- Science03

VECTORS AND SCALARS

Quantities are divided into two groups (types).

Scalar quantities which have magnitude or size only.

Example

Distance
Speed
Mass
Temperature
Energy
Time
Area
Volume
Density
Electric current
Specific heat capacity

2. Vector quantities which have both magnitude and direction.

Example

Displacement
Velocity
Acceleration
Force
Temperature rise and fall
Momentum

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 arrow on it to give its direction

A particle displaced thre to the north, this describes a displacement suppose our scale is 1cm well then draw a 3cm line pointing north ward

NB;

1. The direction of vector can be represented by using compass direction.

2. Two vectors are equal if magnitude and direction are the same

3. Resultant vector can be added by mathematical or graphical/drawing

Vector Arithmetic

Scalar quantity can be added, multiplied, divided or subtracted. Example, if you have two liquid in different measuring cylinder let say fist one contain 10 cm3 and second contain 20cm3 if you asked to find total volume you must add to obtain total volume

Vector addition

A vector which is a sum of a vectors 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.

Adding by mathematical method

When adding two or more vectors by mathematical method mathematical formula used to sum up vectors. Example pythagoras’s theorem, trigonometrically e.t.c

Adding by graphical method

The following are the steps followed when adding two or more vectors by graphical method

(i) choose a suitable scale and write it down on a graph paper

(ii) Pick starting and draw the first vector to scale direction stated (indicate the magnitude and direction)

(iii) Starting from the head of the first vector, draw the second vector to scale to scale in the started direction until all given vector finished

(iv) Draw the line to connect tail of the first drawn vector and the head of the last vector. This is called resultant vector

(v) Measure the length of the resultant vector and convert to actual unit

(vi) Determine the direction of vector Example,

Suppose a man walks starting from point A, a distance of 20m due north and then walks 15m due east. Find his new position from A

Solution

i. Using a scale of 1cm to represent 5 m

ii. Draw a vector AB 4cm due north.

iii. From B draw BD 3cm due east.

iv. Join A and D point

The resultant diagram is a triangle as shown below

v. Measure the length of AD

AD = 5 cm

Change to actual unit

1cm = 5 m

5 cm = ?

Cross multiplication you get 25 m

vi. Determine the direction of vector

Tanθ = 15/20 = 2/4 = 0.75

Θ = 360 51’

Therefore position of D is represented by vector AD of magnitude 25 m at an angle of 360 51’ east of north

There are two methods of vector addition by graphical

The triangle law of vector addition
The parallelogram law of vector addition

THE TRIANGLE LAW OF VECTOR ADDITION
States that " If two vectors in sequence represent the two sides of triangle 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:

Find the resultant force of the two forces 15N and 9N acting on a body making an angle of 60⁰between 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 60⁰using your protractor and placing the tail of the 2^nd forces draw a 3cm line

Example,

A brick is pulled by a force of 4N acting northward and another force of 3N acting north-east. Find the resultant of these two forces.

Solution

I/ Using a scale of 1cm to represent 1 N

II/ Draw a vector AB 4cm due north.

III/ From B draw BD 3cm at 45⁰

IV/ Join A and D point

The resultant diagram is a triangle as shown below

V/ Measure the length of AD

AD = 6.5 cm

Change to actual unit

1cm = 1 N

6.5 cm = ?

Cross multiplication you get 6.5 N

Therefore the resultant of these two forces 6.5 N

Example,

Two forces, one 8 N and the other 6 N, are acting on a body. Given that the two forces are acting perpendicularly to each other, find the magnitude of the third force which would just counter the two forces.

Solution

I/ Using a scale of 1cm to represent 2 N

II/ Draw a vector AB 4cm due north.

III/ From B draw BC 3cm at 900

IV/ Join A and C point

The resultant diagram is a triangle as shown below

V/ Measure the length of AC

AD = 5 cm

Change to actual unit

1cm = 2 N

5 cm = ?

