Theta
th
Iota
i
Kappa
k
Lambda
l
Mu
m
Nu
n
Xi
x
Omicron
o
Pi
p
Rho
r
Sigma
s
Tau
t
Upsilon
u
Phi
ph
Chi
ch
Psi
ps
Omega
o
217