Chapter 5-Solid and Fluid

8/13/2019 Chapter 5-Solid and Fluid

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CHAPTER 5: SOLID AND FLUID

BB101ENGINEERING SCIENCE

5.1 Character ist ics Of Sol ids, Liquids And Gases

Individual atoms can combine with other atoms to form molecules.

Regardless of the type of molecule, matter normally exists as either a solid, a liquid, ora gas.

We call this property of matter is the state of the matter The three normal states of matter have unique characteristics which are listed on the

table.

ASPECT SOLIDS LIQUIDS GASES

Shape Fixed volume and shape Assumes the shape

of the part of the

container

Assumes the shape

and volume of its

container

Compressible Not easily compressible Not easilycompressible

Compressible

Flows of particles Does not flow easily Flow easily Flow easily

Microscopic view

Tightly packed, usually

in a regular pattern

Vibrate (jiggle) but

generally do not move

from place to place

Close together with

no regular

arrangement

Vibrate, move about

and slide past each

other

g

Well separated with

no regular

arrangement

Vibrate and move

freely at high speed


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5.2 Density

Example 1:

A mass of the block is 45 kg and its volume is 0.5 m3. Calculate the density of the block.

Solut ion :

Example 2:A concrete block with a density of 2100 kg/m

3. Calculate the relative density of it.

Solut ion :

v

m

=3

5.0

45

m

kg

= 90 kg/m3

Definition:as the mass (m) per unit volume (v)

of a substance

Formula:

densi ty = mass

volume

= m ( kg /m3)

v ( m3)

Unit: kg/m3@ kgm-3

Densit

Relative Density

Definition:- Relative density is defined as the ratio of thedensity of substance to the density of water

Formula:

Relat ive densi ty = Densi ty of substance(kg/m3)

Densi ty of w ater(kg/m3

)

( water = 1000 kg/m3)

No unit.

Relative density is also expressed as a SPECIFIC

GRAVITY

Density of substance ,

relative =

water

cesubs

tan

=3

3

/1000

/2100

mkg

mkg

= 2.1


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Example 3:

Calculate the volume of a solid with a density of 2.65 kg/m3and its mass is 39.24N.

Solut ion :

v

v

m

kgm

m

mgW

465.2

481.9

24.39

81.924.39

v = 1.51 m-3

Exercise 1:

Calculate the densityfor a rectangular block with mass 0.72kg.The volume of the block is0.00297m

-3. (Ans : 242.4kgm

-3)

Exercise 2:

Aluminium has a density of 2710kgm-3

.Calculate its relative density.(Ans : 2.71)

Exercise 3:

A block with weight 20 N and its volume is 30 cm x 15 cm x 20 cm. Calculate the value ofdensity of substance.

(Ans : 226.7kgm-3

)


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P =A

F

2002/ mN =

28m

F

F = 1600 N

5.3 Pressure

Definition:- Pressure is defined as the normal force per unit area acting on asurface.

The S.I unitof pressure is Pascal (Pa)or Newton per meter square (N/ m2or Nm

2)

Example 4:A car has a weight of 1800N and rests on four tyres, each having a surface area ofcontact with the ground of 0.006m

2.What is the pressure exerted by the tyres on the

ground?

Solut ion :

2006.0

1800

m

N

A

FP

= 30 000 Pa

Example 5:The wind pressure on the wall is 200 Pa. If the wall has an area 0f 8 m

2, what is the force

on it?

Solut ion :

Pressure = Normal Force

Area

P = F ( N )A ( m2)


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Example 6:

A fruit seller uses a knife with a sharp edge and a cross-sectional area of 0.5 cm2to cut

open a watermelon.If the force applied on the knife is 18 N, what is the pressure exertedby the knife on the watermelon?

Solut ion :

24

242

105.0

18

100.55.0

m

N

A

FP

mcm

= 360000 Pa

Exercise 4:

In a hydraulic brake, a force of 50N is applied to a piston of area 5.0 x 10-4

m2.What is the

pressure transmitted throughout the liquid? ( Ans : 100,000 Pa @ 10 kPa )

Exercise 5:

A fruit seller uses a knife with a sharp edge and a cross-sectional area of 0.5 cm2to cut

open a watermelon.a) If the force applied on the knife is 18 N, what is the pressure exerted by the knife

on the watermelon? [ans: 3.6 x 105Pa]

b) After that, he cuts open a papaya using the same knife by exerting a pressure of2.7 10

5Pa. Calculate the magnitude of force applied to cut the papaya.

