1.1 Resultant Force (Force & Motion II)

PHYSICS F5
Theme 1 Newtonian Mechanics

CHAPTER 7: FORCE & MOTION II

7.1 RESULTANT FORCE

Daya Paduan

FORCE & MOTION II
7.1 RESULTANT FORCE

FORCE & MOTION II
7.1 RESULTANT FORCE

CHAPTER 7: FORCE & MOTION II

You will learn:

1.1 Resultant Force (Daya Paduan)
1.2 Resolution of Forces (Leraian Daya)
1.3 Forces in Equilibrium (Keseimbangan Daya)
1.4 Elasticity (Kekenyalan)

7.1 Resultant Force FORCE & MOTION II
7.1 RESULTANT FORCE
Force pull
to the left Force pull
to the right

Tug-of-war competition

Participant A and participant B exert forces FA and FB
respectively on the rope.

What determines whether the rope is at rest, moves to the
left or moves to the right?

7.1 Resultant Force FORCE & MOTION II
7.1 RESULTANT FORCE
Force pull
to the left Force pull
to the right

Obviously, two forces are fighting each other.
The final situation, either to the left or right is depending on the
final result of the two forces.

Assuming the two opposing forces have different magnitudes, the
object will move in the direction of the larger force.

FORCE & MOTION II
7.1 RESULTANT FORCE

7.1 Resultant Force vector sum meaning
plus minus by following

the direction.

The resultant force is the single force that represents the vector

sum of two or more forces acting on an object.

Force A ± Force B = Resultant Force

Maksud lebih satu daya bertindak,

the final result of the vector sum

of forces is named as Resultant
Force.

The meaning of ‘Resultant’
like total money left after

addition or deduction.

Force is vector quantity, FORCE & MOTION II
so to add up or deduct 7.1 RESULTANT FORCE
the values of two forces
depend on the direction 7.1 Resultant Force

of the forces.

The resultant force is the single force that represents the
vector sum of two or more forces acting on an object.

Object to the Left Object Stationary Object to the Right

FORCE & MOTION II
7.1 RESULTANT FORCE

7.1 Resultant Force

Force is vector quantity.

To determine the Resultant force, the direction and the
magnitude of the forces must be involved.

How do we determine the resultant
force of two forces acting on a point?

1.1 Resultant Force FORCE & MOTION II
7.1 RESULTANT FORCE

FORCE & MOTION II
7.1 RESULTANT FORCE

Determining a Resultant Force :

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

The direction of the two Assume that the force to the
forces are same. right is positive:
(i) Magnitude of the

resultant force,
F = 6 + 8 = 14 N
(i) Resultant force, F = 14 N
to the right

Situation 2: Two forces acting on an object in the opposite directions.

The direction of the two Assume that the force to the
forces are different. right is positive:
(i) Magnitude of the

resultant force,
F=-8+6=-2N
(i) Resultant force, F = - 2 N
to the left

Generally: FORCE & MOTION II
7.1 RESULTANT FORCE
Only 1
force Resultant Force, F:
involved
F1 = 5N F = 5N

2 forces F1 = 5N F = F1 + F2
involved F2 = 3N = 5N + 3N
=8N
F1 = 5N
F = F1 + F2
F2 = 3N = 5N - 3N

=2N

FORCE & MOTION II
7.1 RESULTANT FORCE

Determining a Resultant Force :

Situation 3: Two forces acting on an object perpendicular to each other.

8N 2 forces that are
6N perpendicular to each other.

Cannot added or deducted
directly.

FORCE & MOTION II
7.1 RESULTANT FORCE

Determining a Resultant Force :

Situation 3: Two forces acting on an object perpendicular to each other.

8N 6N 2 forces that are
perpendicular to each other。

Cannot added or deducted
directly.

