5.1
5.1 State CharaState Characteristics of cteristics of solid , solid , liquid aliquid and gasnd gas
SOLID
SOLID LIQUID LIQUID GASGAS
Microscopic Microscopic view view Particle Particle Arrangement Arrangement
Tightly packed, usually Tightly packed, usually
in a regular pattern. in a regular pattern.
Close together with no Close together with no regular arrangement. regular arrangement.
Well separated with no Well separated with no regular arrangement. regular arrangement. Shape
Shape Fixed Fixed shape shape Follow Follow container container Undefined Undefined ShapeShape Volume
Volume Fixed volume Fixed volume Fixed Fixed volume volume Undefined Undefined volumevolume Kinetic Energy
Kinetic Energy Content
Content Low Low Average Average HighHigh
Compressibility
Compressibility Hard Hard to to compress compress CompressibleCompressible Easily to compressEasily to compress
Flow
Flow Ability Ability Does Does not not flow flow Flows Flows easily easily Flows Flows easilyeasily
Movement Movement
Vibrate (jiggle) but Vibrate (jiggle) but generally do not move generally do not move from place to place. from place to place.
Vibrate, move about, Vibrate, move about, and slide past each and slide past each
other. other.
Vibrate and move freely Vibrate and move freely
at high speeds. at high speeds.
5.2 Define the Density and Pressure 5.2 Define the Density and Pressure
DEFINITION DEFINITION
The density of a material is defined as its
The density of a material is defined as its mass per unitmass per unit volume.
volume.
SYMBOL SYMBOL •
•The symbol of density isThe symbol of density is ρ (rho).ρ (rho). FORMULA
FORMULA Where: Where:
•ρ (rho) is the density, •ρ (rho) is the density, •
•m is the mass in kg,m is the mass in kg, •
•V is the volume in mV is the volume in m33..
SI UNIT SI UNIT
DENSITY
DENSITY
V
V
m
m
Different Different materials materials usually usually have have different different densities.densities.
Example 1: Example 1:
Calculate the density and relative of wooden block which has dimensions
Calculate the density and relative of wooden block which has dimensions and a and a mass of mass of .. Solution: Solution:
3 3 2 2 2 2 2 2 3 3 4000 4000 10 10 5 5 10 10 2 2 10 10 1 1 10 10 40 40 m m kg kg vv m m DEFINITION DEFINITION RelativRelative density is e density is thethe ratio of the densityratio of the density (mass of a unit(mass of a unit volume)
volume) of a substance to the density of a givenof a substance to the density of a given refer
reference ence material.material.
Specific gravity
Specific gravityusually means relative density withusually means relative density with respect to water.
respect to water.
SYMBOL SYMBOL
•
•The symbol of relative density isThe symbol of relative density is RDRD..
FORMULA: FORMULA: SI UNIT: SI UNIT:
No Unit
No Unit
R
R
e
e
l
l
a
a
t
t
i
i
v
v
e
e
D
D
e
e
n
n
s
s
i
i
t
t
y
y
A solid metal cyl
A solid metal cylinder has radiuinder has radiuss cm and length cm and length . Its mass is. Its mass is . Find the. Find the density of the metal and its relative density.
density of the metal and its relative density.
Solution: Solution:
3 3 6 6 2 2 2 2 2 2 10 10 93 93 .. 3 3 10 10 5 5 .. 0 0 m m l l r r Volume Volume 3 3 6 6 3 3 4 4 .. 16539 16539 10 10 93 93 .. 3 3 10 10 65 65 m m kg kg vv m m 539 539 16 16 1000 1000 4 4 16539 16539 RD RD Density, Density, Relative Relative .. .. water water density of density of material material density of density of DEFINITION: DEFINITION: TheThe pressurepressure,, PP, is defined as the , is defined as the ratio of force to arearatio of force to area
SYMBOL: SYMBOL: •
•The symbol of Pressure isThe symbol of Pressure is PP.. FORMULA:
FORMULA: Where: Where: •
•P is the Pressure,P is the Pressure, •
•F is the Force in newton,F is the Force in newton, •
•A is the Area im mA is the Area im m22..