[ans: 13.5 N]


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Exercise 6:The mass of an elephant is 5000 kg and the area of one its foot is 600 m

2. If the elephant

stands on with two feet and four feet, which one will exert the greatest pressure on theground?

- stands on two feet - stands on fou r feet

The elephant stands on _________ will exert the _____________ on the ground

5.3.1 Variation Of Pressu re With Depth

The deeperyou go into a liquid or gas, the greaterthe pressureon you from thesurrounding fluid will be. The reason for the increased pressure is that the deeperinto a fluid you go, the more fluid, and thus the more weight, you have over top ofyou.

We can calculate the variation of pressure with depth by considering a volume of fluidof height hand cross-sectional areaA(see Fig. 5.3.1).

Figure 5.3.1:Variation of Pressure with Depth

If this volume of fluid is to be in equilibrium, the net force acting on the volume must bezero. There are three external forces acting on this volume of fluid. These forces are:

P =A

F P =

A

F


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1. The force PTAdue to the pressure on top of the volume of fluid. If the fluid is open to the air,PT= PO= 1.01 x 10

5Pa, which is atmospheric pressure.

2. The weight of the volume of fluid, W= mg. Remembering the definition of density,= m/v, and that the volume of the fluid may be calculated as V=Ah, we can writethe weight of the fluid as W= ghA.

3. The force pushing up on the bottom of the volume of fluid, PBA, due to the fluid below thevolume under consideration. If we take the up direction to be positive and add the forces weget

PBA- ghA- PTA= 0,

which gives

PB= PT+ gh.

This provides the general formula relating the pressures at two different points in a fluidseparated by a depth h.

Note: Onlythe density of the fluid and the difference in depthaffects the pressure. The shapeand size of the container are irrelevant.

5.3.2 Pressur e In A Liqu id

The pressure exerted by a static fluid depends only upon the depth of the fluid, thedensity of the fluid, and the acceleration of gravity.

The pressure from the weight of a column of liquid of area, Aand height, his

Pressure, P = g h

where, in SI Units: = fluid density, measured in kg/m3g = acceleration of gravity, measured in measured in N/kg

h = depth of fluid, measured in m


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Example 7:A coin sinks to a depth of 8 m below the surface of the sea. What is the water pressureon the coin? [ water density = 1000 kg/m

3]

Solut ion :

hgP

)8()81.9()1000( 78480 Pa

Example 8:A building is supplied with water from a water tank situated on its roof. If the waterpressure at the ground floor is 1.96 10

5Pa, what is the height of the building?

(Density of water = 1.0 103kg m

-3)

Solut ion :

9810

1096.1

))(81.9)(100.1(1096.15

35

h

h

ghP

= 19.98 m

Example 9:

The figure shows a container with water. What is thepressure of water exerted on fish? [ water density = 1000 kg/m

3]

Solut ion :

)35.0)(81.9)(1000(

ghP

= 3433.5 Pa

Exercise 7:

A small rock sinks to a depth of 3 m in a liquid of density 700 kg/m3. What is the pressure

of the rock?[Ans: 20601 Pa]

5 cm

40 cm


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Exercise 8:

A coin of density 500 kgm-3

sinks in a liquid. At what depth, will the coin be subjected to the the

pressure of 88290 Pa?

[Ans: 18 m]

Exercise 9:The figure below shows a reservoir on a hill. The reservoir supplies water to a water tankon a tall building. Calculate the water pressure at X. ( Density of water = 1000 kg/m

3).

5.4 Pascals Principle

Pascals principlestates that pressure exerted on an enclosed fluid (liquid) istransmitted equally to every part of the fluid (liquid)

Pascals Principle is also known as Pascals principle of transmission of pressure in fluids.

Figure below shows how a simple hydraulic system works based on Pascals principle.

This system consists of two cylinders of cross-sectional area A1and A2respectively. Eachof the cylinders is fitted with a piston and is filled with a liquid such as water or oil.

When a small force, F1is applied to the smaller piston X, a large force, F2is produced onthe larger piston Y.

Piston YOf Area, A2

d2d1

F2F1

Piston XOf Area, A1


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F2is produced by pressure exerting on the larger piston Y due to the pressure transmittedfrom piston X through the liquid.

Based on Pascals principle, the pressure acting at both pistons is the same pressure.Thus;

If the piston Xis moved by a distance d1, the piston Ywould move by a distance d2. Sincethe volume pf liquid transmitted is constant, thus;

Example 10:

Diagram below shows a hydraulic system. The cross-sectional areas of piston P is 0.002 m

2

..If a force of 15 N is applied to a smaller piston, what will be the pressure exerted on thePiston Q?

