PHYS2001 – Physics 1
1.1 The Nature of Physics
From the very big to the very small, Physics covers it all!
This semester we will focus on Classical Mechanics – big
things moving slowly (wrt the speed of light).
Next semester you will study Electricity and Magnetism and
Quantum Mechanics.
Physics is the most fundamental of all the natural sciences.
In science, there is only physics. All the rest is
stamp collecting.
- Earnest Rutherford
The strength of Physics as a science is that all of the physical
laws are based on experimental results.
So why should you be required to take Physics???
Physics has practical applications, but that is not the main
reason you do it.
- Richard Feynman
Our goal is to describe and explain how
and why the world works the way it does.
Through this process you will become analytical thinkers and effective
problem solvers. Physics is the perfect vehicle for doing this. These are
the skills you should take with you when you leave this course.
What do we get with this knowledge???
All technology has its roots in Physics!
Lasers
Rockets
Transistors
Telecommunications
Etc.
1.2 Units
Accurate measurements require precisely defined quantitative units.
We are going to want to measure things like length, mass, and time.
There are several different systems of units used for measurement:
SI – System International or MKS units (meter, kilogram, second)
CGS (centimeter, gram, second)
BE – British Engineering units (ft, slug, second)
In this course we will use SI (MKS) units.
Length = Meter (m) = Distance light travels in a vacuum in 3.3 ns (3.3 × 10-9 s)
Mass = Kilogram (kg) (1 kg ≡ 2.2 lbs. on the Earth’s surface)
Time = Second (s) = The time it takes for 9,192,631,770 vibrations of a Cs-133
atom to occur
1.3 Converting Between Different Units and Base Units
1 km = 1000 m = 1 × 103 m
5.2 cm = 0.052 m = 5.2 × 10-2 m
If you need to brush up on powers of 10 or scientific notation, see Appendix A.
To Do Unit Conversions:
- Treat the units as algebraic quantities
- Multiply by factors of “1”
Example
- Use the “mol method”
dead
My cat, Socola, is fat. He has a 17 in. waist. What is Socola’s waist
size in m?
17 in. 2.54 cm
1 in.
1m
100 cm
Factors of 1
=
(17)(2.54)
100
= 0.43 m
All of your answers must have
appropriate units!!!
Mars Climate Orbiter (1999)!!!
Significant Digits
Every quantity (number) has a certain number of significant digits.
These are digits in the number whose values are known with certainty.
For example, Socola is 27.25 cm tall, with the measurement uncertainty
occurring in the 3rd decimal place.
Thus, all of these digits are significant (known with certainty), and 27.25 cm
has 4 significant digits.
• If a zero is given as the last digit to the right of the decimal
point, it is also significant.
Example: 7.290 m has 4 significant digits.
• However, zeros immediately to the left of an unexpressed
decimal point are not significant.
Example: 1500 m only has 2 significant digits.
• Finally, zeros located between significant digits are
significant.
Example: 1502 m has 4 significant digits.
Example: How many significant digits are there in 150,200.0?
There are 7!
Why do significant digits matter???
Let’s say Socola’s top speed is 7.0 mi/h, and I want to calculate his speed in
m/s. Thus, I need to convert mi to m and h to s.
7.0 mi. 5280 ft.
1h
=
1 mi.
(7.0)(5280)(12)(2.54)
(100)(60)(60)
12 in.
2.54 cm
1 ft.
1 in.
1m
1h
1 min.
100 cm
60 min.
60 s
= 3.12928 m/s
Does it make sense to represent his speed with this accuracy???
No!!!
We only know his speed, 7.0 mi/h, to an accuracy of 2 significant digits.
Thus, 3.12928 m/s → 3.1 m/s
In general, when numbers are multiplied or divided, the number of significant
digits in the answer should be equal to the smallest number of significant digits
in any of the original factors.
Example: Convert 3.141592 m to in.
3.141592 m
100 cm
1 in.
1m
2.54 cm
=
(3.141592)(100)
2.54
= 123.684724409 in.
How many significant digits should this final answer have?? 1? 3? 7?
It should have 7!
123.6847 in.
The 100 cm and the 2.54 cm, etc. are conversion factors. They
are known precisely!
See Appendix B of your text for more on
Significant Digits.
1.4 Trigonometry
Take a look at a right triangle.
We can label the sides relative to the angle q.
opp.
a
hyp.
c
q
b
adj.
sin q =
a
opposite
= c
hypotenuse
cos q =
b
adjacent
= c
hypotenuse
tan q =
opposite
adjacent
a
= b
These Trigonometric Functions are unitless, since they represent the
ratio of two lengths (sides of the triangle).
Example: Michael Jordan casts a shadow of length 1.39 m. If the angle between
the sun’s rays and the ground is 55o, what is Michael’s height, h?
