CHAPTER 3: FORCES BB101 ENGINEERING SCIENCE. LEARNING OUTCOME CLO 1 Identify the basic concept of force. (C1) CLO 1 Identify the basic concept of force

CHAPTER 3:CHAPTER 3:FORCESFORCES

BB101 BB101 ENGINEERING SCIENCEENGINEERING SCIENCE

LEARNING OUTCOMELEARNING OUTCOME

CLO 1 Identify the basic concept of CLO 1 Identify the basic concept of force. (C1)force. (C1)

CLO 3 Apply concept of force in real CLO 3 Apply concept of force in real basic engineering problems. (C2, A1)basic engineering problems. (C2, A1)

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UNDERSTANDING OFFORCE

Apply the concept of forceApply the concept of force Define force and its units.Define force and its units.

Differentiate between weight and massDifferentiate between weight and mass Define Newton’s Second Law.Define Newton’s Second Law. Define forces in equilibrium.Define forces in equilibrium. Apply the concept of force in solving problems.Apply the concept of force in solving problems. Calculate Resultant Force.Calculate Resultant Force.

Understand the concept of Moment ForceUnderstand the concept of Moment Force Define Moment ForceDefine Moment Force Describe principle of momentDescribe principle of moment Apply the concept and formula of moment force in solving the Apply the concept and formula of moment force in solving the

related problemsrelated problems

CONCEPTUAL MAPCONCEPTUAL MAP

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FORCE

Definition

Effect of force

Type of force

Weight and mass

INTRODUCTION

Force in equilibrium

Resultant force using resolution

force

Newton Second Law

MOMENT

Formula of moment

Definition

Concept

Force Moment

Resultant Moment

Application Application Force in our life…Force in our life…

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http://lmspsp.cidos.edu.my/course/view.php?id=333

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INTRODUCTIONINTRODUCTION Definition of Definition of Force Force and its and its units.units.

ForceForce is the action of is the action of pushingpushing or or pullingpulling on an objects. on an objects. The SI unit of force is The SI unit of force is Newton (N)Newton (N) or or kgmskgms-2-2..

The effect of force on an object:The effect of force on an object: A stationary object to move.A stationary object to move. A moving object to change its speed.A moving object to change its speed. A moving object to change its direction of motion.A moving object to change its direction of motion. An object to change its size and shape.An object to change its size and shape.

Application of forceApplication of force Using in mechanical systemUsing in mechanical system Transferred using mechanical instrument, eg: gear, pulley, screw Transferred using mechanical instrument, eg: gear, pulley, screw

and piston.and piston.

How to measure the force?How to measure the force? using a using a spring balancingspring balancing together with together with Newton scale (N). Newton scale (N).

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THE TYPES OF FORCESTHE TYPES OF FORCES

The balance force isThe balance force is::

When two or more external forces acting on a body produce no net no net forceforce, ,

Pulling forces = pushing forcesPulling forces = pushing forces

The following are some of the situation where forces are balanced on a body :

a pile of book resting on a hard surface a car moving at constant velocity along a straight road an airplane is flying horizontally at a constant height with a

constant velocity

10 N 10 N

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Unbalanced force isUnbalanced force is : :

When two or more forces acting on a body are not balanced, there When two or more forces acting on a body are not balanced, there must must be a net force onbe a net force on it. This net force is known as the unbalanced force or it. This net force is known as the unbalanced force or the the resultant forceresultant force..

Pulling forces Pulling forces ≠≠ pushing forces pushing forces

The effects of unbalanced forces acting on an object are shown in the The effects of unbalanced forces acting on an object are shown in the following examplesfollowing examples

Golfer hits a stationary golf ballGolfer hits a stationary golf ball A footballer kicks a fast moving ball towards himA footballer kicks a fast moving ball towards him When the engine of a moving car is shut downWhen the engine of a moving car is shut down

8 N 12 N

THE TYPE OF FORCESTHE TYPE OF FORCES

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TheThe weightweight of an object is defined as the gravitational force acting on the of an object is defined as the gravitational force acting on the object.object.

