1. FORCES

2. What is a force? Intuitively, a force is like a push or a pull
which produces or tends to produce motion

3. Forces experienced in Daily Life
Weight
Normal reaction
Friction
Viscous force
Tension
Upthrust
Lift
Electrical force
Magnetic force

4. Weight W weight is not the same as mass; it is a force
it is the gravitational force exerted by the Earth
it passes through the centre of gravity of the body

5. Normal reaction N two bodies in contact with each other
perpendicular to the surface of contact

6. Friction friction is exerted two surfaces slide across one another
direction is along the surface of contact

7. Cause of friction F hollows and humps all over the surface
movement F
hollows and humps all over the surface
actual contact area only a fraction 1/10000 of total area
extreme high pressure at contact points causes welding of surfaces
forces are needed to overcome these adhesive forces when trying to slide over the surface

8. Static and kinetic friction
It is harder to move a stationary object than to move the object while it is moving
Static friction is the friction exerted by the ground in order to prevent the object from moving
Kinetic friction is the friction exerted by the ground to oppose the motion of the object while it is moving

9. Limiting friction
Static friction is not constant; it varies in magnitude
Suppose a force P is applied trying to move the object P F
2 N
1 N
2 N
1 N
If P is 1 N, F will also be 1 N to prevent object moving
If P increased to 2 N, F also increased to 2 N
But there is limit to how much F can increase to
Maximum possible static friction is called limiting friction
P must exceed limiting friction in order to move object

10. Example 1
In Fig 1.1, an object was moving to the right on a rough surface. In Fig 1.2, an object rests in equilibrium on a rough slope. In both cases, draw the friction force acting on each object.
Fig 1.1
Fig 1.2
friction
friction

11. Viscous force When body moves in fluid, it experiences resistance
such resistance is known as viscous force
examples: air resistance and water resistance
viscous force depends on the speed of the body
the greater the speed, the greater the viscous force

12. Terminal velocity release F W gathering speed v F W
gathering more speed W F V
finally reaches constant terminal velocity W F VT

13. Tension Tension is exerted by a stretched rope, string or spring.
When a body is attached to a string, the tension in the taut string would tend to pull the body.

14. Hooke’s Law F  (l - lo) => F  e => F = ke
where k is force constant
(elastic constant, spring constant or stiffness F constant) F

15. Strain energy in a Deformed wire
Assume that Hooke’s Law is obeyed.
=>For a force-extension graph, it will be a straight line.
In general, work done by a force F in extending a wire from x1 to x2 is the area under the force-extension graph.
=>Work done in extension or strain energy stored in wire, W = ½ Fe = ½ ke2 F x e

16. Example 2
A vertical wire suspended from one end is stretched by attaching weight of 20 N to the lower end. If the extension is 1 x 10-3 m, what is
(a) the force constant;
(b) the energy stored in the wire;
(c) the gravitational potential energy loss by the weight in dropping a distance of
1 x 10-3 m?

17. Assuming Hooke’s law is obeyed,
(a) F = ke
k = F/e = 20/(1x10-3) Nm-1 = 2 x 104 Nm-1
(b) energy stored in wire,
W = ½ Fe
= ½ (20)(1x10 -3)
= 1 x10-2 J

18. Gravitational potential energy lost by weight
= mgh = 2 x 10-2 J
By conservation of energy,
P.E. lost = Energy stored in wire +
heat dissipated when weight at end of wire comes to rest after
vibrating.

19. Upthrust
upthrust
Upthrust is an upward push on a body when it is immersed in a fluid (gas or liquid)
Upthrust is exerted by the fluid
Upthrust is due to pressure difference of fluid at the top and bottom of immersed portion of the body

20. Example 3
Consider an object partially immersed in a fluid of density . The area of the top surface of the object is A and the immersed depth is h. h

21. (c) What is the volume of fluid displaced by the object? h A
(a) What is the pressure difference across the immersed portion of the object?
h  g
(b) Hence write down the expression for the upthrust acting on the object.
h  g A
(c) What is the volume of fluid displaced by the object?
h A
(d) Hence write down the expression for the weight of fluid displaced.
h A  g
(e) Comment on your answers to (b) and (d).
They are the same.

22. Example 3 shows that
Upthrust = weight of fluid displaced
This is actually the Achimedes’ Principle
Archimedes’ Principle states that
the upthrust on a body in a fluid
is equal and opposite to the
weight of the fluid displaced by the body.

