VECTORS AND SCALARS
Quantities are divided into two groups (types).
1. Scalar quantities which have magnitude or size only.
Example
• Distance
• Speed
• Mass
• Temperature
1. Vector quantities which have both magnitude and direction.
Example
• Displacement
• Velocity
• Acceleration
• Force etc
VECTOR DIAGRAM
Since a vector has a magnitude and direction it can be represented by a vector diagram i.e. line which is drawn to scale to represent the magnitude of the vector and an air on it to give its direction i.e. a particle displaced three to the north, this describes a displacement suppose our scale is 1cm well then draw a 3cm line pointing north ward
Since vectors can be represented by a vector diagram. It is very possible to add two vector or more. A vector which is a sum of a vector is called a resultant vector. It represents a number of vector in magnitude and direction. The effect of the single vector will be the same as that of a number of vectors.
Example
Let two ropes pull the same body in different directions. The resultant force will be the net effect of the two ropes acting on a body. There are two methods of vector addition
1. The triangle law of vector addition
2. The parallelogram law of vector addition usually we do use the head to tail of the second vector is placed at the head of the 1st vector and the tail of the 3rd vector at the resultant vector will then be formed by joining the tail of the 1st vector to the head of the last vector. Always drawing the vector to scale imagining to do and the proper direction.
THE TRIANGLE LAW OF VECTOR ADDITION
States that ” If two vectors in sequence represent the two sides of triangle in magnitude then the third side of the triangle will give the resultant vector”

When drawing the head of the first vector should be followed by the tail of the second vector . The third side which represents the resultant vector should be drawn from the tail of one vector and the head of another vector .

The two vectors should be drawn in such away that, tail of one vector joins the tail of another vector and the resultant vector should be drawn from the common point .
Example :
1. Find the resultant force of the two forces 15n and 9n acting on a body making an angle of 600 between them
Solution
Choose scale
Let one centimeter represent 3n
The 15N force will be represented by 5cm and the 9N force by 3cm
Draw a 5cm line using your ruler and pencil at its head measure an angle of 600 using your protractor and placing the tail of the 2nd forces draw a 3cm line

By Pythagoras theorem
F2=62+82
F2=36+64
F=100 ans
If properly drawn the length F in the diagram is 5cm
According to our scale the actual force will be 5 x 2=10N
So to count the two forces a force of 10N acting in the other direction F would be required
THE PARALLELOGRAM LAW OF VECTORS ADDITION
This law is also applicable on adding two velocities just like the triangle law it states that
If two adjacent sides of a parallelogram represent two vectors then the diagonal line of the parallelogram will represent the resultant vector through their common point.
Example
If two forces of 20N and 40N are acting on a body such that they make an angle of 450 between them find their resultant force by using the parallelogram law
Solution
As usual choose a scale 1st
Let 1cm=5N
The 20N force will be represented by 4cm and the 40N force by a line of 80N
Then draw these two forces as adjacent side of the parallelogram with an angle of 450 between them.
Then diagonal lines will represent the resultant force
Class work
Find the resultant force when two forces 8N and 16N from the following angles below
1. 450
2. 600
3. 1200
Solution

The resultant vector length in 3cm
The actual forces is = 3cm x 4N = 12N
The long resultant vector length is 3.5cm
The actual force is 3.5cm x 4N = 12.20N
The diagonal line of length 11.2cm and per our scale actual force(F)
=11.2×5
=56.0N
So the resultant force is 56N
RELATIVE VELOCITY
Definition
This is the velocity of a body with respect to another moving body as observed by a stationary observer
Example
1. Speed of an air plane may be observed by a person observer on the ground to be increased by a tail wind or reduced by head wind. So the wind and the plane are both moving related to one another but the observer is stationary.
1. The speed of a boat in a river may also be observed by an observer at the river bank to be increased downstream or decrease up stream. Again the boat and the water are moving relative to one another but the observer stationary.
Example Suppose a plane is flying at a velocity of 100 km/hr and wind is blowing at a velocity of 25 km/hr if the blowing wind is a)head a) tail
Find the resultant plane velocity relative of to an observer on the grounds
Solution
1. Head wind is “opposing” and so will reduce the velocity
Result velocity = 100 km/hr 25km/hr
= 75 km/h
1. Tail wind “adding” or pushing agent so will increase the velocity
Resulting velocity = 100 km/hr + 25km/hr
These velocity = 125 km/hr
These velocities are of the plane relative to an observer on the ground
By Pythagoras theorem
R2 = (100)2 + (25)2
=10000 + 625
R2 = 10,625
= 103.1 KM/HR
From the diagram to get directions of the resultant velocity were
Cos= 0.9708
θ =14
0
It will make an angel of 140, with the south ward direction
RESOLVING VECTORS
Vectors can be resolved in the vertical and horizontal components
This is because sometimes only part of the vector is used
Example
1. Suppose a lawn mower is pushed with a force missing at a direction of 300 to the ground. That of the force which pushes the mover horizontally is called its resolved horizontal component while the other part which presses the mover vertically into the ground is called its resolved vertical components.
The resolved components can be found by using the parallelogram law.
Force components of CB are represented by OA and OB, since OCAB is a rectangle so OB has no effect in the direction of the force component OC, likewise OC has no effect in the direction of OB.
The resolved components of OA, the vertical direction is OC and a horizontal direction is OB
From cos 300 = OB/OA
since OBA Is a right angled triangle
OB = OA cos 300
OA = OB/cos 30o
Note:
This downward force is balanced by the upward reaction of the ground
Example
Consider picture of weight 5N that is supported by a string on a nail at Cos 500
The tension T at A and B together support the weight. Let the string at A and B make angles of 500 with the vertical

The vertical component of T at A = Tcos 50o

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• ### omirambe Stephen, June 9, 2024 @ 2:18 pmReply

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