SI UNI SI UNITT::
Application of pressure: cutting tools, injection
Application of pressure: cutting tools, injection needleneedle and tip of thumbtack.
and tip of thumbtack.
PRESSURE
PRESSURE
A
A
F
F
P
P
Example 3: Example 3:
A hammer suppl
A hammer supplies a force ofies a force of . . The hammer The hammer head has head has an area an area ofof . . What iWhat is thes the pressure? pressure? Solution: Solution: Pa Pa 10 10 9.86 9.86 or or m m N N 10 10 9.86 9.86 10 10 0 0 .. 7 7 700 700 5 5 2 2 5 5 4 4 A A F F P P
5.3 Variation Of Pressure With Depth 5.3 Variation Of Pressure With Depth
Relating Pressure
Relating Pressure in a liquid to the in a liquid to the DepthDepth and Density and Density of the liquid: of the liquid:
Consider a cylindrical container oh
Consider a cylindrical container oh height,height, and cross-sectional area, and cross-sectional area, which is filled with a liquid which is filled with a liquid of density
of density
Volume
Volume of of liquid liquid in in the the container container ::
V
V
Ah
Ah
MassMass of of liquid liquid in in the the containecontainer r :: mmV V Ah Ah
Force
Force on on point point X X ::
g g Ah Ah mg mg F F
weightweightof of thetheliquidliquid
Pressure
Pressure on on point point X X ::
g g h h A A g g Ah Ah A A F F P P
Therefore Therefore ::h
h
X
X
What will be the: What will be the:
(a) the gauge pressure and (a) the gauge pressure and
(b) the absolute pressure of water
(b) the absolute pressure of water at depthat depth below the surface? (Given that: below the surface? (Given that: , and, and ).). Solution: Solution: a) a) 2 2 2 2 72 72 .. 117 117 117720 117720 12 12 81 81 .. 9 9 1000 1000 m m kkN N m m N N gh gh P
P ga ga ugug ee water water
b) b) 2 2 72 72 .. 218 218 101 101 72 72 .. 117 117 Pr Pr m m kN kN P P P P essure essure AbsolAbsoluteute ga ga ugug ee atmosphereatmosphere
5.4
5.4 Pascal’s PrinciplePascal’s Principle
Pascal’s principle states that pressure exerted on an enclosed fluid (liquid) is
Pascal’s principle states that pressure exerted on an enclosed fluid (liquid) is transmitted equallytransmitted equally to every part of the f
to every part of the f luid (liquidluid (liquid).).
Hydraulic systems can be used to obtain a large force by the application of a much smaller Hydraulic systems can be used to obtain a large force by the application of a much smaller force.
force.
We can turn this phenomenon to our advantage if we alter the areas exposed to equal We can turn this phenomenon to our advantage if we alter the areas exposed to equal pressures, as in an
pressures, as in an hydraulhydraulic lift:ic lift:
Since the pressure must be t
Since the pressure must be the same everywhere:he same everywhere: Pressure is; Pressure is; A A F F P P So; So;
11 22
2 2 2 2 2 2 1 1 1 1 2 2 1 1 A A A A F F F F A A F F A A F F P P P P atat atat
2 2 1 1This says that the Force at the
This says that the Force at the outlet (at 2) is augmented by the size of the area of outlet (at 2) is augmented by the size of the area of the outlet.the outlet. So if we make the area 1000 times
So if we make the area 1000 times larger, we can lift 1000 times the force we apply at Flarger, we can lift 1000 times the force we apply at F11..
By applying Pascal’s principle on a simple hydraulic system, By applying Pascal’s principle on a simple hydraulic system,
Applicatio
Applications of Pascal’s prinns of Pascal’s principle includciple include the hydraulie the hydraulic jack, hydraulic jack, hydraulic lift and hydrauc lift and hydraulic brakes.lic brakes.
Example of Applications of Pascal’s Principle (Hydraulic Example of Applications of Pascal’s Principle (Hydraulic Lift). Lift).
