ARCHIMEDES’ PRINCIPLECh9 (Ch11 Hons.)

Pressure

• Pressure = Force (N) / Area (m2)

• Unit is pascal (Pa) which is VERY small.

• Also use atm, or psi (pounds/inch2)

• Pressure is useful as it expresses the strength of materials (Ex. a container can withstand 150 psi.

• As usual, knowing two variables allows you to calculate the third. ( P = F/a )

Pressure at depth• FLUID – a liquid or a gas.

• Pressure is higher:lower at the bottom of a fluid.

P = P0 + ρ g hWhere P0 is atmospheric pressure

P is pressure at depth,

ρ is density of fluid,

g is acc. Due to gravity

h is the height below the fluid top surface

• Archimedes (287-212 BC), pre-eminent Greek mathematician and inventor, who wrote important works on plane and solid geometry, arithmetic, and mechanics.

Who is Archimedes?

Reason for buoyant force….

The law• Archimedes' Principle, states that

when an object is totally or partially immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced.

The principle is most frequently applied to the behaviour of objects in water, and helps to explain floating and sinking, and why objects seem lighter in water. It also applies to balloons in the air.

The key word in the principle is “upthrust” (or buoyant force), which refers to the force acting upward to reduce the apparent weight of the object when it is in the water or under water.

for example, a metal block with a volume of 100 cm3

is dipped in water, it displaces an equal volume of water, which has a weight of approximately 1 N. The block therefore seems to weigh about 1 N less.

UPTHRUST AND BUOYANT FORCE

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

Density and Buoyancy

From Archimedes’s Principle :Buoyant Force = Weight of fluid displaced = mg (note : F = ma) = Vg (note : = m ) V

Thus FB = V gWhere ……FB = Buoyant Force or Upthrust

= Density of fluid V = Volume of fluid displaced or the volume of the object if it is totally submerged in

the fluid.

Buoyant Force and Flotation

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 Flotation

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 modeling clay 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

Archimedes Principle problems• Use fact that weight of displaced fluid = Fb

• Remember ρ = m / V

• Fg (object) / Fb = ρo / ρ f

• Where Fg (object) is weight of object (or weight in air)

• Fb is buoyant force

• ρ is density (of object or fluid)

1. The weight of the rock in air is 0.85N. When it is completely submerged in water, its weight is 0.45N. What is the buoyant force acting on the rock when it is completely submerged in the water ?

Solution :

Buoyant force = Actual weight – Apparent weight

= 0.85 – 0.45

= 0.4N or 0.4 kg m/s2

Example

2. A concrete slab weight 180N. When it is fully submerged under the sea its apparent weight is 105N.

Calculate the density of the sea water if the volume of the sea water displaced by the concrete slab is 4800 cm3. [ g = 9.8 m/s2 ]

Solution :

Buoyant force = actual weight – apparent weight

= 180 – 105

= 75N

According to Archimedes’s principle

Buoyant force = weight of sea water

displaced

Therefore,

F = ρVg

so…. ρ = F / Vg

= 75kg m/s2 / (4800 x 10-6 m2 x 9.8m/s2 )

= 1530.61 kg / m-3

ExampleExample A rectangular ferry boat is 15m long by 6m A rectangular ferry boat is 15m long by 6m

wide. A 2500kg truck drives onto the ferry. If wide. A 2500kg truck drives onto the ferry. If the density of water is 1000kg/mthe density of water is 1000kg/m33 . Find out . Find out how far down the hull of the ferry sinks.how far down the hull of the ferry sinks.

Δ FΔ Fbb = = ΔΔ weight of water displaced. weight of water displaced.

mmtt g = V g = Vww ρρww g g

mmtt = V = Vww ρρww = (l w = (l w hh) ) ρρww , we need to solve h, we need to solve h

h = mh = mtt / l w / l w ρρw w = 2500kg/ 15m 6m 1000kg/m= 2500kg/ 15m 6m 1000kg/m33

= =

HomeworkHomework

Regular: P343 Concepts Q1-7 Write out in full Regular: P343 Concepts Q1-7 Write out in full sentences. You must read the section on sentences. You must read the section on Archimedes principle and use the ideas within.Archimedes principle and use the ideas within.

