These notes are to supplement the reading of Chapter Two in the Holt Text Physics Text Book by Serway and Faughn.
Additionally the following are helpful.
A java animation for vector addition http://mcasco.com/p1agva.html
A java animation for vector addition http://www.walter-fendt.de/ph14e/resultant.htm
Video on vector components http://www.youtube.com/watch?v=eMVzfoA9dnA&feature=youtube_gdata
Video on inclined plane
http://www.youtube.com/watch?v=VU0lr5KQ79s&feature=SeriesPlayList&p=AD5B880806EBE0A4
VA Sols
PH.1
The student will plan and conduct investigations in which
a) the components of a system are defined;
b) instruments are selected and used to extend observations and measurements of mass, volume, temperature,
heat exchange, energy transformations, motion, fields, and electric charge;
c) information is recorded and presented in an organized format;
d) metric units are used in all measurements and calculations;
e) the limitations of the experimental apparatus and design are recognized;
f) the limitations of measured quantities are recognized through the appropriate use of significant figures or
error ranges;
g) data gathered from non-SI instruments are incorporated through appropriate conversions; and
h) appropriate technology including computers, graphing calculators, and probeware, is used for gathering and
analyzing data and communicating results.
PH.5
The student will investigate and understand the interrelationships among mass, distance, force, and time through
mathematical and experimental processes. Key concepts include
a) linear motion;
d) Newton’s laws of motion;
Vectors
Vectors are numbers with a direction. They are represented with an arrow.
The arrow has two parts a, the front is called the head and the back is called the
tail.
The length of the vector is its magnitude (how big the number is) and it points in
the direction of the vector.
Acceleration, velocity, displacement, and force are all vectors.
3 ways to name a vector
The first way it to use degree notation. Since this is 35 deg less than the
90 deg. We can name this vector
40N @ 55⁰.
The next two way involve treating the axis like a compass rose. We can
name the vector by how far away it is from an axis.
So this becomes
40N @35⁰ E of N
Because the vector is at the location, where if you start at North and go
35 deg to the East of it. Furthermore We can also find the angle from
East. There are 90 degrees between East and North. Since the vector is
35 degrees to the East of North we can subtract from 90 degrees and we
get 55 degrees. So we can name the vector
40N at 55⁰ N of E.
Another Example:
50 m/s @ 25⁰
50 m/s @ 25⁰ N of E
50 m/s @ 65⁰ E of N
45 m/s @ 240⁰
45 m/s @ 30⁰ W of S
45 m/s @ 60⁰ S of W
Diagrams on this page are not to scale.
Vectors are always added head to tail.
The sum or two or more vectors is the resultant.
Ways to add vectors
1.
2.
3.
4.
5.
Graphical Method
Pythagorean theorem
Components
Matrix – not explained in these notes
Law of Sines and Cosines - not explained in these notes
Using the Graphical Method
With the graphical method you use a ruler and protractor. Start with your principle axis and making a scale, and
then graph the vectors head to tail.
For example: Find the resultant velocity of a bird flying 9.0m/s @ 30⁰N of E in a wind that is blowing 2m/s @ 20⁰N of W.
First, we draw our principle axis.
Then we make a scale. For our purpose we will use
2m/s =1cm
I will then use a protractor to find 30deg North of East
and put a mark there. (Highlighted for emphasis)
Now I will use my scale to determine how far to draw
the line. 9.0m/s (
1𝑐𝑚
2𝑚/𝑠
)= 4.5 cm
Next, we use a ruler to draw a 4.5cm line toward the 30
degree mark.
Now, at the head of the first vector, we make a
secondary axis and use the protractor and mark 20
degrees N of W. (Highlighted here.)
1𝑐𝑚
The next vector is drawn in that direction, 2m/s (2𝑚/𝑠)=
1cm long
Then the resultant, shown in green here, is measured
and the angle found. (I got 3.9 cm @ 42⁰ N of E)
Using our scale this becomes
3.9 cm(
2𝑚/𝑠
1𝑐𝑚
)=7.8m/s @ 42⁰
Using the Pythagorean Theorem
Vectors that are at right angles from each other can be added using the
Pythagorean Theorem. Since you are using math to determine the
length of the vectors, all of the rest of the images will not be to scale.
For example: A persons walks so they have a displacement 30m due N
then they turn and walk again so they have a displacement of 20 m due
East, what is their total displacement for both time periods.
Start by free handing the vectors on the principle axis head to tail.
Next I sketch in the resultant (in green) and use the
Pythagorean Theorem to calculate the length.
A2 +B2 = C2
(30m)2 + (20m)2 =C2
C=√900 + 400
C=36m
Now we need to find the angle Ѳ. To do this we use SOHCAH-TOA
TOA means Tan Ѳ = opp/adj side on a triangle. For this situation we get
Tan Ѳ =
20 𝑚
30 𝑚
20 𝑚
)
30 𝑚
Ѳ = Tan – (
(Note: Your calculator must be in deg mode.)
Ѳ = 33.7 degrees
So our answer is 36m @ 33.7⁰ E of N or 36m at 54.3⁰ N of E
Using vector Components
Components are used when you have most of your vectors will add like a 1-D problem. You can break a vector into
pieces (components) that go along the axis and simple add and subtract them.
Example: A twenty kg box is pulled with 100 N@ a 30 degree angle from the horizontal across a floor, which has a
coefficient of friction of .45 with the box. What will the acceleration be?
First, make a sketch with a free body
diagram.
Now, we can start putting the
numbers into our diagram. Force of
weight is just mg.
The force applied was given,
however, we need to find the
component of this vector that are
along the x and y axis.
We will use SOH-AH-TOA for this.
Once the Fax and Fay are calculated, we will work on the Fn and Ff. Remember that Fn is the force pushing
perpendicularly out of the surface. Therefore we will calculate it by subtracting Fay from the absolute value of the force
of weight.
Now we are ready to solve for the acceleration using Newton’s 2nd Law. Bear in mind, Fn + Fay + Fw cancel each other
out. All of the movement will be horizontal, no movement will be vertical.
∑F=ma
Ff + Fax = ma
-65.7 N + 86.6 N = 20kg (a)
a = 1.05 m/s2
It is interesting to note here. That because the Fa is pulling up as well as forward that force normal is reduced. This is
important, because if force normal is reduced so is friction. What this means is for items with a lot of friction, it is easier
to pull at an angle than straight across.
Using vector Components On an Incline
When you encounter inclined plane problems it is often easiest to solve the problem, by turning your axis along the
inclined plane, and using vector components to deal with the force of weight.
Normal Axis
Tilted Axis
An Example.
A .5kg box is on a inclined plane that has been tilted at 15 degrees above the horizontal. If the coefficient of friction is
.1, what will the acceleration be?
We start with a sketch and a free body diagram.
Next we break the Fw into its components that are
parallel (||) and perpendicular (┴) to the inclined
plane.
Then we put the numbers into the diagram.
First calculate the force of weight.
Next use SOH-CAH-TOA to calculate F||
and F ┴. Bear in mind that F┴ is
negative and F|| is positive.
Force normal is the opposite of F┴ here.
Force friction is also in the negative dir.
Finally we can calculate the
acceleration by using Newton’s 2nd
law.
Bear in mind Fw is not used and its components F|| and F┴ are used in its place. F┴ and Fn cancel each
other out.
∑F=ma
Ff +F|| =ma
-0.473 N + 1.27 N = (.5kg)(a)
a=1.59m/s2