2 2 2 2 1 1 1 1
A
A
F
F
A
A
F
F
2 2 2 2 1 1 1 1A
A
x
x
x
x
A
A
Where: Where: F F 11 = force at 1 = force at 1 F F 2 2 = force at 2 = force at 2 AA11 = cross sectional Area at 1 = cross sectional Area at 1
A
A2 2 = cross sectional Area at 2 = cross sectional Area at 2
x
x 11 = distance moved at 1 = distance moved at 1
x
A hydraulic car
A hydraulic car lift has lift has a pump a pump piston with radiuspiston with radius . The resultant piston has a radius. The resultant piston has a radius of
of . The total weight of the car and plunger is. The total weight of the car and plunger is . If the bottom ends of the. If the bottom ends of the piston and plunger are at the same height, what input force is required to stabilize the car and piston and plunger are at the same height, what input force is required to stabilize the car and output plunger?
output plunger?
Solution: Solution:
We need to use the area for circular objects,
We need to use the area for circular objects, for both the piston and plunger. Apply for both the piston and plunger. Apply Pascal's Principle: Pascal's Principle:
N N r r r r F F A A A A F F F F B B A A B B B B A A B B A A 20 20 .. 131 131 150 150 .. 0 0 0120 0120 .. 0 0 20500 20500 22 2 2 2 2 2 2
5.5 Archimedes' Principle 5.5 Archimedes' Principle Archimedes’ principleArchimedes’ principle states that an object which is partially or wholly immersed in a states that an object which is partially or wholly immersed in a fluid (liquid or gas) is acted upon by an upward buoyant force equal to the weight of the fluid (liquid or gas) is acted upon by an upward buoyant force equal to the weight of the fluid it displaces.
fluid it displaces.
An object wei
An object weighs less in wghs less in water than it does in the ater than it does in the air.air.
This loss of weight is due to the upthrust of the water acting upon it and is equal to the weight of This loss of weight is due to the upthrust of the water acting upon it and is equal to the weight of the liquid displaced.
the liquid displaced.
If the weight of the water displaced is less than the weight of the object, the object will sink. If the weight of the water displaced is less than the weight of the object, the object will sink.
Otherwise the object will float, with the weight of the water displaced equal to the weight of the Otherwise the object will float, with the weight of the water displaced equal to the weight of the object.
object.
Archimedes Principle: Archimedes Principle:
The buoyant force is equal to the The buoyant force is equal to the weight of the displaced water. weight of the displaced water.
Floatation: Floatation:
The principle of floatation states that a floating body displaces its own weight of the liquid in which it The principle of floatation states that a floating body displaces its own weight of the liquid in which it floats.
floats.
According to Archi
According to Archimedes’ Princimedes’ Principle:ple:
Therefore; Therefore;
Figure below show four situations of object in a liquid: Figure below show four situations of object in a liquid:
S S i i t t u u a a t t i i o o n n 1 1 F F W W oo
Buoyant
Buoyant
force
force
object
object
of
of
Weight
Weight
However; However;displaced
displaced
liquid
liquid
of
of
Weight
Weight
force
force
Buoyant
Buoyant
l l l l o o o o l l l l o o o o l l o o l l o o V V V V g g V V g g V V g g m m g g m m W W W W For totally submerged object; For totally submerged object;
l l o o V V V V S S i i t t u u a a t t i i o o n n 2 2 F F W W oo
Buoyant
Buoyant
force
force
object
object
of
of
Weight
Weight
However; However;displaced
displaced
liquid
liquid
of
of
Weight
Weight
force
force
Buoyant
Buoyant
l l l l o o o o l l l l o o o o l l o o l l o o V V V V g g V V g g V V g g m m g g m m W W W W
For totally submerged object; For totally submerged object;
Buoyant force Buoyant force Weight Weight Rising Rising
<
<
Buoyant force Buoyant force Weight Weight Rising Rising>
>
Buoyant force = WBuoyant force = Weight of liquid displacedeight of liquid displaced
V Vgg mg mg F F displaced displaced liquid liquid of of Weight Weight = = force force Buoyant Buoyant
S S i i t t u u a a t t i i o o n n 3 3 However; However;
displaced
displaced
liquid
liquid
of
of
Weight
Weight
force
force
Buoyant
Buoyant
l l l l o o o o l l l l o o o o l l o o l l o o V V V V g g V V g g V V g g m m g g m m W W W W
For totally submerged object; For totally submerged object;
l l o o V V V V S S i i t t u u a a t t i i o o n n 4 4 F F W W oo
Buoyant
Buoyant
force
force
object
object
of
of
Weight
Weight
However; However;displaced
displaced
liquid
liquid
of
of
Weight
Weight
force
force
Buoyant
Buoyant
l l l l o o o o l l l l o o o o l l o o l l o o V V V V g g V V g g V V g g m m g g m m W W W W
For totally submerged object; For totally submerged object;
l l o o V V V V NOTE :
NOTE : ooobject, l object, l liquid disliquid dis pla placeced d
Archimede
Archimedes' Principle explains why s' Principle explains why steel ships float.steel ships float.