Pressure on piston X,P1= F1

A1

Pressure on piston Y,P2= F2

A2

Pressure on Piston X = Pressure on Piston Y

P1 = P2

F1 = F2A1 A2

A1d1 = A2d2

Input

Piston P

Output

Piston Q


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Solut ion :

QPiston

P

Pa

m

NP

QPistonessurePPistonessure

PP

2

21

21

7500

002.0

15

PrPr

Example 11:

The cylinder piston of a hydraulic jack has a cross-sectional area of 0.06 m2and the plunger

has across-sectional area of 0.002 m2.

i) The upward force for lifting a load placed on top the large piston is 9000 N.

calculate the downward force on the plunger required to lift this load.

Solut ion :

A1= 0.002 m

F1F2 = 9000 N

plunger

A2= 0.06 m

1P = 2P

1

1

A

F =

2

2

A

F

2

1

002.0 m

F =

206.0

9000

m

N

21

002.0 m

F = 2/150000 mN

1F =

22 002.0/150000 mxmN

= 300 N


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ii) If the distance moved by the plunger is 75 cm, what is the distance moved by

the large piston in cm?

Solut ion :

Example 12:

The diagrams shows a hydraulic jack.A force F of 20N applied on the piston P is required to lifta load of weight W by the piston Q. The cross sectional-area of input and output pistons arerespectively 64 cm2and 192 cm2.What is the mass of the load?

Solut ion :

22

21

192

)81.9(

64

20

PrPr

cm

m

cm

N

A

F

A

F

QPistonessurePPistonessure

PP

Q

Q

P

P

m=6.11kg

11dA = 22dA

cmxm 75002.0 2 = 2206.0 dxm

2d =

2

2

06.0

75002.0

m

cmxm

= 2.5 cm

W

F

P Q


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Exercise 10:

In a hydraulic brake, a force of 50N is applied to a piston of area 2.0 x 10-4

m2.What is the

pressure transmitted throughout the liquid?

Exercise 11:

Figure as below show a balanced hydraulic pressure pump. Given F1=20N, A1=0.2m2

and A2=0.3m2. Calculate the force at F2.

Exercise 12:Figure below shows a hydraulic jack used to lift a 3600 kg load. The cross-sectionalareas of input piston R and output piston S are 100 cm

2and 450 cm

2respectively.

[g =9.81 N kg-1

]

i) What is the weight of the load? ii) Calculate the pressure acting on the oil.

A1

F A2

F1 F2


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iii)What is the pressure exerted by the iv) Calculate the force, F, required to supportoil on piston R? the load.

v) If the displacement of piston R is 36 cm, calculate the displacement of piston S.

5.5 Archimedes Principle

Archimedes principlestates that an object which is partially or wholly immersed in a fluid(liquid or gas) is acted upon by an upward buoyant force equals to the weight of thefluid it displaces.

The buoyant force is created by the increase of pressurewith depth.

Consider an object of base area A, immersed in a liquid of density .

Using the formula of pressure, P = ghand P = F / A

The buoyant force = the resulting force acting upwards

= F2 - F1

= ( h2gA ) - ( h1gA )

h1

Base area = A

h2

F1= h1gA

F2= h2gA


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= ( h2h1 ) A g

= (Vliquid displaced) g

= (mliquid displaced) g

= weight of liquid displaced by the water

The principle of floatationstates that a floating body displaces its own weight of the

liquid in which it floats.

Hence;

Example 13:

If the submarine was floating on the sea surface,its volume of body under the sea surface is850m

3.The density of sea water is 1030kgm

-3,

( i ) Calculate the weight of water displaced.

Solut ion :

2

33

81.9875500

)81.9)(1030)(850(

mskg

kgmm

gVdisplacedwaterofweight

= 8588655 N

Hence;

Buoyant Force = weight of liquid displaced by the object

The weight of liquid displaced = weight of the body(i.e the buoyant force)

The volume of liquid displaced = volume of the body(i.e the buoyant force)

The buoyant / = the weight of liquid displaced ( w )

Upthrust Force = the mass of liquid ( m ) x gravitational acceleration ( g )

= [ density of liquid () x volume of liquid displaced (Vliquid) ] x gravitational acceleration (g)

= [ density of liquid () x volume of body submerged (Vbody) ] x gravitational acceleration (g)


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( ii ) Calculate the mass of the submarine

Solut ion :

Because the submarine is floating freely,therefore weight of submarine = weight of waterDisplaced.