Weight (N) = mass (kg) x gravity (msWeight (N) = mass (kg) x gravity (ms-2-2)) W = mgW = mg

where, W = weight, m = mass, gwhere, W = weight, m = mass, g = gravitational field (9.81ms= gravitational field (9.81ms-2-2))

The SI unit for weight is The SI unit for weight is Newton (N)Newton (N) and it is a and it is a vector quantityvector quantity

The The massmass of an object is a of an object is a measure of its inertiameasure of its inertia

Mass is Mass is constant quantityconstant quantity and it is a and it is a scalar quantityscalar quantity. It is the same . It is the same irrespective of where the object is.irrespective of where the object is.

WEIGHT & MASSWEIGHT & MASS

Mass (kg) ≠ weight (N)

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WEIGHT & MASSWEIGHT & MASS

• The difference between weight and mass is summarized in below table

WeightWeight MassMass

Dependent on the acceleration Dependent on the acceleration due to gravitydue to gravity Is a constant quantityIs a constant quantity

Is a vector quantityIs a vector quantity Is a scalar quantityIs a scalar quantity

Is measured in Newton (N)Is measured in Newton (N) Is measured in kilogram (kg)Is measured in kilogram (kg)

Please think of ……..Please think of ……..

Why we say, we weigh our body, Why we say, we weigh our body, while the given unit is in ‘while the given unit is in ‘kgkg’. ’. Remember “Remember “kgkg” is unit of ” is unit of massmass, , not the unit of not the unit of weightweight. Therefore, . Therefore, why we didn’t say we scale the why we didn’t say we scale the mass. So, please think of …..mass. So, please think of …..

11Unit in ‘kg’

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NEWTON’S LAWNEWTON’S LAW

Definition : Definition :

When net force acting on an object is not zero, the object will accelerate at the When net force acting on an object is not zero, the object will accelerate at the direction of exerted forcedirection of exerted force

F F a a

F = maF = ma

Where,Where, F = Force F = Force

m = massm = mass

a = acceleration (9.81msa = acceleration (9.81ms-2-2))

GROUP ACTIVITYGROUP ACTIVITY

Gallery walkGallery walk

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Exercise 1Exercise 11) Determine the direction of situations below and give the explanation.

a)6 N 15 N

c)10 N

5 N

b)8 N

4 N

4 N

4 N2) A certain force is applied to a 2.0 kg mass. The mass is accelerated at

1.5 ms-2. if the same force is applied to a 5.0 kg mass, what is the acceleration og the mass?

3) A car of mass 700 kg accelerates from rest to 105 km h-1 in 10 s. what is the accelerating force developed by the car engine?

4) A toy car of mass 800 g is pulled along a level runway with a constant speed by a force of 2 N.

a) what is the friction on the toy car?

b) when the force is increased to 6 N, what is

i) the unbalanced force acting on it?

ii) the acceleration of the toy car?

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EXERCISE 2EXERCISE 2 An astronaut has mass of 70 kg. what is his weight ifAn astronaut has mass of 70 kg. what is his weight if

(a) he is on the surface of the Earth where the gravitational field (a) he is on the surface of the Earth where the gravitational field strength strength is 9.8 N kg is 9.8 N kg-1-1??

(b) he is on the surface of the moon where the gravitational field (b) he is on the surface of the moon where the gravitational field strength strength is 1/6 of that on the surface of the Earth? is 1/6 of that on the surface of the Earth?

A spacecraft of mass 800 kg is orbiting above the Earth’s surface at a height A spacecraft of mass 800 kg is orbiting above the Earth’s surface at a height where its gravitational field strength is 2.4 N kgwhere its gravitational field strength is 2.4 N kg-1-1..