Applications o
Applications of Archimedes’ prinf Archimedes’ principle can be ciple can be founfound in ships, d in ships, submarines, hot-air ballonsubmarines, hot-air ballons and thes and the hydrometer. hydrometer. Buoyant force Buoyant force Weight Weight Floating Floating
=
=
=
=
Buoyant force Buoyant force Weight Weight Floating Floating •• Displaced water weight < ball weightDisplaced water weight < ball weight Sink
Sink
•
• Displaced water weight = hull Displaced water weight = hull weightweight Float
Example 6: Example 6:
The buoyant force acting on t
The buoyant force acting on the object will decrease when the:he object will decrease when the:
weight weight of of the the object object decreasedecrease
Example 7: Example 7:
A
A concrete concrete slab slab weight weight isis , when it is fully submerged under the sea, its apparent weight is, when it is fully submerged under the sea, its apparent weight is
. Calculate the density of the sea water if the volume of the sea water displaced by the. Calculate the density of the sea water if the volume of the sea water displaced by the concrete slab is concrete slab is .. Solution: Solution: N N 48 48 102 102 150 150 ight ight Apprent we Apprent we weight weight Actual Actual force force Buoyant Buoyant According to Archi
According to Archimedes’ Princimedes’ Principle:ple:
3 3 6 6 6 6 1019 1019 81 81 .. 9 9 10 10 4800 4800 48 48 81 81 .. 9 9 10 10 4800 4800 48 48 displaced displaced water water sea sea of of weight weight force force Buoyant Buoyant m m kg kg Vg Vg mg mg F F B B Object Object Liquid Liquid Buoyant Force Buoyant ForceFigure below shows a boat loaded with some goods floating on the sea. The density of the sea is Figure below shows a boat loaded with some goods floating on the sea. The density of the sea is
..
(a) Calculate the weight of the boat. (a) Calculate the weight of the boat.
(b)
(b) Figure below shows the Figure below shows the situation of maximum loading of situation of maximum loading of the boat.the boat.
Calculate the additional weight of goods that has to be added to the boat to reach this Calculate the additional weight of goods that has to be added to the boat to reach this situation. situation. Solution: Solution: (a) (a) N N Vg Vg 14994 14994 8 8 .. 9 9 5 5 .. 1 1 1020 1020 water water sea sea of of Weight Weight boat boat the the of of Weight Weight (b)
(b) For For maximum maximum loading;loading;
N N Vg Vg 44982 44982 8 8 .. 9 9 5 5 .. 4 4 1020 1020 displaced displaced water water sea sea of of Weight Weight weight weight Additional Additional boat boat of of Weight Weight Therefore; Therefore; N N
29988
29988
14994
14994
44982
44982
added
added
be
be
to
to
goods
goods
of
of
weight
weight
Additional
Additional
Tutorial 5a (Density and
Tutorial 5a (Density and Pressure)Pressure)
1.
1. An An object has object has a a mass mass ofof and a volume of and a volume of . What is the . What is the density of the object?density of the object? 2.
