Operator Generic Fundamentals
Physics
© Copyright 2014
Operator Generic Fundamentals
2
Physics Introduction
• Physics is a science based upon exact measurement of physical
quantities that are dependent upon three fundamental dimensions
– Mass
– Length
– Time
• These units must be understood in order to lay the foundation for the
concepts and principles presented in classical and atomic physics,
this information complies with ACAD 10-001
© Copyright 2014
Intro
Operator Generic Fundamentals
3
Terminal Learning Objectives
At the completion of this training session, the trainee will demonstrate
mastery of this topic by passing a written exam with a grade of ≥ 80
percent on the following Terminal Learning Objectives (TLOs):
1. Convert between units of measure associated with the English and
System Internationale (SI) measuring systems.
2. Describe vector quantities, and how they are represented.
3. Solve resultant vector problems.
4. Describe the measurement of force and its relationship to free body
diagrams.
5. Apply Newton's laws of motion to a body at rest.
6. Calculate the change in velocity when two objects collide, with
respect to the conservation of momentum.
7. Describe energy, work, and power in mechanical systems.
© Copyright 2014
Intro
Operator Generic Fundamentals
4
Enabling Learning Objectives for TLO 1
1. Describe the three fundamental dimensions: length, mass, and
time.
2. List standard units of the fundamental dimensions for each of the
following systems:
• International System of Units (SI)
• English System
3. Differentiate between fundamental and derived measurements.
4. Convert between English and SI units of mass and length.
5. Convert time measurements between the following:
• Years, weeks, days, hours, minutes, seconds
© Copyright 2014
TLO 1
Operator Generic Fundamentals
5
Three Fundamental Dimensions
ELO 1.1 – Describe the three fundamental dimensions: length, mass, and
time.
•
•
Mass: amount of material present in an object
–
Description of "how much" material makes up an object
–
Not the same thing as weight, even though units are similar
Weight (derived unit): measurement that describes the force of
gravity on "mass" of an object
© Copyright 2014
ELO 1.1
Operator Generic Fundamentals
6
Three Fundamental Dimensions
•
Length: distance between two points
–
Needed to locate position of a point in space ⇒ describe size of a
physical object or system
–
When measuring a length of pipe, ends of pipe are two points ⇒
distance between two points is length
• Time: duration between two instants
–
Typically measured in: seconds, minutes, or hours
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
7
Three Fundamental Dimensions
Knowledge Check
Which of the following is NOT a fundamental dimension?
A. Weight
B. Length
C. Mass
D. Time
Correct answer is A.
© Copyright 2014
ELO 1.1
Operator Generic Fundamentals
8
Units of Measure
ELO 1.2 – List standard units of the fundamental dimensions for each of
the following systems: International System of Units (SI) and English
System
• Units define magnitude of a measurement
• For example: length
– Could be centimeter or meter, each of which describes a different
magnitude of length
– Could also be inches, miles, fathoms, or other units
• Units must be specified when same physical quantity may be
measured using a variety of different units
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
9
Unit Systems
• Two unit systems in use
– English units
– International System of Units (SI)
English System
• Used in U.S.
• Various units for the fundamental dimensions or measurements
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
10
Unit Systems
English Units of Measure
Length
Mass
Time
Inch
Ounce
Second*
Foot*
Pound*
Minute
Yard
Ton
Hour
Mile
Day
Month
Year
* denotes standard unit of measure
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
11
Unit Systems
System Internationale (SI)
• SI system is made up of two related systems
– Meter-kilogram-second (MKS) system
– Centimeter-gram-second (CGS) system
• Simpler to use than English system because they use a decimalbased system in which prefixes are used to denote powers of ten
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
12
SI versus English System
• SI based on powers of ten:
– One kilometer = 1,000 meters
– One centimeter = one one-hundredth of a meter
• English system has odd units of conversion:
– Mile = 5,280 feet
– Inch = one twelfth of a foot
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
13
MKS Units of Measure
MKS system used primarily for calculations in Physics
MKS Units of Measure
Length
Mass
Time
Millimeter
Milligram
Second*
Meter*
Gram
Minute
Kilometer
Kilogram*
Hour
Day
Month
Year
* denotes standard unit of measure
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
14
CGS Units of Measure
Both MKS and CGS systems are used in Chemistry
CGS Units of Measure
Length
Mass
Time
Millimeter
Milligram
Second*
Meter*
Gram
Minute
Kilometer
Kilogram*
Hour
Day
Month
Year
* denotes standard unit of measure
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
15
Dimensions Of Familiar Objects and
Events
Approximate Lengths of Familiar Objects
Object
Length (meters)
Diameter of earth's orbit around the sun
2 x 1011
Football field
0.91 x 102 (100 yards)
Diameter of a dime
2 x 10-2
Thickness of a window pane
1 x 10-3
Thickness of paper
1 x 10-4
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
16
Dimensions Of Familiar Objects and
Events
Approximate Masses of Familiar Objects
Object
Mass (kilograms)
Earth
6 x 1024
House
2 x 105
Car
2 x 103
Dime
3 x 10-3
Postage stamp
5 x 10-8
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
17
Dimensions Of Familiar Objects and
Events
Approximate Time Durations of Familiar Events
Event
Time (seconds)
Age of earth
2 x 1017
Human life span
2 x 109
Earth's rotation around the sun
3 x 107
Earth's rotation on its axis
8.64 x 104
Time between heart beats
1
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
18
Units of Measure
Knowledge Check
The measurement system most used in the United States when dealing
with physics is:
A. foot – pound – second
B. mks
C. cgs
D. foot – pound – minute
Correct answer is A.
© Copyright 2014
ELO 1.2
Operator Generic Fundamentals
19
Derived Measurements
ELO 1.3 – Differentiate between fundamental and derived
measurements.
• Most physical quantities have units that are combinations of the three
fundamental dimensions (length, mass, time)
• When these dimensions or measurements are combined, they
produce derived units
– Derived from one or more fundamental measurements
– Can be combination of same or different units
© Copyright 2014
ELO 1.3
Operator Generic Fundamentals
20
Area and Volume
• Area
– Product of two lengths (e.g., width x length for a rectangle)
– Units of length squared ⇒ in.2 or m2
1 m x 1 m = 1 m2
4 in. x 2 in. = 8 in.2
• Volume
– Product of three lengths (e.g., length x width x depth for a
rectangular solid)
– Units of length cubed, ⇒ in.3 or m3
– MKS and CGS unit systems have a specific unit for volume ⇒ liter
1 liter = 1,000 cm3
2 in. X 3 in. X 5 in. = 30 in.3
© Copyright 2014
ELO 1.3
Operator Generic Fundamentals
21
Density, Velocity and Acceleration
• Density
– Measure of the mass of an object per unit volume
– Units of mass divided by length cubed ⇒ kg/m3 or lbs/ft3
• Velocity
– Change in length per unit time
– Units ⇒ km/h or ft/s
• Acceleration
– Measure of change in velocity or velocity per unit time
– Units ⇒cm/s2 or ft/s2
© Copyright 2014
ELO 1.3
Operator Generic Fundamentals
22
Fundamental and Derived Measurements
Knowledge Check
Match the following terms with the correct dimensions of measurement.
1. A measure of the change in velocity per unit time is
centimeters per second per second (cm/s2) or feet per
second per second (ft/s2)
A. Acceleration
2. The product of three lengths (e.g., length x width x depth B. Volume
for a rectangular solid) is (in3) or cubic meters (m3)
3. Mass of an object per unit volume is kilograms per cubic
meter (kg/m3) or pounds per cubic foot (lbs/ft3)
C. Density
4. Length per unit time is kilometers per hour (km/hr) or
feet per second (ft/s)
D. Velocity
Correct answers are: 1-A, 2-B, 3-C, 4-D
© Copyright 2014
ELO 1.3
Operator Generic Fundamentals
23
Need for Unit Conversions
ELO 1.4 – Convert between English and SI Systems of mass and length.
• Personnel may encounter both English and SI systems of units in
their work
– Measurements taken or read from an instrument may be different
from those required by procedure, requiring conversion of
measurements
– Documentation provided by industrial equipment vendors may
use English system of measurement and may need to be
converted to SI measurements
• In order to apply measurements from the SI system to the English
system and vice versa, it is necessary to develop relationships of
known equivalents (Conversion Factors)
• Equivalents can then be used to convert from given units to desired
units
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
24
Conversion Factors
• Conversion Factors
– Based on relationships of equivalents from different measurement
systems
– Applied to given measurement to convert it to required units
– Equivalent relationships between different units of measurement
are defined in Conversion Tables
• Example Conversion Factors:
1 yard = 0.9144 meters
1 kilogram = 2.205 pounds mass (lbm)
1 hour = 3,600 seconds
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
25
Typical Conversion Factors
Converting Table
© Copyright 2014
Measurement
Unit of Measurement Conversion Measurement
Time
60 seconds
60 minutes
= 1 minute
= 1 hour
Length
1 yard
12 inches
5,280 feet
1 meter
1 inch
= 0.9144 meter
= 1 foot
= 1 mile
= 3.281 feet
= 0.0254 meter
Mass
1 lbm
2.205 lbm
1 kg
= 0.4535 kg
= 1 kg
= 1,000 g
Area
1 ft2
10.764 ft2
1 yd2
1 mile2
= 144 in2
= 1 m2
= 9 ft2
= 3.098 x 106 yd2
Volume
7.48 gal
1 gal
1l
= 1 ft3
= 3.785 l
= 1,000 cm3
ELO 1.4
Operator Generic Fundamentals
26
Performing Unit Conversions
Convert 5 feet to inches
• First select appropriate equivalent relationship from the conversion
table on slide 28:
1 𝑓𝑜𝑜𝑡 = 12 𝑖𝑛𝑐ℎ𝑒𝑠
• Conversion is basically a multiplication by 1
• Divide both sides of equation 1 ft = 12 inches by 1 foot to obtain the
following:
1 𝑓𝑡
12 𝑖𝑛𝑐ℎ𝑒𝑠
=
1 𝑓𝑡
1 𝑓𝑜𝑜𝑡
5 𝑓𝑒𝑒𝑡 = 5 𝑓𝑒𝑒𝑡 ×
© Copyright 2014
or
12 𝑖𝑛𝑐ℎ𝑒𝑠
1=
1 𝑓𝑜𝑜𝑡
12 𝑖𝑛𝑐ℎ𝑒𝑠
= 5 × 12 𝑖𝑛𝑐ℎ𝑒𝑠 = 60 𝑖𝑛𝑐ℎ𝑒𝑠
1 𝑓𝑜𝑜𝑡
ELO 1.4
Operator Generic Fundamentals
27
Unit Conversion - Examples
• Example 1: Convert 795 m to ft.
• Solution 1:
– Step 1: Select the equivalent relationship from the conversion
table (See slide 28)
1 𝑚𝑒𝑡𝑒𝑟 (𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑢𝑛𝑖𝑡𝑠) = 3.281 𝑓𝑡 (𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑢𝑛𝑖𝑡𝑠)
– Step 2: Divide to obtain the factor 1 as a ratio
𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑢𝑛𝑖𝑡𝑠
𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑢𝑛𝑖𝑡𝑠
3.281 𝑓𝑡
1=
1𝑚
– Step 3: Multiply the quantity by the ratio:
3.281 𝑓𝑡
795 𝑚
3.281 𝑓𝑡
795 𝑚 ×
=
×
= 795 × 3.281 𝑓𝑡
1𝑚
1
1𝑚
= 2,608.395 𝑓𝑡
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
28
Unit Conversion - Multiple Steps
• If an equivalent relationship between given units and desired units
cannot be found in conversion tables, multiple conversion factors
must be used
• Conversion is performed in several steps until measurement is in
desired units
• Given measurement must be multiplied by each conversion factor
(ratio)
• After common units have been canceled out, answer will be in
desired units
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
29
Unit Conversion - Examples
• Example 2: Convert 2.91 square miles to square meters
• Solution 2:
– Step 1: Select equivalent relationship from conversion table on
slide 28.
o There is no direct conversion shown for square miles to square
meters ⇒ multiple conversions will be necessary
o Following conversions will be used:
1 𝑠𝑞 𝑚𝑖𝑙𝑒 = 3.098 × 106 𝑠𝑞 𝑦𝑑
1 𝑠𝑞 𝑦𝑑 = 9 𝑠𝑞 𝑓𝑡
10.764 𝑠𝑞 𝑓𝑡 = 1 𝑠𝑞 𝑚
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
30
Unit Conversion - Examples
• Solution 2 (continued):
– Step 2: Express relationship as a ratio (desired unit/present unit):
3.098 × 106 𝑠𝑞 𝑦𝑑
1=
1 𝑠𝑞 𝑚𝑖𝑙𝑒
– Step 3: Multiply quantity by ratio:
3.098 × 106 𝑠𝑞 𝑦𝑑
2.91 𝑠𝑞 𝑚𝑖𝑙𝑒𝑠 ×
= 9.015 × 106 𝑠𝑞 𝑦𝑑
1 𝑠𝑞 𝑚𝑖𝑙𝑒
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
31
Unit Conversion - Examples
• Solution 2 (continued):
– Step 4: Repeat steps until value is in desired units
9 𝑓𝑡 2
1=
1 𝑦𝑑 2
9.015 × 106 𝑦𝑑 2
9 𝑓𝑡 2
×
= 8.114 × 107 𝑓𝑡 2
2
1 𝑦𝑑
1 𝑚2
1=
10.764 𝑓𝑡 2
8.114 ×
107
𝑓𝑡 2
1 𝑚2
(8.114 × 107 𝑚2 )(1 𝑚2 ) 8.114 × 107 𝑚2
×
=
=
10.764 𝑓𝑡 2
10.764
10.764
= 7.538 × 106 𝑚2
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
32
Unit Conversion - Examples
It is possible to perform all conversions in a single equation as long as
all appropriate conversion factors are included:
2.91
𝑚𝑖 2
3.098 × 106 𝑦𝑑 2 9 𝑓𝑡 2
1 𝑚2
×
×
×
2
2
1 𝑚𝑖
1 𝑦𝑑
10.764 𝑓𝑡 2
2.91 × (3.098 × 106 )(9)(1 𝑚2 )
=
10.764
8.114 × 107 𝑚2
=
10.764
= 7.538 × 106 𝑚2
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
33
Unit Conversion - Examples
• Example 3: A Swedish firm is producing a valve that is to be used by
an American supplier. The Swedish firm uses the MKS system for all
machining. To conform to the MKS system, how will the following
measurements be listed?
Valve stem = 57.20 in.
Valve inlet and outlet
I.D. = 22.00 in.
O.D. = 27.50 in.
