CHAPTER 2

HIDROSTATICS

OUTLINE 2.1 DENSITY 2.2 PRESSURE 2.3 PASCAL’S LAW 2.4 ARCHIMEDES PRINCIPLE

•mass, m is measured in kilograms (kg)•volume, V is measured in cubic metres (m3)density, ρ is measured in kilograms per cubic metres (kg/m3)

2.1 DENSITY

m

ρ V

What is density?

)(

)(,

vvolume

mmassdensity

2.1.1 SPECIFIC GRAVITY, s.g

Is defined as:

s.g= Density of substanceDensity of water, 1000kgm-3

• the unit is : no unit (dimensionless)

EXAMPLE 2.1 Find the mass and weight of the air in a

classroom at 24°C with 10.0m x 8.0 m floor and ceiling 3.0 m high.

Solution: given density of air = 1.20kg/m3

m air= ρair x V classroom

= (1.20kg/m3)((10x8x3)m3)

=288kg Hence, weight of air: w=mg

=(288kg)(9.8m/s2)=2822.4N

QUESTION 1

Calculate the volume of a liquid of mass 45 kg and density 900 kg/m3.

ρ = m / V

becomes:V = m / ρ= 45 kg ÷ 900 kg/m3 volume of liquid = 0.05 m3

QUESTION 2When a small stone is immersed into the water inside a measuring cylinder the level increases from 20.0 to 27.5 ml. Calculate the density of the stone in g/cm3 if its mass is 60g.

Volume of stone = (27.5 – 20.0) ml= 7.5 cm3

ρ = m / V= 60g / 7.5cm3 density of the stone = 8.0 g/cm3

QUESTION 5

Calculate the density in g/cm3 and kg/m3 of a metal cylinder of radius 2cm, height 3cm and mass 400g.

Volume of a cylinder = π x r2 x h= π x (2cm)2 x 3cm= 3.142 x 4 x 3= 37.7 cm3

ρ = m / V= 400 g / 37.7 cm3 metal density = 10.6 g/cm3

= 10 600 kg/m3

9

QUESTION 65. (II) A bottle has a mass of 35.00 g when empty

and 98.44 g when filled with water. When filled with another fluid, the mass is 88.78 g. What is the specific gravity of this other fluid?

Solution: Take the ratio of the density of the fluid to that of water, noting that the same volume is used for both liquids.

fluid fluid fluidfluid

water waterwater

88.78 g 35.00 g0.8477

98.44 g 35.00 g

m V mSJ

m V m

SGfluid

ANSWERSdensity mass volume

240 g 40 cm3

3000 kg/m3 4500 kg

0.80 g/cm3 80 cm3

9 kg 0.003 m3

6 g/cm3

3 g/cm3

1.5 m3

64 g

Complete:

Choose appropriate words to fill in the gaps below:

Density is equal to ______ divided by _________ and can be measured in kilograms per ______ metres.

A density of _______kg/m3 is the same as a density of 1 g/cm3. This is the density of ________.

The ________ of a stone can be measured by immersing the stone into water. The volume of water ________ by the stone is equal to the volume of the stone. The volume of the water displaced is found using a _________ cylinder.

displaced

cubic

1000

mass water

volume

WORD SELECTION:densit

ymeasuring

displaced

cubic

1000

mass

water

volume

density

measuring

2.2 PRESSURE, P (IN A FLUID)

When in fluid (either in liquid or gas) is at rest, it exerts a force perpendicular to any surface in contact with it, such as a cointaner wall or the surface of a body immersed in it.

Consider a small surface of area, A and a normal force, Fn exerted by the fluid perpecdicular to the surface:

Pressure, P at any point is defined as the normal force per unit are: P=F/A

The SI unit: N/m2 or Pascal, Pa1 atm = 1.013 x 105 Pa or 15 lbs/in2

*One atmosphere is the pressure exerted on us every day by the earth’s atmosphere.

PRESSURE, Ppressure = force area

p = F A

units:force, F – newtons (N)area, A – metres squared (m2)pressure, p – pascals (Pa)

14

The pressure is the same in every direction in a fluid at a given depth.

Pressure varies with depth.

P = F = rAhg so P = rgh

A A

15

A FLUID AT REST EXERTS PRESSURE PERPENDICULAR TO ANY SURFACE THAT IT CONTACTS. THERE IS NO PARALLEL COMPONENT THAT WOULD CAUSE A FLUID AT REST TO FLOW.

16

PROBLEM 79. (I) (a) Calculate the total force of the

atmosphere acting on the top of a table that measures (b) What is the total force acting upward on the underside of the table?

9. (a) The total force of the atmosphere on the table will be the air pressure times the area of the table.

5 2 51.013 10 N m 1.6 m 2.9 m 4.7 10 NF PA (b) Since the atmospheric pressure is the same on the underside of the

table (the height difference is minimal), the upward force of air pressure is the same as the downward force of air on the top of the table,

54.7 10 N

m. 2.9m 6.1

2.2.1 THE EFFECT OF GRAVITY ON FLUID PRESSURE(ATMOSPHERIC PRESSURE AND GAUGE PRESSURE)

The pressure P1 on the surface of the water is 1 atm, or 1.013 x 105 Pa. If we go down to a depth h below the surface, the pressure becomes greater by the product of the density of the water , the acceleration due to gravity g, and the depth h. Thus the pressure P2 at this depth is

h h hp2 p2

p2

p1 p1p1

ghpp 12

18

In this case, p2 is called the absolute pressure -- the total static pressure at a certain depth in a fluid, including the pressure at the surface of the fluid

The difference in pressure between the surface and the depth h is gauge pressure

ghpp 12

Note that the pressure at any depth does not depend of the shape of the container, only the pressure at some reference level (like the surface) and the vertical distance below that level.

h h hp2 p2 p2

p1 p1p1

EXERCISE 8 14.(II) What are the total force and the

absolute pressure on the bottom of a swimming pool 22.0 m by 8.5 m whose uniform depth is 2.0 m? (b) What will be the pressure against the side of the pool near the bottom?

