SUBJECT: PHYSICS

CLASS: SS 2

DATE:

TERM: 1st TERM

REFERENCE TEXT

New School Physics by M.W Anyakoha
SSCE WAEC Past Questions
UTME Past Questions

WEEK SIX AND SEVEN

TOPIC: EQUILIBRIUM OF FORCES

CONTENT

Principles of moment
Conditions for equilibrium of a rigid body
Centre of gravity and stability
Couple

A body is said to be in equilibrium if under the action of several forces, it does accelerate uniformly, rotate with uniform angular velocity or remain at rest. Example a stone at rest, the earth rotating round the sun, a body moving along a path at uniform velocity.

FOR WEEKS SIX AND SEVEN

MOMENT OF A FORCE

The moment of a force is the product of the force and the perpendicular distance

Moment = Force x perpendicular distance = F X d Unit =Nm

CONDITIONS FOR EQUILIBRIUM

1.The sum of the upward forces must be equal to the sum of the downward forces.

2.The sum of the clockwise moment above a point must be equal to the sum of anticlockwise moment about the same point

F1 F2 F3

F4 F5

Upward forces [ UF] = F1 + F2 + F3

Downward forces [ DF] = F4 + F5

For a body at equilibrium,

UF = DF

F1 + F2 + F3 = F4 + F5

F1 F2

X1 X2

A B

X3 X4

F3 F4

For a body at equilibrium,

Upward forces = downward force,

F1 + F2 = F3 + F4

(F1+F2) – (F3+F4)=0

Clockwise moment = F2 X2 + F4X4

Anticlockwise moment = F1X1+ F3X3

(F1X1+ F3X3) –(F2X2 + F4X4)= 0

sum of clockwise moment =sum of anticlockwise moment

COUPLE

A couple is a system of two parallel, equal and opposite forces acting along the same line. The effect of a couple is to rotate the body.

r F

Fig I F

d F

fig ii

The moment of a couple is the product of one of the forces and the perpendicular distance between the lines of action of the two forces

In fig (i), M = f x 2r

In fig (ii), M = f x d

The distance between the two equal forces is called the arm of the couple, the moment of a couple is also called a torque

APPLICATION OF THE EFFECT OF COUPLES

It is easier to turn a tap on or off by applying couple
It is easier to turn a steering wheel of a vehicle by applying a couple with our two hands instead of a single force with one arm.

EXAMPLES 1: A light beam AB sits on two pivots C and D . A load of 1ON hangs at O,2m from the support at c. Find the value of the reaction forces P and Q at C and D.

P O Q

A B

4m 2m 6m

C D

10N( weight)

P + Q = 10N

X 2 = Q (2 + 6 )

20 = 8Q

Q = 20/8 =2.5 N

Taking moment about D

P x8 = 10 x6

P = 60/8

=7.5N

Q = 10 -7.5

= 2.5 N

EXAMPLE 2: A pole AB of length 10m and weight 600N has its centre of gravity 4m from the end A, and lies on horizontal ground. Draw a diagram to show the the forces acting on the pole when the end B is lift this end. prove that this force applied at the end A will not be sufficient to lift the end A from the ground.

4m 6m

A Ground level

600N

Clockwise moment =600 x 4 =2400Nm

Anticlockwise moment =p x 10 = 10pNm

P =240Nm

If this force of 240Nm is applied at A, we have

P= 240Nm

4m 6m

600N

Taking moment about B, we have

clockwise moment =240 x 10 =2400Nm

Anticlockwise moment =600 x 6 =3600 Nm

The anticlockwise moment is greater than the clockwise moment .

Therefore , the 240N force A will not be sufficient to lift the end A because the turning effect due to the 600N force far exceeds that due to the 240N force

EXAMPLE 3:

60O

20N

Find the moment of the force of 20N in the diagram above about A and B

Taking moment about A

Cos 60 =d/3m

D= 3 cos 60

D = 1.5m

Moment about A =F x d

M = 20 x 1.5= 30 Nm

The Moment about B= 0

EXAMPLE 4: A uniform rod lm long weighing 100N is supported horizontally on two knife edges placed 10cm from its ends. What will be the reaction at the support when a 40N load is suspended 10cm from the midpoint of the rod.