• Solution 3:
– Valve stem:
𝑚
57.20 𝑖𝑛.× 0.0254
= 1.453 𝑚
𝑖𝑛.
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
34
Unit Conversion - Examples
• Solution 3 (continued):
– Valve inlet and outlet:
𝑚
𝐼. 𝐷. 22.00 𝑖𝑛. × 0.0254
= 0.559 𝑚
𝑖𝑛.
𝑂. 𝐷. 27.50 𝑖𝑛. × 0.0254
© Copyright 2014
𝑚
= 0.699 𝑚
𝑖𝑛.
ELO 1.4
Operator Generic Fundamentals
35
Common Conversion Factors
Common Conversion Factors for Units of Mass
g
kg
t
lbm
1 gram
1
0.001
10-5
2.2046 x 10-3
1 kg
1,000
1
0.001
2.2046
1 metric ton (t)
106
1,000
1
2,204.6
1 pound-mass
(lbm)
453.59
0.45359
4.5359 x 10-4
1
1 slug
14,594
14.594
0.014594
32.174
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
36
Common Conversion Factors
Conversion Factors for Common Units of Length
cm
m
km
in.
ft
mi
1
0.01
10-5
0.3937
0.032808
6.2137 x 10-6
1 meter
100
1
0.001
39.37
3.2808
6.2137 x 10-4
1 kilometer
105
1,000
1
39,370
3,208.8
0.62137
1 inch
2.5400
0.025400
2.54 x 10-5
1
0.083333
1.5783 x 10-5
1 foot
30.480
0.30480
3.0480 x 10-4
12.000
1
1.8939 x 10-4
1 mile
1.6093 x 105
1,609.3
1.6093
63,360
5,280
1
1 centimeter
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
37
Conversion between English and SI
Systems
Knowledge Check
Match the following dimensions to their equivalents.
1. 3,280.8 feet
A. 1 mile per hour
2. 453,590 grams
B. 1 lbm
3. 0.017 hours
C. 1 minute
4. 1.47 feet per
second
D. 1 kilometer
Correct answers are: 1-D, 2-B, 3-C, 4-A
© Copyright 2014
ELO 1.4
Operator Generic Fundamentals
38
Common Conversion Factors
ELO 1.5 – Convert time measurements between the following: years,
weeks, days, hours, minutes, and seconds.
Conversion Factors for Common Units of Time
sec
min
1 second
1
0.017
1 minute
60
1
0.017
1 hour
3,600
60
1
1 day
86,400
1,440
24
1
2.74 x 10-3
8,760
365
1
1 year
© Copyright 2014
hr
day
Year
2.7 x 10-4 1.19 x 10-5
3.1 x 10-8
6.9 x 10-4
1.9 x 10-6
3.15 x 107 5.26 x 105
ELO 1.5
4.16 x 10-2 1.14 x 10-4
Operator Generic Fundamentals
39
Converting Time Units
Converting Time Units Demonstration
Step
Convert three years into
seconds.
Result
The conversion factor goes directly
from years to seconds.
7
1 𝑦𝑒𝑎𝑟 = 3.15 × 10 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
1.
Select the appropriate
relationship from the
conversion table.
2.
Express the relationship as a
ratio (desired units/present
units).
7
3.15 × 10 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
1 𝑦𝑒𝑎𝑟
7
3.
Multiply the quantity by the
ratio.
Repeat the steps until the
Correctvalue
answer
is inis:
theDdesired units.
4.
© Copyright 2014
3.15 × 10 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
3 𝑦𝑒𝑎𝑟𝑠
1 𝑦𝑒𝑎𝑟
7
= 9.45 × 10 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
Multiple steps are not necessary in
this case.
ELO 1.5
Operator Generic Fundamentals
40
Converting Time Units
Knowledge Check
Calculate the number of minutes in 3.2 years.
A. 164,000
B. 168,000
C. 1,640,000
D. 1,680,000
Correct answer is D.
© Copyright 2014
ELO 1.5
Operator Generic Fundamentals
41
TLO 1 Summary
• The three fundamental measurements are:
– Length: distance between two points
– Mass: amount of material in an object
– Time: duration between two instants
• The English system of units consists of the following standard units:
– Foot, Pound, Second
• The SI system of measurement consists of the following standard
units:
– Meter, Kilogram, Second, Centimeter, Gram, and Second
© Copyright 2014
TLO 1
Operator Generic Fundamentals
42
TLO 1 Summary
• Combinations of units comprise derived units to describe various
physical quantities. For example:
– Mass - pounds-mass (lbm) or kilograms (kg)
– Volume - cubic inches (in3) or liters (l)
– Density - mass per unit volume (lbm/in3 or kg/l)
• Unit Conversion: conversion tables list equivalent relationships.
• Unit Conversion Steps:
– Step 1: Select the equivalent relationship from the conversion
table.
– Step 2: Express the relationship as a conversion factor.
– Step 3: Multiply the given quantity by the conversion factor.
© Copyright 2014
TLO 1
Operator Generic Fundamentals
43
TLO 1 Summary
Now that you have completed this lesson, you should be able to:
1. Describe the three fundamental dimensions: length, mass, and
time.
2. List standard units of the fundamental dimensions for each of the
following systems: a. International System of Units (SI) b. English
System.
3. Differentiate between fundamental and derived measurements.
4. Convert between English and SI units of mass and length.
5. Convert time measurements between the following: a. Years,
b. Weeks, c. Days, d. Hours, e. Minutes, f. Seconds
© Copyright 2014
TLO 1
Operator Generic Fundamentals
44
Vector Quantities
TLO 2 – Describe vector quantities and how they are represented.
• Vectors are quantities with
– Magnitude
– Direction
• They are useful in
– Understanding physical concepts
– Solving various problems involving motion
– Understanding the interrelationship of force and motion
© Copyright 2014
TLO 2
Operator Generic Fundamentals
45
Enabling Learning Objectives for TLO 2
1. Define the following as they relate to vectors:
a. Scalar quantity
b. Vector quantity
c. Vector component
d. Resultant
2. Describe methods for identifying vectors in written material.
© Copyright 2014
TLO 2
Operator Generic Fundamentals
46
Vector Terminology
ELO 2.1 – Define the following as they relate to vectors: scalar quantity,
vector quantity, vector component, and resultant.
• Scalars - quantities that have magnitude only ⇒ independent of
direction
• Vectors - have both magnitude and direction
– Indicated graphically using an arrow
o Length of a vector (arrow) represents magnitude
o Arrow shows direction of vector
• Resultant - single vector which represents combined effect of two or
more other vectors (called components)
• Component – two or more vectors that contribute to a net result, or
“resultant” vector. Can be determined either graphically or by using
trigonometry
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
47
Scalar Quantities
• Scalar Quantity
– Quantity that has magnitude only
– No directional component
• Include
– Time
– Speed
– Temperature
– Volume
– Density
– Mass
– Energy
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
48
Vector Quantities
• Vector quantity has both
– Magnitude
– Direction
• Magnitude
– Size of a vector
– Referred to as a vector's displacement
– Can be thought of as the scalar portion of vector
– Represented by length of vector
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
49
Vector Quantities
• Direction of vector indicates
how vector is oriented relative
to some reference axis
• Giving direction to scalar A
makes it a vector
– Vector A is oriented in the
NE quadrant with a direction
of 45° north of the E-W axis
• Length of A is representative of
its magnitude or displacement
Figure: Vector Reference Axis
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
50
Scalar vs. Vector Example
• Car moving at 80 kilometers per hour
– 80 km/hour only refers to car's speed; no indication of direction
car is moving
– Scalar quantity
• Car traveling at 80 km/hour due east
– Indicates velocity of car, magnitude (80 km/hour) and direction
(due east)
– Vector quantity
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
51
Diagramming a Vector Quantity
• Straight line drawn to show unit of length
– Length represents magnitude of vector
• Arrow drawn on one end of line
– Arrow represents direction of vector
• Description of Vector
– Simple straight lines used to illustrate direction and magnitude of
quantities
– Have a starting point at one end (tail) and an arrow at opposite
end (head)
Figure: Simple Vector
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
52
Vector Quantities
• All have magnitude and direction, includes:
– Displacement
– Velocity
– Acceleration
– Force
– Momentum
– Magnetic field strength
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
53
Vector Quantities
Knowledge Check
Select all that are true.
A. Vectors have direction and magnitude.
B. Scalars have direction and magnitude.
C. Temperature is a vector quantity.
D. Force is a vector quantity.
Correct answers are A and D.
© Copyright 2014
ELO 2.1
Operator Generic Fundamentals
54
Vector Identification
ELO 2.2 – Describe methods for identifying vectors in written material.
• Vector quantities are often represented by boldfaced letter
– For example: A, B, C, R
– Particular quantities may be predefined: F - force, V - velocity, and
A - acceleration
• Sometimes represented as: 𝑨 𝑩 𝑪 𝑹
• Written vector quantities must include specific magnitude and
direction
– 50 km/hour due north
– 50 lbf at 90°
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
55
Identifying Vectors
Knowledge Check
Which of the following is not a way to represent a vector?
A. As a quantity with both magnitude and direction (50 km/hr, due
north)
B. As a quantity (30 mi/hr)
C. As a bold, capital letter with an arrow over it
D. As a bold capital letter for predefined items, such as F for force
Correct answer is B.
© Copyright 2014
ELO 2.2
Operator Generic Fundamentals
56
TLO 2 Summary
• Scalar quantities:
– Magnitude only
– Independent of direction
– Examples of scalars include time, volume, and temperature.
• Vector quantities:
– Both magnitude and direction
– Length of arrow signifies magnitude
– Arrow shows direction
– Examples of vectors include force, velocity, and acceleration.
• Vector identification:
– Boldfaced capital letters (A, F, R) with arrows over them
– Graphically, or (x, y) coordinates
– Directional Coordinates, or Degrees
© Copyright 2014
TLO 2
Operator Generic Fundamentals
57
TLO 2 Summary
Important Facts
• A resultant is a single vector that can replace two or more vectors.
• You can obtain components for any two non-parallel directions if the
vectors are in the same plane.
• Restricting the treatment to perpendicular directions and twodimensional space, the components of a vector are the two vectors in
the x and y (or east-west and north-south) directions which produce
the same effect as the original vector (or add to produce the original
vector).
• Components are determined from data, graphically or analytically.
© Copyright 2014
TLO 2
Operator Generic Fundamentals
58
TLO 2 Summary
Now that you have completed this lesson, you should be able to:
1. Define the following as they relate to vectors: scalar quantity, vector
quantity, vector component, and resultant.
2. Describe methods for identifying vectors in written material.
© Copyright 2014
TLO 2
Operator Generic Fundamentals
59
Solving Vector Problems
TLO 3 – Solve resultant vector problems.
• Now that you know how to distinguish between a scalar and a vector,
you can solve vector problems
• These problems range from
– Graphing vectors
– Determining a vector’s components
– Adding vectors by various methods
• Vectors are a necessary tool for understanding the interrelationship of
force and motion
© Copyright 2014
TLO 2
Operator Generic Fundamentals
60
Enabling Learning Objectives for TLO 3
1. Given a magnitude and direction, graph a vector using the
rectangular coordinate system.
2. Determine components of a vector from a resultant vector.
3. Add vectors using the following methods:
a. Graphical
b. Component addition
c. Analytical
© Copyright 2014
TLO 3
Operator Generic Fundamentals
61
Graphing Vectors
ELO 3.1 – Given a magnitude and direction, graph a vector using the
rectangular coordinate system.
• Vector quantities graphically
represented using rectangular
coordinate system
• Two-dimensional system that
uses an x-axis and a y-axis
– x-axis is horizontal straight
line
– y-axis is a vertical straight
line, perpendicular to
x-axis
Figure: Rectangular Coordinate System
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
62
Rectangular Coordinate System
• Point of origin - intersection of
axes
• Each axis marked off in equal
divisions in all four directions
from point of origin
• On horizontal axis (x),
– Values to right of origin are
positive (+)
– Values to left of origin are
negative (-)
• Very important to use same
units (divisions) on both axes
© Copyright 2014
Figure: Rectangular Coordinate System
ELO 3.1
Operator Generic Fundamentals
63
Rectangular Coordinate System
• Rectangular coordinate system
creates four infinite quadrants
– Quadrant I - located above
and to right of origin
– Quadrant II - located above
and to left of origin
– Quadrant III - located to left
and below origin
– Quadrant IV - located below
and to right of origin
Figure: Rectangular Coordinate System
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
64
Displaying Vectors Graphically
• Using a ruler and graph paper, a rectangular coordinate system
should be laid out
• Label x- and y-axes
• Equal divisions should be marked off in all four directions
– Divisions to right and above point of origin labeled positive (+)
– Divisions to left and below point of origin labeled negative (-)
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
65
Graphing Vectors
• Beginning at point of origin a
line segment of proper length is
drawn along x-axis, in positive
direction
– Line segment represents
vector magnitude, or
displacement
• Arrow is placed at head of
vector to indicate direction
• Tail of vector located at point of
origin
© Copyright 2014
Figure: Displaying Vectors Graphically - Magnitude
ELO 3.1
Operator Generic Fundamentals
66
Displaying Vectors That Do Not Fall on X
or Y-axes
• Tail located at point of origin
• Head location:
– If coordinates (x, y) are
given, values can be plotted
to locate vector head
– If vector described in
degrees, line segment can
be rotated counterclockwise
from x-axis to proper
orientation
© Copyright 2014
ELO 3.1
Figure: Displaying Vectors Graphically - Direction
Operator Generic Fundamentals
67
Using Directional Coordinates
• Because x- and y-axes define direction, conventional directional
coordinates may be used to identify x- and y-axes
• Degrees may also be used to identify x- and y-axes
Figure: Degree Coordinates
Figure: Directional Coordinates
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
68
Vector Resultants and Components
• Resultant - single vector which represents combined effect of two or
more other vectors
– Result or sum of vector addition
• Components - can be determined either graphically or by using
trigonometry
• Vector addition different from addition of pure numbers unless
addition takes place long a straight line
– Reduces to number line of standards or scale addition
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
69
Graphing Vectors
Knowledge Check
The graph of a vector is:
A. a straight line with an arrow is drawn on one end. The length of
the line represents the magnitude of the vector, and the arrow
represents the direction of the vector.
B. intended to convey direction only. Notes are needed to convey
magnitude.
C. indicative of a quantity with magnitude but no direction.
D. a means of showing the direction of the force, but is not drawn
to scale.
Correct answer is A.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
70
Vector Components
ELO 3.2 – Determine components of a vector from a resultant vector.
• Components are vectors, when
added, yield the resultant vector
• Shown in previous example,
traveling 3 km north and then 4
km east yields a resultant
displacement of 5 km 37° north
of east
• Shows that component vectors
of any two non-parallel
directions can be obtained for
any resultant vector in the same
plane
© Copyright 2014
ELO 3.2
Figure: Vector Addition – Not in Straight Line
Operator Generic Fundamentals
71
Plotting Component Vectors
• Component vectors determined by plotting on a rectangular
coordinate system
• Example - Resultant vector of 5 units at 53° can be broken down into
its respective x and y magnitudes
– x value of 3 and y value of 4 can be determined using
trigonometry or graphically
– Magnitudes and position can be expressed by one of several
conventions including:
o (3,4)
o (x = 3, y = 4)
o (3 at 0° , 4 at 90°) and (5 at 53°)
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
72
Plotting Component Vectors
• If resultant vector given (instead
of component coordinates),
components can be determined
graphically
– Resultant vector plotted first
on rectangular coordinates
– Vector coordinates projected
to axis (dashed lines)
– Length along x-axis is Fx
– Length along y-axis is Fy
© Copyright 2014
ELO 3.2
Figure: Vector Components
Operator Generic Fundamentals
73
Component Vector Example
For resultant vector shown, determine component vectors given:
FR = 50 N at 53°
Figure: Component Vectors
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
74
Component Vector Example: Solution
• First, project a perpendicular
line from the head of FR to xaxis and a similar line to y-axis
• Where projected lines meet
axes determines magnitude of
component vectors:
30 N at 0° (Fx)
40 N at 90° (Fy)
Figure: Component Vectors
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
75
Determining Component Vectors Using
Trigonometry
• Trigonometry may also be used to determine vector components
• Relationship between an acute angle of a right triangle and its sides
is given by three ratios:
– Sine
– Cosine
– Tangent
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
76
Right Triangle Relationships
sin 𝜃 =
𝑎
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
=
𝑐 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝜃 =
𝑏
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
=
𝑐 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
tan 𝜃 = =
𝑐 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
Figure: Right Triangle
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
77
Calculating Component Vectors
Example 1
Determine component vectors, Fx
and Fy, for
FR = 50 N at 53°, using
trigonometric functions
Figure: Component Vectors
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
78
Calculating Component Vectors
Example 1
Fx is calculated as follows:
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
cos 𝜃 =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝐹𝑥
cos 𝜃 =
𝐹𝑅
or
𝐹𝑥 = 𝐹𝑅 cos 𝜃
𝐹𝑥 = (50 𝑁)(cos 53°)
𝐹𝑥 = (50 𝑁)(0.6018)
𝐹𝑥 = 30 𝑁 𝑜𝑛 𝑥– 𝑎𝑥𝑖𝑠
© Copyright 2014
Figure: Component Vectors
ELO 3.2
Operator Generic Fundamentals
79
Calculating Component Vectors
Example 1
Fy is calculated as follows:
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
sin 𝜃 =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝐹𝑦
sin 𝜃 =
𝐹𝑅
or
𝐹𝑦 = 𝐹𝑅 sin 𝜃
𝐹𝑦 = (𝐹𝑅 )(sin 𝜃)
𝐹𝑦 = (50 𝑁)(sin 53°)
𝐹𝑦 = (50 𝑁)(0.7986)
𝐹𝑦 = 40 𝑁 𝑜𝑛 𝑦– 𝑎𝑥𝑖𝑠
© Copyright 2014
Figure: Component Vectors
ELO 3.2
Operator Generic Fundamentals
80
Calculating Component Vectors
Example 1
• Components for FR are
– Fx = 30 N at 0°
– Fy = 40 N at 90°
• This result is identical to result
obtained using graphic method
Figure: Component Vectors
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
81
Calculating Component Vectors
Example 2
What are the component vectors,
given FR = 80 N at 220°?
Figure: FR = 80 N at 220°
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
82
Calculating Component Vectors
Example 2
Fx is calculated as follows:
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝑐𝑜𝑠 𝜃 =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝐹𝑥
cos 𝜃 =
𝐹𝑅
or
𝐹𝑥 = 𝐹𝑅 cos 𝜃
𝐹𝑥 = (80 𝑁)(cos 220° )
𝐹𝑥 = (80 𝑁)(−0.766)
𝐹𝑥 = −61 𝑁 𝑎𝑡 0° or 61 𝑁 𝑎𝑡 180°
© Copyright 2014
ELO 3.2
Figure: FR = 80 N at 220°
Operator Generic Fundamentals
83
Calculating Component Vectors
Example 2
Fy is calculated as follows:
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
sin 𝜃 =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
sin 𝜃 =
𝐹𝑦
𝐹𝑅
or
𝐹𝑦
𝐹𝑦
𝐹𝑦
𝐹𝑦
=
=
=
=
© Copyright 2014
𝐹𝑅 sin 𝜃
(80 𝑁)(sin 220°)
80 𝑁 −0.6428
−51 𝑁 𝑎𝑡 90° or 51 𝑁 𝑎𝑡 270°
ELO 3.2
Figure: FR = 80 N at 220°
Operator Generic Fundamentals
84
Calculating Component Vectors
Example 2
Components for FR:
𝐹𝑥 = 61 𝑁 𝑎𝑡 180°
𝐹𝑦 = 51 𝑁 𝑎𝑡 270°
Figure: FR = 80 N at 220°
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
85
Determining Components of a Vector
Knowledge Check
A vector begins at (2, 3) and ends at (-1, 7). Select all of the
statements about this vector that are true.
A. The vector has a magnitude of 5.
B. The vector has a direction of 120°.
C. The vector has a direction of 127°.
D. The vector has a horizontal component of magnitude 3, direction
180°, and a vertical component of magnitude 4, direction 90°.
Correct answers are A, C, and D.
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
86
Methods Used To Add Vectors
ELO 3.3 – Add vectors using the following methods: graphical,
component addition, and analytical.
• Several methods have been developed to add vectors
• In this presentation, the graphical, component addition and analytical
methods will be explained
– Either the graphical method or the component addition method
will provide a fairly accurate result
– If a higher degree of accuracy is required, an analytical method
using geometric and trigonometric functions is required
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
87
Graphic Method Of Vector Addition
• Equipment needed to use Graphic Method of Vector Addition:
– Standard linear (non-log) graph paper
– Ruler
– Protractor
– Writing instrument (pencil)
• Method utilizes a five-step process
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
88
Graphic Method Of Vector Addition
• Step 1 - Plot first vector (F1) on
rectangular (x-y) axes
– Ensure same scale used on
both axes
– Place tail (beginning) of first
vector at origin of axes
Figure: Plotting Vector on Rectangular Axes
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
89
Graphic Method Of Vector Addition
• Step 2 - Draw second vector
(F2) connected to end of first
vector
– Start tail of second vector at
head of first vector
– Ensure second vector drawn
to scale
– Ensure proper angular
orientation of second vector
with respect to axes of
graph
© Copyright 2014
ELO 3.3
Figure: Plotting Vector F2
Operator Generic Fundamentals
90
Graphic Method Of Vector Addition
• Step 3 - Add other vectors sequentially
– Add one vector at a time
– Always start tail of new vector at head of previous vector
– Draw all vectors to scale and with proper angular orientation
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
91
Graphic Method Of Vector Addition
• Step 4 - When all given vectors
have been drawn, draw and
label a resultant vector, FR ,
from point of origin of axes to
head of final vector
– Tail of resultant vector is tail
of first vector drawn
– Head of resultant vector is
at head of last vector drawn
Figure: Plotting Resultant Vector
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
92
Graphic Method Of Vector Addition
• Step 5 - Determine magnitude
and direction of resultant vector
– Measure displacement and
angle directly from graph
using ruler and protractor
– Determine components of
resultant by projection onto
x- and y-axes
Figure: Plotting Resultant Vector
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
93
Component Addition Method
• Component Addition Method - addition of vector coordinates on a
rectangular (x, y) coordinate system
• Vector components added along each axis to determine magnitude
and direction of resultant
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
94
Component Addition Method
• Coordinates locate specific
point in system
– Specific point is head of
vector
– Head can be located by
counting units along x-axis
and units along y-axis
– Example: Point has
coordinates (4,3)
• x component = +4
• y component = +3
Figure: Vector Component Addition Method
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
95
Component Addition Method
• Four-step method:
– Step 1 - Determine x- and y-axes components of all original
vectors
– Step 2 - Mathematically combine all x-axis components
(+𝑥 𝑎𝑡 180° = −𝑥 𝑎𝑡 0° )
– Step 3 - Mathematically combine all y-axis components
(+𝑦 𝑎𝑡 270° = −𝑦 𝑎𝑡 90°)
– Step 4 - Resulting (x, y) components are the (x, y) components of
the resulting vector
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
96
Component Addition Method
Example 1
• Find the resultant vector:
– 𝐹1 = (4,10)
– 𝐹2 = (−6,4)
– 𝐹3 = (2, −4)
– 𝐹4 = (10, −2)
• Step 1: Determine x- and y-axis components of all four original
vectors
𝑥– 𝑎𝑥𝑖𝑠 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 = 4, −6, 2, 10
𝑦– 𝑎𝑥𝑖𝑠 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 = 10, 4, −4, −2
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
97
Component Addition Method
Example 1 (continued)
• Step 2: Mathematically combine all x-axis components
𝐹𝑥 = 4 + (−6) + 2 + 10
𝐹𝑥 = 4 − 6 + 2 + 10
𝐹𝑥 = 10
• Step 3: Mathematically combine all y-axis components
𝐹𝑦 = 10 + 4 + (−4) + (−2)
𝐹𝑦 = 10 + 4 − 4 − 2
𝐹𝑦 = 8
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
98
Component Addition Method
Example 1 (continued)
• Step 4: Express resultant vector
– Resultant components from previous additions are coordinates of
resultant
𝐹𝑥 = 10, 𝐹𝑦 = 8
𝐹𝑅 = (10, 8)
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
99
Component Addition Method
Example 2
• Determine the resultant, FR, given:
– 𝐹1 = 30 𝑁 𝑎𝑡 0°, 10 𝑁 𝑎𝑡 90°
– 𝐹2 = 50 𝑁 𝑎𝑡 0°, 50 𝑁 𝑎𝑡 90°
– 𝐹3 = 45 𝑁 𝑎𝑡 180°, 30 𝑁 𝑎𝑡 90°
– 𝐹4 = 15 𝑁 𝑎𝑡 0°, 50 𝑁 𝑎𝑡 270°
• Follow the sequence used in first example, remembering that x at
180° is -x at 0°, and y at 270° is -y at 90°
𝐹𝑥 = 30 + 50 + (−45) + 15 = 50 𝑁
𝐹𝑦 = 10 + 50 + 30 + (−50) = 40 𝑁
𝐹𝑅 = 50 𝑁 𝑎𝑡 0°, 40 𝑁 𝑎𝑡 90°
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
100
Analytical Method Of Vector Addition
• When vector components are added to determine magnitude and
direction of resultant, calculations using trigonometric functions are
most accurate method for making this determination.
– Graphic and component addition methods of obtaining resultant
described previously can be hard to use and time consuming.
– Accuracy is a function of the scale used in making the diagram
and how carefully the vectors are drawn.
• Analytical method can be simpler and far more accurate.
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
101
Review of Mathematical Functions
• Pythagorean Theorem is used
to relate lengths sides of right
triangles
– In any right triangle, square
of length of hypotenuse
equals sum of squares of
lengths of other two sides
Figure: Right Triangle – Pythagorean Theorem
𝑐2 = 𝑎2 + 𝑏2 𝑜𝑟
𝑐 = 𝑎2 + 𝑏 2
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
102
Review of Mathematical Functions
𝑎
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
sin 𝜃 = =
𝑐 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑏
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
cos 𝜃 = =
𝑐 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
tan 𝜃 = =
𝑐 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
• Trigonometric functions:
– Cos used to solve for Fx
– Sin used to solve for Fy
– Tan normally used to solve
for θ
o Sin and cos may also be
used
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
103
Review of Mathematical Functions
• On a rectangular coordinate system, sine values of θ are positive (+)
in quadrants I and II and negative (-) in quadrants III and IV
• Cosine values of θ are positive (+) in quadrants I and IV and negative
(-) in quadrants II and III
• Tangent values are positive (+) in quadrants I and III and negative (-)
in quadrants II and IV
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
104
Review of Mathematical Functions
• When mathematically solving for tan θ, calculators will specify angles
in quadrants I and IV only
• Actual angles may be in quadrants II and III
• Each problem should be analyzed graphically in order to ensure a
realistic solution
• Quadrant II and III angles may be obtained by adding or subtracting
180° from value calculated
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
105
Using the Analytical Method
• A man walks 3 km in one direction, then turns 90° and continues to
walk for an additional 4 km
• In what direction and how far is he from his starting point?
– First, draw a simple sketch
Figure: Hypotenuse and Angle
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
106
Using the Analytical Method
• His net displacement is found
using:
𝑅=
𝑎2 + 𝑏2
𝑅=
32 + 42
• His direction (angle of
displacement) is found using tan:
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝑎
tan 𝜃 =
𝑏
4
tan 𝜃 =
3
tan 𝜃 = 1.33
𝜃 = tan−1 1.33
tan 𝜃 =
𝑅 = 25
𝑅 = 5 𝑘𝑚
𝜃 = 53°
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
107
Using the Analytical Method
• His new location is 5 km at 53° from his starting point
Figure: Hypotenuse and Angle
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
108
Resultant of Several Vectors
Example 1
• Three forces,
(F1, F2, and F3) act on an object
• Find resultant force (FR)
Figure: Force Acting on an Atomic Nucleus
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
109
Resultant of Several Vectors
Example 1 (continued)
• Step 1: Draw x and y
coordinates and three forces
from point of origin or center of
object
– Component vectors and
angles have been added to
drawing to aid in discussion
Figure: Individual Force Vectors – Nucleus Example
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
110
Resultant of Several Vectors
Example 1 (continued)
• Step 2: Resolve each vector
into its rectangular components:
Vector Angle x component y component
𝐹1𝑦 = 𝐹1 sin 𝜃1
F1
θ1
𝐹1𝑥 = 𝐹1 cos 𝜃1
F2
θ2
𝐹2𝑥 = 𝐹2 cos 𝜃2 𝐹2𝑦 = 𝐹2 sin 𝜃2
F3
θ3
𝐹3𝑥 = 𝐹3 cos 𝜃3 𝐹3𝑦 = 𝐹3 sin 𝜃3
© Copyright 2014
ELO 3.3
Figure: Individual Force Vectors – Nucleus Example
Operator Generic Fundamentals
111
Resultant of Several Vectors
Example 1 (continued)
• Step 3: Sum the x and y components.
𝐹𝑅𝑥 = Σ𝐹𝑥 = 𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥
𝐹𝑅𝑦 = Σ𝐹𝑦 = 𝐹1𝑦 + 𝐹2𝑦 + 𝐹3𝑦
(“Σ” means summation)
• Step 4: Calculate magnitude of FR
𝐹𝑅 =
2
2
𝐹𝑅𝑥
+ 𝐹𝑅𝑦
• Step 5: Calculate angle of displacement
𝐹𝑅𝑦
𝐹𝑅𝑥
𝐹𝑅𝑦
−1
𝜃 = tan
𝐹𝑅𝑥
tan 𝜃 =
© Copyright 2014
Figure: Individual Force Vectors – Nucleus Example
ELO 3.3
Operator Generic Fundamentals
112
Resultant of Several Vectors
Example 2
• Given three forces acting on an object, determine magnitude and
direction of resultant force FR
– 𝐹1 = 90 𝑁 𝑎𝑡 39°
– 𝐹2 = 50 𝑁 𝑎𝑡 120°
– 𝐹3 = 125 𝑁 𝑎𝑡 250°
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
113
Resultant of Several Vectors
Example 2 (continued)
• Step 1: Draw vectors
– Draw x and y coordinate axes
on a sheet of paper
– Draw F1, F2, and F3 from point of
origin
o Not necessary to be totally
accurate in placing vectors in
drawing
o Approximate location in
correct quadrant is all that is
necessary
Figure: Individual Force Vectors – Nucleus Example
– Label drawing
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
114
Resultant of Several Vectors
Example 2 (continued)
• Step 2: Resolve each force into its rectangular coordinates
Force Magnitude Angle
F1
F2
F3
© Copyright 2014
90 N
50 N
125 N
39°
120°
250°
x component
y component
= 90 sin 39° 𝐹1𝑦
𝐹1𝑥 = 90 cos 39° 𝐹1𝑥
= (90) (0.777)
𝐹1𝑦
𝐹1𝑥 = 69.9 𝑁
𝐹1𝑦 = 56.6 𝑁
𝐹2𝑥 = 50 cos 120°
𝐹2𝑦 = (50) sin 120°
𝐹2𝑥 = (50)(−0.5)
𝐹2𝑦 = (50)(0.866)
𝐹2𝑥 = −25 𝑁
𝐹2𝑦 = 43.3 𝑁
𝐹3𝑥 = (125) cos 250°
𝐹3𝑦 = (125) sin 250°
𝐹3𝑥 = 125)(−0.342)
𝐹3𝑦 = (125)(−0.94)
𝐹3𝑥 = −42.8 𝑁
𝐹3𝑦 = −117.5 𝑁
ELO 3.3
= (90)(0.629)
Operator Generic Fundamentals
115
Resultant of Several Vectors
Example 2 (continued)
• Step 3: Sum the x and y components
𝐹𝑅𝑥 = 𝐹1𝑥 + 𝐹2𝑥 + 𝐹3𝑥
𝐹𝑅𝑥 = 69.9 𝑁 + (−25 𝑁) + (−42.8 𝑁)
𝐹𝑅𝑥 = 2.1 𝑁
𝐹𝑅𝑦 = 𝐹1𝑦 + 𝐹2𝑦 + 𝐹3𝑦
𝐹𝑅𝑦 = 56.6 𝑁 + 43.3 𝑁 + (−117.5𝑁)
𝐹𝑅𝑦 = −17.6 𝑁
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
116
Resultant of Several Vectors
Example 2 (continued)
• Step 4: Calculate magnitude of FR.
𝐹𝑅 =
𝐹𝑅 =
𝐹𝑥2 + 𝐹𝑦2
2.1
2
+ −17.6
2
𝐹𝑅 = 314.2
𝐹𝑅 = 17.7 𝑁
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
117
Resultant of Several Vectors
Example 2 (continued)
• Step 5: Calculate angle of displacement
𝐹𝑅𝑦
tan 𝜃 =
𝐹𝑅𝑥
−17.6
tan 𝜃 =
2.1
tan 𝜃 = −8.381
𝜃 = tan−1 −8.381
𝜃 = −83.2°
• 𝐹𝑅 = 17.7 𝑁 at −83.2° 𝑜𝑟 276.8°
Note: A negative angle means a clockwise rotation from the zero axis.
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
118
Adding Vectors
Knowledge Check
Determine the resultant of the following vectors:
F1, magnitude 6, angle 150°
F2, magnitude 11, angle 240°
F3, magnitude 5, 0°
A. Magnitude 8.7, 228°
B. Magnitude 7.9, 210°
C. Magnitude 11.1, 196°
D. Magnitude 7.9, 196°
Correct answer is A.
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
119
TLO 3 Summary
Graphic Method Summary:
• Draw rectangular coordinates, Draw first vector, Draw second vector
connected to the end (head) of first vector with proper angular
orientation, Draw remaining vectors, starting at the head of the
preceding vector, Draw resultant vector from the origin of axes to
head of final vector, Measure the length of the resultant vector,
Measure the angle of the resultant vector addition
Figure: Vector, Magnitude 7, 30°
© Copyright 2014
TLO 3
Operator Generic Fundamentals
120
TLO 3 Summary
Component-Addition Method Summary:
• Determine the x- and y- axes of all original vectors, Mathematically
combine all x-axis components, Mathematically combine all y-axis
components, The results are the components of the resultant vector.
Figure: Graphing Vector Demonstration
© Copyright 2014
Operator Generic Fundamentals
121
TLO 3 Summary
Analytical Method Summary:
• Draw x and y coordinate axes, Draw component vectors from point of
origin, Resolve each vector into rectangular components, Sum the x
and the y components, Calculate magnitude of FR, Calculate angle of
displacement.
𝑎2 + 𝑏 2 = 𝑐 2
• The cosine will be used to solve for Fx. Sine will be used to solve for
Fy. Tangent will normally be used to solve for θ, although sine and
cosine may also be used, depending on which vector characteristics
are known.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
122
TLO 3 Summary
Now that you have completed this lesson, you should be able to:
1. Given a magnitude and direction graph a vector using the
rectangular coordinate system.
2. Determine components of a vector from a resultant vector.
3. Add vectors using the following methods:
a. Graphical
b. Component addition
c. Analytical
© Copyright 2014
TLO 3
Operator Generic Fundamentals
123
Measurement of Force
TLO 4 – Describe the measurement of force and its relationship to free
body diagrams.
• When determining how an object reacts to a force or forces, it is
important to understand the different types of forces that may act on
the object
• Force is defined as a vector quantity that tends to produce an
acceleration of a body in the direction of its application
• Must be acquainted with the various types of forces that exist to
– Construct a correct free-body diagram
– Apply the appropriate equation in order to solve force-related
problems
© Copyright 2014
TLO 4
Operator Generic Fundamentals
124
Enabling Learning Objectives for TLO 4
1.
2.
3.
4.
5.
6.
7.
Describe the following types of forces: tensile, compressive,
frictional, centripetal, and centrifugal.
Describe the factors that affect the magnitude of the static and
kinetic friction force.
Describe force as it applies to an object and its velocity.
Describe weight as it applies to an object and its velocity.
Given the mass of an object and a value for gravity, calculate the
weight of the object.
Describe the purpose of a free-body diagram, and given all
necessary information, construct a free-body diagram.
Describe the conditions necessary for a body to be in force
equilibrium.
© Copyright 2014
TLO 4
Operator Generic Fundamentals
125
Tensile and Compressive Forces
ELO 4.1 – Describe the following types of forces: tensile, compressive,
frictional, centripetal, and centrifugal.
• Applied force on an object tends to pull the object apart, defines
tension
• Applied force tends to compress the object, it is in compression
• Ropes, cables, etc, attached to bodies only support tensile loads, and
therefore exhibit tension when placed free-body diagrams
• Fluid forces are almost always compressive forces
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
126
Centripetal Force
• An object moving at constant speed in a circle is not in equilibrium
• Although magnitude of linear velocity is not changing, direction of
velocity is continually changing
• A change in direction requires acceleration, an object moving in a
circular path has a constant acceleration towards center of circular
path
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
127
Centripetal Force
• Force is required to cause
acceleration (Newton's second
law of motion)
𝐹 = 𝑚𝑎
• To have constant acceleration
towards center of circular path,
there must be a net force acting
towards center
• Force is known as Centripetal
force
• Without centripetal force object
will move in a straight line
© Copyright 2014
Figure: Centripetal Force
ELO 4.1
Operator Generic Fundamentals
128
Centrifugal Force
• Appears to oppose direction of
motion, acts on an object that
follows a curved path
– Appears to be a force
directed away from center of
circular path
– Actually a fictitious force
– An apparent force that is
used to describe forces
present due to an object's
rotation
© Copyright 2014
ELO 4.1
Figure: Apparent Centrifugal Force
Operator Generic Fundamentals
129
Centrifugal Force
• Not an actual force
• Can be proven not an actual
force by cutting string
– Plane will fly off in a straight
line that is tangent to circle
at velocity it had the
moment string was cut
– If an actual centrifugal force
was present, plane would
not fly away in a line tangent
to circle, but would fly
directly away from the circle
© Copyright 2014
ELO 4.1
Figure: Loss of Centripetal Force
Operator Generic Fundamentals
130
Types of Forces
Knowledge Check
Match the following terms to the appropriate definitions.
1. Tensile forces only
A. Ropes and cables
2. Cause of circular motion
B. Fluid forces
3. Contact between objects
C. Friction
4. Compressive forces only
D. Centripetal
Correct answers are 1-A, 2-D, 3-C, 4-B.
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
131
Friction
ELO 4.2 – Describe the factors that affect the magnitude of the static and
kinetic friction force.
• Results from two surfaces in contact, where one of the surfaces is
attempting to move parallel to or over other surface
– Dry friction (sometimes called Coulomb friction)
– Fluid friction
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
132
Fluid Friction
• Develops between layers of fluid moving at different velocities
– Used in considering problems involving flow of fluids through
pipes
– Important to thermodynamics and fluid flow
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
133
Dry Friction
• Laws of dry friction are best understood by the following experiment
– A block of weight W is placed on a horizontal plane surface
– Forces acting on block are its weight (W) and the normal force (N)
of surface
Figure: Frictional Forces
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
134
Dry Friction
• Weight has no horizontal component
• Normal force of surface also has no horizontal component
– Reaction is therefore normal to surface and is represented by N in
part (a) of previous figure
Figure: Frictional Forces
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
135
Dry Friction
• Horizontal force P is applied to block as shown in part (b) of previous
figure
– If P is small, block will not move
– Some other horizontal force must therefore exist that balances P
Figure: Frictional Forces
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
136
Dry Friction
• This balancing force is the static-friction force (F)
– Resultant of a great number of forces acting over entire surface of
contact between block and plane
• Nature of these forces is not known exactly
– Assumed that forces are due to irregularities of surfaces in
contact and to molecular action
Figure: Frictional Forces
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
137
Dry Friction
• If force P is increased, friction force F also increases, continuing to oppose
P, until its magnitude reaches a certain maximum value FM
• If P is further increased, friction force cannot balance it anymore, and
block starts sliding
• As soon as block has been set in motion, magnitude of F drops FM to a
lower value FK because of less interpenetration between irregularities of
surfaces in contact when these surfaces move with respect to one another
• Block continues sliding with increasing velocity (i.e., it accelerates)
– Friction force, FK – kinetic-friction force, remains approximately
constant
Figure: Frictional Forces
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
138
Static-Friction and Kinetic-Friction Force
Static-Friction Force
• Maximum value FM of static-friction force is proportional to normal
component N of reaction of surface
𝐹𝑀 = µ𝑆 𝑁
• µS - coefficient of static friction (a constant)
Kinetic-Friction Force
• Magnitude FK of kinetic-friction force may be expressed as:
𝐹𝐾 = µ𝐾𝑁
• µK - coefficient of kinetic friction (a constant)
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
139
Magnitudes of Coefficients of Friction
• µS and µK do not depend upon area of surfaces in contact
– Both coefficients depend strongly on nature of surfaces in contact
– Also depend upon the exact condition of surfaces
o Value is seldom known with accuracy greater than 5 percent
Note: Frictional forces are always opposite in direction to motion (or
impending motion) of object
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
140
Friction Force
Knowledge Check
Select all statements about the friction force that are true.
A. The direction of the friction force is opposite the direction of
motion.
B. The coefficient of kinetic friction between two surfaces is greater
than the coefficient of static friction for the same two surfaces.
C. Friction force is proportional to the normal force.
D. When an object is not moving, the friction force is zero.
Correct answers are A and C.
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
141
Force and Velocity
ELO 4.3 – Describe weight as it applies to an object and its velocity.
Force - vector quantity that tends to produce an acceleration of a body
in the direction of its application
• Changing body's velocity causes body to accelerate
• Mathematically defined by Newton's second law of motion:
𝐹 = 𝑚𝑎
Where
F = force on object (Newton or lbf)
m = mass of object (kg or lbm)
a = acceleration of object (m/sec2 or ft/sec2)
© Copyright 2014
ELO 4.3
Operator Generic Fundamentals
142
Force
• Characterized by its point of application, its magnitude, and its
direction
• Two or more forces may act upon an object without affecting its state
of motion
• For example, a book resting upon a table has a downward force
acting on it caused by gravity and an upward force exerted on it from
table top
– These two forces cancel and net force on book is zero
– Can be verified by observing that no change in state of motion
has occurred
© Copyright 2014
ELO 4.3
Operator Generic Fundamentals
143
Force and Velocity
Knowledge Check
Select all the true statements.
A. Application of force always causes acceleration.
B. Force is a vector quantity.
C. When an object remains at rest, there is no force applied to the
object.
D. When an object remains at rest, the net force on the object is
zero.
Correct answers are B and C.
© Copyright 2014
ELO 4.3
Operator Generic Fundamentals
144
Weight
ELO 4.4 – Describe weight as it applies to an object and its velocity.
• Force can be thought of simply as a push or pull, but is more clearly
defined as any action on a body that tends to change the velocity of
the body.
• Weight is a force exerted on an object due to the object's position in a
gravitational field.
• Newton's laws of motion are some of the fundamental concepts used
in the study of force and weight.
© Copyright 2014
ELO 4.4
Operator Generic Fundamentals
145
Weight
• Special application of concept of force
– Defined as force exerted on an object by gravitational field of
earth, or more specifically, pull of the earth on body
𝑊 = 𝑚𝑔 or 𝑊 =
𝑚𝑔
𝑔𝑐
(English system)
• W = weight (N or lbf)
• m = mass (kg or lbm) of object
• g = local acceleration of gravity
– on earth = 9.8 m/sec2 or 32.17 ft/sec2
• gc = a conversion constant employed to facilitate the use of Newton's
second law of motion with English system of units, equal to 32.17 ftlbm/lbf-sec2
– gc has same numerical value as acceleration of gravity at sea level
© Copyright 2014
ELO 4.4
Operator Generic Fundamentals
146
Weight
• Mass of a body is same regardless of whether on the moon, earth,
etc.
• Weight of a body depends upon local acceleration of gravity
– Weight of an object is less on the moon than on earth because
local acceleration of gravity is less on the moon than on earth
© Copyright 2014
ELO 4.4
Operator Generic Fundamentals
147
Weight – Examples
• Calculate the weight of a person
with a mass of 84 kg.
𝑊 = 𝑚𝑔
• Calculate the weight of a person
with a mass of 84 kg on the
moon. Gravity on the moon is
1.63 m/sec.
𝑚
= 84 𝑘𝑔 9.81
𝑠𝑒𝑐 2
𝑊 = 𝑚𝑔
𝑚
𝑊 = 84 𝑘𝑔 1.63
𝑠𝑒𝑐 2
2
= 8.24 × 10 𝑁
𝑊 = 1.37 × 103 𝑁
© Copyright 2014
ELO 4.4
Operator Generic Fundamentals
148
Gravitational Force Review
• Any object dropped will accelerate as it falls, not in physical contact
with any other body
• Gravitational force
– Concept that one body, exerts a force on another body, even
though they are far apart
– Gravitational attraction of two objects depends upon mass of each
and distance between them (Newton's Law of Gravitation)
© Copyright 2014
ELO 4.4
Operator Generic Fundamentals
149
Weight
Knowledge Check
Select all the true statements.
A. The mass of a body is the same, wherever the body is located.
The weight of a body, however, depends upon the local
acceleration due to gravity.
B. Weight is a function of mass only, and does not vary based on
location.
C. Weight is a vector quantity.
D. Weight is a scalar quantity.
Correct answers are A and C.
© Copyright 2014
ELO 4.4
Operator Generic Fundamentals
150
Calculating Weight
ELO 4.5 – Given the mass of an object and a value for gravity, calculate
the weight of the object.
• Calculate the weight of an object by the following four steps:
– Determine which system (English or SI) is used, and select the
appropriate formula
– Determine mass of the object
– Determine local acceleration due to gravity
– Complete formula and calculate the weight
© Copyright 2014
ELO 4.5
Operator Generic Fundamentals
151
Calculating Weight - Example
Calculate the weight of an object with mass of 262 kg.
• Step 1: Determine which system is used (English or metric) and
choose the correct formula
– Mass is given in kg, metric system is being used, and the formula
is 𝑊 = 𝑚𝑔
• Step 2: Determine the mass of the object
– Mass is 262 kg
• Step 3: Determine the local acceleration due to gravity
– Local acceleration due to gravity is 9.8 m/sec2
© Copyright 2014
ELO 4.5
Operator Generic Fundamentals
152
Calculating Weight - Example
• Step 4: Complete the formula and calculate the weight of the object.
𝑊 = 𝑚𝑔 = 262 𝑘𝑔
𝑚
9.8
𝑠𝑒𝑐 2
𝑘𝑔– 𝑚
= 2.6 × 10
𝑠𝑒𝑐 2
3
3
= 2.2 × 10 𝑁
© Copyright 2014
ELO 4.5
Operator Generic Fundamentals
153
Calculating Weight
Knowledge Check
An object weighs 144 pounds-mass on earth. What would it weigh on a
planet with acceleration due to gravity of 14.9 ft/sec2?
A. 66.6 pounds-mass
B. 66.6 kilograms
C. 2.19 x 102 pounds-mass
D. 2.19 x 102 kilograms
Correct answer is A.
© Copyright 2014
ELO 4.5
Operator Generic Fundamentals
154
Free-Body Diagrams Introduction
ELO 4.6 – Describe the purpose of a free-body diagram, and given all
necessary information, construct a free-body diagram.
• To study the effect of forces on a body, isolate the body and
determine all forces acting upon it
• The diagram with the representation of all external forces acting on it
is called a Free-Body Diagram
• It has long been established that the free-body diagram method is the
key to the understanding of engineering problems
• The isolation of a body is the tool that clearly separates cause and
effect and focuses attention to the application of a selected principle
© Copyright 2014
ELO 4.6
Operator Generic Fundamentals
155
Simple Free-Body Diagram
• Book resting on table
• Two forces are acting on book
to keep it stationary
– Weight (W) of book exerts a
force downward on table
– Normal force (N) is exerted
upward on book by table
⇒equal to weight of book
– A normal force is defined as
any perpendicular force with
which any two surfaces are
pressed against each other
© Copyright 2014
Figure: Book on a Table
ELO 4.6
Operator Generic Fundamentals
156
Constructing a Free-Body Diagram
• Step 1. Determine which body/bodies to be isolated
– Body chosen will usually involve one or more of desired unknown
quantities
• Step 2. Isolate body or combination of bodies chosen with a diagram
that represents its complete external boundaries
• Step 3. Represent all forces that act on isolated body as applied in
their proper positions in diagram of isolated body
– Do not show forces that object exerts on anything else, since
these forces do not affect object itself
• Step 4. Indicate the choice of coordinate axes directly on diagram.
Pertinent dimensions may also be represented for convenience
© Copyright 2014
ELO 4.6
Operator Generic Fundamentals
157
Constructing a Free-Body Diagram
• Free-body diagram serves purpose of focusing accurate attention on
action of external forces; therefore, diagram should not be cluttered
with excessive information
• Force arrows should be clearly distinguished from other arrows to
avoid confusion
© Copyright 2014
ELO 4.6
Operator Generic Fundamentals
158
Free-Body Diagram – Example 1
• Car is being towed by a force of some magnitude
• Construct a free-body diagram showing all forces acting on car
Figure: Car
© Copyright 2014
ELO 4.6
Operator Generic Fundamentals
159
Free-Body Diagram – Solution 1
• Car is chosen and isolated
• All forces acting on car are
represented with proper
coordinate axes
– Fapp – Force applied to tow car
– Fk – Frictional force that
opposes applied force due to
weight of car and nature of
surfaces (car's tires and road
surface)
– W – Weight of car
– N – Normal force acting on car
© Copyright 2014
ELO 4.6
Figure: Free-Body Diagram: Car Under Tow
Operator Generic Fundamentals
160
Calculating Weight
Knowledge Check
Select all the true statements for the diagram:
Figure: Hanging Object
A. If the hanging object is at rest, the net forces applied to it must
be zero.
B. Cable T2 has no force applied to it.
C. Cable T2 has a horizontal force pulling the hanging object to the
left but no vertical force applied.
D. Cable T1 has a vertical force to match the weight of the hanging
object, and a horizontal force pulling the hanging object to the
right.
Correct answers are A, C, and D.
© Copyright 2014
ELO 4.6
Operator Generic Fundamentals
161
Force Equilibrium
ELO 4.7 – Describe the conditions necessary for a body to be in force
equilibrium.
• Knowledge of the forces required to maintain an object in equilibrium
is essential in understanding the nature of bodies at rest and in
motion
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
162
Net Force
• When forces act on an object, result may be a change in object's
state of motion
• If certain conditions exist, forces may combine to maintain a state of
equilibrium or balance.
• To determine if a body is in equilibrium, overall effect of all forces
acting on it must be assessed
• All forces that act on an object result in essentially one force that
influences object's motion
• Force which results from all forces acting on body defined as Net
Force
• Important to remember that forces are vector quantities
• When analyzing various forces, must account for both magnitude
(displacement) as well as direction in which force is applied
– Best done using a free-body diagram
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
163
Book Resting on Table
• Book remains stationary resting on table because table exerts normal
force upward equal to weight of book ⇒ net force on book is zero
• If force applied to book and effect of friction is neglected, net force
will be equal to applied force
– Book will move in direction of applied force
Figure: Net Force Acting on Book on Table
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
164
Book Resting on Table
• Free-body diagram (a) below shows that weight (W) of book is
canceled by normal force (N) of table since they are equal in
magnitude but opposite in direction
• Resultant (net) force is therefore equal to applied force (FAPP)
Figure: Net Force Acting on Book on Table
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
165
Equilibrium and Inertia
• An object in equilibrium is considered to be in a state of balance ⇒
net force on object is equal to zero
– If vector sum of all forces acting on an object is equal to zero,
then object is in equilibrium
• Newton's first law of motion describes equilibrium and the effect of
force on a body that is in equilibrium
– "An object remains at rest (if originally at rest) or moves in a
straight line with a constant velocity if the net force on it is zero.“
• Newton's first law of motion is also called the “Law of Inertia”
• Inertia - tendency of a body to resist a change in its state of motion
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
166
First Condition of Equilibrium
• Consequence of Newton's first law
• May be written in vector form:
– "A body will be in translational equilibrium if and only if the vector
sum of forces exerted on a body by the environment equals zero.“
• Example – if three forces act on a body, it is necessary for the
following to be true for body to be in equilibrium:
𝐹1 + 𝐹2 + 𝐹3 = 0
• Equation may also be written:
Σ𝐹 = 0
– Sum includes all forces exerted on body by its environment
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
167
First Condition of Equilibrium
• The vanishing of this vector sum is a necessary condition, called the
First Condition of Equilibrium
– Must be satisfied in order to ensure translational equilibrium
• In three dimensions (x, y, z), component equations of First Condition
of Equilibrium are:
Σ𝐹𝑥 = 0 Σ𝐹𝑦 = 0 Σ𝐹𝑧 = 0
• Condition applies to objects in motion with constant velocity and to
bodies at rest or in static equilibrium
(referred to as Statics)
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
168
Equilibrium Example
• Object has a weight of 125 kg
• Object is suspended by cables
as shown
• Calculate tension (T1 ) in cable
at 30° with the horizontal
• Tension in a cable is force
transmitted by cable
• Tension at any point in cable
can be measured by removing a
suitable length of cable and
inserting a spring scale
© Copyright 2014
ELO 4.7
Figure: Hanging Object
Operator Generic Fundamentals
169
Equilibrium Example
• Object and its supporting cables are motionless (in equilibrium) ⇒ net
force acting on intersection of cables is zero
– Sum of x-components of T1, T2 and T3 is zero
Σ𝐹𝑥 = 𝑇1𝑥 + 𝑇2𝑥 + 𝑇3𝑥 = 0
– Sum of y-components also zero
Σ𝐹𝑦 = 𝑇1𝑦 + 𝑇2𝑦 + 𝑇3𝑦 = 0
• Tension T3 equal to weight of the object - 125 N
Figure: Free-Body Diagram: Hanging Object
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
170
Equilibrium Example
• x and y components of tensions can be found using trigonometry
(sine function)
• Substituting known values into equation:
𝛴𝑭𝑦 = 𝑻1 sin 30° + 𝑻2 sin 180° + 𝑻3 sin 270°
= 0 𝛴 𝑻1 0.5 + 𝑻2 0 + 125 𝑁 −1 = 0
0.5 𝑻1 − 125 𝑁 = 0
0.5 𝑻1 = 125 𝑁
𝑻1 = 250 𝑁
Figure: Free-Body Diagram: Hanging Object
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
171
Equilibrium Example
• A simpler method to solve this problem involves assigning a sign
convention to free-body diagram and examining direction of forces
• By choosing (+) as upward direction and (-) as downward direction:
– Upward component of T1 is
+ T1 sin 30°
– Tension T3 is -125 N
– T2 has no y-component
Figure: Free-Body Diagram: Hanging Object
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
172
Equilibrium Example
• Using same equation as before:
Σ𝐹𝑦 = (𝑇1 sin 30°) − 125 𝑁 = 0
0.5 𝑇1 = 125 𝑁
𝑇1 = 250 𝑁
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
173
Equilibrant
• If the sum of all forces acting upon a body is equal to zero, that body
is said to be in force equilibrium
• If the sum of all the forces is not equal to zero, any force or system of
forces capable of balancing the system is defined as an equilibrant
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
174
Equilibrant Example
• A 900 kg car is accelerating (on
a frictionless surface) at a rate
of 2 m/sec2
• What force must be applied to
the car to act as an equilibrant
for this system?
• First, a free-body diagram is
drawn
• A force, F2 , MUST be applied in
opposite direction to F1 such
that sum of all forces acting on
car is zero
Figure: Free-Body Diagram: Accelerating Car
Σ 𝐹𝑜𝑟𝑐𝑒𝑠 = 𝐹1 + 𝐹2 + 𝑁 + 𝑊 = 0
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
175
Equilibrant Example
• Since car remains on surface,
forces N and W are in equal
and opposite directions
• Force F2 must be applied in an
equal and opposite direction to
F1 in order for forces to be in
equilibrium
Figure: Free-Body Diagram: Accelerating Car
𝐹2 = 𝐹1 = 𝑚𝑎 = (900 𝑘𝑔 × 2 𝑚/sec2)
= 1,800 𝑘𝑔– 𝑚/sec2
= 1,800 𝑁𝑒𝑤𝑡𝑜𝑛𝑠
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
176
Force Equilibrium
Knowledge Check
A 900-kg car is accelerating (on a frictionless surface) at a rate of 2
m/sec2. What force must be applied to the car to act as an equilibrant
for this system?
A. 1,800 N
B. 1,800 foot-pounds
C. 8,820 N
D. 8,820 foot-pounds
Correct answer is A.
© Copyright 2014
ELO 4.7
Operator Generic Fundamentals
177
TLO 4 Summary
Measurements of Force
• A tensile force is a force that tends to pull an object apart,
Compressive force is a force that tends to compress an object.
• Frictional force is the force resulting from two surfaces in contact,
where one of the surfaces is attempting to move with respect to the
other surface. The magnitude of the frictional force is affected by the
following:
– Weight of the object being moved
– Type of surface on the object being moved
– Type of surface on which the object is moving
• Static-frictional forces are those frictional forces present when an
object is stationary, whereas kinetic-frictional forces are those
frictional forces present between two objects that are moving.
© Copyright 2014
TLO 4
Operator Generic Fundamentals
178
TLO 4 Summary
• Centripetal force is the force on an object moving in a circular path
that is directed towards the center of the path.
• Centrifugal force is the fictitious force that appears to be directed
away from the center of the circular path.
• Force is a vector quantity that tends to produce an acceleration of a
body in the direction of its application.
𝐹 = 𝑚𝑎
• Weight is the force exerted on an object due to gravity. (On the earth,
it is the gravitational pull of the earth on the body.)
𝑚𝑔
𝑊 = 𝑚𝑔 𝑜𝑟 𝑊 =
𝑔𝑐
© Copyright 2014
TLO 4
Operator Generic Fundamentals
179
TLO 4 Summary
Free-Body Diagrams
• A free-body diagram isolates a body and illustrates all the forces that
act on the body so that a complete and accurate account of all of
those forces may be considered. Constructing a free-body diagram
requires the following four steps:
• Determine the body or combination of bodies to be isolated.
• Isolate the body or combination of bodies with a diagram that
represents the complete external boundaries.
• Represent all forces that act on the isolated body in their proper
positions within the diagram.
• Indicate the choice of coordinate axes directly on the diagram.
© Copyright 2014
Operator Generic Fundamentals
180
TLO 4 Summary
Net Force, Inertia, and Equilibrium
• The force that is the resultant force of all forces acting on a body is
defined as the net force.
• If the vector sum of all the forces acting on an object is equal to zero,
then the object is in equilibrium.
• Inertia is the tendency of a body to resist a change in its state of
motion.
• The First Condition of Equilibrium is stated as follows: "A body will be
in translational equilibrium if and only if the vector sum of forces
exerted on a body by the environment equals zero, or 𝐹1 + 𝐹2 + 𝐹3 = 0
or, 𝛴𝐹 = 0
• Any force or system of forces capable of balancing a system so that
the net force is zero is defined as an equilibrant.
© Copyright 2014
Operator Generic Fundamentals
181
TLO 4 Summary
Now that you have completed this lesson, you should be able to:
1. Describe the following types of forces: tensile, compressive,
frictional, centripetal, and centrifugal.
2. Describe the factors that affect the magnitude of the static and
kinetic friction force.
3. Describe force as it applies to an object and its velocity.
4. Describe weight as it applies to an object and its velocity.
5. Given the mass of an object and a value for gravity, calculate the
weight of the object.
6. Describe the purpose of a free-body diagram, and given all
necessary information, construct a free-body diagram.
7. Describe the conditions necessary for a body to be in force
equilibrium.
© Copyright 2014
TLO 4
Operator Generic Fundamentals
182
Newton’s Laws of Motion
TLO 5 – Apply Newton's laws of motion to a body at rest.
• The basis for modern mechanics was developed in the seventeenth
century by Sir Isaac Newton
• From his studies of objects in motion, he formulated three
fundamental laws
• The study of Newton's laws of motion allows us to understand and
accurately describe the motion of objects and the forces that act on
those objects
© Copyright 2014
TLO 5
Operator Generic Fundamentals
183
Enabling Learning Objectives for TLO 5
1. Describe Newton’s 1st, 2nd, and 3rd laws of motion.
2. Describe Newton’s law of universal gravitation.
© Copyright 2014
TLO 5
Operator Generic Fundamentals
184
Newton’s Laws of Motion
ELO 5.1 – Describe Newton's 1st, 2nd, and 3rd laws of motion.
• Newton's First Law of Motion - "an object remains at rest (if
originally at rest) or moves in a straight line with constant velocity if the
net force acting on it is zero.“
• Newton’s Second Law - "the acceleration of a body is proportional to
the net (i.e., sum or resultant) force acting on it and in the direction of
that net force.“
– Establishes the relationship between force, mass, and acceleration
– Mathematically: 𝐹 = 𝑚𝑎
F = force (Newton = 1 kg-m/sec2, or lbf)
m = mass (kg or lbm)
a = acceleration (m/sec2 or ft/sec2)
© Copyright 2014
ELO 5.1
Operator Generic Fundamentals
185
Newton’s Second Law
• 𝐹 = 𝑚𝑎 can be used to calculate an object’s weight at the earth’s
surface
– F is force, or weight, caused by gravitational acceleration of earth
acting on mass, m, of object
– Acceleration designated, g, which equals
9.8 m/sec2 or 32.17 ft/sec2
(g - gravitational acceleration constant)
• For weight: 𝐹 = 𝑚𝑔
© Copyright 2014
ELO 5.1
Operator Generic Fundamentals
186
Newton’s Third Law
• Newton's Third Law of Motion - “if a body exerts a force on a
second body, the second body exerts an equal and opposite force on
the first”
– “For every action there is an equal and opposite reaction"
• Forces always occur in pairs of equal and opposite forces
– Example: Downward force exerted on desk by pencil is
accompanied by upward force of equal magnitude exerted on
pencil by desk
• Principle holds for all forces, variable or constant, regardless of their
source
© Copyright 2014
ELO 5.1
Operator Generic Fundamentals
187
Newton’s Laws of Motion
Knowledge Check
State Newton’s Second Law of Motion.
A. A particle with a force acting on it has an acceleration
proportional to the magnitude of the force and in the direction of
that force.
B. For every action there is an equal but opposite reaction.
C. An object remains at rest (if originally at rest) or moves in a
straight line with constant velocity if the net force on it is zero.
D. Motion of an object is determined by the size and shape of the
object, not the mass of the object.
Correct answer is A.
© Copyright 2014
ELO 5.1
Operator Generic Fundamentals
188
Newton’s Universal Law of Gravitation
ELO 5.2 – Describe Newton's law of universal gravitation.
• Newton’s Universal Law Of Gravitation - "Each and every mass in
the universe exerts a mutual, attractive gravitational force on every
other mass in the universe. For any two masses, the force is directly
proportional to the product of the two masses and is inversely
proportional to the square of the distance between them."
© Copyright 2014
ELO 5.2
Operator Generic Fundamentals
189
Universal Law Of Gravitation
• Newton expressed the universal law of gravitation using equation:
𝑚1 𝑚2
𝐹 = 𝐺
𝑟2
– 𝐹 = force of attraction (Newton = 1kg-m/sec2 or lbf)
– 𝐺 = universal constant of gravitation
(6.67 x 10-11 m3/kg-sec2 or 3.44 x 10-8 ft3/slug-sec2)
– 𝑚1 = mass of first object (kg or lbm)
– 𝑚2 = mass of second object (kg or lbm)
– r = distance between centers of the two objects (m or ft)
© Copyright 2014
ELO 5.2
Operator Generic Fundamentals
190
Universal Law Of Gravitation
• Value of g (gravitational acceleration constant), at surface of earth
can be determined using universal law of gravitation
– Value is known to be 9.8 m/sec2
(or 32.17 ft/sec2)
– Can be calculated using: 𝐹 = 𝐺
© Copyright 2014
ELO 5.2
𝑚1 𝑚2
𝑟2
Operator Generic Fundamentals
191
Calculating the Gravitational Acceleration
Constant
• First, assume that earth is much larger than an object and that the
object resides on surface of earth
– Value of r will be equal to radius of earth
• Second, understand that force of attraction (F) for the object is equal
to object's weight (F)
• Setting Equations equal to each other yields: 𝐹 = 𝐺
𝑀𝑒 𝑚1
= 𝑚1 𝑎
𝑟2
– Me = mass of the earth (5.95 x 1024 kg)
– m1 = mass of the object
– r = radius of the earth (6.367 x 106 m)
© Copyright 2014
ELO 5.2
Operator Generic Fundamentals
192
Calculating the Gravitational Acceleration
Constant
• Mass (m1) of object cancels, and value of (g) can be determined as
follows since a = g by substituting (g) for (a) in previous equation
𝑀𝑒
𝑔=𝐺 2
𝑟
3
𝑚
𝑔 = 6.67 × 10−11
𝑘𝑔– 𝑠𝑒𝑐 2
5.97 × 1024 𝑘𝑔
6.371 × 106 𝑚 2
𝑚
𝑔 = 9.81
𝑠𝑒𝑐 2
© Copyright 2014
ELO 5.2
Operator Generic Fundamentals
193
Calculating the Gravitational Acceleration
Constant
• If object is significant distance from earth, can demonstrate that (g) is
not a constant value but varies with distance (altitude) from earth
• Object at an altitude of 30 km (18.63 mi). Value of (g) is:
𝑟 = 30,000 𝑚 + 6.371 × 106 𝑚
= 6.401 × 106 𝑚
𝑔 = 6.67
× 10−11
𝑚3
𝑘𝑔– 𝑠𝑒𝑐 2
5.97 × 1024 𝑘𝑔
6.401 × 106 𝑚 2
𝑚
𝑔 = 9.72
𝑠𝑒𝑐 2
© Copyright 2014
ELO 5.2
Operator Generic Fundamentals
194
Calculating the Gravitational Acceleration
Constant
• Height of 30 km only changes (g) from
9.8 m/sec2 to 9.7 m/sec2
– Even smaller change for objects closer to earth
• (g) is normally considered constant value since most calculations
involve objects close to surface of earth
© Copyright 2014
ELO 5.2
Operator Generic Fundamentals
195
Universal Gravitation
Knowledge Check
Select all statements about the Law of Universal Gravitation that are
true.
A. The law applies to every mass in the universe.
B. The force due to gravity is mutual.
C. The force due to gravity is directly proportional to the distance
between the bodies.
D. The force due to gravity is attractive.
Correct answers are A, B, and D.
© Copyright 2014
ELO 5.2
Operator Generic Fundamentals
196
TLO 5 Summary
• Newton's First Law of Motion – an object at rest will remain at rest
or an object moving in a straight line with a constant velocity will
remain so, if the net force on it is zero.
• Newton's Second Law of Motion – the acceleration of a body is
inversely proportional to its mass and directly proportional to the net
(i.e., resultant) force acting on it; the acceleration acts in the direction
of the net force. F=ma
• Newton's Third Law of Motion – the forces of action and reaction
between interacting bodies are equal in magnitude and opposite in
direction. For every action there is an equal and opposite reaction.
• Newton's Universal Law of Gravitation – each and every mass in
the universe exerts a mutual, attractive gravitational force on every
other mass in the universe. For any two masses, the force is directly
proportional to the product of the two masses and is inversely
proportional to the square of the distance between them.
F = G(m1m2/r2)
© Copyright 2014
TLO 5
Operator Generic Fundamentals
197
TLO 5 Summary
Now that you have completed this lesson, you should be able to:
1. Describe Newton's 1st, 2nd, and 3rd laws of motion.
2. Describe Newton's law of universal gravitation.
© Copyright 2014
TLO 5
Operator Generic Fundamentals
198
Momentum
TLO 6 – Calculate the change in velocity when two objects collide, with
respect to the conservation of momentum.
• Learning the laws governing moving objects will enable you to
understand
– Interactions of particles and atoms in the fission process
– Way turbines turn steam expansion into mechanical energy
– Causes of water hammer
– Other important physical phenomena important to power plant
operation
© Copyright 2014
TLO 6
Operator Generic Fundamentals
199
Enabling Learning Objectives for TLO 6
1.
Describe momentum and conservation of momentum.
2.
Using the conservation of momentum, calculate the velocity of an
object (or objects) following a collision of two objects.
© Copyright 2014
TLO 6
Operator Generic Fundamentals
200
Momentum
ELO 6.1 – Describe momentum and conservation of momentum.
• A measure of the motion of a moving body
• An understanding of momentum and the conservation of momentum
are essential tools in solving physics problems
© Copyright 2014
ELO 6.1
Operator Generic Fundamentals
201
Momentum
• Momentum is a basic and widely applicable concept of physics
– Measure of motion of a moving body
– Result of product of body's mass and velocity: 𝑃 = 𝑚𝑣
P = momentum of the object (kg-m/sec or ft-lbm/sec)
𝑚 = mass of the object (kg or lbm)
𝑣 = velocity of the object (m/sec or ft/sec)
• Momentum is a vector quantity, results from velocity of object
• If different momentum quantities are to be added as vectors, direction
of each momentum must be taken into account
• To simplify understanding of momentum, only straight line motions
considered in this presentation
© Copyright 2014
ELO 6.1
Operator Generic Fundamentals
202
Momentum Example
• Calculate momentum of 7 kg bowling ball rolling down lane at 7
m/sec
𝑃 = 𝑚𝑣
𝑝 = 7.0 𝑘𝑔
𝑚
7.0
𝑠
𝑘𝑔– 𝑚
𝑝 = 49
𝑠
© Copyright 2014
Operator Generic Fundamentals
203
Force and Momentum
• Direct relationship between force and momentum:
– Rate at which momentum changes with time is equal to net force
applied to an object
– Directly from Newton's second law of motion: 𝐹 = 𝑚𝑎
– Special case of Newton's second law for a constant force which
gives rise to a constant acceleration
• Acceleration is rate at which velocity changes with time ⇒
We know that, 𝐹 = 𝑚𝑎
and since, 𝑎 =
then, 𝐹 = 𝑚
© Copyright 2014
∆𝑣
∆𝑡
∆𝒗
∆𝑡
ELO 6.1
Operator Generic Fundamentals
204
Force and Momentum
• Acceleration is rate at which velocity changes with time
– 𝐹 = 𝑚𝑎
𝑎=
∆𝑣
∆𝑡
• Substituting P for mv and P0 for
mv0
𝐹=
∆𝑣
𝐹=𝑚
∆𝑡
or
𝑚𝑣 − 𝑚𝑣𝑜
𝐹=
𝑡 − 𝑡𝑜
© Copyright 2014
𝑃 − 𝑃𝑜
𝑡 − 𝑡𝑜
𝐹=
ELO 6.1
∆𝑝
∆𝑡
Operator Generic Fundamentals
205
Force and Momentum Example
The velocity of a rocket must be increased by 35 m/sec to achieve
proper orbit around the earth. If the rocket has a mass of 5,000 kg and
it takes 9 seconds to reach orbit, calculate the required thrust (force) to
achieve this orbit.
• Initial velocity (vo) and final velocity (v) are unknown
• Do know change in velocity (v-vo) - 35 m/sec
• Equation used can be used to find the solution
∆𝑣
𝐹=𝑚
∆𝑡
𝑚
35 𝑠
𝐹 = (5.00 × 10 𝑘𝑔)
9𝑠
4 𝑘𝑔– 𝑚
𝐹 = 1.94 × 10
2
𝑠𝑒𝑐
4
𝐹 = 1.94 × 10 𝑁
3
© Copyright 2014
ELO 6.1
Operator Generic Fundamentals
206
Force and Momentum Example
• Initial velocity (vo) and final
velocity (v) are unknown
∆𝑣
𝐹=𝑚
∆𝑡
• Do know change in velocity (vvo) - 35 m/sec
𝑚
35
3
𝑠
𝐹 = (5.00 × 10 𝑘𝑔)
9𝑠
• Equation used can be used to
find the solution
4
𝐹 = 1.94 × 10
𝑘𝑔– 𝑚
2
𝑠𝑒𝑐
4
𝐹 = 1.94 × 10 𝑁
© Copyright 2014
ELO 6.1
Operator Generic Fundamentals
207
Conservation of Momentum
• One of most useful properties of momentum is that it is conserved
– Means that if no net external force acts upon an object,
momentum of object remains constant
– Using 𝐹 =
∆𝑃 = 0
∆𝑝
,
∆𝑡
it can be seen that if force (F) is equal to zero, then
• Important when solving problems involving collisions, explosions,
etc., where external force is negligible, and momentum before event
(collision, explosion) equals momentum following event
© Copyright 2014
ELO 6.1
Operator Generic Fundamentals
208
Conservation of Momentum Examples
• Applies when a bullet is fired from a gun
– Prior to firing the gun, both the gun and the bullet are at rest (i.e.,
VG and VB are zero) ⇒ total momentum is zero:
𝑚𝐺 𝑣 𝐺 + 𝑚 𝐵 𝑣 𝐵 = 0
or
𝑚𝐺𝑣𝐺 = −𝑚𝐵𝑣𝐵
• When gun is fired, momentum of recoiling gun is equal and opposite
to momentum of bullet
– In other words, momentum of bullet (mBvB) is equal to momentum
of gun (mGvG), but of opposite direction
© Copyright 2014
ELO 6.1
Operator Generic Fundamentals
209
Law of Conservation of Momentum
• Sum of a system's initial momentum is equal to sum of a system's
final momentum: Σ 𝑃𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = Σ 𝑃𝑓𝑖𝑛𝑎𝑙
• Where a collision of two objects occurs, conservation of momentum
can be stated as:
𝑃1 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑃2 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑃1 𝑓𝑖𝑛𝑎𝑙 + 𝑃2 𝑓𝑖𝑛𝑎𝑙
or
(𝑚1𝑣1)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + (𝑚2𝑣2)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑚1𝑣1)𝑓𝑖𝑛𝑎𝑙 + (𝑚2𝑣2)𝑓𝑖𝑛𝑎𝑙
• In a case where two bodies collide and have identical final velocities,
this equation applies: 𝑚1𝑣1 + 𝑚2𝑣2 = (𝑚1 + 𝑚2)𝑣𝑓
© Copyright 2014
ELO 6.1
Operator Generic Fundamentals
210
Law of Conservation of Momentum
• Consider two railroad cars rolling on a level, frictionless track
• Cars collide, become coupled, and roll together at a final velocity (vf)
• Momentum before and after the collision is expressed with
(𝑚1𝑣1)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + (𝑚2𝑣2)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑚1𝑣1)𝑓𝑖𝑛𝑎𝑙 + (𝑚2𝑣2)𝑓𝑖𝑛𝑎𝑙
• If initial velocities of two objects (v1 and v2 ) are known, then final
velocity (vf ) can be calculated by rearranging the equation
𝑚1 𝑣1 + 𝑚2 𝑣2 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑣𝑓𝑖𝑛𝑎𝑙 =
𝑚1 + 𝑚2
Figure: Momentum
© Copyright 2014
ELO 6.1
Operator Generic Fundamentals
211
Law of Conservation of Momentum
Example
Railroad cars in figure below have masses of m1 = 1,000 kg and
m2 = 1,300 kg. First car (m1) is moving at a velocity of 9 m/sec and
second car (m2) is moving at a velocity of 4 m/sec. First car overtakes
second car and couples with it.
Calculate final velocity of two cars.
Figure: Momentum
© Copyright 2014
ELO 6.1
Operator Generic Fundamentals
212
Law of Conservation of Momentum
Solution
Final velocity (vf ) can be easily calculated using the equation below
𝑣𝑓𝑖𝑛𝑎𝑙
𝑚1 𝑣1 + 𝑚2 𝑣2 𝑖𝑛𝑖𝑡𝑖𝑎𝑙
=
𝑚1 + 𝑚2
𝑚
1,000 𝑘𝑔 9.0 𝑠 + 1,300 𝑘𝑔
=
1,000 𝑘𝑔 + 1,300 𝑘𝑔
𝑣𝑓𝑖𝑛𝑎𝑙
𝑚
= 6.17
𝑠𝑒𝑐
𝑣𝑓𝑖𝑛𝑎𝑙
𝑚
4𝑠
Figure: Momentum
© Copyright 2014
ELO 6.1
Operator Generic Fundamentals
213
Momentum
Knowledge Check
Select all statements about momentum that are true.
A. Momentum is conserved in all collisions.
B. Momentum is conserved in elastic collisions, but not in inelastic
collisions.
C. Momentum of an object may be changed by applying force to
the object.
D. Momentum is a vector quantity.
Correct answers are A, C, and D.
© Copyright 2014
ELO 6.1
Operator Generic Fundamentals
214
Elastic and Inelastic Collisions
ELO 6.2 – Using the conservation of momentum, calculate the velocity of
an object (or objects) following a collision of two objects.
Law of conservation of momentum does not consider whether collision
is elastic or inelastic.
• Elastic collision - both momentum and kinetic energy are conserved
– Example - head-on collision of two billiard balls of equal mass
• Inelastic collision - momentum is conserved, but system kinetic
energy is not conserved
– Example - head-on collision of two automobiles where part of
initial kinetic energy is lost as metal crumples during the impact
© Copyright 2014
ELO 6.2
Operator Generic Fundamentals
215
Collisions and Velocity
• To calculate the velocity of objects after elastic collisions:
– Draw the interaction, including known masses and velocities of all
objects, before and after the collision
– Determine which directions (e.g. up and to the right) are positive in
this solution.
– Organize the known information, including masses and velocities
of all objects, before and after the collision
– Determine the unknown that you must solve for
– Use the formulae given below to determine the solution.
– Formula to use if objects are together after collision:
𝑚1𝑣1 + 𝑚2𝑣2
𝑣𝑓 =
𝑚1 + 𝑚2
– Formula to use if objects are separate after collision:
(𝑚1𝑣1)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + (𝑚2𝑣2)𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = (𝑚1𝑣1)𝑓𝑖𝑛𝑎𝑙 + (𝑚2𝑣2)𝑓𝑖𝑛𝑎𝑙
© Copyright 2014
ELO 6.2
Operator Generic Fundamentals
216
Collisions and Velocity Example
• Consider two railroad cars that have masses of m1 = 1,000 kg and m2
= 1,300 kg. The first car (m1) is moving at a velocity of 9 m/sec and
the second car (m2) is moving at a velocity of 4 m/sec. The first car
overtakes the second car and couples with it.
• Calculate the final velocity of the two cars.
• The final velocity (vf ) can be calculated as follows:
𝑚1𝑣1 + 𝑚2𝑣2
𝑣𝑓 =
𝑚1 + 𝑚2
𝑣𝑓 = ( 1,000𝑘𝑔
9
𝑚
+ 1,300 𝑘𝑔
sec
4
𝑚
)/(1,000𝑘𝑔 + 1,300 𝑘𝑔)
sec
𝑣𝑓 = 6.17 𝑚/sec
© Copyright 2014
ELO 6.2
Operator Generic Fundamentals
217
Collisions and Velocity
Knowledge Check
An object with mass of 1.0 x 102 grams is traveling at 50 m/s when it
strikes an object with mass of 10 kg, which is at rest. Assuming that
the first object is at rest after the elastic collision, what is the velocity
and direction of the second object?
A. 0.50 m/s, in the same direction that the first object was moving.
B. 0.50 m/s, in the opposite direction that the first object was
moving.
C. 5.0 m/s, in the same direction that the first object was moving.
D. 5.0 m/s in the opposite direction that the first object was moving.
Correct answer is A.
© Copyright 2014
ELO 6.2
Operator Generic Fundamentals
218
TLO 6 Summary
Momentum
• Momentum is the measure of the motion of a moving body. It is equal
to the body's mass multiplied by the velocity at which it is moving.
• Momentum can be defined as: 𝑝 = 𝑚𝑣
The Law of Conservation of Momentum
• The Law of Conservation of Momentum states that if no net external
force acts upon a system, the momentum of the system remains
constant. If the net force (F) is equal to zero, then ∆p = 0.
• Momentum before and after a collision can be calculated using:
𝑚1 𝑣1
© Copyright 2014
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
+ 𝑚2 𝑣2
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
= 𝑚1 𝑣1
TLO 6
𝑓𝑖𝑛𝑎𝑙
+ 𝑚2 𝑣2
𝑓𝑖𝑛𝑎𝑙
Operator Generic Fundamentals
219
TLO 6 Summary
Now that you have completed this lesson, you should be able to:
1. Describe momentum and conservation of momentum.
2. Using the conservation of momentum, calculate the velocity of an
object (or objects) following a collision of two objects.
© Copyright 2014
TLO 6
Operator Generic Fundamentals
220
Energy, Work and Power
TLO 7 – Describe energy, work, and power in mechanical systems.
• Energy is the measure of the ability to do work or cause a change
• Energy does not simply appear and disappear
• Energy is transferred from one position to another or transformed
from one type of energy to another
• Work is a measure of the amount of energy required to move an
object
© Copyright 2014
TLO 7
Operator Generic Fundamentals
221
Enabling Learning Objectives for TLO 7
1.
2.
3.
4.
5.
Describe the following terms:
a. Energy
b. Potential energy
c. Kinetic energy
d. Work
Calculate energy and work for a mechanical system.
State the First Law of Thermodynamics, "Conservation of Energy.“
State the mathematical expression for power.
Calculate power in a mechanical system.
© Copyright 2014
TLO 7
Operator Generic Fundamentals
222
Energy
ELO 7.1 – Describe the following terms: energy, potential energy, kinetic
energy, and work.
• Energy - measure of the ability to do work
• Energy determines capacity of a system to perform work and may be
stored in various forms
• Basic mechanical systems involve concepts of potential and kinetic
energy
• Both thermal and mechanical energy can be separated into two
categories:
– Transient Energy - energy in motion; energy being transferred
from one place to another
– Stored Energy - energy contained within a substance or object
© Copyright 2014
ELO 7.1
Operator Generic Fundamentals
223
Potential Energy
• Potential Energy - energy stored in an object because of its position
– Example: potential energy of an object above surface of earth in
gravitational field
• PE also applies to energy due to separation of electrical charge and
to energy stored in a spring
– Energy due to position of any force field
• Example: Energy stored in hydrogen and oxygen as PE to be
released upon burning
– Relative separation distance associated with elemental forms of
hydrogen and oxygen represents PE
– Compound water is formed during burning, resulting in a release
of PE of separation
© Copyright 2014
ELO 7.1
Operator Generic Fundamentals
224
Potential Energy
• When discussing mechanical PE, we look at the position of an object,
relative to other objects or to a reference point
– Measure of an object's position is its vertical distance above a
reference point
– Reference point is normally earth's surface, but can be any point
© Copyright 2014
ELO 7.1
Operator Generic Fundamentals
225
Potential Energy
• PE of object represents work required to elevate object to that
position from reference point
𝑃𝐸 = 𝑤𝑜𝑟𝑘 𝑡𝑜 𝑒𝑙𝑒𝑣𝑎𝑡𝑒
= 𝑤𝑒𝑖𝑔ℎ𝑡 × ℎ𝑒𝑖𝑔ℎ𝑡
=
𝑚𝑔𝑧
𝑔𝑐
Where:
𝑃𝐸 = potential energy in N-m (Joules) or (ft-lbf)
𝑚 = mass in kg (lbm)
𝑔 = 9.8 m/sec2 (32.17 ft/sec2)
𝑔𝑐 = 32.17 (lbm-ft)/(lbf-sec2)
𝑧 = height above a reference in m (ft)
Note: 𝑔𝑐 is used only when using the English system of measurement
© Copyright 2014
ELO 7.1
Operator Generic Fundamentals
226
Kinetic Energy
• Kinetic Energy - energy stored in object due to its motion
• KE of an object represents amount of energy required to increase
velocity of object from rest (v = 0) to its final velocity
• Can also represent work an object it can do as it pushes against
something in slowing down (waterwheel or turbine)
© Copyright 2014
ELO 7.1
Operator Generic Fundamentals
227
Kinetic Energy
Because KE is due to motion of object, and motion is measured by
velocity, KE for an object can be calculated in terms of its velocity:
2
1
𝑚𝑣
𝐾𝐸 = 𝑚𝑣 2 𝑜𝑟 𝐾𝐸 =
2
2𝑔𝑐
Where:
𝐾𝐸 = kinetic energy in N-m (Joules) or (ft-lbf)
𝑚 = mass in kg (lbm)
𝑣 = velocity in m/sec (ft/sec)
𝑔𝑐 = (32.17 lbm-ft)/(lbf-sec2)
Note: 𝑔𝑐 is used only when using the English system of measurement
© Copyright 2014
ELO 7.1
Operator Generic Fundamentals
228
Thermal and Internal Energy
• Thermal Energy - energy related to temperature (higher
temperature, greater molecular movement, and greater energy)
– If an object has more thermal energy than adjacent substance,
substance at higher temperature will transfer thermal energy (at a
molecular level) to cooler substance
– Note: This energy is moving from one place to another (energy in
motion) referred to as transient energy, more commonly – heat
• Internal Energy - energy stored in a substance because due to
motion and position of particles of substance
‒ Only stored energy in a solid material
‒ Important to heat transfer and fluid flow
© Copyright 2014
ELO 7.1
Operator Generic Fundamentals
229
Mechanical Energy and Work
• Mechanical Energy - energy related to motion or position
– Transient mechanical energy - commonly referred to as work
– Stored mechanical energy exists in two forms: KE or PE
– Both KE and PE can be found in fluids and solid objects
• Work - commonly thought of as activity requiring exertion
– Physics Definition of Work - work done by a force acting on a
moving object if the object has component of motion in the
direction of the force
– Work is done by a person, a machine, or an object by applying a
force and causing something to move
© Copyright 2014
ELO 7.1
Operator Generic Fundamentals
230
Work
• If you push on a box (apply a force) and it moves three feet, work has
been performed by you to the box, while work has been performed on
the box
– If you push on the box and it does not move, then work, by our
definition, has not been accomplished
• Work is defined mathematically as: 𝑊 = 𝐹 × 𝑑
– 𝑊 = work done in J (ft-lbf)
– 𝐹 = force applied to the object in N (lbf)
– 𝑑 = distance the object is moved with the force applied in m (ft)
© Copyright 2014
ELO 7.1
Operator Generic Fundamentals
231
Energy Definitions
Knowledge Check
Match the following terms to the appropriate definitions
1. Energy stored in an object because
of its motion
A. Work
2. The energy stored in an object
because of its position
B. Thermal energy
3. Force applied through a distance
C. Kinetic energy
4. That energy related to temperature
D. Potential energy
Correct answers are1-C, 2-D, 3-A, 4-B.
© Copyright 2014
ELO 7.1
Operator Generic Fundamentals
232
Calculating Energy and Work
ELO 7.2 – Calculate energy and work for a mechanical system.
• To calculate potential energy, use the following steps:
– Determine whether the English or metric system is being used
– Determine the height above reference point and mass of the
object in question.
– Ensure all units are consistent. Convert to appropriate units if
necessary.
– Apply the appropriate formula
𝑃𝐸 = 𝑚𝑔𝑧 (SI system)
𝑃𝐸 =
𝑚𝑔𝑧
𝑔𝑐
(English system)
– Calculate the potential energy.
© Copyright 2014
ELO 7.2
Operator Generic Fundamentals
233
PE Example
• What is the potential energy of a 50 kg object suspended 10 meters
above the ground?
𝑚
𝑃𝐸 = (50 𝑘𝑔)(9.81 2 )(10.0 𝑚)
𝑠
2
2
𝑃𝐸 = 4.90 × 10 𝑁– 𝑚 or 4.90 × 10 𝐽
© Copyright 2014
ELO 7.2
Operator Generic Fundamentals
234
Calculating Kinetic Energy
• To calculate kinetic energy, use the following steps:
– Determine the velocity and mass of the object in question.
– Ensure all units are consistent. Convert to appropriate units if
necessary.
– Use the formula to calculate the kinetic energy.
© Copyright 2014
ELO 7.2
Operator Generic Fundamentals
235
KE Example
• What is the kinetic energy of a 10 kg object that has a velocity of 8
m/sec?
1
𝑘𝑔
2
𝐾𝐸 = 𝑚𝑣 = 10
2
2
𝐾𝐸 = 5 𝑘𝑔
𝑚
8.0
𝑠𝑒𝑐
2
64 𝑚2
sec 2
2
𝐾𝐸 = 3.2 × 10 𝐽
© Copyright 2014
ELO 7.2
Operator Generic Fundamentals
236
Calculating Work
• To calculate work use the following steps:
– Determine the force applied and the distance through which the
force acts on the object
– Ensure all units are consistent. Convert to appropriate units if
necessary.
– Use the formula (𝑊 = 𝐹 × 𝑑) to calculate the work done.
© Copyright 2014
ELO 7.2
Operator Generic Fundamentals
237
Work Examples
• An individual pushes a large box for three minutes. During that time,
a constant force of 200 N is exerted to the box, but it does not move.
How much work has been accomplished?
𝑊 = 𝐹 × 𝑑
𝑊 = 200 𝑁 𝑥 0 𝑚
𝑊 = 0 𝐽 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
• The same box is pushed again. A horizontal force of 200 N is applied
to the box, and the box moves five meters horizontally. How much
work has been done?
𝑊 = 𝐹𝑥𝑑
𝑊 = 200 𝑁 𝑥 5 𝑚
𝑊 = 1,000 𝐽
© Copyright 2014
ELO 7.2
Operator Generic Fundamentals
238
Calculating Energy and Work
Knowledge Check
Match the terms below to the appropriate formula.
1.
𝟏
𝒎𝒗𝟐
𝟐
A. Work
2. mgz
B. Momentum
3. Fd
C. Kinetic energy
4. mv
D. Potential energy
Correct answers are1-C, 2-D, 3-A, 4-B.
© Copyright 2014
ELO 7.2
Operator Generic Fundamentals
239
Law of Conservation of Energy
ELO 7.3 – State the First Law of Thermodynamics, "Conservation of
Energy."
• The First Law of Thermodynamics - "energy cannot be created or
destroyed, only altered in form."
– PE - measure of force applied to an object, in order to raise it from
the point of origin to some height
– Energy expended in raising object is equivalent to PE gained by
object because of its height
– Example of transfer of energy as well as alteration of type of
energy
© Copyright 2014
ELO 7.3
Operator Generic Fundamentals
240
Law of Conservation of Energy
• Example: Person throwing a baseball
– While ball is in person’s hand, it contains no KE
– Force must be applied to ball in order to throw it
– Ball leaves hand of person throwing it with a velocity, giving KE
equal to work applied by person’s hand
• Mathematically transfer of energy can be described by following
simplified equation:
𝐸𝑛𝑒𝑟𝑔𝑦𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝐸𝑛𝑒𝑟𝑔𝑦𝑎𝑑𝑑𝑒𝑑 – 𝐸𝑛𝑒𝑟𝑔𝑦𝑟𝑒𝑚𝑜𝑣𝑒𝑑 = 𝐸𝑛𝑒𝑟𝑔𝑦𝑓𝑖𝑛𝑎𝑙
© Copyright 2014
ELO 7.3
Operator Generic Fundamentals
241
Law of Conservation of Energy
• Energy initial - energy initially stored in an object/substance
– Energy can exist in various combinations of KE and PE
• Energy added - energy added to object/substance:
– Heat can be added
o Heat is energy gained or lost at a microscopic level
– Energy can be added in form of stored energy in any mass
added, such as water to a fluid system
– Work can be done on a system
o Energy gained at macroscopic level
© Copyright 2014
ELO 7.3
Operator Generic Fundamentals
242
Law of Conservation of Energy
• Energy removed - energy removed from an object/substance
– Heat can be rejected
– Work can be done by the system
– Energy can be in form of energy stored in any mass removed
• Energy final - energy remaining within object/substance after all
energy transfers and transformations occur
– Energy can exist in various combinations of KE, PE, flow, and
internal energy
© Copyright 2014
ELO 7.3
Operator Generic Fundamentals
243
Simplified Energy Balance
• Each component of Energy Balance can be broken down:
– 𝐸𝑛𝑒𝑟𝑔𝑦𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐾𝐸1 + 𝑃𝐸1
– 𝐸𝑛𝑒𝑟𝑔𝑦𝑎𝑑𝑑𝑒𝑑 = work done on and heat added to system
– 𝐸𝑛𝑒𝑟𝑔𝑦𝑟𝑒𝑚𝑜𝑣𝑒𝑑 = work done by and heat removed from system
– 𝐸𝑛𝑒𝑟𝑔𝑦𝑓𝑖𝑛𝑎𝑙 = 𝐾𝐸2 + 𝑃𝐸2
• Resulting energy balance:
𝐾𝐸1 + 𝑃𝐸1 + 𝐸𝑎𝑑𝑑𝑒𝑑 − 𝐸𝑟𝑒𝑚𝑜𝑣𝑒𝑑 = 𝐾𝐸2 + 𝑃𝐸2
• Neglecting any heat removed or added to system, can replace Eadded
and Eremoved with associated work terms:
𝐾𝐸1 + 𝑃𝐸1 + 𝑊𝑜𝑛 = 𝐾𝐸2 + 𝑃𝐸2 + 𝑊𝑏𝑦
© Copyright 2014
ELO 7.3
Operator Generic Fundamentals
244
Conservation of Energy
Knowledge Check
Match the terms below to the appropriate definitions.
1. Energy cannot be created or
destroyed, only altered in form.
A. Kinetic energy
2. 𝐾𝐸1 + 𝑃𝐸1 + 𝐸𝑎𝑑𝑑𝑒𝑑 = 𝐾𝐸2 + 𝑃𝐸2 + 𝐸𝑟𝑒𝑚𝑜𝑣𝑒𝑑
B. Conservation of Energy
3. Energy due to position
C. Potential energy
4. Energy due to motion
D. Energy balance
Correct answers are 1-B, 2-D, 3-C, 4-A.
© Copyright 2014
ELO 7.3
Operator Generic Fundamentals
245
Power
ELO 7.4 – State the mathematical expression for power.
• Power: a measure of the rate at which energy is used.
• Thermal power: term used to refer to the transfer of heat.
• Mechanical power: term used to describe when work is being done.
• Power - amount of energy used per unit time or rate of doing work
– Measured in units of:
– Joules/sec (also known as watts)
– British Thermal Units (BTUs)
– Horsepower
– ft-lbf/sec
© Copyright 2014
ELO 7.4
Operator Generic Fundamentals
246
Thermal Power
• Thermal Power - measure of thermal energy used per unit time
– Rate of heat transfer or heat flow rate
– Examples of thermal power units:
o British Thermal Units (BTU)
o Kilowatts (kW)
• Thermal power is calculated by:
𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 =
© Copyright 2014
ℎ𝑒𝑎𝑡 𝑢𝑠𝑒𝑑
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
ELO 7.4
Operator Generic Fundamentals
247
Mechanical Power
• Mechanical Power - Mechanical energy used per unit time ⇒ rate at
which work is done
• Expressed in units of:
– Joules/sec (joules/s) or watt (W) in mks
– Feet-pounds force per second (ft-lbf/s) or horsepower (hp) in
English system
• Mechanical power can be calculated using the following
mathematical expression:
𝑃𝑜𝑤𝑒𝑟 =
© Copyright 2014
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
ELO 7.4
Operator Generic Fundamentals
248
Mechanical Power
• Because work can be defined as force multiplied by distance, can
also use:
𝐹𝑑
𝑃=
𝑡
Where:
𝑃 = Power (W or ft-lbf/s)
𝐹 = Force (N or lbf)
𝑑 = distance (m or ft)
𝑡 = time (sec)
• In English system of measurement, horsepower is commonly used
for describing power ratings of equipment such as pumps, motors
and engines
• One horsepower is equivalent to 550 ft-lbf/s and 745.7 watts
© Copyright 2014
ELO 7.4
Operator Generic Fundamentals
249
Mechanical Power
𝐹𝑑
In the equation, 𝑃 = , d divided by t (distance per unit time) is same
𝑡
as velocity ⇒ alternate description of power is:
𝐹𝑣
𝑃=
550
Where:
𝑃 = power (hp)
𝐹 = force (lbf)
𝑣 = velocity (ft/s)
© Copyright 2014
ELO 7.4
Operator Generic Fundamentals
250
Power
Knowledge Check
Match the terms below to the appropriate definitions.
1. Energy used per unit time
A. Horsepower
2. J/s
B. Thermal power
3. 550 ft-lbf/s
C. Watt
4. Heat used per unit time
D. Power
Correct answers are 1-D, 2-A, 3-C, 4-B.
© Copyright 2014
ELO 7.4
Operator Generic Fundamentals
251
Calculating Power
ELO 7.5 – Calculate power in a mechanical system.
• To calculate power use the following steps:
– Determine the information known and the information needed
– Ensure that units are consistent, convert units as necessary
– Choose the appropriate formula based on the information
available and the terms being used
• Formulae for calculating power:
– 𝑃 =
𝐹𝑣
550
– 𝑃 =
𝐹𝑑
𝑡
ℎ𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟
– 𝑃𝑜𝑤𝑒𝑟 =
© Copyright 2014
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
ELO 7.5
Operator Generic Fundamentals
252
Power Example 1
A pump provides a flow rate of 10,000 lpm. The pump does 1.5 x 108
Joules of work every 100 minutes. What is the power of the pump?
Answer:
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑃𝑜𝑤𝑒𝑟 =
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
1.5 × 108 𝐽𝑜𝑢𝑙𝑒𝑠
𝑃=
100 𝑚𝑖𝑛
1 𝑚𝑖𝑛
60 𝑠𝑒𝑐
𝑃 = 2.5 × 104 𝑊𝑎𝑡𝑡𝑠
© Copyright 2014
ELO 7.5
Operator Generic Fundamentals
253
Power Example 2
A boy rolls a ball with a steady force of 1 lbf, giving the ball a constant
velocity of 5 ft/s. What is the power used by the boy in rolling the ball?
Answer:
𝐹𝑣
𝑃=
550
1 𝑙𝑏𝑓
𝑃=
5
𝑓𝑡
𝑠𝑒𝑐
550
𝑃 = 9.0 × 10−3 ℎ𝑝
© Copyright 2014
ELO 7.5
Operator Generic Fundamentals
254
Power Example 3
A race car traveling at constant velocity can go one quarter mile (1,320
ft) in 5 seconds. If the motor is generating a force of 1,890 lbf pushing
the car, what is the power of the motor in hp? Assume the car is
already at full speed at t=0.
Answer:
𝑃=
𝐹𝑑
𝑡
𝑃=
1,890 𝑙𝑏𝑓 1,320 𝑓𝑡
5 𝑠𝑒𝑐
1 ℎ𝑝
𝑓𝑡– 𝑙𝑏𝑓
550
𝑠𝑒𝑐
𝑃 = 907 ℎ𝑝
© Copyright 2014
ELO 7.5
Operator Generic Fundamentals
255
Calculating Power
Knowledge Check
A man constantly exerts 50.0 lbf to move a crate 50 feet across a level
floor. The move requires 10 seconds. What is the power expended by
the man pushing the crate?
A. 0.450 horsepower
B. 2.5 x 103 foot-pounds force
C. 4.50 horsepower
D. 2.5 x 103 watts
Correct answer is A.
© Copyright 2014
ELO 7.5
Operator Generic Fundamentals
256
TLO 7 Summary
• Energy is the ability to do work. The work done on an object is the
product of the force on the object and the distance the object moves
(in the direction of the force).
• Kinetic energy is the energy an object has, due to its motion.
• Potential energy is the energy an object has, due to its position.
• 𝑃𝐸 = 𝑚𝑔𝑧 or 𝑃𝐸 =
𝑚𝑔𝑧
𝑔
𝑐
• The law of conservation of energy: Energy cannot be created or
destroyed, only altered in form.
• Simplified energy balance:
• Power is the amount of energy used per unit time.
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑃𝑜𝑤𝑒𝑟 =
𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
© Copyright 2014
TLO 7
Operator Generic Fundamentals
257
TLO 7 Summary
Now that you have completed this lesson, you should be able to:
1. Describe the following terms:
a. Energy
b. Potential energy
c. Kinetic energy
d. Work
2. Calculate energy and work for a mechanical system.
3. State the First Law of Thermodynamics, "Conservation of Energy."
4. State the mathematical expression for power.
5. Calculate power in a mechanical system.
© Copyright 2014
TLO 7
Operator Generic Fundamentals
258
Module Summary
• The physics module covered the concepts of force, motion,
momentum, energy, work, and power
– These concepts will be used in future modules
• It is necessary for you to understand these concepts in order to
understand power plant equipment
• These concepts are also key to your ability to operate power-plant
equipment safely and efficiently
© Copyright 2014
Summary
Operator Generic Fundamentals
259
Module Summary
Now that you have completed this module, you should be able to:
1. Convert between units of measure associated with the English and
System Internationale (SI) measuring systems.
2. Describe vector quantities, and how they are represented.
3. Solve resultant vector problems.
4. Describe the measurement of force and its relationship to free body
diagrams.
5. Apply Newton's laws of motion to a body at rest.
6. Calculate the change in velocity when two objects collide, with
respect to the conservation of momentum.
7. Describe energy, work, and power in mechanical systems.
© Copyright 2014
Summary
Operator Generic Fundamentals