PASCAL’S PRINCIPLE

Expected Learning Outcomes

Student should be able to:

state Pascal’s Principle.explain hydraulic systems.describe applications of Pascal’s Principle.solve problems involving Pascal’s Principle.

PASCAL’S PRINCIPLE

Transmission of Pressure in a Liquid

1. Liquids are practically incompressible.

2. The compression force causes pressure to act on the surface of the water.

Pressure = Force (compression) Surface area of liquid

3. Pascal’s principle states that in a confined fluid, an externally applied pressure is transmitted uniformly in all direction.

4. In a hydraulic system, Pascal’s principle is applied as a force multiplier. The force multiplier of a hydraulic system can be represented by the equation:

Output force = Output piston area Input force Input piston area

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Transmission of pressure

Pascal’s principleUses of Pascal’s principle in

everyday lifeSolving problems involving Pascal’s

principle

FLUIDS

Applications of Pascal’s Principle

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Solve the problems below.

Example 1:

In hydraulic brake, a force of 80 N is applied to a piston with area of 4 cm2.

(a) What is the pressure transmitted throughout the liquid?(b) If the piston at the wheel cylinder has an area of 8 cm2, what is the force exerted on it?

Solution:

(a) P = F/A = 80 N/4 cm2

= 20 N cm-2

(b) F = P x A = 20 N cm-2 x 8 cm2

= 160 N

Example 2:

The figure shows a 10 N weight balancing a X N weight placed on a bigger syringe.

What is the value of X ?

Solution:

F1/ A1 = F2/ A2

10 N / 1.5 cm2 = X N / 4.5 cm2

Therefore X = 1.0 / 1.5 x 4.5 = 30 N

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Example 3:

The mass of X is 2 kg. It is placed at a piston A. The cross section areas of A and B are 5 cm2 and 80 cm2

respectively.

(a) Calculate the force which acts on piston A.

F = mg = 2 x 10 = 20 N

(b) Find the pressure which is exerted on piston B.

P = F/A = 20 N / 5 x 10-4 m2

= 40 000 N m-2

(c) Find the mass of Y which can be lifted by piston B.

F1/A1 = F2/A2

2 x 10/ 5 x 10-4 = m x 10 / 80 x 10-4

m = 32 kg

(d) If piston A moves down by 20 cm, then piston B will go up by

5 x 20 cm = 80 x l cm3

l = 1.25 cm

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Text Book : Mastery Practice 3.4 (Page 95)Challenge Yourself (Page 96)

ARCHIMEDES’ PRINCIPLE

Expected Learning Outcomes

Student should be able to:

explain buoyant force. relate buoyant force to the weight of the liquid displaced. state Archimedes’ principle. describe applications of Archimedes’ principle. solve problem involving Archimedes’ principle.

ARCHIMEDES’ PRINCIPLE

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ARCHIMEDES’ PRINCIPLE

Buoyant force = Weight of fluid displaced Principle of floatation

Applicationsa) Hydrometer c) Shipsb) Hot air balloons d) Submarines

A floating object displaces fluid that is equivalent to its own weight.

Practical Book : Experiment 3.1 (Page 62)

Applications of Archimedes’ Principle

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Solve the problems below.

Example 1:

What is the buoyant force acting on the nut when immersed in water?

Solution:

Actual weight = 0.50 NApparent weight = 0.40 NBouyant force = Actual weight – Apparent weight = 0.50 – 0.40 = 0.10 N

Example 2:

A stone weight 2.5 N. When it 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:

Bouyant force = Actual weight – Apparent weight = 2.5 – 2.2 = 0.3 N

Bouyant force = weight of the solution displaced = ρVg

0.3 = ρ x ( 25 x 10-6 ) x 9.8 where ρ is the density of the solution.