= upward buoyant force= 8588655 N

281.9

8588655

,

ms

N

g

Wmsubmarineofmass

= 875500kg

( iii ) What is the principle applied in the calculation above

Archimedes Principle

Example 14:

Figure 5.5

Figure above shows a rectangular block of density 3200kgm-3

and dimensions 30cm x 45cm x25cm is hung to a spring balance and the block is fully submerged in water.[Density of water =1000kgm

-3]

(a) ( i ) Calculate the pressure by the water acting on the upper surface of the block.Solut ion :

5.081.91000

ghP

= 4905 Pa


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( ii ) Calculate the pressure by the water acting on the lower surface of the block.

Solut ion :

75.081.91000

ghP

= 7357.5 Pa

( iii ) What is the force acting on the upper surface of the block by the water?

Solut ion :

135.04905

APF

A

FP

= 662.2 N

( iv ) What is the force acting on the lower surface of the block by the water?

Solut ion :

135.05.7357

APF

A

FP

= 993.3 N

( v ) Calculate the net force acting on the block.

Nett Force = 338N

(vi) Determine the volume of water displaced by the block.

Solut ion :

2.045.03.0 displacedwaterofVolume

= 3.375 x 10-2

m3


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(vii) Calculate the mass of water displaced by the block.

Solut ion :

210375.31000

VdisplacedwaterofMass

= 33.75 kg

(viii) What is the weight of water displaced by the block?

Solut ion :

281.9)75.33(

mskg

mgW

= 331.1 N

( v ) What is the buoyant force on the block.

Solut ion :

The buoyant Force = The Weight of water displaced= 331.1 N

Example 15:

A boat of mass 200kg floats on lake. If the density of the water in the lake is 1000kgm -3,whatis the volume of the displaced water ?

Solut ion :

31000

200

kgm

kg

mv

v

m

= 0.2 m-3


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Exercise 13:An object of mass 50 kg and density 4 x 10

3kg/m

3. resting at the bottom of a swimming pool.

What is the buoyant force acting on the object [water density = 1000 kgm-3

and g = 9.81N/kg ]

Exercise 14:A man throws a box of mass 26kg in a sea.The box floats on the surface of the sea water.Ifthe volume of the box that immersed in the sea water is 0.025m

-3,what is the density of the

sea water ?

Exercise 15:

Figure as below show a wood block measuring 0.1m x 0.2 m x 0.2 m is floating on thesurface of the water. Determine:

a) The related principle for the floating wood block as in the figure [ans: Archimedes]b) The mass of the block if its density is 60 kg/m

3 [ans: 0.24 kg]

c) The weight of the water displaced caused by the wood block [ans: 2.3544 N]

water

Wood block


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20 cm

END OF CHAPTER

1. Give the definition, formula and S.I unit for:

(a)Density of substance b) Relative density / Specific gravity

2. A container as show as figure below is filled with oil. If the mass of the oil is 3500 g. Calculatethedensity of oil, the specific gravity of oiland the pressureof container if 5 kN force isapplied on it surface. [ans: 145.833 kg/m3, 0.148, 83.333 kPa]

3. A small rock sinks to a depth of 3 m in a liquid of density 700 kg/m3. At what depth, indifferent liquid of density 500 kg/m

3, will the rock experience the same pressure? [ans: 4.2 m]

4. A block of metal shows a reading of 120N when it is placed on a weighing machine in the air.However, it shows a reading of 50N when it is submerged in the oil.(Density of oil = 850 kgm

-3)

(a) Find

(i) the true weight of the metal

(ii) The apparent weight of the metal

(iii) The buoyant force acting on metal

(iv) The volume of the metal

(v) The density of the metal

(b) State and define the principle applied

30 cm

40 cm


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REFERENCES

1.) Physics Online,unwinnipegtheory,http://theory.uwinnipeg.ca

2.) Lim Peng Chew,Total Reference Physics,Form 4,Sasbadi Sdn. Bhd,pp 126.

3) Thong Kum Soon,You Kok Yeow,Progressive Assesment Physics,Form 4,PustakaYakin Pelajar Sdn. Bhd. Malaysia.
http://theory.uwinnipeg.ca/http://theory.uwinnipeg.ca/http://theory.uwinnipeg.ca/http://theory.uwinnipeg.ca/

Chapter 5-Solid and Fluid

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