è
h
55o
55o
1.39 m
1.39 m
tan q =
opposite
adjacent
=
h
h
1.39 m
è h = (1.39 m)(tan 55o) = 1.98 m
(6’ 6”)
Other Trig. Stuff
Also for right triangles, we will use the Pythagorean Theorem:
a2 + b2 = c2
c
a
b
For all triangles, we can use the following:
Law of Sines
a
b
=
=
sin a
sin b
c
sin g
c
a
Law of Cosines
c2 = a2 + b2 -2ab∙cosg
a
b
g
b
1.5 Vectors and Scalars
Scalars are quantities (including units) that give the size or magnitude of
something.
Examples:
Mass = 45 kg
Length = 16.2 m
Speed = 15.0 m/s
Vectors are objects that require both magnitude AND direction to completely
describe them.
Examples:
Displacement = 10 m, 20o N of E
Velocity = 25 m/s, 35o above the horizontal
Vectors are usually represented as arrows. The arrow points along the
direction of the vector, and the length of the vector arrow is proportional to
its magnitude.
Remember → Vectors have direction, scalars do not!
In the book, the vectors are represented in bold face with an arrow over
them. In class and on homework, quizzes, etc., you should represent them
with an arrow above the quantity.
→
For example, a tension force: T
→
Example: Represent a position vector r that is 10 m in length and
directed at 20o N of E.
N
→
r
20o
W
m
10
E
Vectors are extremely important!!! They occur
everywhere in physics. Many objects are in
motion, moving in all directions!!!
S
1.6 Vector Addition and Subtraction
The simplest situation occurs when the vectors are collinear, i.e. they lie
along the same direction.
Example: A car moves due east with a displacement vector A = 275 m. It then
continues to move due east with
→ a displacement vector B = 125 m. What is
the total displacement vector, R, of the car?
→ → →
What we want to find is R = A + B.
N
A = 275 m
B = 125 m
E
R = 400 m
→
R = 275 m due east + 125 m due east = 400 m due east
When adding two vectors, always place the tail of the second vector on the
head of the first.
What if the two vectors to be added are not collinear???
Example: A car first moves with a displacement vector A = 275 m due
o N of
east. It then moves with a displacement vector B = 125 m
at
55
→ → →
W. What is the total displacement of the car? i.e. what is R = A + B.
N
Since these vectors are head to tail, the resultant
vector is drawn from the origin to the head of the
second vector.
→
First, how do I find the magnitude of R, i.e. its
length?
→
R
A = 275 m
I can’t use Pythagorean Thm., since I don’t have a
right triangle.
But we do know the lengths of
two sides and the angle
opposite the unknown side →
B = 125 m
55o
We use Law of Cosines!
E
R 2 = A 2 + B2 – 2ABcos55o ⇒ R = 2752 + 1252 – 2(275)(125)cos55o
R = 228 m
We are only half done, since R is a vector, we need both
magnitude and direction to completely describe it.
So how do we represent its direction???
N
So now I know the lengths of all three sides and
one angle, so I can use Law of Sines to find θ.
R
m
8
2
=2
θ
A = 275 m
B = 125 m
55o
E
o
R
B
⎡
⎤
B sin 55o
125
sin
55
−1
o
=
⇒
sin
θ
=
⇒
θ
=
sin
=
26
.
7
⎢
⎥
sin 55o sin θ
R
228
⎣
⎦
→
So, to describe R, we would say the following:
→
R = 228 m, directed at 26.7o N of E.
Vector Subtraction
→
→
Let’s say I have two vectors A and B, and I know the
→
→ → →
resultant vector C, such that C = A + B.
We can use the following property of vectors: When a vector
is multiplied by (-1), its magnitude remains unchanged, but its
direction is reversed.
→ → →
C=A+B
Now let’s solve the above vector
→
equation for A:
C
B
A
→ → →
A=C-B
→
→
Vector (-B) is just vector B with its
direction reversed!
And now we recognize that vector
subtraction is just vector addition:
C
A
(-B)
→ → →
A = C + (-B)
→
→
Vector C and vector (-B) are now head
→
to tail, so A is the resultant vector.
Many times vectors won’t be drawn head to tail, but will start from the same
point.
This occurs quite often in problems involving multiple forces.
→
→
→ →
Let’s say we have vectors A and B as shown. What is A + B?
Since they both begin from the
same point, we can use the
parallelogram method.
y
First, complete the parallelogram
formed by the two vectors.
→
B
ltant
u
s
e
R
→
A
x
The resultant vector is then
drawn in from the origin of the
two vectors to the opposite
vertex of the parallelogram.
1.7 Vector Components
A vector can be “broken down” into components which add vectorially to
equal the original vector.