Cross multiplication you get 10 N

Therefore magnitude of third force is 10 N

THE PARALLELOGRAM LAW OF VECTORS ADDITION

This law is also applicable on adding two vectors 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’’

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 1^st

Let 1cm=5N

The 20N force will be represented by 4cm and the 40N force by a line of 8cm

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

The diagonal line of length 11.2cm and per our scale actual force(F)

=11.2x5

=56.0N

So the resultant force is 56N

The diagonal line of length 11.2cm and per our scale actual force(F)

=11.2x5

=56.0N

So the resultant force is 56N

Example,

Two forces AB and AD of magnitude 40 N and 60 N respective are pulling a body on horizontal table. If the two forces makes an angle 300 between them, find the resultant force on the, body.

Solution

I/ Using a scale of 1cm to represent 10 N

II/ Draw a vector AD 6 cm horizontal from point A

III/ From point A draw AB 3 cm at 300 from vector AD

IV/ Complete the parallelogram ABCD

V/ Join A and c point

The resultant diagram is a triangle as shown below

VI/ Measure the length of Ac

Ac = 9.7 cm

Change to actual unit

1cm = 10 N

9.7 cm = ?

Cross multiplication you get 97 N

Therefore the resultant of these two forces 6.5 N

Example,

Two ropes of 3 m and 6 m long are tied to a ceiling and their free ends are pulled by a force of 100 N as shown in the figure below. Find the tensions in each rope if they make angle 30° between them.

Diagram:

Solution

I/ Using a scale of 1cm to represent 1 m

II/ Draw a vector AD 6 cm horizontal from point A

III/ From point A draw AB 3 cm at 300 from vector AD

IV/ Complete the parallelogram ABCD

V/ Join A and c point

The resultant diagram is a triangle as shown below

VI/ Measure the length of Ac

Ac = 8.7 cm

AC is the equal to 100 N because action is equal to opposite reaction, Ac = 8.7 cm =100 N

Now:

Tension at 3 cm calculated by:

8.7 cm = 100 N

3 cm = ?

Cross multiplication you get 34.5 N

Therefore the Tension at 3 cm is 34.50N

Then:

Tension at 6 cm calculated by:

8.7 cm = 100 N

6 cm = ?

Cross multiplication you get 69 N

Therefore the Tension at 6 cm is 69 N

Example,

Find the resultant force when two forces act as shown in the figure below.

Join lines to get resultant force

I/ Using a scale of 1cm to represent 1 N

II/ Draw a vector AD 8 cm horizontal from point A

III/ From point A draw AB 6 cm at

IV/ Complete the parallelogram ABCD

V/ Measure the length of Ac

Ac = 10 cm

Change to actual unit

1cm = 1 N

10 cm = ?

Cross multiplication you get 10 N

Therefore the resultant of these two forces 10 N

Example,

Find the resultant force, F, when two forces,

9 N and 15 N, act on an object with an angle of 60⁰ between them.

Solution

I/ Using a scale of 1cm to represent 3 N

II/ Draw a vector AD 5 cm horizontal from point A

III/ From point A draw AB 3cm at 60⁰ from vector AD

IV/ Complete the parallelogram ABCD

V/ Join A and c point

The resultant diagram is a triangle as shown below

VI/ Measure the length of Ac

Ac = 7 cm

Change to actual unit

1cm = 3 N

7 cm = ?

Cross multiplication you get 21 N

Therefore the resultant force, F is 21 N

Class work

Find the resultant force when two forces 8N and 16N from the following angles below

90⁰
45⁰
60⁰
120⁰

1. 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

ABSOLUTE VELOCITY

Definition

Is the velocity observed seen as same in every inertial frame of reference.

Relative Velocity

Defn: Relative velocity is the velocity of a body with respective to another moving or stationary body.

Defn: Relative velocity is the velocity of a body with relative to the moving observer.

Relative velocity of an object is the velocity of the object with respect to any other frames of reference.