2. A A substancsubstance e having having a a density density ofof . What is . What is the volume of the substance if the mass isthe volume of the substance if the mass is
?? 3.
3. A A room room with with a a dimension ofdimension of is filled with is filled with of air. What is the mass of of air. What is the mass of the air?
the air? 4.
4. A A measuring cylinder measuring cylinder is is filled up filled up with with aa liquid having a mass of liquid having a mass of . What is the density. What is the density of the liquid?
of the liquid? 5.
5. A A liquid having liquid having a a density ofdensity of . If 1cm. If 1cm33 of the liquid turn into of the liquid turn into vapors, find the vapors, find the density of the vapor.
density of the vapor. 6. Liquid
6. Liquid and and having a density of having a density of and and respectively. Without any respectively. Without any changeschanges of volume, a
of volume, a liquid liquid is added to is added to liquid liquid . Calculate the additional density of the liquid.. Calculate the additional density of the liquid. 7. A
7. A pressure is exerted on the pressure is exerted on the floor. Calculate the force acting on the floor. floor. Calculate the force acting on the floor. 8.
8. A A wood block wood block with a with a dimension ofdimension of height having a mass of height having a mass of
. Determine the pressure exerted by the wood block.. Determine the pressure exerted by the wood block. 9.
9. A A blade wblade with a ith a dimension ofdimension of having a force of having a force of to cut a meat. What is the to cut a meat. What is the pressure exerted by the blade?
pressure exerted by the blade? 10.
10. A A tank tank with with a a dimension dimension ofof is filled with paraffin. is filled with paraffin. (( ))
Calculate : Calculate :
a) The pressure exerted on the base of the tank a) The pressure exerted on the base of the tank b) The force that acted on the
b) The force that acted on the base of the tankbase of the tank 11. A density of seawater is
11. A density of seawater is . What is the pressure exerted by the seawater at a vertical. What is the pressure exerted by the seawater at a vertical depth of
depth of . ( Given:. ( Given: ))
Tutorial 5b
Tutorial 5b (Pascal’ Principle)(Pascal’ Principle)
1.
1. Figure shows a simple hydraulic system. PisFigure shows a simple hydraulic system. Piston A & B ton A & B has a cross sectional area ofhas a cross sectional area of
and and respectively. respectively.
a)
a) What is What is the pressure the pressure at pat piston A, iston A, whenwhen load is placed on it? load is placed on it? b)
b) What is What is the the pressure at pressure at piston B.piston B. c)
c) If load W placed aIf load W placed at piston B, determine t piston B, determine the load W that cathe load W that can be pushed by n be pushed by piston B.piston B.
2.
2. Figure shows a Figure shows a pipe system that pipe system that is filled wis filled with oil. If ith oil. If the piston A the piston A is pushed is pushed withwith force: force: (a) What is the pressure exerted by the oil
(a) What is the pressure exerted by the oil (b)
Cross sectional area of
Cross sectional area of ,, and and Cross sectional Cross sectional area area ofof
a)
a) What What is is the the pressure at pressure at pistonpiston , if, if force acting on a piston force acting on a piston ?? b)
b) If If loadload placed at piston placed at piston , determine the load, determine the load that can be pushed by piston that can be pushed by piston .. c)
c) If the distance moveIf the distance moved by the piston A is d by the piston A is 1.5m, what is the d1.5m, what is the distance moved istance moved by piston B?by piston B? d)
d) If If loadload is being replaced with 210kg load, what is the force acted on piston is being replaced with 210kg load, what is the force acted on piston in order to in order to support the load at piston
support the load at piston ??
Tutorial 5c
Tutorial 5c (Archimede(Archimedes’ s’ Principle)Principle)
1. An empty boat having a weight of
1. An empty boat having a weight of is floating statically . is floating statically . ((
))a)
a) What What is is the the buoyant buoyant force?force? b)
b) What is What is the volume the volume of of the displaced the displaced water?water? 2.
2. A A cube of cube of metal having metal having a a volume ofvolume of is completely submerged in ….is completely submerged in …. a) a) water water ((
)) b) b) oil oil ((
)) c) c) Oxygen Oxygen ((
)) According to theAccording to the following, wfollowing, what is the buoyant hat is the buoyant force?force? 3.
