1. SOLID & FLUIDS
BB101 – ENGINEERING
SCIENCE

2. Introduction

3. MATTER
Definition :Matter is anything that has
mass and occupies space (volume).
Example of matter : Chair, books,
water, wood & others

4. Objective:
State the characteristics
of solid, liquid and
gas.


5. Characteristic
of Matter
Objective:
State the characteristics
of solid, liquid and
gas.

CHARACTERIS
TIC
Arrangement of
particles
SOLID
LIQUID
Very closely
packed
Closely Packed
Shapes & Volume
Fix Shape &
Volume
Force between
Particles
Very strong
forces
GAS
Widely Spaced
Fixed volume, but
not fixed shape
Takes the shapes
& volume of its
container
Not have fixed shape
and volume
Takes the shape &
volume of its container
Weak forces
Very weak forces or
negligible
 Move freely and
Movement
Vibrate & spin
around their
position
Vibrate & move
randomly but not
freely
randomly in all
directions.
 High speed & colliding

6. Objective:
State the characteristics of solid, liquid and gas.

7. Objective:
State the characteristics of solid, liquid and gas.

8. Objective:
State the characteristics of solid, liquid and gas.

9. Objective:
State the characteristics of solid, liquid and gas.

10. Objective:
State the characteristics of solid, liquid and gas.

11. Objective:
State the characteristics of solid, liquid and gas.

12. Objective:
State the characteristics of solid, liquid and gas.

13. DENSITY OF MATTER
Definition of density : is defined as mass per
unit volume
Formulae
The SI unit is kg/m3 and g/cm3 (special case)

14. Example 1
An Object has a mass of 750g and a volume of
5.0 x 10-4m3 .
Solution
m : 750g
mass 0.75kg
ρ=
volume
0.75
=
5 × −4
10
=1500kg / m 3
@
m : 750g
mass
ρ=
volume
750
=
5 × −4
10
=0.015 g / m 3

15. Watch it
 Sinking Objects.flv

16. Relative Density
 Relative density also known as a specific gravity of
matter .
 To compare the densities of two materials, we
compare each with the densities of water.
 Formulae Relative Density
ρ material
σ =
ρ water
Relative Density didn’t have SI unit.

17. Substances
Density(kg/m3) Subtances
Solids
Densities
(kg /m3)
Gases
Copper
8890
Air
1.29
Iron
7800
Carbon Dioxide
1.96
Lead
11300
Carbon
Monoxide
1.25
Aluminium
2700
Helium
0.178
Ice
917
Hydrogen
Wood,white pain
420
Oxygen
1.43
2300
Nitrogen
1.25
Concrete
Cork
240
Liquids
Propane
Water
1000
Seawater
1025
Oil
Ammonium
870
Mercury
13600
Alcohol
790
0.0899
0.760
2.02
Objective: Define density and
its unit

18. Example 2
Find the relative density if Copper is
8890 kgm-3 and water is 1000 kgm-3 .
Relative Density
ρ material
σ=
ρ water
8890kg / m −3
σ=
1000kg / m −3
= 8.89
( NoUnit )

19. Pressure

20. Situation of Pressure
Using hand
Using nail
Situation 1
(Increasing the pressure by
reducing the area)
WHICH BALLOON POP EASIER?

21. Example 3
Objective:
Define
pressure
and its unit

22. Situation 2
Formula
The gauge pressure at any depth
from the surface of a fluid;
Pressure in Liquids.
Pressure, P = ρ g h
whereas
:
ρ = density of liquid
h = depth
P =ρ g h also known as the
hydrostatic pressure.

23. Pressure depends on depth &
density
Have you ever noticed?
That’s why; the dams are built
much thicker at the base than at
the top, because, the pressure
exerted by the water increases
with depth.

