1.1 Resultant Force

1.1 resultant
force

CONTENTS

01 02 03 04

• Types of Forces Resultant Applying Problem
• Free Body Force Newton’s Solving
Diagram Second Law

TYPES OF FORCES

FRICTIONAL FORCE/ GRAVITATIONAL FORCE/
DRAG WEIGHT

TYPES OF FORCES

TENSION, T NORMAL, N

Resultant force

When two forces of the same magnitude acting in
opposite directions are applied on a stationary object,
the object remain at rest.

FORCES ACTED ON THE CEILING FAN
Tension, T

Weight, W

FORCES ACTED ON THE VASE ON THE TABLE
Normal, N

?

ASSUMING THE TWO OPPOSING FORCES HAVE
DIFFERENT MAGNITUDES, THE OBJECT WILL MOVES IN
THE DIRECTION OF THE LARGER FORCE

ASSUMING THE TWO OPPOSING FORCES HAVE
DIFFERENT MAGNITUDES, THE OBJECT WILL MOVES IN
THE DIRECTION OF THE LARGER FORCE

An aeroplane has same magnitude of
forces in all directions
Which direction does the aeroplane moves?

1. DRAG = THRUST, LIFT = WEIGHT
2. RESULTANT FORCE = 0
3. THE AEROPLANE MOVES IN CONSTANT VELOCITY

There is Resultant Force!

DETERMINE A RESULTANT FORCE

A resultant force is the
vector sum of the forces
acting on a point. How
do you determine the
resultant forces acting
on a point?

Situation 1: two forces acting on an object in
the same direction

● Magnitude
● Direction
● Resultant Force

Situation 1: two forces acting on an object in
the same direction

● Magnitude

F = 3N + 8N = 7 N

● Direction
to the right

● Resultant Force

F = 7 N to the right

Situation 2: two forces acting on an object in
the opposite direction

● Magnitude

● Direction

● Resultant Force

Situation 2: two forces acting on an object in
the opposite direction

● Magnitude

F = +5N -3N = +2 N

● Direction
to the right

● Resultant Force

F = 2 N to the right

or

F=+2N

Situation 3: two forces acting on an object
PERPENDICULAR TO EACH OTHER

Use Pythagoras’ Theorem
OR

Trigonometry
OR

Drawing in the scale
diagram

= +

= +

F

= +
=
= 10 N

=



= 0.75

F = − 0.75

= 36.9 ⁰

Thus,

cos 36.9 ⁰ =

SOH
CAH F = ⁰
TOA
cos 36.9
= __________ N



Situation 4: two forces acting on an object in
directions that are not perpendicular to each other

There are two ways to solve You need to
this problem. have these
measuring
First, triangle of forces instruments; a
method and second,
parallelogram of forces ruler, a
protractor and
method a compasses

Triangle method

example Choose a suitable
scale and use ruler
700 N
& protractor

600 N
65 ⁰

6 cm

Scale 1 cm = 100 N

example

600 N Remember starts
65 ⁰ F2 from the head of

F1

700 N

7 cm

65 ⁰

6 cm
Scale 1 cm = 100 N

600 N example
65 ⁰
Remember starts F starts
from the tail of F1 to the head

of F2. Then, measure

700 N

F

F2 = 7 cm

65 ⁰
F1 = 6 cm
Scale 1 cm = 100 N

Parallelogram method

500 N example
65 ⁰
Choose a suitable scale
and use ruler & compasses

800 N

5 cm

65 ⁰

8 cm

Scale 1 cm = 100 N

500 N example
65 ⁰
With the aid of
compasses, complete

the parallelogram

800 N

5 cm

65 ⁰

8 cm

Scale 1 cm = 100 N

example

Draw and measure the
diagonal line, F. Measure

500 N 800 N
65 ⁰

5 cm F


8 cm

Scale 1 cm = 100 N

Newton’s First Law

Newton’s Second Law







Problem-solving

A. Free Body Diagram

b. Weight of the trolley

Weight of the trolley, W = Normal Reaction, R

c. Resultant force acting on the trolley

Force, F = ma Given information;
mass of the trolley, m = 1.2 kg
Acceleration of the trolley, a = 4.0 −2

F = ma
= 1.2 kg x 4.0 −2
= 4.8 N

D. Tension in the string pulling the trolley

Forces acted on the trolley at the
x-axis are Tension, T and
Frictional Force.
From the information given, the
trolley is accelerating,
Thus;
Resultant Force, F = T -

Tension, T = F +
T = 4.8 N + 6 N
T = 10.4 N

E. Find the Mass of the Load Acceleration of
the mass is
To find m, the same as
acceleration of
the trolley

"Truth is ever to be found in
simplicity, and not in the

multiplicity and confusion of
things."

ISAAC NEWTON

THANKS!

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Cikgu Fazilah.blogspot.

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