Nb:

 Velocity of one object let say VA respect another object let say VB is denoted by symbol VAB.

 if all object moving to the same direction, it seems to observe low speed, therefore we minus two velocity of moving body, (-VB)

VAB = VA + (-VB)

VAB = VA - VB

 if all object moving to the opposite direction, it seems to observe high speed, therefore we plus two velocity of moving body, (+VB)

VAB = VA + (+VB)

VAB = VA + VB

 relative velocity also can be calculated by triangle method and by parallelogram methods

Example

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.

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

Head wind is “opposing” and so will reduce the velocity

Result velocity = 100 km/hr-25km/hr

= 75 km/h

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

Since: Resultant vector is measured as an anticlockwise angle of rotation from due east

θ = 270⁰ - 14⁰– anticlockwise to the east

Example,

Car A is moving with a velocity of 20 m/s while car B is moving with a velocity of 30 m/s.

Calculate the velocity of car B relative to car A if:

{a} they are moving in the same direction

{b} They are moving in the opposite directions.

Data given

Velocity of Car A, VA = 20 m/s

Velocity of Car B, VB = 30 m/s

Relative velocity, VBA = ?

Solution:

{a} they are moving in the same direction

From: VBA = VB – VA

VBA = 30 – 20

VBA = 10 m/s

{b} They are moving in the opposite directions.

From: VBA = VB + VA

VBA = 30 + 20

VBA = 50 m/s

RESOLUTION OF VECTOR

As we study at trigonometrically ration when we have values of hypotenuse and angle formed with horizontal we can calculate vertical component and horizontal component.

Consider the diagram below where the toy car pulls at a certain angle but it seems to move horizontally due to horizontal force/vector, not only that but

Vertical force/vector = y is formed

From the diagram:

Horizontal force/vector = x

Vertical force/vector = y

Extract the triangle from the above diagram

Horizontal force/vector is given by the formula

From:

Cos θ = X/F – multiply for F both sides you get

X = FCos θ

Vertical force/vector is given by the formula

From:

Sin θ = Y/F – multiply for F both sides you get

Y = FSin θ

Example,

A nail is being pulled using a string from a wall. The string forms an angle of 30° with the normal. If the force being used is 10 N, part of the force will tend to bend the nail while the other part will try to pull it out.

FIGURE;

What the magnitude of the force:

{a} Tend to bend the nails?

{b} Tend to pull the nails out?

Solution:

Kept the information above into vector form

{a} Force tends to bend the nails, f1 = ?

f1 = 10 x cos 30°

f1 = 10 x 0.866

f1 = 8.66 N

{b} Force tends to pull the nails out, f2 = ?

F2 = 10 x sin 300

F2 = 10 x 0.5

F2 = 5.0 N

Example,

A body is being acted on by two forces: F1 = 18 N acting at an angle of 25° and F2 = 30 N acting at 140° from due East. Find the resultant of the two forces, F, by separating the forces into x- and y- components.

Solution:

Draw the diagram first

First find F1X and F2X

Where:

F1 = 18 N

F2 = 30 N

From: X = F. Cos θ

F1X = F1. Cos 25

F1X = 18 x Cos 25

F1X = 18 x 0.9063

F1X = 16.31 N - toward east

Then:

F2X = F1. Cos 40

F2X = 30 x Cos 40

F2X = 30 x 0.7660

F2X = 22.98 N - toward west

Assume the wanted direction is east so the direction of force to west will be negative. Find their net force, FX = ?

FX = F1X + F2X

FX = 16.31 + (-22.98)

FX = 16.31 - 22.98

FX = - 6.67 N - toward west

Second find F1y and F2y

Where:

F1 = 18 N

F2 = 30 N

From: Y = F. Sin θ

F1Y = F1. Sin 25

F1Y = 18 x Sin 25

F1Y = 18 x 0.4226

F1Y = 7.6 N - toward north

Then:

F2Y = F1. Sin 40

F2Y = 30 x Sin 40

F2Y = 30 x 0.6428

F2Y = 19.28 N - toward north

Assume the wanted direction is north. Find their net force, FY = ?

FY = F1Y + F2Y

FY = 7.6 + 19.28

FY = 26.88 N - toward north

Modify the vector diagram

Lastly find the resultant of the two forces, F = ?

