Unformatted text preview: Physics for Scientists and
Engineers I
PHY 2048, Section 4 Dr. Beatriz Roldán Cuenya University of Central Florida, Physics Department, Orlando, FL Chapter 0  Introduction
I. General II. International System of Units
III. Conversion of units
IV. Dimensional Analysis
V. Problem Solving Strategies 1 I. Objectives of Physics
 Find the limited number of fundamental laws that govern natural
phenomena.
 Use these laws to develop theories that can predict the results of future
experiments.
Express the laws in the language of mathematics.
 Physics is divided into six major areas:
1. Classical Mechanics (PHY2048)
2. Relativity
3. Thermodynamics
4. Electromagnetism (PHY2049)
5. Optics (PHY2049)
6. Quantum Mechanics II. International System of Units
POWER PREFIX ABBREVIATION 15 peta P 1012 tera T 109 giga G mega M kilo k 10
QUANTITY UNIT NAME UNIT SYMBOL Length meter m Time second s Mass kilogram kg Speed 10 6 103 m/s hecto h m/s 101 deka da N 101 deci D 102 centi c 103 milli m micro µ nano n pico p femto f 2 Acceleration
Force 2 Newton Pressure Pascal Pa = N/m2 Energy Joule 10 J = Nm
10 Power Watt 6 W = J/s
109 Temperature Kelvin K 10 12 1015 2 III. Conversion of units Chainlink conversion method: The original data are multiplied successively
by conversion factors written as unity. Units can be treated like algebraic
quantities that can cancel each other out.
Example: 316 feet/h m/s feet 1 h 1 m ⋅ = 0.027 m/ s
316
⋅ h 3600s 3.28 feet IV. Dimensional Analysis
Dimension of a quantity: indicates the type of quantity it is; length [L],
mass [M], time [T]
Dimensional consistency: both sides of the equation must have the same
dimensions.
Example: x=x0+v0t+at2/2 [L] = [L ] + [L] [T ] + [L2] [T 2 ] = [L] + [L] + [L ]
[T ]
[T ] Note: There are no dimensions for the constant (1/2) Significant figure one that is reliably known. Zeros may or may not be significant:
 Those used to position the decimal point are not significant.
 To remove ambiguity, use scientific notation.
Ex: 2.56 m/s has 3 significant figures, 2 decimal places.
0.000256 m/s has 3 significant figures and 6 decimal places.
10.0 m has 3 significant figures.
1500 m is ambiguous
1.5 x 103 (2 figures), 1.50 x 103 (3 fig.),
1.500 x 103 (4 figs.) Order of magnitude the power of 10 that applies. 3 V. Problem solving tactics
• Explain the problem with your own words.
• Make a good picture describing the problem.
• Write down the given data with their units. Convert all data into S.I. system.
• Identify the unknowns.
• Find the connections between the unknowns and the data.
• Write the physical equations that can be applied to the problem.
• Solve those equations.
• Always include units for every quantity. Carry the units through the entire
calculation.
• Check if the values obtained are reasonable
units. order of magnitude and Chapter 1  Vectors
I. Definition II. Arithmetic operations involving vectors
A) Addition and subtraction
 Graphical method
 Analytical method Vector components B) Multiplication 4 Review of angle reference system
90º
0º<θ1<90º
90º<θ2<180º
θ2 θ1 180º
θ3
180º<θ3<270º 0º Origin of angle reference system θ4
270º<θ4<360º 270º
Angle origin Θ4=300º=60º I. Definition
Vector quantity: quantity with a magnitude and a direction. It can be
represented by a vector.
Examples: displacement, velocity, acceleration. Same displacement
Displacement does not describe the object’s path. Scalar quantity: quantity with magnitude, no direction.
Examples: temperature, pressure 5 II. Arithmetic operations involving vectors
Vector addition: s = a +b b
 Geometrical method a
s = a +b Rules:
a +b =b +a (commutative law) (a + b ) + c = a + (b + c ) Vector subtraction: (3.1) (associative law) (3.2) d = a − b = a + ( −b ) (3.3) Vector component: projection of the vector on an axis.
a x = a cos
(3.4)
a y = a sin θ
2
a = ax + a 2
y Scalar components of a Vector magnitude
(3.5) tan θ = ay Vector direction ax 6 Unit vector: Vector with magnitude 1.
No dimensions, no units.
ˆjˆ
i , ˆ, k → unit vectors in positive direction of x, y, z axes ˆ
a = axi + a y ˆ
j (3.6) Vector component Vector addition:
 Analytical method: adding vectors by components.
