Figure 4.2: An example of two vectors being added to give a resultant
4.7.2
Vector addition using components
In Fig 4.3 two vectors are added in a slightly different way to the methods discussed so fa
Image - located at infinity Ob ject
between f and P diag
Image - located behind the mirror - erect (right side up) - magnified (increased in size, larger)
- virtual
3.5.2
Convex Mirrors
diag base ray
8
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
8
Using our known trigonometric ratios we can calculate the value of ;
6 tan =
8
= arctan
= 36.8o .
Step 6 : Quote the final answer
Our final answer is a resultant of 1
Step 4 : Apply the scale conversion
We now use the scale to convert the length of the resultant in the scale diagram to the actual
displacement in the problem. Since we have chosen a scale of 1cm = 1k
In a rough sketch one should include all of the information given in the problem. All of the
magnitudes of the displacements are shown and a compass has been included as a reference
direction.
Step 2
The parallelogram method is restricted to the addition of just two vectors. However, it is arguably the
most intuitive way of adding two forces acting at a point.
4.6.2
Algebraic Addition and Subtract
+ =
=
Now you have discovered one use for vectors; describing resultant displacement how far and in what
direction you have travelled after a series of movements.
Although vector addition here has bee
We know which way the man is running around the track and we know his
speed. His velocity at point B will be his speed (the magnitude of the
velocity) plus his direction of motion (the direction of hi
Step 3 : Choose a positive direction
Lets make to the right the positive direction. This means that to the left becomes the
negative direction.
Step 4 : Now define our vectors algebraically
With right
sin
8
0
=
sin 135
18.5
o
= 8 sin 135
18.5
= arcsin(0.3058)
= 17.8o
sin
Thus, F A C = 62.8o .
Step 4 : Quote the resultant
Our final answer is then:
Resultant Displacement: 18.5km on a bearing of 62.
Finish
(Shop)
Start
(House)
Figure 4.1: Illustration of Displacement
OR
Definition: Displacement is a vector with direction
pointing from some initial (starting) point to some final (end) point and wh
Step 2 : Choose a suitable scale
In this problem a scale of 1cm = 0.5N would be appropriate, since then the vector diagram
would take up a reasonable fraction of the page. We can now begin the accurat
Step 4 : Now define our vectors algebraically
With right positive:
v initial
=
+3m.s1
and
v f inal
=
2m.s1
Step 5 : Subtract the vectors
Thus, the change in velocity of the ball is:
v
= (2m.s1 ) (+3m.
N
W
E
S
40m
Step 2 : Determine the length of the resultant
Note that the triangle formed by his separate displacement vectors and his resultant
displacement vector is a right-angle triangle.
We can th
and east). Algebraic techniques, however, are not limited to cases where the vectors to be combined
are along the same straight line or at right angles to one another. The following example illustrate
Note that the angle of incidence i is between the incident ray and the normal 2 to the
not between the incident ray and the surface of the mirror.
There are two forms of an image formed by reflection:
a from
b gives a new vector
c:
c = b
a
=
b + (
a)
This clearly shows that subtracting vector
a from
b is the same as adding (
a ) to
b
In mathematical form, subtracting
.
Look at the following examp
Take the next vector and draw it as an arrow starting from the arrowhead of the first vector in the
correct direction and of the correct length.
Continue until you have drawn each vector each time s
(NOTE TO SELF: This is actually average velocity. For instantaneous s change to dif- ferentials.
Explain that if is large then we have average velocity else for infinitesimal time interval instantaneo
and still function. However since total internal reflection only occurs when light is going from a medium to a
less dense medium, it is necessary ti coat each fibre with glass of a lower refractive in
4
3
2
1
0
-1
-2
-3
-4
0
1
2
3
4
5
d-5iag
de-fi4ning -a3ngles -i2, r
an-d1 N
Light is refracted according to the laws of refraction:
3.5.4
Laws of Refraction
Laws of Refraction:
sin i /sin r is a const
3.3
Introduction
Light is at first, something we feel incredibly familiar with. It can make us feel warm, it allows us to see,
allows mirrors and lenses to work, allows for .
Under more careful study
Let us test the first one. It says one step forward and then another step forward is the same as an arrow
twice as long two steps forward.
It is possible that you end up back where you started. In thi
Step 2 : Now we determine speed from the distance and time.
We know that speed is distance covered per unit time. So if we divide the distance
covered by the time it took we will know how much distanc
We now know the lengths of the sides of the triangle for which our vector is the hypotenuse.
If you look at these sides we can assign them directions given by the dotted arrows. Then our
original red
and
sE
=
250 cos 30o
= 216.5 km
and sE are the magnitudes of the components they are in the directions north and east respectively.
Remember sN
(NOTE TO SELF: SW: alternatively these results can be ar
3.2.4
Compound Microscope
This type of microscope uses two convex lenses. The first creates a real magnified image of the object that
is in turn used by the second lens to create the final image. This
tangent can be used as an illustration. Else defer until chapter on Graphs and Equations of Motion.
Instantaneous velocity: reading on the speedometer in a direction tangent to the path. Instantaneous
Figure 4.3: Components of vectors can be added as well as the vectors themselves
Step 2 : Resolve the red vector into components
Let us start with the bottom vector. If you are told that this vector h
The Parallelogram Method
When needing to find the resultant of two vectors another graphical technique can be applied- the
parallelogram method. The following strategy is employed:
Choose a scale and