Lesson Notes By Weeks and Term - Senior Secondary School 2







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






  • 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.




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



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 



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                   


  1. It is easier to turn a tap on or off by applying couple
  2. 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



     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


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



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


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 








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 



 R1 + R2 = 140N

 Taking moment about R1

 R2 x 80 = (100 x 40 ) + (40 x 50 )

80R2 = 4000 + 2000

      R2 = 6000/80


       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 = 6640 – 720x

        600x+ 720x = 6640 

        x = 6640/1320

        x = 20. 18cm



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.



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



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

  1. 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.
  2. 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.
  3. 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.



Students project.

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



  1. 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)
  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?



  1. The S.I unit of moment is (a) Jm (b)Wm (c)Nm
  2. 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.
  3. 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.



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

                12m        P




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



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

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