→
Let’s say that I have some vector A in the xy-plane:
→
We can break A down into its x- and y-components.
y
Properties of the components:
→
A
→
1. They add vectorially to equal A.
→
Ay
→
Ax
2. They are perpendicular to each other.
x
3. They run parallel to the x- and y-axis,
respectively.
Vectors vs. Scalar Components
The components of a vector are vectors themselves, i.e. they have both
magnitude and direction.
→
Let’s say that I have the following vector, A, in the xy-plane:
y
→
The components of A would look
like this:
→
Ax = 10 m
→
A
→
x
Ay = 15 m
Let’s say the magnitude (length) of
→
→
Ax is 10 m, and the magnitude of Ay
is 15 m.
Scalar components are scalars. They
are just positive or negative numbers.
So, for the above example, we would have the following:
Vector Components
Scalar Components
→
Ax = 10 m in the positive x-direction
Ax = 10 m
→
Ay = 15 m in the negative y-direction
Ay = -15 m
1.8 The Component Method of Vector Addition
Many times, using Law of Sines or Law of Cosines will work when doing
vector addition.
However, there are times when the geometry can get complicated, and a
better way presents itself.
This is the component method of vector addition. It works every time and
for any number of vectors.
→
→
Let’s say I have two vectors A and B, and I know the
→
→ → →
resultant vector C, such that C = A + B.
Let’s say the vectors are in the xy-plane and they look like the following:
y
→
C
→
B
→
A
x
y
→
→
Let’s go ahead and break A and B down into
their x- and y-components.
→
C
→
By
→
Cy
→
Bx
→
Ay
→
B
→
A
→
Ax
→
Cx
’s add the x- and y-components of
Now let→
vector C to the plot.
x
→ → →
→ → →
Now it’s easy to see that Ax + Bx = Cx and Ay + By = Cy
y
y
→
C
→
By
→
C
→
Cy
è
→
Ay
→
Ax
→
Bx
x
→
Cx
x
→
Once you have the two components of vector C, then how do you calculate its
magnitude???
Use the Pythagorean Theorem!!!
C2 = Cx2 + Cy2
Remember, this method works for any number of vectors, so in general, given n
vectors, the scalar components of the resultant vector Rx and Ry will be:
Rx = Ax + Bx +Cx + Dx + ∙ ∙ ∙ + nx
Ry = Ay + By +Cy + Dy + ∙ ∙ ∙ + ny
Once the scalar components Rx and Ry are calculated, then the magnitude
of R is found by Pythagorean Theorem.
R2 = Rx2 + Ry2
Example: Leaving a late movie at 2 AM on a Friday night, you can’t remember
where you parked your car. From the theater door you walk 30 m due north,
stop, and walk 10 m due east. Here you stop again and then walk 15 m at 30o
south of east. Finally you reach your car. What is the magnitude and direction of
the resultant vector (called your displacement vector)? Give the direction relative
to east.
y, north
The resultant vector, R, would
then be:
B = 10 m
30o
C
=1
5m
Notice that the following vector
equation would be correct:
Car
A = 30 m
R
→ → → →
R=A+B+C
x, east
Theater
So how do I find the magnitude
→
of R???
We can use the component method of vector addition!!!
y, north
B = 10 m
Cx
30o
C
=1
Cy
5m
A = 30 m
Let’s begin by breaking down the
vectors A, B, and C into their x- and
y-components.
A and B are very easy, since they
already line up along one of our
axes directions.
A is along the y-direction, so it has
no x-component, and Ay = A.
B is along the x-direction, so it has
no y-component, and Bx = B.
x, east
Theater
So the only vector we really need to break down is C.
Now we make a table of values showing the x- and y-components for each vector.
→
A
→
B
→
C
x-component
y-component
0
30 m
10 m
0
Ccos30o = 15cos30o = 13 m
-Csin30o = -15sin30o = -7.5 m
y, north
So, from the component method of vector
addition, the following must be true:
Cx
Bx = 10 m
C
30o
=1
5m
Cy
Rx = Ax + Bx + Cx
Ry = Ay + By + Cy
Rx = 0 + 10 m + 13 m = 23 m
Ay = 30 m
R
Ry
Ry = 30 m + 0 – 7.5 m = 22.5 m
R = R 2x + R 2y
θ
Rx
Theater
x, east
How about the direction of R?
R = (23) 2 + (22.5) 2 = 32.2 m
From the diagram tanθ = Ry/Rx.
So, θ = tan-1(22.5/23) = 44.4o
→
A
→
B
→
C
→
Finally, x-component
the complete description of the resultant
y-component
vector, R, is:
0
10 m
30 m
32.2 m directed at 44.4o N of E.
0
or
o
o = 13 m
= -15sin30
-7.5 m
Ccos30o =32.2
15cos30
o aboveo the
m directed
at 44.4-Csin30
positive=x-axis.