ˆ
r = a + b = (a x + bx )i + (a y + by ) ˆ
j (3.7) Vectors & Physics:
The relationships among vectors do not depend on the location of the origin of
the coordinate system or on the orientation of the axes.
 The laws of physics are independent of the choice of coordinate system.
2
a = a x + a 2 = a '2 + a '2
y
x
y (3.8) θ = θ '+φ Multiplying vectors:
 Vector by a scalar: f = s⋅a  Vector by a vector:
Scalar product = scalar quantity
(dot product) a ⋅ b = ab cos φ = a x bx + a y by + a z bz (3.9) 7 Rule: a ⋅b = b ⋅ a a ⋅ b = ab ← cos φ = 1 (φ = 0 ) (3.10) a ⋅ b = 0 ← cos φ = 0 (φ = 90 )
i ⋅ i = j ⋅ j = k ⋅ k = 1⋅1 ⋅ cos 0 = 1
i ⋅ j = j ⋅ i = i ⋅ k = k ⋅ i = j ⋅ k = k ⋅ j = 1 ⋅1 ⋅ cos 90 = 0 cos ϕ = Angle between two vectors: a ⋅b
a ⋅b Multiplying vectors:
 Vector by a vector
Vector product = vector (cross product) ˆ
ˆ
a × b = c = (a y bz − by a z )i − (bz a x − az bx ) ˆ + (a x by − bx a y )k
j
c = ab sin φ a × b = 0 ← sin φ = 0 (φ = 0 )
a × b = ab ← sin φ = 1 (φ = 90 ) Magnitude Vector product
Direction Rule: right hand rule b × a = − (a × b ) (3.12) c perpendicular to plane containing a , b 1) Place a and b tail to tail without altering their orientations.
2) c will be along a line perpendicular to the plane that contains a and b
where they meet.
3) Sweep a into b through the smallest angle between them. 8 Righthanded coordinate system
z
k
i j y x Lefthanded coordinate system
z
k
j i x y i ×i = j × j = k ×k = 0 i × i = j × j = k × k = 1⋅1⋅ sin 0 = 0 i × j = −( j × i ) = k
j × k = − (k × j ) = i
k × i = − (i × k ) = j 9 P1: If B is added to C = 3i + 4j, the result is a vector in the positive direction of the y axis, with a
ˆˆ
magnitude equal to that of C. What is the magnitude of B?
Method 2
Method 1
Isosceles triangle
B + C = B + (3iˆ + 4 ˆ) = D = Dˆ
j
j θ C = D = 32 + 4 2 = 5 C tan θ = (3 / 4) → θ = 36.9 D ˆ
ˆj
B + (3i + 4 ˆ) = 5 ˆ → B = −3i + ˆ → B = 9 + 1 = 3.2
j
j θ B / 2
θ sin =
→ B = 2 D sin = 3.2
D 2 2 B/2 B P2: A fire ant goes through three displacements along level ground: d1 for 0.4m SW, d2 0.5m E, d3=0.6m at
60º North of East. Let the positive x direction be East and the positive y direction be North. (a) What are the
x and y components of d1, d2 and d3? (b) What are the x and the y components, the magnitude and the direction
of the ant’s net displacement? (c) If the ant is to return directly to the starting point, how far and in what direction
should it move?
(a)
d1x = −0.4 cos 45 = −0.28m N
D d1 y = −0.4 sin 45 = −0.28m E d 2 x = 0.5m 45º d4 d3 D = 0.52 2 + 0.242 = 0.57 m 0.24 = 24.8 0.52 North of East d3 x = 0.6 cos 60 = 0.30m d2 P2 ˆ
D = d 4 + d 3 = (0.22iˆ − 0.28 ˆ) + (0.3i + 0.52 ˆ) = (0.52iˆ + 0.24 ˆ)m
j
j
j θ = tan −1 d2 y = 0 d1 (b)
d 4 = d1 + d 2 = (−0.28iˆ − 0.28 ˆ) + 0.5iˆ = (0.22iˆ − 0.28 ˆ)m
j
j d3 y = 0.6 sin 60 = 0.52m ˆ
d1 = 4iˆ + 5 ˆ − 6k
j
ˆ
ˆ
d = −i + 2 ˆ + 3k
j
2 ˆ
ˆ
d 3 = 4i + 3 ˆ + 2 k
j (c) Return vector
negative of net displacement,
D=0.57m, directed 25º South of West ( a ) r = d1 − d 2 + d 3 ?
(b) Angle between r and + z ?
(c) Component of d1 along d 2 ?