(a) what is meant by gravitational field strength at a point in the (a) what is meant by gravitational field strength at a point in the gravitational field? gravitational field?

(b) Calculate the gravitational force experienced by the spacecraft(b) Calculate the gravitational force experienced by the spacecraft

A rock has a mass of 20.0 kg and weight of 90.0 N on the surface of a planet.A rock has a mass of 20.0 kg and weight of 90.0 N on the surface of a planet.

(a) What is the gravitational field strength on the surface of the (a) What is the gravitational field strength on the surface of the planet?planet?

(b) what are the mass and the weight of the rock on the surface of the (b) what are the mass and the weight of the rock on the surface of the Earth where its gravitational field strength is 9.8 N kg Earth where its gravitational field strength is 9.8 N kg-1-1??

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PROPERTIES OF VECTORPROPERTIES OF VECTOR

Showed by symbol and arrowShowed by symbol and arrow

magnitude

direction

Write as vector AB

A B

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VECTOR OF FORCESVECTOR OF FORCES

Showed by arrow.Showed by arrow. The length of this diagonal represents the magnitude resultant force, The length of this diagonal represents the magnitude resultant force,

F and its direction.F and its direction.

magnitude direction

F = 30 N

Vertical vector F

F = 15 N

Force at 30 °

30°

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FORCES IN EQUILIBRIUMFORCES IN EQUILIBRIUM

An object at rest is in An object at rest is in equilibrium. This is because the . This is because the forcesforces acting on it are balanced and the acting on it are balanced and the resultant forceresultant force is is zerozero..

Fx 2

Fy 1

Fy 2

Fx 1

Horizontal force Fx1 = Fx2

Therefore Fx1 – Fx2 = 0

Vertical force Fy1 = Fy2

Therefore Fy1 – Fy2 = 0

Since the resultant force on an object in equilibrium is zero, if the forcesare resolved into horizontal and vertical components, then

(a)The sum of all the horizontal components of the forces = 0(b)The sum of all the vertical components of the forces = 0

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ADDITION FORCES :ADDITION FORCES :RESULTANT FORCESRESULTANT FORCES

A A resultant forceresultant force is a single force that represents the combined is a single force that represents the combined effect of two or more forces with effect of two or more forces with magnitudemagnitude and and directiondirection

Vertical force = The forces acting at y-axialVertical force = The forces acting at y-axial Horizontal force = The forces acting at x-axialHorizontal force = The forces acting at x-axial

The effects of force are depends on:The effects of force are depends on:1.1. The magnitude – the value of forces in The magnitude – the value of forces in Newton’sNewton’s unit unit2.2. The directions – left, right, up and downThe directions – left, right, up and down

A B

C

AB + BC = AC

AC is resultant force

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A A resolution forcesresolution forces is a single force that can be is a single force that can be resolvedresolved into into two perpendicular componentstwo perpendicular components

Fx is the Vertical component of force whereas Fy is the Horizontal component of force

ADDITION FORCES:ADDITION FORCES:RESOLUTION FORCERESOLUTION FORCE

F

θ

Fy

Fx

Figure shows a force F is resolved into two perpendicular components Fx and Fy.

With that,Fx = F cos Fy = F sin

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PROBLEM SOLVINGPROBLEM SOLVING

Example 1Example 1:: The table being pulled by two forces with the magnitude of 6N and 8N The table being pulled by two forces with the magnitude of 6N and 8N

respectively. The angle between the two forces is 60respectively. The angle between the two forces is 60°°

SolutionSolution

Method 1 : can be determined by using the parallelogram of Method 1 : can be determined by using the parallelogram of forcesforces

Method 2 : can be resolved into two perpendicular Method 2 : can be resolved into two perpendicular components (using formula)components (using formula)

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Method 1 : ParallelogramMethod 1 : Parallelogram Steps:Steps:

Choose the scale . Eg: 1 cm = 1m. Using the graph paper and set the point Draw the forces F1 and F2 from a point with an angle of with

each other.

Draw another two lines to complete the parallelogramDraw another two lines to complete the parallelogram Draw the diagonal of the parallelogram. The length of this Draw the diagonal of the parallelogram. The length of this

diagonal represents the magnitude resultant force, F and its diagonal represents the magnitude resultant force, F and its direction, direction, αα can be determined by measuring the angle can be determined by measuring the angle between the diagonal with either one side of the parallelogrambetween the diagonal with either one side of the parallelogram

F1

F2

F

F1

F2

F

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Method 1 : ParallelogramMethod 1 : Parallelogram StepsSteps::

FFxx = ______ N = ______ N FFy y = ______ N= ______ N

Fx and Fy are the vertical and horizontal components of the force :

Magnitude of Fx = 6.0 cos 60° = 3.0 N

Magnitude of Fy = 8.0 sin 60° = 7.0 N

6.0 N

60°

8.0 N

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METHOD 2 : FORMULAMETHOD 2 : FORMULA

Using formula :Using formula :

R R = √ F= √ FY Y 22 + F + FX X

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= √ 6= √ 622 + 8 + 822

= √ 36 + 64= √ 36 + 64

= √ 100= √ 100

= 10 m= 10 m

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Exercise 3Exercise 3

1.1. Who will win?Who will win?

2.2. Calculate the total net force between the following interaction:Calculate the total net force between the following interaction:

5 N10 N

25 N

a)

b) 30 N

55 N

8 N

25 N32 N

c)

d) 45 N

25 N

25 N25 N

45 N 15 N 20 N

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1.1. The Total of force between two or more interactionThe Total of force between two or more interaction

Example 1:Example 1:

F1

F2

Total net force

F = F1 + F2

Total net force

F = (F2 + F3 ) - F1

F2

F3

F1

THE TOTAL OF FORCETHE TOTAL OF FORCE

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5 N10 N

15 N

Total net force

= ( 5 + 10 ) – 15 N

= 0 (equilibrium state)

28 N24 N

Total net force

= 28 – 24 N

= 4 N (move to right side)

1.1. The Total of force between two or more interactionThe Total of force between two or more interaction

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Method 2: Scale drawing :

• Using paper graph to get accurate reading / value

•Scale 1cm = 1N

•Use the protractor to measure the angle of 30°

10 N

30°

FX

30°

10 cm

Fx

Fx = _________ cm

= _________ N


2. 2. The total of force acting at on angle The total of force acting at on angle

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2.2. The total of force acting at on angle The total of force acting at on angle


Method 1 : Analysis

FX = 10 cos θ

= 10 cos 30°

= 8.67 N

10 N

30°

FX

30


Example 4Example 4

Method 1: Analysis

Fx = FX1 + FX2

FX1 = 10 cos 35° = 8.19 N

Fx2 = 12 cos 40° = 9.19 N

Fx = 8.19 + 9.19

= 17.38 N

10 N

35° FX40°

12 N


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10 N

35°

FX

40°

12 N

35°

10 cm

12 cm

40°

Fx

Fx = _________ cm

= _________ N

Method 2: Scale drawing

• Using paper graph to get accurate reading/value

•Scale 1cm = 1N

•Use the protractor to measure the angle of 35°


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MOMENTMOMENT The moment of a force can be worked out using the The moment of a force can be worked out using the

formula: moment = force applied × perpendicular formula: moment = force applied × perpendicular distance from the pivot. If the magnitude of the distance from the pivot. If the magnitude of the force is force is FF newtons and the perpendicular distance is newtons and the perpendicular distance is dd metres then: metres then:

MOMENT = Force X Perpendicular distance (arm)MOMENT = Force X Perpendicular distance (arm)= F X d= F X d= (Newton) x (meter)= (Newton) x (meter)

SI unit SI unit = Nm = Nm Force (F)

Distance (d)

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MOMENT OF FORCESMOMENT OF FORCES For an object to be in For an object to be in static equilibrium,,

1.1. the sum of the forces must be zero, but also the sum of the torques (moments) about any point. For a two-the sum of the forces must be zero, but also the sum of the torques (moments) about any point. For a two-dimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: dimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations:

The total of anti-clockwise moment = the total of clockwise momentThe total of anti-clockwise moment = the total of clockwise moment

2.2. The total of normal force = the total of interaction forceThe total of normal force = the total of interaction force

F1 F2

RF

d1 m d2 m

F1 d1 = F2 d2 Nm

F1 + F2 = RF N

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For an object to be in For an object to be in static equilibrium, , The centre of gravity can be determine using The centre of gravity can be determine using moment resultantmoment resultant

methodmethod

Resultant moment = the total of force moment Resultant moment = the total of force moment

Given, the centre of gravity at x is A

Then, label the reference point of moment at A

Resultant moment = ( F1 + F2 + F3 ) x

The total of force moment = F1 (0) + F2 (x1) + F3 (x1 + x2)

So, ( F1 + F2 + F3 ) x = F1 (0) + F2 (x1) + F3 (x1 + x2)

F1 (0) + F2 (X1) + F3 (X1 + X2)

( F1 + F2 + F3 )

X =

F3

A B

F1

x2

x

F2

x1

MOMENT OF FORCESMOMENT OF FORCES

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Determine the centre of gravity for force action, so that the bar Determine the centre of gravity for force action, so that the bar remains in horizontal equilibriumremains in horizontal equilibrium

A B

20 N 50 N4 m

x


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Example 5 : Example 5 :

SolutionSolution


Method 2:

20 ( 0 ) + 50 ( 4 )

( 20 + 50 )

200

70

= 2.86 m

=

x =

Method 1:

Given the centre of gravity at x form A is,

The total of anti-clockwise moment = the total of clockwise moment

20 ( x ) = 50 ( 4 - x )

20 ( x ) = 200 – 50 x

20 x + 50 x = 200 x

70 x = 200

x = 2.86 m

x

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P Q

50 N 100 N4 m

x

25 N1 m



Determine the centre of gravity for force action, so that Determine the centre of gravity for force action, so that the bar the bar remains in horizontal equilibriumremains in horizontal equilibrium

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MOMENT OF FORCESMOMENT OF FORCES Example 5 : Example 5 :

SolutionSolution

Method 2:

50 ( 0 ) + 25 ( 1 ) + 100 ( 5 )

50 + 25 + 100

525

175

= 3.00 m

x =

=

Method 1:

Given the centre of gravity at x form P is,

The total of anti-clockwise moment = the total of clockwise moment

50 + 25 ( - 1 ) = 100 ( 5 - )

50 + 25 - 25 = 500 – 100

175 = 525

= 3.00 m

xxx

x

x

xx x

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EXERCISE 4EXERCISE 4

5 N 8 N 15 N2 m

8 m

8 N 12 N 10 N

1 m 4 m 2 m

15 N

10 N 15 N 55 N

1 m

12 N

1 m 6 m

16 N

12 m

45°

25 N

a)

b)

c)

d)

1.1. Determine the centre of gravity for force action, so that the Determine the centre of gravity for force action, so that the bar bar remains in horizontal equilibriumremains in horizontal equilibrium

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EXERCISE 5 EXERCISE 5

1.1. Find the interaction for both of following points, RFind the interaction for both of following points, RAA and R and RBB : :

a)a)

b)b)

0.5m 1m0.5m

RB

3N2NRA

1m 1m

RB2kg

RA

1m 1m

1kg 1kg

THE ENDTHE END

Documents

CHAPTER 3: FORCES BB101 ENGINEERING SCIENCE. LEARNING OUTCOME CLO 1 Identify the basic concept of force. (C1) CLO 1 Identify the basic concept of force