19

5 2 3 3 2

0

5 2

5 2 7

1.013 10 N m 1.00 10 kg m 9.80 m s 2.0 m

1.21 10 N m

1.21 10 N m 22.0 m 8.5 m 2.3 10 N

P P gh

F PA

20

(b) The pressure against the side of the pool, near the bottom, will be the same as the pressure at the

bottom,

5 21.21 10 N mP

21

2.3 PASCAL’S PRINCIPLE Pascal’s Principle - if an external pressure is

applied to a confined fluid, the pressure at every point within the fluid increases by that amount. Applications: hydraulic lift and brakes

Pout = Pin

And since P = F/a

Fout = Fin

Aout Ain

Mechanical Advantage:

Fout = Aout

Fin Ain

22

2.4 Buoyancy and Archimedes’ Principle

This is an object submerged in a fluid. There is a net force on the object because the pressures at the top and bottom of it are different.

The buoyant force is found to be the upward force on the same volume of water:

23

The net force on the object is then the difference between the buoyant force and the gravitational force.

24

If the object’s density is less than that of water, there will be an upward net force on it, and it will rise until it is partially out of the water.

25

For a floating object, the fraction that is submerged is given by the ratio of the object’s density to that of the fluid.

SINKING AND FLOATING OBJECTS

The reading of spring balance is 2.7 N

The reading of spring balance is 1.7 N

What is the reading of spring balance if the wood is attached to it ? ZERO

BUOYANT FORCE AND FLOATATION

Buoyant force = weight the object floats and stationary

Buoyant force > weight the object moves up

Buoyant force < weight the object moves down

THE LAW OF FLOATATION

A floating object displaces its own weight of fluid in which it floats.

warm fresh water

cold fresh water

warm sea water

cold sea water

THINK!!!!!

1. Why the depth of ship immersed in the water different?

Fresh water less dense than sea water and warm water less dense than coldwater so warm fresh water need to be displaced more to keep the uptrust force equal with weight of the boat so it still can float.

2. If the plasticine is formed into a ball, it will sink.But when it is formed into a hull it will float. Why?

-

BECAUSE…..

APPLICATIONS

Hot air balloon1. rises upwards (Upthrust > Weight of hot air (helium gas) +

weight of airship fabric + weight of gondola + weight of passengers.)( balloon expand)

2..descends(Upthrust < Weight of hot air (helium gas) +

weight of airship fabric + weight of gondola + weight of passengers.)(balloon shrinks)

3. stationary (Upthrust = Weight of hot air (helium gas) +

weight of airship fabric + weight of gondola + weight of passengers.)( balloon size uncanged)

The density of sea water varies with location and season. To ensure that a ship is loaded within safe limits , the Plimsoll line marked on the body of the ship acts as a guide.

PLIMSOLL LINE OF THE SHIP

If ballast tanks empty Upthrust > weight submarine rises to surfaceIf ballast tanks full Upthrust < weight submarine sinks to bottom

SUBMARINE

HYDROMETER

An hydrometer is an instrument used to measure the density of a liquid.

In a liquid of lesser density , the hydrometer is more submerged.The hydrometer floats higher in a liquid of higher density.

lead shot to make it float upright

39

22. (I) A geologist finds that a Moon rock whose mass is 9.28 kg has an apparent mass of 6.18 kg when submerged in water. What is the density of the rock?

22. The difference in the actual mass and the apparent mass is the mass of the water displaced by the rock. The mass of the water displaced is the volume of the rock times the density of water, and the volume of the rock is the mass of the rock divided by its density. Combining these relationships yields an expression for the density of the rock.

rockactual apparent water rock water

rock

3 3 3 3rockrock water

9.28 kg1.00 10 kg m 2.99 10 kg m

9.28 kg 6.18 kg

mm m m V

m

m

40

24.(II) A crane lifts the 18,000-kg steel hull of a ship out of the water. Determine (a) the tension in the crane’s cable when the hull is submerged in the water, and (b) the tension when the hull is completely out of the water.

24.(a) When the hull is submerged, both the buoyant force and the tension force act upward on the hull, and so their sum is equal to the weight of the hull. The buoyant force is the weight of the water displaced.

buoyant

hull waterbuoyant hull water sub hull water hull

hull hull

3 34 2 5 5

3 3

1

1.00 10 kg m 1.8 10 kg 9.80 m s 1 1.538 10 N 1.5 10 N

7.8 10 kg m

T F mg

mT mg F m g V g m g g m g

41

24. (b)When the hull is completely out of the water, the tension in the crane’s cable must be equal to the weight of the hull.

4 2 5 51.8 10 kg 9.80 m s 1.764 10 N 1.8 10 NT mg

42

34.(III) A 5.25-kg piece of wood floats on water. What minimum mass of lead, hung

from the wood by a string, will cause it to sink?34.For the combination to just barely sink, the total

weight of the wood and lead must be equal to the total buoyant force on the wood and the lead.

0.50SG

weight buoyant wood Pb wood water Pb water

wood Pb water waterwood Pb water water Pb wood

wood Pb Pb wood

1 1

F F m g m g V g V g

m mm m m m

water

wood woodPb wood wood

water

Pb Pb

1 11 1 10.50

5.25 kg 5.76 kg11 11 1

11.3

SGm m m

SG