### PHYSICS

# Vectors and scalar quantities in physics

By

Addition of two vectors (compounding two vectors)

The resultant of more than two vectors

# Vectors and scalar

In physics physical quantities are divided
into two; **vectors and scalar quantities**

**What is Vector quantity? ** Vector quantities are physical quantities that have both magnitudes and directions. Examples of vector quantities are displacement, velocity, force, displacement and many more.

**What is scalar quantity? ** Scalar quantities are physical quantities that have magnitudes, but no directions. Examples are, distance, speed, mass, temperature, etc, .

Scalar quantities unlike vector quantities can be added and subtracted using ordinary algebra.

## Representations of vectors

A vector can be represented by line OA, as show in the diagram below. The magnitude of vector is represented by the length of the line OA. If a vector is refers to be OA, it means that the vector have a magnitude equivalent to the length of OA, and act from O to A

A vector of 40 units can be represented on paper by an equivalent length, for example let say, 5cm, it means that 1cm, represent 8units.

The direction of the vector is represented by angle which it makes with a give line OX, as shown on the diagram, by the arrow.

### Parallelogram law of vector

** **

The **parallelogram law of vector, **states that if two vectors are represented in magnitudes and directions, by the adjacent sides of a parallelogram, the resultant is represented in magnitude and direction by the diagonal of the parallelogram, drawn from the common points.

#### Addition of two vectors (compounding two vectors)

Two or more vectors acting on a body in given direction can
be added or combined to give a single vector, which produce the same effect.
The single vector produced after combining two or more vectors is called the **resultant.**

Consider two forces of magnitude **P=30N **and **Q=40N, **acting
on a body **O **in any of the following
directions:

(A).**P
**and **Q **act in the same direction

In this case the resultant R, of the forces, **P **and **Q , ** is given by the algebraic sum of the two force, that is

**R= P + Q = 30 + 40 + 70N along OX**

(B).**P** and **Q **act in opposite direction

For this number (B), the resultant of the two forces is obtained by subtraction, since they are acting in opposite direction to each other; therefore the resultant is given as

**R = Q – P = 40 – 30 = 10N along OX**

(C).**P** and **Q **act at right angle to each other

In the case where the forces are inclined to each other at an angle, the resultant of the two forces cannot be obtained by simple algebraic subtraction**, or addition, **but by what is called **vector addition.**

To get the resultant of the vectors inclined to each other,
we can use the **parallelogram law of vector,
trigonometry ratio, or scalar drawing **to obtain the magnitude and
directions of the resultant vector.

So from the Pythagoras’s theorem the solution of (c), is given as;

**R ^{2} = P^{2} + Q^{2} **

The direction of the resultant with respect to a given line **OX** is the angle which **R** makes with **OQ**. It is given by:

The HTML Clipboard

= , assuming that P=3 and Q =4, then

** **** **

In alternative method, the **resultant R** and the **direction α** of the vector can also be found using **scale drawing. **The ** P **and ** Q **are represented using any suitable scale , for example **1cm **to represent **10N, so that P (30N), ** is represented by **3cm ** and ** Q (40N) ** is represented by ** 4cm. ** The rectangle is completed as shown below.

Now you, measure, the diagonal and convert it into Newton, using
the same scale of **1cm=10N **to obtain
the **resultant R.**

The direction of the **resultant
vector, **which is the angle **α, **can
also be measured using a **protractor. **

If you have done your scale drawing well, you will get approximately the same value like the one you got using Pythagoras theorem.

(D).**P **and **Q **act at an acute angle Ѳ =60^{o}

If ** p **and ** Q ** are inclined at angle 60^{o}, . In this case, you can use cosine rule to obtain
the resultant that is using

**R ^{2} =P^{2 }+ Q^{2} – 2PQcos (180-Ѳ)**

And because cos** (180-Ѳ**) = – **cosѲ,** then we have

**R ^{2} = P^{2} + Q^{2 }+ 2PQcosѲ**

Where **Ѳ** is the
angle between **P** and **Q,**

And α, can be found using the sine rule

**But sine 120 ^{o} = sin (180^{o} -120^{o}) = sin60^{o}**, then we have

You can also use scale drawing to obtain, the **resultant **and the direction, by using scale of **1cm = 10N, **as shown, below. Again we use scale of 1cm=10N. we draw the length 4cm to represent Q ( 40N), using protractor , we draw the angle 60^{o}, then we draw length 3cm to represent P (30N). Then we complete the parallelogram, as shown with dotted lines. The resultant and its direction can now be obtained by measuring the resultant R, with meter and measuring the direction of it which is the angle α between OA, and OQ using protractor

(E).**P **and **Q **act at an obtuse angle O-120^{o}

If **P **and **Q **are inclined to at 120^{o},
the resultant can also be solved using **cosine
rule, **that is

**R ^{2} = P^{2}
+ Q^{2} + 2PQ cos120^{o}, **but

**cos120**therefore,

^{o}= – cos60^{o},**R ^{2} = P^{2}
+ Q^{2} – 2PQcos60^{o} =36N**

Alternatively we can also use scale drawing to solve or obtain resultant and its direction.

## Resolution of vectors

A vector, such as force OA =10N, acting at angle 30^{o}, with direction of OX, can be broken or resolved into two perpendicular parts. This component or parts can be determined by drawing a rectangle for which the force will be represented by diagonal

Using Pythagoras’s theorem, we can obtain

OX =10cos30^{o }(horizontal component) and

OY=10sin30^{o} (vertical component)

These two components called horizontal and vertical
component can also be resolved to produce original result, by using **vector addition**

#### The resultant of more than two vectors

To find the resultant
of several forces, f_{1},
f_{2},f_{3} f_{4}…… acting on a point O, we
first reduce the whole system to two perpendicular forces and then compound the
force to obtain the resultant

To reduce the system to two perpendicular forces, we simple find the component of each forces, f1, f2,f3,…along two perpendicular direction OX and OY, and then add to obtain ** P ** and ** Q**

Let look at the example below, Find the resultant of the system of forces given in diagram below

Stage A, is the given problem needed to be solved, so from stage A, it was reduced to stage B, and to stage C, where it is finally reduced to two perpendicular force of 4N and 2N. From stage C, the vectors can now be solved using Pythagoras theorem.

Where R = resultant

, the direction
is given by **tan- ^{1}2**=

**63**with horizontal

^{o}## Relative motion

When two bodies A and B are in motion are moving in a straight line , the velocity of A in relative to B is found by adding the velocity of ** **reversed to the velocity of A.

Example1: A car traveling in a straight road at 100km/h passes a bus going in the same direction at 60km/h. What is the velocity of the car in relative to the velocity of the bus?

Answer: the velocity of the car in relative to the bus is **100 – 60 = 40km/h**

Example 2: If a car in motion with velocity of 100km/h and a bus with 60km/h are moving in opposite direction, what will be the velocity of the car in relative to bus?

Answer: The relative velocity of the car to that of the bus is
**100 + 60 =160km/h**

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lawal

May 22, 2019 at 8:26 pm

quite educative. thanks alot

admin

May 22, 2019 at 9:38 pm

You are welcome