Solid and fluid

SOLID & FLUIDS
BB101 – ENGINEERING
SCIENCE

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

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

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

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

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)

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

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.

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

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 )

Situation of Pressure

Using hand

Using nail

Situation 1
(Increasing the pressure by
reducing the area)
WHICH BALLOON POP EASIER?

Example 3

Objective:
Define
pressure
and its unit

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.

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.

Example 4

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

76cm
P atm
mercury

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

Watch it

 ..My DocumentsMy VideosRealPlayer DownloadsPrinciple of hyd

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

Applications of Pascal’s
Principle

 Hydraulic Brakes of a
car

 Hydraulic Brake

 Hydraulic Jack

Hydraulic system
Shows a simple hydraulics system built according to
Pascal Principle
Input Force

Area large
piston

Area small
piston

Fluid
Output Force

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?

solution

Objective:
Application
of
Pascal
Principle

Archimedes Principle

Displaced Volume

Watch it

 ..My DocumentsRealPlayer DownloadsConceptual
Physics Demo of Archimedes' principle - YouTube.flv

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

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

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

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

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

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.

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

Conclusion of this solid & fluid

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

Solid and fluid

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