(d ) Component of d1 perpendicular to d 2 and in plane of d1, d 2 ? ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(a) r = d1 − d 2 + d 3 = (4i + 5 ˆ − 6k ) − (−iˆ + 2 ˆ + 3k ) + (4i + 3 ˆ + 2k ) = 9iˆ + 6 ˆ − 7 k
j
j
j
j −7 ˆ
(b) r ⋅ k = r ⋅1 ⋅ cosθ = −7 → θ = cos −1 = 123 12.88 d1perp r = 9 2 + 6 2 + 7 2 = 12.88m (c) d1 ⋅ d 2 = −4 + 10 − 18 = −12 = d1d 2 cos θ → cos θ =
d1// d1 ⋅ d 2
d1d 2 d1
θ d1//
d2 d ⋅d
− 12
= d1 cos θ = d1 1 2 =
= −3.2m
d1d 2
3.74 d 2 = 12 + 22 + 32 = 3.74m
(d ) d1 = d12// + d12perp → d1 perp = 8.77 2 − 3.2 2 = 8.16m
d1 = 4 2 + 52 + 6 2 = 8.77 m P3 ˆ
ˆ
If d1 = 3i − 2 ˆ + 4k
j (d1 + d 2 ) ⋅ (d1 × 4d 2 ) ? ˆ
d 2 = −5i + 2 ˆ − k
jˆ (d1 + d 2 ) = a → contained in (d1 , d 2 ) plane
(d1 × 4d 2 ) = 4(d1 × d 2 ) = 4b → perpendicular to (d1, d 2 ) plane
a perpendicular to b → cos 90 = 0 → 4a ⋅ b = 0 Tip: Think before calculate !!! 10 P4: Vectors A and B lie in an xy plane. A has a magnitude 8.00 and angle 130º; B has
components Bx= 7.72, By= 9.20. What are the angles between the negative direction of
ˆ
the y axis and (a) the direction of A, (b) the direction of AxB, (c) the direction of Ax(B+3k)?
y A
130º (a) Angle between − y and A = 90 + 50 = 140 x (b) Angle − y , ( A × B ) = C → angle − ˆ , k because C perpendicular
jˆ B plane ( A, B ) = ( xy ) → 90 ˆ
(c) Direction A × ( B + 3k ) = D
ˆ
ˆ
E = B + 3k = −7.72iˆ − 9.2 ˆ + 3k
j
ˆ
ˆ
iˆ
j
k
ˆ
D = A × E = − 5.14 6.13 0 = 18.39iˆ + 15.42 ˆ + 94.61k
j
− 7.72 − 9.20 3 D = 18.392 + 15.42 2 + 94.612 = 97.61
ˆ
ˆ
j
j
j
− ˆ ⋅ D = − ˆ ⋅ (18.39i + 15.42 ˆ + 94.61k ) = −15.42 − ˆ ⋅ D − 15.42 j
cos θ = =
→ θ = 99 1 ⋅ D 97.61 P5: A wheel with a radius of 45 cm rolls without sleeping along a horizontal floor. At time t1 the dot P painted
on the rim of the wheel is at the point of contact between the wheel and the floor. At a later time t2, the
wheel has rolled through onehalf of a revolution. What are (a) the magnitude and (b) the angle (relative
to the floor) of the displacement P during this interval?
y
Vertical displacement: 2 R = 0.9m
Horizontal displacement: 1
(2πR ) = 1.41m
2 d ˆ
r = (1.41m)i + (0.9m) ˆ
j
r = 1.412 + 0.9 2 = 1.68m x 2R tan θ = → θ = 32.5 πR P6: Vector a has a magnitude of 5.0 m and is directed East. Vector b has a magnitude of 4.0 m and is directed
35º West of North. What are (a) the magnitude and direction of (a+b)?. (b) What are the magnitude and
direction of (ba)?. (c) Draw a vector diagram for each combination. a b a = 5iˆ
ˆ
b = −4 sin 35 i + 4 cos 35 ˆ = −2.29iˆ + 3.28 ˆ
j
j N
125º ba
a W a+b ˆ
(b) b − a = b + (− a ) = −7.29i + 3.28 ˆ
j (a) a + b = 2.71iˆ + 3.28 ˆ
j
a + b = 2.71 + 3.28 = 4.25m b − a = 7.29 2 + 3.28 2 = 8m 3.28 tan θ = → θ = 50.43 2.71 3.28 tan θ = − → θ = −24.2 7.29 2 E 2 or 180 + (−24.2 ) = 155.8 S 180 − 155.8 = 24.2 North of West 11 ...
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This note was uploaded on 07/30/2011 for the course PHY 2049 taught by Professor Saha during the Spring '08 term at University of Central Florida.
 Spring '08
 SAHA
 Physics

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