23. Lift What helps birds and aeroplanes maintain its flight?
The answer is the upward lift force exerted on their wings when in motion

24. Electric force
Electric force is exerted between two electric charges + + -
like charges repel
unlike charges attract

25. Magnetic force
Magnetic force is exerted between two magnetic materials or between electric currents N N S
like poles repel
unlike poles attract

26. Forces experienced in Daily Life
Weight
Normal reaction
Friction
Viscous force
Tension
Upthrust
Lift
Electrical force
Magnetic force

27. Different forces upthrust normal reaction normal reaction weight

28. Different forces
lift
tension
weight
weight

29. Different forces How did this ‘forward force’ come about?
normal reaction
speed
air resistance
friction
forwardforce
weight

30. Different forces How did this ‘thrust’ come about? lift air resistance
weight

31. Who exerts on who Friction exerted by Gases expelled by rocket
A force is always exerted by some body on some other body.
What makes a car move?
What makes a rocket fly?
Friction exerted by
ground on tires
Gases expelled by rocket

32. Test Yourself. Identify the forces
weight
normal reaction
weight
weight
FGM

33. Fundamental types of force
When scientists examined all the forces,
they found that many of them are similar in nature.
Scientists have identified 4 fundamental types of force:
gravitational force
electromagnetic force
nuclear force
weak force

34. All forces in our daily life can be classified into one of the fundamental types. In the following table, identify the nature of each force:
gravitational
electromagnetic
electromagnetic
electromagnetic
electromagnetic
electromagnetic
electromagnetic

35. 2 Addition of Vectors Parallelogram RuleB R A B

36. Triangle Rule A B B R A

37. Finding resultant force
The magnitude of resultant force can be found by
drawing vector diagram to scale
calculation (pythagoras theorem, cosine rule, etc)
resolution

38. Example 4 Two forces are given below: 70º 5 N 4 N 30º
Find the magnitude of the resultant force.

39. Method 1 Drawing vector diagram to scale
Scale used is 1 cm : 1 N
5 N
(5 cm)
4 N
(4 cm) R
(5.8 cm)
From the vector diagram,
magnitude of resultant R is 5.8 N
What is missing in the answer?

40. Method 2: By calculation
Using Cosine rule: x2 = (4) (5) cos =>x =5.84 N
Using Sine rule:
5 N
30
4 N
70
80 x

41. Method 3 By resolving vectors
4 cos 70°
70º
5 N
5 sin 30°
30º
4 sin 70°
5 cos 30°
4 N
Rx = 5 cos 30° + 4 cos 70° = N
Ry = 5 sin 30° - 4 sin 70° = N
Rx N Ry
1.26 N R
Magnitude of resultant R is given by
R2 = (5.70)2 + (1.26)2  R = 5.8 N

42. Example 5
Two horizontal forces act at a point to produce a resultant force of magnitude 40 N in the eastward direction. Given that one of the forces is in the northward direction and has a magnitude of 30 N, find the magnitude and direction of the second force. N E F 
30 N
40 N
Magnitude of second force F =  = 50 N
Angle  = tan-1 (40/30) = 53°,
direction of F is 53° east of south (or bearing 127°)

43. Centre of gravity and Free body diagram Centre of Gravity
The centre of gravity of a body is the single point at which the entire weight of the body can be considered to act.

44. Free body diagram (Important)
is a diagram showing all the forces acting on a particular object
is an important tool for solving problems

45. Example 6
An object A of weight w rests on top of another object B of weight W placed on the ground, as shown.
Draw separate free body diagrams showing forces acting on
(a) A only
(b) B only, and
(c) A and B together.

46. Answer N1 N2 w N3 N3 W
W+w
N1 = normal reaction exerted by B on A
N2 = normal reaction exerted by A on B
N3 = normal reaction exerted by ground on B
N1 is numerically equal to N2 (action / reaction pair)

47. Common forces in free body diagrams

48. Force exerted by surface (only)
Total force R exerted by surface on moving
object consists of two components
- normal reaction N
- frictional force F
R is also known as the contact force N R
Motion F

49. A Non lecture Note Example
An object of weight W, resting on a rough surface, is connected to a suspended object of weight w by a string over a smooth pulley. Draw and label the forces acting on each object.
Normal reaction
tension
tension
friction w W

50. Turning effect of a force
Consider a water wheel which is free to rotate about its centre.
Water flowing to the right exerts force on lower blades.
This force causes the wheel to rotate about its centre.
We say that the force has a turning effect.
Turning effect of a force is also known as its moment.
Amount of moment depends on force and distance away.