By using Pythagoras’ theorem,

R²= 26.88² + (-6.67²)

R = 27.70 N

Get the direction

Tan θ = Fy/Fx

Tan θ = 26.88/6.67

Tan θ = 4.03

θ = 76.060 – to the west or θ = 103.940– to the east

Therefore resultant force is 27.70 N at an angle of 103.940 to west θ = 76.060 – to the west or θ = 103.940– to the east

Therefore resultant force is 27.70 N at an angle of 103.940 to west

PRODUCED BY PHYSICIST MAYUNGA CONTACT 0745884799

VECTORS AND SCALARS

Quantities are divided into two groups (types).

Scalar quantities which have magnitude or size only.

Example

Distance
Speed
Mass
Temperature
Energy
Time
Area
Volume
Density
Electric current
Specific heat capacity

2. Vector quantities which have both magnitude and direction.

Example

Displacement
Velocity
Acceleration
Force
Temperature rise and fall
Momentum

VECTOR DIAGRAM

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

NB;

1. The direction of vector can be represented by using compass direction.

2. Two vectors are equal if magnitude and direction are the same

3. Resultant vector can be added by mathematical or graphical/drawing

Vector Arithmetic

Vector addition

Adding by mathematical method

When adding two or more vectors by mathematical method mathematical formula used to sum up vectors. Example pythagoras’s theorem, trigonometrically e.t.c

Adding by graphical method

The following are the steps followed when adding two or more vectors by graphical method

(i) choose a suitable scale and write it down on a graph paper

(ii) Pick starting and draw the first vector to scale direction stated (indicate the magnitude and direction)

(iii) Starting from the head of the first vector, draw the second vector to scale to scale in the started direction until all given vector finished

(iv) Draw the line to connect tail of the first drawn vector and the head of the last vector. This is called resultant vector

(v) Measure the length of the resultant vector and convert to actual unit

(vi) Determine the direction of vector Example,

Suppose a man walks starting from point A, a distance of 20m due north and then walks 15m due east. Find his new position from A

Solution

i. Using a scale of 1cm to represent 5 m

ii. Draw a vector AB 4cm due north.

iii. From B draw BD 3cm due east.

iv. Join A and D point

The resultant diagram is a triangle as shown below

v. Measure the length of AD

AD = 5 cm

Change to actual unit

1cm = 5 m

5 cm = ?

Cross multiplication you get 25 m

vi. Determine the direction of vector

Tanθ = 15/20 = 2/4 = 0.75

Θ = 360 51’

Therefore position of D is represented by vector AD of magnitude 25 m at an angle of 360 51’ east of north

There are two methods of vector addition by graphical

The triangle law of vector addition
The parallelogram law of vector addition

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:

Find the resultant force of the two forces 15N and 9N acting on a body making an angle of 60⁰between 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 60⁰using your protractor and placing the tail of the 2^nd forces draw a 3cm line

Example,

A brick is pulled by a force of 4N acting northward and another force of 3N acting north-east. Find the resultant of these two forces.

Solution

I/ Using a scale of 1cm to represent 1 N

II/ Draw a vector AB 4cm due north.

III/ From B draw BD 3cm at 45⁰

IV/ Join A and D point

The resultant diagram is a triangle as shown below

V/ Measure the length of AD

AD = 6.5 cm

Change to actual unit

1cm = 1 N

6.5 cm = ?

Cross multiplication you get 6.5 N

Therefore the resultant of these two forces 6.5 N

Example,

Solution

I/ Using a scale of 1cm to represent 2 N

II/ Draw a vector AB 4cm due north.

III/ From B draw BC 3cm at 900

IV/ Join A and C point

The resultant diagram is a triangle as shown below

V/ Measure the length of AC

AD = 5 cm

Change to actual unit

1cm = 2 N

5 cm = ?