Honors: Question 11 – 17 P352 (Concepts)Honors: Question 11 – 17 P352 (Concepts)

22ndnd day homework day homework HONORS: P 353 Q2 (8300Lb), 5 (317mHONORS: P 353 Q2 (8300Lb), 5 (317m22) )

(density), 10 (1.3 million lb),12 (32N),13 (density), 10 (1.3 million lb),12 (32N),13 (4.76 X10(4.76 X1044),14 (24) (pressure), 19 (1.26x ),14 (24) (pressure), 19 (1.26x 101055Pa), 21 (3.5x 10Pa), 21 (3.5x 1066 N), 23 (1.2X10 N), 23 (1.2X10 55 Pa) Pa) (pressure with depth)(pressure with depth)

Regular : P327 Q 2 (2.7X10Regular : P327 Q 2 (2.7X1022), 3 (1.2X10), 3 (1.2X10 3 3

Pa, 6.0 X10Pa, 6.0 X10 -2 -2 N . P330 Q1 (1.11X10 N . P330 Q1 (1.11X108 8 Pa),Pa),

2 (8.25X10 2 (8.25X10 5 5 Pa), 3 (2.7X10 Pa), 3 (2.7X10 44 Pa) , 4 (1.03 Pa) , 4 (1.03 X10X1055 Pa, 1.05X10 Pa, 1.05X1055 Pa). P331 Q1,2,3,4 Pa). P331 Q1,2,3,4

Pascal’s principlePascal’s principle Honors P329 , reg. P326Honors P329 , reg. P326 Any change in pressure in an enclosed, Any change in pressure in an enclosed,

static fluid is transmitted equally to all parts static fluid is transmitted equally to all parts of the fluid and the container walls.of the fluid and the container walls.

What does static mean?What does static mean? Examples of Pascal’s principle include a Examples of Pascal’s principle include a

hydraulic car jack, a hydraulic ram etc. hydraulic car jack, a hydraulic ram etc.

–

Ex: car jack has small piston area = 1.0 X10Ex: car jack has small piston area = 1.0 X10 -6-6 m m22

and a large piston of 0.50 X10and a large piston of 0.50 X10-4-4 m m22. . – How much can it lift if the small piston is pushed How much can it lift if the small piston is pushed

down with a force of 180N? down with a force of 180N?

– Given:Given: Pascal says PPascal says P1 1 = P= P22 ,So F ,So F1 1 / A/ A11 = F = F22 / A / A22

– Rearrange to get FRearrange to get F2 2 on its own Fon its own F2 2 = A= A2 2 FF11 / A / A11

– So substitute to solve…. So substitute to solve….

– Tonight for homework do P Tonight for homework do P

P1P2

Important general rule…..Important general rule…..

Fluids flow from areas of ______ pressure Fluids flow from areas of ______ pressure to areas of ____ pressure.to areas of ____ pressure.

Pressure in a fluid at any point pushes Pressure in a fluid at any point pushes equally in all directionsequally in all directions

Bernoulli’s PrincipleBernoulli’s Principle Daniel Bernoulli (Swiss) in the early 1700’s Daniel Bernoulli (Swiss) in the early 1700’s

observed that air and other fluids do not observed that air and other fluids do not behave as one would expect. Specifically:behave as one would expect. Specifically:

““A fluid moving at a high velocity exhibits A fluid moving at a high velocity exhibits ____ pressure . A fluid moving at a low ____ pressure . A fluid moving at a low velocity exhibits ____ pressure.”velocity exhibits ____ pressure.”

This principle explains many phenomena: This principle explains many phenomena: Being pushed into path of speeding vehicle, Being pushed into path of speeding vehicle,

a carburetor and many more….. a carburetor and many more…..

Shower curtain pushed inShower curtain pushed in

House roof coming off in tornadoHouse roof coming off in tornado

Curve balls in any ball sportCurve balls in any ball sport

Lift from an airfoil (wing)Lift from an airfoil (wing)

Where to find Bernoulli’s Where to find Bernoulli’s Principle in your book..Principle in your book..

Honors: P 343- P344 Honors: P 343- P344

Regular : P335Regular : P335