3. A A metal block metal block having having a a weight ofweight of is completely being submerged in to the water. The weight is completely being submerged in to the water. The weight of the block when it completely submerged is
of the block when it completely submerged is a)
a) What iWhat is the s the volume volume of thof the bloe blockck b)
b) What is What is the dethe density onsity of the f the block (block (
).).4.
4. Figure (a) shows an object is weighed in air and found to have a weight of 2.0N. While Figure (b)Figure (a) shows an object is weighed in air and found to have a weight of 2.0N. While Figure (b) shows the object is
shows the object is completelcompletely submerged into the water.y submerged into the water. a)
a) What iWhat is the s the mass mass of the of the object?object? b)
b) What What is is the the buoyant buoyant force?force? c)
c) What is What is the mass the mass of the of the displaced displaced water?water? d)
Answer 5a: Answer 5a: 1) 1) 1500 1500 kg/mkg/m33 2) 2) 0.0003 0.0003 mm33 3) 3) 97.5 97.5 kg kg 4) 4) 800 800 kg/mkg/m33 5) 5) 2 2 kg/mkg/m33 6) 840kg/m6) 840kg/m33 7) 7) 20 20 N N 8) 8) 159.6159.6 PaPa 9) 9) 8.3 8.3 x x 101066 Pa Pa 10a) 15696 10a) 15696 Pa Pa 10b) 10b) 188352 188352 N N 11) 11) 303129 303129 PaPa Answer 5b: Answer 5b: 1a)
1a) 200kPa 200kPa b) b) 200kPa 200kPa c)100 c)100 kgkg 2a)
2a) 66.67 66.67 kPa kPa b) b) 13.33N13.33N 3a)
3a) 500 500 Pa Pa b) b) 600N 600N c) c) 0.1 0.1 m m d) d) 140N140N
Answer 5c: Answer 5c: 1a)
1a) 2000 2000 N N 1b) 1b) 0.2 0.2 mm33 2a) 2a) 7.848 7.848 N N 2b) 2b) 6.28 6.28 N N 2c) 2c) 0.12 0.12 NN 3a)
3a) 2 2 x x 1010-5-5 m m33 3b) 3b) 5000 5000 kg/mkg/m33 4a)
4a) 0.2 0.2 kg kg 4b) 4b) 0.2 0.2 N N 4c) 4c) 0.2 0.2 kg kg 4d) 4d) 2 2 x x 1010-4-4 m m33
Minimum requirement assessment task for this topic: Minimum requirement assessment task for this topic: 1 Theory Test & 1 End-of-Chapter
1 Theory Test & 1 End-of-Chapter
Specification of Theory Test: CLO1- C1 & CLO3-C2, A1 Specification of Theory Test: CLO1- C1 & CLO3-C2, A1 Specification of Labwork: CLO2- C2, P1
Specification of Labwork: CLO2- C2, P1 ****************
**************************************************************************************************************************************************************************************** COURSE LEARNING OUTCOME (CLO)
COURSE LEARNING OUTCOME (CLO)
Upon completion of this topic, students should be
Upon completion of this topic, students should be able to:able to: 1.
1. Identify Identify the bathe basic sic concept concept of of solid solid and and fluid (fluid (C1)C1) 2.
2. Apply coApply concept oncept of solid f solid and fluid and fluid to prto prove reove related physilated physics prics principles. nciples. (C2,P1)(C2,P1) 3.
3. Apply the Apply the concept oconcept of solid f solid and fluid and fluid in real in real basic enbasic engineering pgineering problems. roblems. (C2, A1(C2, A1)) ****************
**************************************************************************************************************************************************************************************** Compliance to PLO
Compliance to PLO
PLO 1, LD1 (Knowledge)-Test 2PLO 1, LD1 (Knowledge)-Test 2
PLO 2, LD2 PLO 2, LD2 (Practical Skills)- Experiment 3(Practical Skills)- Experiment 3