Cross multiplication you get 10 N

Therefore magnitude of third force is 10 N

THE PARALLELOGRAM LAW OF VECTORS ADDITION

This law is also applicable on adding two vectors 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’’

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 1^st

Let 1cm=5N

The 20N force will be represented by 4cm and the 40N force by a line of 8cm

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

The diagonal line of length 11.2cm and per our scale actual force(F)

=11.2x5

=56.0N

So the resultant force is 56N

The diagonal line of length 11.2cm and per our scale actual force(F)

=11.2x5

=56.0N

So the resultant force is 56N

Example,

Two forces AB and AD of magnitude 40 N and 60 N respective are pulling a body on horizontal table. If the two forces makes an angle 300 between them, find the resultant force on the, body.

Solution

I/ Using a scale of 1cm to represent 10 N

II/ Draw a vector AD 6 cm horizontal from point A

III/ From point A draw AB 3 cm at 300 from vector AD

IV/ Complete the parallelogram ABCD

V/ Join A and c point

The resultant diagram is a triangle as shown below

VI/ Measure the length of Ac

Ac = 9.7 cm

Change to actual unit

1cm = 10 N

9.7 cm = ?

Cross multiplication you get 97 N

Therefore the resultant of these two forces 6.5 N

Example,

Diagram:

Solution

I/ Using a scale of 1cm to represent 1 m

II/ Draw a vector AD 6 cm horizontal from point A

III/ From point A draw AB 3 cm at 300 from vector AD

IV/ Complete the parallelogram ABCD

V/ Join A and c point

The resultant diagram is a triangle as shown below

VI/ Measure the length of Ac

Ac = 8.7 cm

AC is the equal to 100 N because action is equal to opposite reaction, Ac = 8.7 cm =100 N

Now:

Tension at 3 cm calculated by:

8.7 cm = 100 N

3 cm = ?

Cross multiplication you get 34.5 N

Therefore the Tension at 3 cm is 34.50N

Then:

Tension at 6 cm calculated by:

8.7 cm = 100 N

6 cm = ?

Cross multiplication you get 69 N

Therefore the Tension at 6 cm is 69 N

Example,

Find the resultant force when two forces act as shown in the figure below.

Join lines to get resultant force

I/ Using a scale of 1cm to represent 1 N

II/ Draw a vector AD 8 cm horizontal from point A

III/ From point A draw AB 6 cm at

IV/ Complete the parallelogram ABCD

V/ Measure the length of Ac

Ac = 10 cm

Change to actual unit

1cm = 1 N

10 cm = ?

Cross multiplication you get 10 N

Therefore the resultant of these two forces 10 N

Example,

Find the resultant force, F, when two forces,

9 N and 15 N, act on an object with an angle of 60⁰ between them.

Solution

I/ Using a scale of 1cm to represent 3 N

II/ Draw a vector AD 5 cm horizontal from point A

III/ From point A draw AB 3cm at 60⁰ from vector AD

IV/ Complete the parallelogram ABCD

V/ Join A and c point

The resultant diagram is a triangle as shown below

VI/ Measure the length of Ac

Ac = 7 cm

Change to actual unit

1cm = 3 N

7 cm = ?

Cross multiplication you get 21 N

Therefore the resultant force, F is 21 N

Class work

Find the resultant force when two forces 8N and 16N from the following angles below

90⁰
45⁰
60⁰
120⁰

1. 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

ABSOLUTE VELOCITY

Definition

Is the velocity observed seen as same in every inertial frame of reference.

Relative Velocity

Defn: Relative velocity is the velocity of a body with respective to another moving or stationary body.

Defn: Relative velocity is the velocity of a body with relative to the moving observer.

Relative velocity of an object is the velocity of the object with respect to any other frames of reference.

Nb:

 Velocity of one object let say VA respect another object let say VB is denoted by symbol VAB.

 if all object moving to the same direction, it seems to observe low speed, therefore we minus two velocity of moving body, (-VB)

VAB = VA + (-VB)

VAB = VA - VB

 if all object moving to the opposite direction, it seems to observe high speed, therefore we plus two velocity of moving body, (+VB)

VAB = VA + (+VB)

VAB = VA + VB

 relative velocity also can be calculated by triangle method and by parallelogram methods

Example

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.

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

Head wind is “opposing” and so will reduce the velocity

Result velocity = 100 km/hr-25km/hr

= 75 km/h

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

Since: Resultant vector is measured as an anticlockwise angle of rotation from due east

θ = 270⁰ - 14⁰– anticlockwise to the east

Example,

Car A is moving with a velocity of 20 m/s while car B is moving with a velocity of 30 m/s.

Calculate the velocity of car B relative to car A if:

{a} they are moving in the same direction

{b} They are moving in the opposite directions.

Data given

Velocity of Car A, VA = 20 m/s

Velocity of Car B, VB = 30 m/s

Relative velocity, VBA = ?

Solution:

{a} they are moving in the same direction

From: VBA = VB – VA

VBA = 30 – 20

VBA = 10 m/s

{b} They are moving in the opposite directions.

From: VBA = VB + VA

VBA = 30 + 20

VBA = 50 m/s

RESOLUTION OF VECTOR

As we study at trigonometrically ration when we have values of hypotenuse and angle formed with horizontal we can calculate vertical component and horizontal component.

Consider the diagram below where the toy car pulls at a certain angle but it seems to move horizontally due to horizontal force/vector, not only that but

Vertical force/vector = y is formed

From the diagram:

Horizontal force/vector = x

Vertical force/vector = y

Extract the triangle from the above diagram

Horizontal force/vector is given by the formula

From:

Cos θ = X/F – multiply for F both sides you get

X = FCos θ

Vertical force/vector is given by the formula

From:

Sin θ = Y/F – multiply for F both sides you get

Y = FSin θ

Example,

FIGURE;

What the magnitude of the force:

{a} Tend to bend the nails?

{b} Tend to pull the nails out?

Solution:

Kept the information above into vector form

{a} Force tends to bend the nails, f1 = ?

f1 = 10 x cos 300

f1 = 10 x 0.866

f1 = 8.66 N

{b} Force tends to pull the nails out, f2 = ?

F2 = 10 x sin 300

F2 = 10 x 0.5

F2 = 5.0 N

Example,

Solution:

Draw the diagram first

First find F1X and F2X

Where:

F1 = 18 N

F2 = 30 N

From: X = F. Cos θ

F1X = F1. Cos 25

F1X = 18 x Cos 25

F1X = 18 x 0.9063

F1X = 16.31 N - toward east

Then:

F2X = F1. Cos 40

F2X = 30 x Cos 40

F2X = 30 x 0.7660

F2X = 22.98 N - toward west

Assume the wanted direction is east so the direction of force to west will be negative. Find their net force, FX = ?

FX = F1X + F2X

FX = 16.31 + (-22.98)

FX = 16.31 - 22.98

FX = - 6.67 N - toward west

Second find F1y and F2y

Where:

F1 = 18 N

F2 = 30 N

From: Y = F. Sin θ

F1Y = F1. Sin 25

F1Y = 18 x Sin 25

F1Y = 18 x 0.4226

F1Y = 7.6 N - toward north

Then:

F2Y = F1. Sin 40

F2Y = 30 x Sin 40

F2Y = 30 x 0.6428

F2Y = 19.28 N - toward north

Assume the wanted direction is north. Find their net force, FY = ?

FY = F1Y + F2Y

FY = 7.6 + 19.28

FY = 26.88 N - toward north

Modify the vector diagram

Lastly find the resultant of the two forces, F = ?

By using Pythagoras’ theorem,

R²= 26.88² + (-6.67²)

R = 27.70 N

Get the direction

Tan θ = Fy/Fx

Tan θ = 26.88/6.67

Tan θ = 4.03

θ = 76.060 – to the west or θ = 103.940– to the east

Therefore resultant force is 27.70 N at an angle of 103.940 to west θ = 76.060 – to the west or θ = 103.940– to the east

Therefore resultant force is 27.70 N at an angle of 103.940 to west

PRODUCED BY PHYSICIST MAYUNGA CONTACT 0745884799

Tuesday 22 December 2020

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