51. Moment of a force l A F Moments about A = F l (clockwise)
The moment of a force about an axis is defined as the product of the force and the perpendicular distance between the axis and the line of action of the forces.
The moment of a force is also known as the torque. l
Moments about A
= F l (clockwise) A F

52. Moment of a force l l A A F F l A F Example: Not in lecture notes
300 A l
300
Mtd 1
F sin 300
F cos 300 F A l F
300
Mtd 2
Moments about A
= F l sin 30 0 (clockwise)
= 1/2 F l
l sin 30 0
lsin 30 0

53. Example 7 Find the moments of the following forces about point A. 5 m
40
3 m
60
30 N
40 N
Moment of 30 N about A =
30 × 4 = 120 N m (anticlockwise)
Moment of 40 N about A =
40 × 3 sin 60
= 104 N m (clockwise)
Moment of 20 N about A =
20 × 5 cos 40
= 77 N m (clockwise)

54. Torque of a couple
The torque of a couple is equal to the product of one of its forces and the perpendicular distance between the lines of action of the two forces.
Couple = pair of equal and opposite forces whose lines of action do not coincide F A x d F
Taking moment about any arbitrary point, say A,
total anticlockwise moment = F × (d+x) - F × x
= F d

55. Example 8
Calculate the torque acting on the rod 2.0 m long in Figs 9.1 and 9.2.
10 N
10 N
Fig. 9.1
Fig. 9.2
2.0 m
30º
2.0 cos 30º
10 N
10 N
Fig. 9.1:
Torque = F d = 10 × 2.0 = 20 N m
Fig. 9.2:
Perpendicular distance between 10 N forces = 2.0 cos 30º
Torque = 10 × 2.0 cos 30º = 17 N m

56. 5 System in equilibrium A system is in equilibrium when there is
no resultant force and no resultant torque.
Second condition: Resultant torque is zero
First condition: Resultant force is zero
forces would form a closed triangle or polygon
sum of components resolved in any direction is zero
system is said to be in translational equilibrium
is either at rest or moving with constant velocity
has constant linear momentum
total clockwise moment = total anticlockwise moment
if there are only 3 forces, they would intersect at a common point
system is said to be in rotational equilibrium
is at rest or rotating with constant angular velocity
has constant angular momentum

57. Example 9
A horizontal force F is exerted on the pendulum of weight W, causing the pendulum to be suspended at an angle  to the vertical, as shown. Find F in terms of W and .   T T W F F W
From the vector diagram,
tan  = F / W  F = W tan 

58. Example 10
A body of weight 200 N is suspended by two cords, A and B, as shown in the diagram. Find the tension in each cord.
60º
cord B TB TB
cord A W TA
60º TA W
From the vector diagram,
tan 60º = W / TA  TA = W / tan 60º = 200 / tan 60º = 115 N
sin 60º = W / TB  TB = W / sin 60º = 200 / sin 60º = 231 N

59. Example 11
A uniform rod is supported with the fulcrum exactly at the centre of the rod. Two masses were placed on the rod and the system is in equilibrium. Find m.
0.45 m
0.30 m N
2.0 kg m W
2.0 × g
m g
Taking moments about the fulcrum,
clockwise moments = anticlockwise moments
m g × = 2.0 × g × 0.45
 m = kg

60. Example 12
A uniform rod XY of weight 20 N is freely hinged to a wall at X. It is held horizontal by a string attached at Y at an angle of 20º to the rod, as shown.
string
20º X Y
Find
(a) the tension in the string,
(b) the magnitude of the force exerted by the hinge.

61. Example 12 (continued) string R R T 20 N 20º 70º X Y T 29 N 20 N
(a) Let  be the length of the rod and
T be the tension in the string
Taking moments about X,
anticlockwise moments = clockwise moments
T sin 20º ×  = 20 × ( / 2)
 T = N
(b) Let R be the force exerted by the hinge
From the vector diagram, using cosine rule,
R2 = (20)(29) cos 70º
 R = 29 N

62. Example 13
A heavy uniform beam of length  is supported by two vertical cords as shown.
cord A
cord B TA TB
(3/10) 
(7/10) 
weight
Taking moments about the centre of gravity,
clockwise moments = anticlockwise moments
TA × (2/10)  = TB × (5/10) 
 ratio TA / TB = / 2
tension in cord A
Find the ratio
tension in cord B

63. The End