Resultant 1. Draw the diagonal 2. Calculate the
Force of the rectangle that length of the
diagonal using
represents the Pythagoras’
resultant force, F of Theorem

the two forces

FORCE & MOTION II
7.1 RESULTANT FORCE

Determining a Resultant Force :

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

F1

FORCE & MOTION II
7.1 RESULTANT FORCE

Determining a Resultant Force :

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

Triangle of forces method Parallelogram of forces method

By using a ruler and a protractor, draw the By using a ruler and a protractor, draw the
force F1 followed by force F2 to form two force F1 and force F2 from a point to form

sides of a triangle. two adjacent sides of a parallelogram

Measure the F F Measure the
angle, θ angle, θ
θ
θ

Measure the length of side F With the aid of a pair of compasses,
and calculate the magnitude of complete the parallelogram. Draw the
diagonal from the point of action of the
the resultant force using the
scale you have chosen. forces. The diagonal represents the
resultant force, F

FORCE & MOTION II
7.1 RESULTANT FORCE

7.1 Resultant Force

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

Vector Force Table Kit FORCE & MOTION II
7.1 RESULTANT FORCE

FORCE & MOTION II
7.1 RESULTANT FORCE

Vector Force Table Kit

FORCE & MOTION II
7.1 RESULTANT FORCE

Resultant Force on an Object in Various States of Motion

A free body diagram of an object is a diagram that shows all the forces acting
on that object only.

Force: Normal Force: Normal Force: Normal
reaction from reaction from reaction from the

the table the wedge inclined plane

Force: Weight Force: Weight
of book of bag

Free body diagram of a book Free body diagram of a
on a table. bag on an inclined plane.

Newton's third law: If an object A exerts a force on object B, then object B
must exert a force of equal magnitude and opposite direction back on object A.

FORCE & MOTION II
7.1 RESULTANT FORCE

Resultant Force on an Object in Various States of Motion

A free body diagram of an object is a diagram that shows all the forces acting
on that object only.

Free body diagram of a moving
trailer and a moving rocket

Free body diagram of a moving trailer Force

Force

Force

Force Force
Force

Force

FORCE & MOTION II

When we know the Resultant force of the object, 7.1 RESULTANT FORCE

then we would know acceleration of the object because F = ma.

Object at rest or
moving with
constant velocity,
acceleration = 0

FORCE & MOTION II
7.1 RESULTANT FORCE

State the value of the Acceleration
and Force, either zero or not zero.

Zero Not Zero Zero
Zero
Zero Not Zero
Upward force and
Weight and Normal Upward force weight are equal
reaction are equal larger than weight

Solving Problems Involving Resultant Force, Mass FORCE & MOTION II
7.1 RESULTANT FORCE
and Acceleration of an Object (F = ma)

Solving Problems Involving Resultant Force, Mass FORCE & MOTION II
7.1 RESULTANT FORCE
and Acceleration of an Object (F = ma)

Solving Problems Involving Resultant Force, Mass FORCE & MOTION II
7.1 RESULTANT FORCE
and Acceleration of an Object (F = ma)

Example 2: FORCE & MOTION II
7.1 RESULTANT FORCE

(a) Sketch the free body diagram using the symbol W to represent the weight

of the passenger and symbol R for the normal reaction from the floor of
the lift.
(b) Calculate the magnitude of the normal reaction, R when the lift is:

(i) stationary
(ii) moving upwards with an acceleration of 1.2 m s–2
(iii) moving with a uniform velocity of 8.0 m s–1 [Gravitational

acceleration, g = 9.81 m s–2]

Solution: (b) (ii) The resultant force acts upwards
(i) Resultant force, F = ma
F=0
R=W R – W = ma
R = mg R – 588.6 = 60 × 1.2
R = 72 + 588.6
= 60 × 9.81
= 588.6 N = 660.6 N

(iii) Resultant force, F = 0
R – W = ma
R–W=0
R=W
R = 588.6 N

Example 3： FORCE & MOTION II
7.1 RESULTANT FORCE

Figure 1.10 shows a trolley of mass 1.2 kg being pulled on a table by a load
through a pulley. The trolley moves with an acceleration of 4.0 m s–2
against a friction of 6.0 N.