R1 R2

10cm 40cm 10cm 30cm 10cm

40N

100N

R1 + R2 = 140N

Taking moment about R1

R2 x 80 = (100 x 40 ) + (40 x 50 )

80R2 = 4000 + 2000

R2 = 6000/80

R2=75N

R1 = 140 – 75 =65N

EXAMPLE 5. A metre rule is found to balance horizontally at the 50cm mark, When a body of mass 60kg is suspended at the 6cm mark, the balance point is found to be at the 30cm mark, calculate. The weight of the metre rulethe distances of the balance point to the 60kg mass if the mass is moved to the 13cm mark

6cm 24m2om 50cm

600N W

w x 20 = 24 x 600

w = 14400/20

= 720N

13cm xcm 37cm 50cm

600N 720N

600x(X)=720(37-X)
600x = 6640 – 720x

600x+ 720x = 6640

x = 6640/1320

x = 20. 18cm

CENTRE OF GRAVITY

The centre of gravity of a body is the point through which the line of action of the weight of the body always passes irrespective of the position of the body. It is also the point at which the entire weight of the body appears to be concentrated.

The centre of mass of a body is the point at which the total mass of the body appears to be concentrated. Sometimes, the center of mass may coincides with the centre of gravity for small objects.

EVALUATION

With the aid of diagrams, explain how you can determine the centre of gravity of four named regular uniform bodies.
Describe an experiment to determine the centre of gravity of an irregular lamina.

STABILITY OF OBJECTS

There are three types of equilibrium- stable equilibrium, unstable equilibrium, and neutral equilibrium.

Stable equilibrium: a body is said to be in stable equilibrium if it tends to return to its original position when slightly displaced. A low centre of gravity and wide base will put objects in stable equilibrium e.g. a cone resting on its base ; a racing car with low C.G and wide base; a ball or a sphere in the middle of a bowl.
Unstable equilibrium: a body is said to be in an unstable equilibrium if when slightly displaced it tends to move further away from its original position e.g. a cone or an egg resting on its apex. High C.G. and a narrow base usally causes unstable equilibrium.
Neutral equilibrium: a body is said to be in neutral equilibrium if when slightly displaced, it tends to come to rest in its new position e.g a cone or cylinder or an egg resting on its side.

EVALUATION

Students project.

Each student will make paper model of the three types of equilibrium.

GENERAL EVALUATION

A uniform beam AB of length 6m and mass 20kg rests on support P and Q placed 1m from each end of the beam. Masses of 10kg and 8kg are placed at A and B respectively. Calculate the reactions at P and Q (g = 9.8ms-2)
A box is pushed along a horizontal floor by a horizontal force of 60 N. There is a frictional force between the box and the floor of 50 N.What is the gain in kinetic energy of the box when it moves a distance of 4.0 m?

WEEKEND ASSIGNMENT

The S.I unit of moment is (a) Jm (b)Wm (c)Nm
A uniform metre rule of mass 100g balances at the 40cm mark when a mass X is placed at the 10cm mark.What is the value of X? (a)33.33g (b)43.33g (C) 53.33g.
Two forces each of magnitude 10N act jn opposite directions at the end of a table.If the length of the table is 50cm.Find the moment of the couple on the tab le.(a)0.5Nm (b)5Nm (c) 50Nm.

4 A pole AB of length 5M and weigh 300N has its centre of gravity 2.0M from the end A,and lies on horizontal ground.Calculate the force required to begin to lift this end

(a) 60N (b)120N (c) 240N.

5 When a body is acted upon by several forces and it does not accelerates or rotates, the body is said to be in (a) space (b)equilibrium (C) motion.

THEORY

1 State the conditions necessary for a body to be in equilibrium, mention the three types of equilibrium)

12m P

300

10N

Use the diagram above to calculate the moment of the force of 10N about the point p.

READING ASSIGNMENT

New Sch. Physics FOR SSS –M W ANYAKOHA PAGES 173-182.

Lesson Notes By Weeks and Term - Senior Secondary School 2