24. Example 4
Figure below show a Barometer mercury. Find the
pressure? (p mercury = 13600 kgm-3)
vacuum
P atm
76cm
P atm
mercury

25. Solution

26. Pascal‘s Principle
Piston is
pushed
in
The transmission of pressure in liquid
The figure show that when the plunger is
pushed in, the pressure of water at the
end of the plunger will cause water to
spurt out in directions.
 Pascal’s principle states that the pressure exerted on a confined liquid is
transmitted equal in all direction

27. Watch it
 Pascal's Principle.mp4

28.  In a hydraulic system, pressure on both
piston is equal

29. Applications of Pascal’s
Principle
 Hydraulic Brakes of a
car
 Hydraulic Brake
 Hydraulic Jack

30. Hydraulic system
Shows a simple hydraulics system built according to
Pascal Principle
Input Force
Area large
piston
Area small
piston
Fluid
Output Force

31. Example 5
The cylindrical piston of a hydraulic jack
has a cross sectional area of 0.06m2 and
the plunger has a cross-sectional area of
0.002m2.
a) The upward force for lifting a load
placed on top the large piston 9000N.
Calculate the downward force on the
plunger required to lift this load
assuming a 100% work efficiency.
b) If the distance moved by the plunger
is 75cm,what is the distance moved by
the large piston?

32. solution
Objective:
Application
of
Pascal
Principle

33. Archimedes Principle
Displaced Volume

34. Watch it
 ....Archimedes' principle.mp4

35. Example 6
 Figure below shows the weight of a mass in the air is
15N. The mass is immersed in water which has density of
1000 kg/m³. Calculate:
a) The buoyant force
b) The weight of water displaced
c) The volume of the immersed body

36. Solution

37. Activity in group
 1.Find the volume of copper of mass 200g if density the cooper is
8890kg/m3?Answer
 2. The mass of a proton is 1.67x 10-27kg and it can be considered to be a
sphere of roughly 1.35 x10-15m radius. What its density? Answer
 3.Find the density of alcohol if 307g occupies 855cm3?Answer

38. Activity in group
 4. A fruit seller uses a knife with a sharp edge and a cross-sectional
area of 0.5cm2 to cut open a watermelon. Answer
 a) If the force applied on the knife is 18N,what is the pressure exerted
by the knife on the watermelon
 b) After that, he cuts open a papaya using the same knife by exerting
a pressure 2.7 x 105 Pa. Calculate the magnitude of force applied to
cut the papaya

39. Activity in group
 5.From the figure below, force input is given by 4000 N
and diameter at small piston is 100 cm. If the diameter
at large piston given by 250 cm, find the maximum
mass of the car than can lift by the force input
4000N.Answer

40. Activity in group
 6.A concrete slab weighs is 150N. When it fully
submerged under the sea, its apparent weight is 102N.
answer
 a) Calculate the buoyant force by concrete slab when
immersed in sea water.
 b)Calculate the density of the sea water in kg/m3 if the
volume of the sea water displaced by the concrete slab
is 4800cm3

41. Answer

42. Answer

43. Answer

44. Answer
 5. Force input = 4000 N
 Area input = π (100/100)2 = 3.1416 m2.
 Area input = π (250/100)2 = 19.635 m2.
 (F1/A1) = (F2/A2)
 (4000 / 3.1416) = ( F2 / 19.635 )
 F2 = (4000 / 3.1416) x 19.635 = 25000 N
 Mass of the car = 25000 / 9.81 = 2548.42 kg.

45. Answer
6. a)Buoyant force= Actual weight- Apparent weight

= 150N-102N

=48N
b) Bouyant force= Weight of sea water displaced
 48N = p x (4800 x 10-6) x 9.81 N kg-1

p= 1020kgm-3
F=p x V x g

46. Conclusion of this solid & fluid

47. Reference
 Longman Essential Physics SPM Yap Eng Keat& Khoo
Goh Kow 2012
 Pelangi STPM Physics volume2 Poh Liong Yong 2014
 JMSK Engineering science BB101 Fourth Ed. 2012
 http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html
 http://physics.tutorvista.com/fluiddynamics/archimedes-principle.html

48. Thank you