(a) Sketch the free body diagram of the trolley and the load.
Use W = the weight of the trolley, R = normal reaction on
the trolley, FR = friction, T = tension of the string and B = the
weight of the load.

(b) Compare the weight of the trolley, W with the normal
reaction, R.

(c) Calculate the resultant force acting on the trolley, F.
(d) Calculate the tension in the string pulling the trolley, T.
(e) What is the mass of the load, m? [Gravitational acceleration,

g = 9.81 m s–2]

(b) Weight of trolley,
W = normal reaction, R

For this part, the The load and trolley FORCE & MOTION II
produce the tension, 7.1 RESULTANT FORCE
Resultant Force = T – Fr
(R and W are equal, T at this part, too The T here overcome
the frictional force
cancel each other) and pull the trolleh
moves to the right

The load produces
the tension, T at this

part

For whole system, the For this part, the
load is pulling the Resultant Force = B - T
trolley to move?
The Weight B overcome
Look into the deatil, the T and pull the load
there are two
downward
separated but related
parts of the system

FORCE & MOTION II
7.1 RESULTANT FORCE

Common mistake:

Thinking B against FR.
Resultant force = B - FR

Trolley moves to this
direction ?

Example 3： FORCE & MOTION II
7.1 RESULTANT FORCE

Figure 1.10 shows a trolley of mass 1.2 kg being pulled on a table by a load
through a pulley. The trolley moves with an acceleration of 4.0 m s–2 against
a friction of 6.0 N.

(c) Calculate the resultant force acting on the trolley, F.
(d) Calculate the tension in the string pulling the trolley, T.
(e) What is the mass of the load, m?
[Gravitational acceleration, g = 9.81 m s–2]

(c) Mass of trolley, m = 1.2 kg The trolley moved with a
Acceleration of the trolley, a = 4.0 m s–2 = 4.0 m s–2 which caused
F = ma
by resultant force.
= 1.2 × 4.0
= 4.8 N

(d) Resultant force, F = 4.8 N The tension larger
than frictional force,
Friction, FR = 6.0 N
F = T – FR, so trolley moved.
thus T = F + FR
T = 4.8 + 6.0 = 10.8 N

Example 3： FORCE & MOTION II
7.1 RESULTANT FORCE

Figure 1.10 shows a trolley of mass 1.2 kg being pulled on a table by a load
through a pulley. The trolley moves with an acceleration of 4.0 m s–2 against
a friction of 6.0 N.

(e) What is the mass of the load, m?
[Gravitational acceleration, g = 9.81 m s–2]

(e) Acceleration of the load, a = 4.0 m s–2
Gravitational acceleration, g = 9.81 m s–2

F= B – T

F = ma 4m = 9.81m – 10.8 The load is moving
= m × 4.0 down because B
= 4m 5.81m = 10.8
larger than T.
B = mg
= m × 9.81 m = 10.8
= 9.81m 5.81

= 1.86 kg

Thrust Frictional FORCE & MOTION II
Force Force / 7.1 RESULTANT FORCE
Tension
Weight
Examples
of Force

Buoyant Impulsive
Force Force

FORCE & MOTION II
7.1 RESULTANT FORCE

(a) Resultant Force, (b) Resultant Force,
F = F1 + F2 F = F1 + F2

F = (9N + 12N) – 11N F = (5N + 2N) – 13N
= 15 N = -6 N

FORCE & MOTION II
7.1 RESULTANT FORCE

180 N

240 N

FORCE & MOTION II
7.1 RESULTANT FORCE

（b) （c) The ball moved to the
northeast
180 N
θ tan θ = 180
240
240 N
θ = tan-1 180
Resultant force, F = 2402 + 1802
= 300 N 240

= 36.87o

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