EQUILIBRIUM OF FORCES

Subject: 

Physics

Term:

FIRST TERM

Week:

WEEK 6

Class:

SS 2

Topic:

EQUILIBRIUM OF FORCES

Previous lesson: 

The pupils have previous knowledge of

CONCEPT OF PROJECTILE.

that was taught as a topic in the previous lesson

Behavioural objectives:

At the end of the lesson, the learners will be able to

• Resultant of force
• Equilibrant force
• Equilibrium, types of equilibrium
• Moment of force
• couple
• Conditions for equilibrium for system of parallel forces
• Centre of gravity (types of equilibrium) – Centre of mass

Instructional Materials:

• Wall charts
• Pictures
• Related Online Video
• Flash Cards

Methods of Teaching:

• Class Discussion
• Group Discussion
• Asking Questions
• Explanation
• Role Modelling
• Role Delegation

REFERENCES:

• new school Physics by MW Anyakoha
• New system PHYYSICS for senior secondary schools. Dr. Charles Chew.
• Comprehensive Certificate Physics by Olumuyiwa Awe
• Senior School Physics BY PN Okeke, SF Akande
• STAN Physics.

Content:

WEEK SIX

DATE ………………………………

TOPIC: EQUILIBRIUM OF FORCES

CONTENT

• Resultant of force
• Equilibrant force
• Equilibrium, types of equilibrium
• Moment of force
• couple
• Conditions for equilibrium for system of parallel forces
• Centre of gravity (types of equilibrium) – Centre of mass

PERIOD ONE;

Resultant force

This can be defining a single force which can produce the same effect as the combined force on a system. The addition of two or more force produces the resultant force. The resultant of any system of force can be obtained through any of the process described earlier.

Equilibrant force

This is that force which when added to a system of vectors will make the resultant of the system zero. Equilibrant has the same magnitude as the resultant force but it always acts in a direction opposite to that of the resultant.

R

F

2

F

1

F

2

F

1

For a system in equilibrium, the resultant force is ZERO.

Types of equilibrium

1. Stable equilibrium; a body is in stable equilibrium if it velocity and it resultant force is zero. ( v = 0 and R = 0)
2. Dynamic equilibrium: a body is said to dynamic equilibrium if its velocity is constant or it is rotating with a constant angular velocity. For bodies in dynamic equilibrium, velocity is not zero but the resultant force on it is zero

(i.e, )

1. Translational equilibrium: a body is said to be in translational equilibrium if there is no net force acting on it though it is at rest or moving with constant velocity.

Thus a body is said to be in equilibrium is it resultant force is zero.

Moment of a force

The turning effect of a force is it moment. Moment of a force about a point can be define as the product of the force and it perpendicular distance from the point.

CASE

1

;

O

d

F Moment of the force F about the point O = F x d CASE 2.

O

d

F

The perpendicular component of F is . So moment of F about O is

Moment CASE 3:

The force F will create a translational motion and not a turning effect. Therefore the moment of F in this case is ZERO.

N.B note that the moment of a force is maximum when the force is at right angle.

EVALUATION

1. The product of a force and its perpendicular distance force a point called —- (a) resultant (b) equilibrant (c) moment of force (d) couple
2. If moment M = Fdsin, for what value of will the moment of the force F be zero? (a) 0⁰ (b) 90⁰ (c) 180⁰ (d) 270⁰.
3. The angle between the resultant and the equilibrant of a system of force is —- (a) 0⁰ (b) 90⁰ (c) 180⁰ (d) 270⁰.
4. The type of equilibrium possess by a body falling through a fluid after attaining it terminal velocity is —- equilibrium. (a) stable (b) unstable (c) dynamic (d) neutral
5. The resultant force on a body in translational equilibrium is —-

PERIOD TWO

Principle of moment

This states that for a system in equilibrium, the algebraic sum of moments about any point is zero.

It can also be stated thus, for a system in equilibrium, the sum of the clockwise moment about a point is equal to the sum of the anticlockwise moment about the same point.

Consider the system below,

d

c

b

a

Q

P

W

1

W

2

W

Three downward forces, W, W1, W2.

Reaction act P and Q, constitute the two upward force acting on the body.

Taking moment about P,

Don’t forget moment of a force is Force x perpendicular distance.

Classwork: following the example above, take moment about the point Q and write out the clockwise and the anticlockwise moments.

Examples;

1. A 40cm P B

7N

In the diagram above, AB represent a uniform rod of length 1.50m which is in equilibrium on a pivot at p. if AP = 40cm, calculate the mass of the rod. (g = 10ms^-2

Solution:

A 40cm P B

Since the rod is uniform, it weight act at the

centre of the rod

7N W

the rod is 1.5m long, it centre is taking moment about the pivot clockwise moment = f x d = 7N x 0.4m = 2.8 Nm

Anticlockwise moment = f x d = W x (0.75 – 0.4) = 0.35W

At equilibrium, clockwise moment = anticlockwise moment

0.35W = 2.8

The weight of the rod is 8.0N

But weight W = mg

Mass of the rod is 0.8 kg

2. A metre rule is found to balance at 48 cm marked. When a body of 60 g is suspended at 6 cm, the balance point is found to be 30 cm. i. Calculate the mass of the rule. ii. What is the new balance point if the 60 g is moved to 13 cm mark.

^Solution 0 6 30 48 100

60 g W

W is mass of metre rule

N.B

• a metre rule is 100 cm long
• for uniform metre rule the weight (position of c.g) is 50 cm mark
• for non-uniform metre rule, c.g is at the balance point when no load is on the rod

Moment of a couple M = F x d

M = 10 x 0.5= 5 Nm

Centre of gravity

This can be defined as a point on a body through which the line of action of the resultant weight of the body passes through. It is the point on an object where the resultant weight of the body is acting.

The position of the centre of gravity of an object can be determined through the following methods:

i. Balancing method ii. Plumb line methods

EXPERIMENT 4-5:

• • an experiment to determine the centre of balance of metre rule using the balancing method.
  • An experiment to determine the centre of gravity of a laminar irregular cardboard using the plumb line method

Uniform objects often have their centre of gravity at their midpoint / centre.

The position of the centre of gravity of an object determines the stability of the object.

TYPES OF EQUILIBRIUM / STABILITY

1. Stable equilibrium: bodies in stable equilibrium

i. Have centre of gravity close to their base (low c.g) ii. Have wide base iii. Returns to their original position after a slight tilt.

Wide

base

c.g

.

.

c.g

Examples of bodies in stable equilibrium; a cone sitting on it base, a funnel set upside down on a table

N.B when bodies in neutral equilibrium are slightly tilted, their potential energy increases but the line of action remains within the base.

2. Unable equilibrium: bodies in unstable equilibrium

• • 1. Have centre of gravity high above the base.
    2. Have narrow base
    3. Fall away from their original position when they are slightly tilted

c.g

.

N.B when bodies in neutral equilibrium are slightly tilted, their potential energy decreases and the line of action falls outside the base.

3. Neutral equilibrium:For bodies in neutral equilibrium, the potential energy remains unchanged when they are slightly tilted.

the diagram below typify the types of equilibrium.

43

|

Page

Ball

in

unstable

equilibrium

Ball

in

neutral

equilibrium

(you can easily identify the type of equilibrium by considering how a body will fall off its equilibrium position when it is slightly tilted)

Centre of mass

This can be defined as the point on an object where the application of a force will produce acceleration and not a turning effect.

EVALUATION

• Mention two example each of bodies in (i) stable equilibrium (ii) unstable equilibrium (iii) neutral equilibrium
• Define a couple.