3.5 Archimedes' Principle 73w2g

3.5 Applying Archimedes’ Principle

Learning Outcome:  Explain buoyant force,  Relate buoyant force to the weight of liquid displaced,  State Archimedes’ principle,  Describe applications of Archimedes’ principle,  Solve problems involving Archimedes’ principle.

Idea of buoyancy

Lesson 1

observation : The fishing net is more difficult to pull when more of the net is out of the water surface. Why ? Buoyant force is smaller when the net is less immersed into the water .

Buoyant force Top surface

Pressure = P1, area = A F1

Force, F1 = P1A (acting downward)

Bottom surface Pressure = P2, area = A

F2

Force, F2 = P2A (acting upward) P2 is greater than P1 hence F2 is greater than F1

The net upward force F2- F1 is known as buoyant force

Volume of fluid displaced Volume of object is equal to volume of fluid displaced

If an object is placed in the container, some water will overflow. Liquid is displaced by the object.

Buoyant force

Object above the water level.

The deeper the object is, the more liquid displaced, and

No buoyant force

the stronger the buoyant force

Hands-on activity 3.6 pg 61 Aim : To understand buoyant force. Observation : Weight of stone in air, W1 / N

Weight of stone in water, W2 / N

Discussion : 1.

W2 is less than W1. When an object is immersed in water, the weight becomes lighter.

2. The value of the buoyant force = Weight in air – weight in water = ???

Idea – factors affecting buoyant force

Lesson 2

Big buoyant force

Heavy ferry made of steel plates can float. Do you know why ?

Small buoyant force

A block of steel sink.

Archimedes’s Principle The relationship between buoyant force and weight of the fluid displaced was first discovered in the third century BC by the Greek scientist Archimedes. It is stated as follows : When an object is wholly or partially immersed in a fluid, it experiences a buoyant force equal to the weight of the fluid displaced.

A brief history

1

King of Sicily

2

Goldsmith

3

Archimedes

4

6

5

7 After this, Archimedes investigate the relationship between buoyant force and the weight of fluid displaced

An immersed body is buoyed up by a force equal to the weight of the fluid displaced.

Experiment 3.1 pg 62 Data Weight of plasticine in air , Wa /N

Weight of plasticine in water,Ww /N Weight of empty beaker Wo / N Weight of beaker plus displaced water ,W1/N Analysis of Data:

Buoyan t force

Apparent loss in weight of the plasticine = Wa – Ww Weight of water displaced =W1- Wo

Factors affecting buoyant force

Lesson 3

Buoyant force = weight of liquid displaced = mg Archimede s Principle = Vg Volume of fluid displaced,V A larger volume of the object submerged in the fluid will experience a larger buoyant force.

Buoyant Density of the force fluid,  A denser liquid exerts a larger buoyant force.

Gravitational field strength , g

Buoyant force = weight of fluid displaced (for all cases) F

F floating

F

W W W

Totally immersed Weight of object is greater than buoyant force

Partially submerged

Immersed just under water surface

Weight of object is equal to buoyant force

Weight of object is equal to buoyant force

Object is denser than fluid

Object is less dense than fluid

Density of object = density of fluid

Worked example (Exploring pg 189) A block of wood with a density of 25000 kg m-3 and a mass of 0.3 kg is immersed in a liquid of density 15000 kg m-3. Calculate the buoyant force acting on the block of wood. solution

Volume of liquid displaced = volume of wooden block = Mass density = 0.3 25000 = 1.2 x 10-5 m3

Buoyant = weight of liquid force displaced = Volume x density x g = 1.2 x 10-5 x1.5x104 x 10

= 1.8 N

Worked example (Exploring pg 189) A stone weighs 2.5 N. When it is fully submerged in a solution, its apparent weight is 2.2 N. Calculate the density of the solution if its volume displaced by the stone is 25 cm3. [ g= 9.8 N kg-1] solution

Buoyant force = 0.3 N Buoyant = weight of solution force displaced = Volume x density x g 0.3 = (25 x10-6)x x 9.8

 = 1224 kg m-3 25cm-3

Applications of Archimedes’ Principle Hydrometer – to determine the density of liquid Which liquid is denser ?

Weight of hydrometer = weight of liquid displaced A denser liquid has a smaller volume displaced.

Liquid A

Lead shots

Liquid A is denser than liquid B.

Liquid B

Applications of Archimedes’ Principle Ship – To ensure a ship is loaded within safe limit. Ship can float because the volume of water displaced is sufficiently large and buoyant force is equal to the weight of ship.

A ship will submerged deeper in fresh water because the density of fresh water is less than sea water.

To ensure a ship is loaded within safe limits, the Plimsoll line marked on the body of the ship acts as a guide.

View video

Applications of Archimedes’ Principle Hot air balloon – To rise and float a hot air balloon in the air. A hot air balloon displaces a large volume of air. The balloon ride upwards when buoyant force is greater than its weight. The balloon descends when the buoyant force is less than its weight. The balloon remains stationary in the air when buoyant force is equal to its weight.

View video

Applications of Archimedes’ Principle Submarine – to control the sinking and rising of submarine Submarine rises by forcing compressed air into the ballast tank

Submarine dives by itting water into the ballast tank

SUMMARY 1.When an object is immersed in liquid or gas, it experiences a buoyant force. 2. buoyant force =apparent loss in weight

3. Buoyant force = weight of fluid displaced. = Vg where V = volume of fluid displaced  = density of fluid

4. When an object is floating, buoyant force =weight of the object.

Evaluation A

B

C true It is also equal to the buoyant force

Evaluation

A B C

Evaluation

A B C D

Home work Mastery Practice 3.5

pg 100 Worksheet 3.5