, ρ = 0.3 / ( 25 x 106 ) x 9.8

= 1224 kg m-3

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Example 3:

Figure below shows the cross-section of a submarine. The volume of the submarine is 240 m3. When the ballast tank is empty, the submarine floats at the surface of the sea with 3/4 of its volume below the surface of the sea. The density of sea water is 1200 kg m-3.

a) i) On figure above mark and label two vertical forces acting on the submarine. ii) State the relationship between the two forces in a (i).

The weight of the submarine is equal to the upthrust.

b) i) Calculate the magnitude of one of the two forces in a (i).

Volume of water displaced= ¾ x 240 m3

Upthrust= ρVh= 1200 x (3/4x240) x 10= 2.16 x 106 N

ii) Name the principle used in b (i). Archimedes’ Principle

c) The captain of the submarine observed that the submarine is not strong enough to dive safely to the bottom of the sea. Suggest modifications that can be made to the submarine to make it safer. Justify your suggestions.

- Thicker walls to withstand the higher water pressure. - Wall made of steel. Steel is strong under compression. - Make the cross section circular. A circular shape can withstand higher pressure.

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Upthrust

Weight

Example 4:Diagram below shows an experiment.

(Density of water is 1 000 kg m-3)

(a) Calculate

(i) the apparent reduction in weight of the stone

Apparent loss in weight = 8 – 4 = 4 N

(ii) the buoyant force that acts on the stone

= 4 N

(iii) the mass of water displaced

Weight of water displaced = 4 N mg = 4 m x 10 = 4 m = 0.4

(iv) the volume of the stone

V = m/ρ = 0.4 m3 / 1000 = 4 x 10-4 m3

(v) the density of the stone

ρ = m/V = 1 kg / 4 x 10-4 m3

= 2500 kgm-3

(b) state the principle that you used in (a) above

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Archimedes’ Principle

BERNOULLI’S PRINCIPLE

Expected Learning Outcomes

Student should be able to:

state Bernoulli’s principle. explain that a resultant force exists due to a difference in fluid pressure. describe applications of Bernoulli’s principle. solve problems involving Bernoulli’s principle.

Bernoulli’s principle state that when the velocity of a fluid is high, the pressure is low, and when the velocity is low, the pressure is high.

SITUATION A:

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Applications

a) Aerofoilb) Carburetorsc) Bunsen Burnersd) Insecticide sprayers

Pressure in moving fluids

BERNOULLI’S PRINCIPLE

Text Book : Mastery Practice 3.5 (Page 100)

SITUATION B :

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Solve the problems below.Example 1:Diagram 1 shows a glass tube with uniform cross-section. It is used to study the relationship between velocity and pressure in fluids.

Diagram 1

a) The tube is filled with water and the two ends are stoppered. i) Mark on the above diagram, the water levels at tubes A, B and C.ii) Explain your answers in a (i) above. The water is stagnant. The pressures in the three tubes are the same.

b) Later, the water in the tube is allowed to flow with uniform speed from P to Q.

Diagram 2

i) Mark on the above diagram, the water levels at tubes A, B and C.ii) Explain your answer in (b) (i). The pressure decreases from A to C and the water can flow from P to Q.

c) The experiment in (b) is repeated by replacing the above tube with another tube as shown in Diagram 3 below.

Diagram 3

i) Mark on the above diagram, the water levels at tubes A, B and C.ii) Compare the velocity of water at X, Y and Z. Velocity of water at Y is the greatest. Velocity of water at X is the smallest.

VY > VZ > VX

iii) What is the relationship between the velocity of water and water level in the tubes.The higher the velocity of water, the lower is the water level.

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Example 2:

Diagam 4 shows an experiment. The tube is inverted into basin of water.

Diagram 4

a) The air flows in the tube from A to B.i) Compare the velocity of air at K, L and M. Velocity of air at L is the greatest.

Velocity of air at K is the smallest.VL > VM > VK

ii) Mark the water levels in tubes P, Q and R.iii) Compare the pressure in tubes P, