Unformatted text preview: PHYSICS FOR ENGINEERS 1 (PHY 2043) FALL 2011
Prerequisite course required MAC 2281 or MAC 2311
Instructor: Dr. Grigoriy Kreymerman COURSE TEXT BOOK:
PHYSICS FOR SCIENTIST AND ENGINEERS, Vol1, Sixth Edition
Authors: Paul A. Tipler and Gene Mosca
Publisher: W. H. Freeman and Company (http://whfreeman.com/tipler/)
The biggest part of this book is Mechanics fallowing by Oscillations Waves and Thermodynamics. There will be four tests and four 15 minutes quizzes.
First, second and third tests will be graded as 105 points max. The final cumulative test will be graded as 120 points max. The total points for four tests are 435 max. Quizzes each 21 points total 84 points max.
Homework assignments 60 points total max.Total course points 1
possible are 579. WELCOME TO PHYSICS!
The required textbook
for this course is PHYSICS
For Scientists
and Engineers
Sixth Edition
by Paul A. Tipler and Gene
Mosca
(©2008, W.H. Freeman & Company) 2 1CHAPTER
MEASUREMENT AND VECTORS
Units
Conversion of Units
Dimensions of Physical Quantities
Vectors
General Properties of Vectors 3 To express measurement quantitatively we need introduce units.
Physical unit is the define standard for measurement of physical
quantity or value.
For example Time _____ seconds, minutes, hours, days, ……
Distance_____
meters, kilometers, feet, yards, miles……. Mass_______ kilograms, grams, milligrams, pounds, slugs….
U.S. customary system inches, feet, yards, miles, pounds,….
Foot originally defined as the length of the king’s foot in France.
The basic standard, which use in physics is
The International System of Units
In 1960, an international committee established a set of standards for the scientific community called SI (for System International). 4 There are seven base quantities in the SI system. They are length, mass, time, electric current, thermodynamic temperature, amount of substance (mole), and luminous intensity. For now we define only three base units or fundamental quantities:
Time, Length, and Mass
Time  second (s) Historically it is (1/60)(1/60)(1/24) of the mean solar day Modern etalon of the time define by frequency of E.M. radiation emitted
from a certain level excitation in cesium (9192631770 cycles per second) 5 Lengthmeter (m) Historically length was define as 1/10,000000 10 − 7 or of the distance between the equator and North Pole. Standard 1m between two scratches on platinum –iridium alloy at constant temperature adapted 1889. Today length define by speed of light (299792458 m/s)in empty space. The meter is the distance which light travel
1/ 299792458 second. 6 Masskilogram (kg) Historically was defined as the mass of one liter of distillated water at 4ºC. The kilogram is now defined to be the mass of a specific platinumiridium alloy cylinder, which store in the International Bureau of Weights and Measures in Sèvres, France. 7 The classical example of importance of units is NASA fiasco with Mars probe in 1999. The Jet Propulsion Laboratory in California and Lockhead Martin in Colorado use different units of measurement. Units Prefixes 8 Conversion of Units
Example:
How many centimeters are there in one foot? How many centimeters are there in one mile?
We can use the facts that there are 2.540 centimeters in 1 inch and 12 inches in 1 foot to show that there are 30.48 cm per ft. We can then use the fact that there are 5280 feet in 1 mile to find the number of centimeters in one mile. Multiply 2.540 cm/in by 12 in/ft to find the number of cm per ft:
cm in 2.540 12 = 30.48 cm/ft in ft Multiply 30.48 cm/ft by 5280 ft/mi to find the number of centimeters in one mile: cm ft 30.48 5280 = 1.609 × 105 cm/mi ft mi 9 Dimensions of Physical Quantities We going use next notation : for distance = [L], for time = [T], mass = [M]
We can multiply or divide physical quantities (values)
in deferent units.
Example speed = [L] / [T]
When we adding or subtracting physical quantities (values)
dimensions of this quantities should be same.
Example total length = [L1] + [L2]
Dimensional analysis. Falling time as function of high and ……
10 Example:
Force has dimensions of mass times acceleration. Acceleration has dimensions of speed divided by time. Pressure is defined as force divided by area. What are the dimensions of pressure? Express pressure in terms of the SI base units kilogram, meter and second.
Use the definition of pressure and the dimensions of force and area to obtain: ML [F] = T 2 = M [ P] =
2 LT 2 [ A] [ L ] Express pressure in terms of the SI base units to obtain: m
kg ⋅ 2
N
s = kg
=
m2
m2
m ⋅ s2 11 What combination of force and one other physical quantity has the dimensions of power? Where power is [ P ] = [ F ] [ L] [T ] Let X represent the physical quantity of interest. Express the relationship of X to force and power dimensionally: [ F ] [ X ] = [ P]
Solve for [ X ] [ X ] = [ P]
[F] Substitute the dimensions of force and power and simplify to obtain: [ F ] ⋅ [ L]
[ X ] = [T ] = [ L ]
[ F ] [T ] Because the dimensions of velocity are L/T, we can conclude that:
[ P] = [ F ] [v]
12 True or false:
(a) Two quantities must have the same dimensions in order to be added.
(b) Two quantities must have the same dimensions in order to be multiplied.
(a) True. You cannot add ″apples to oranges″ or a length (distance traveled) to a volume (liters of milk).
(b) False. The distance traveled is the product of speed (length/time) multiplied by the time of travel (time). 13 Example:
In doing a calculation, you end up with m/s in the numerator and m/s^2 in the denominator. What are your final units? (a) m^2/s^3, (b) 1/s, (c) s^3/m^2, (d) s, (e) m/s. 14 Uncertainty of measurement Problem:
A circular hole of radius 8.470 × 10−1 cm must be cut into panel. The ± 1 ⋅ 10−3
tolerance of radius is cm. If the actual hole is larger than the desired radius by the allowed tolerance, what is the difference between the actual area and the desired area of the hole?
Solution: ( ΔA = πr02 − πr 2 = π r02 − r 2 ) ΔA = π( r0 − r ) ( r0 + r ) ( )[ ( ΔA = π 1.0 × 10 −3 cm 8.470 ×10 −1 cm + 8.470 × 10 −1 cm + 1.0 ×10 −3 cm
= 5.3 ×10 −3 cm 2 15 )] Order of Magnitude
Rounding number to the nearest power of 10 called an order of magnitude. 10 5 g
9 ⋅10 g is approximately 4 9 ⋅10 −5 g is approximately ? Problem:
Assume, percentage of Human Body that is Water Up to 60 percent. The mass of a water molecule is 30 × 10^27 kg. If the mass of a person is 60 kg, estimate the number of water molecules in that person.
N represent the number of water molecules in a person of mass m N=
N= mhuman body m water molecule × 0.6 60 kg
30 × 10− 27 kg
molecule × 0.6 = 1.2 × 1027 molecules 16 VECTORS
Vector define by magnitude (displacement or length) and direction. Equal Vectors 17 Addition and Subtraction of Vectors
HeadtoTail method addition. Vector sum, or resultant.
Parallelogram method addition Vector addition obeys the commutative law
18 → → → → → →
Vector addition is associative. A+ B + C = A+ B + C → →
→
→→
C = A+ B
A
B C = A+ B it is not equal to . Except when and are parallel. 19 Subtraction of Vectors 20 A vector points in the +x direction. Show graphically at least three A choices for a vector such that points in the + y direction.
B+ A
B 21 The vector equation that describes the relationship among vectors A, B and C is A.
B.
C.
D.
E. B =C + A B =C − A C = A− B A= B −C A= B +C
22 Component of Vector
Component of vector in define direction is the projection of the vector onto an axis in that direction. 23 The process of finding the x,y, and z component of vector is called resolving the vector We can find angle θ if we know Ax and Ay.
tanθ = Ay
Ax θ = arctan Ay
Ax 2
A = Ax + A 2
y In three dimensions.
2
A = Ax + A2 + Az2
y 24 → → → C = A+ B C X = AX + B X CY = AY + BY 25 Problem:
A vector A 7.00 units long and a vector B 5.50 units long are added. Their sum is a vector C 10.0 units long. Determine the angle between the original two vectors. 26 θ = 180° − α
Apply the law of cosines C 2 = A2 + B 2 − 2 AB cos α A2 + B 2 − C 2 α = cos −1 2 AB −1 7.00 α = cos = 105.6° 2 + 5.50 2 − 10.0 2 2( 7 ) ( 5.5) θ = 180° − 105.6° = 74° 27 Unit Vectors ˆ
A Unit Vector of vector is vector which is in same direction as A
A A
vector but without dimensions and with magnitude equal to 1. ˆ = A where A = A2 + A2 + A2
A
x
y
z
A ˆ
A = Axˆ + Ay ˆ + Az k
i
j ˆ
ˆ
A + B = Axˆ + Ay ˆ + Az k + Bxˆ + B y ˆ + Bz k =
i
j
i
j ( ( ) )( ) ˆ
= ( Ax + Bx )ˆ + Ay + B y ˆ + ( Az + Bz ) k
i
j 28 ˆ
ˆ + 4.7 ˆ
B = (−7.7)i + 3.2 ˆ
j
Given the following vectors: , and
A = 3.4i
j ˆ (a) Find the vector , in unit vector notation, D
C = 5.4i + (−9.1) ˆ
j such that D + 2 A − 3C + 4 B = 0 (b) Express your answer in Part (a) in terms of magnitude.
(c) Find unit vector for D vector.
(a) Solve the vector equation that gives the condition that must satisfied for D
D = −2 A + 3C − 4 B Substitute for A, C and B ( )( )( ˆ
ˆ
ˆ
D = −2 3.4i + 4.7 ˆ + 3 5.4i − 9.1 ˆ − 4 − 7.7 i + 3.2 ˆ
j
j
j
ˆ
= ( − 6.8 + 16.2 + 30.8) i + ( − 9.4 − 27.3 − 12.8) ˆ
j
ˆ
= 40.2 i − 49.5 ˆ
j (b) Use the Pythagorean Theorem 2
D = Dx + D 2
y (c ) D= ( 40.2) 2 + ( − 49.5) 2 ˆ
ˆ
ˆ
ˆ = D = 40.2i − 49.5 j ≈ 0.63i − 0.78 ˆ
D
j
D
63.8 = 63.8 29 ) 30 ...
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 Physics, Sixth Edition, θ, Paul A. Tipler

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