SOLID & FLUIDS BB101 – ENGINEERING SCIENCE
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SOLID & FLUIDS
BB101 – ENGINEERING SCIENCE
Introduction
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
CHARACTERISTIC
SOLID LIQUID GAS
Arrangement of particles
Very closely packed
Closely Packed Widely Spaced
Shapes & Volume Fix Shape & Volume
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
Force between Particles
Very strong forces
Weak forces Very weak forces or negligible
MovementVibrate & spin around their
position
Vibrate & move randomly but not
freely
Move freely and randomly in all directions.
High speed & colliding with another
Density Higher density High density Low density
Objective:State the characteristics of
solid, liquid and gas.
Objective:State the characteristics of solid, liquid and gas.
Objective:State the characteristics of solid, liquid and gas.
Objective:State the characteristics of solid, liquid and gas.
Objective:State the characteristics of solid, liquid and gas.
Objective:State the characteristics of solid, liquid and gas.
Objective:State the characteristics of solid, liquid and gas.
Objective:State the characteristics of solid, liquid and gas.
DENSITY OF MATTERDefinition 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 0.75kg @ m : 750g
3
4
/1500
105
75.0
mkg
volume
mass
3
4
/015.0
105750
mg
volume
mass
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
Relative Density didn’t have SI unit.
water
material
Substances Density(kg/m3) Subtances Densities (kg /m3)
Solids 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 0.0899
Wood,white pain 420 Oxygen 1.43
Concrete 2300 Nitrogen 1.25
Cork 240 Ammonium 0.760
Liquids Propane 2.02
Water 1000 Objective: Define density and its unitSeawater 1025
Oil 870
Mercury 13600
Alcohol 790
Gasoline 680
water
material
Relative Density
Example 2
Find the relative density if Copper is 8890 kgm-3 and water is 1000 kgm-3 .
)(
89.8
/1000
/88903
3
NoUnit
mkg
mkg
Pressure
Situation of Pressure
Situation 1 (Increasing the pressure by reducing the area) WHICH BALLOON POP EASIER?
Using hand Using nail
Example 3
Objective: Define
pressure and its unit
Situation 2
Pressure in Liquids.
Formula The gauge pressure at any depth from the surface of a fluid;
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)
vacuum76cm
mercury
P atm P atm
Solution
Pascal‘s Principle
Pascal’s principle states that the pressure exerted on a confined liquid is transmitted equal in all direction
The transmission of pressure in liquid
Piston is pushed in
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.
Watch it
..\My Documents\My Videos\RealPlayer Downloads\Principle of hydraulic lift - YouTube.flv
In a hydraulic system, pressure on both piston is equal
Applications of Pascal’s Principle
Hydraulic Brake
Hydraulic Jack
Hydraulic Brakes of a car
Hydraulic system
Shows a simple hydraulics system built according to Pascal Principle
Input Force
Output Force
Area small piston
Area large piston
Fluid
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
ofPascal
Principle
Archimedes Principle
Displaced Volume
Watch it
..\My Documents\RealPlayer Downloads\Conceptual 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
Solution
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
Answer
Answer
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 F=p x V x g
48N = p x (4800 x 10-6) x 9.81 N kg-1
p= 1020kgm-3
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/fluid-dynamics/archimedes-principle.html
Thank you
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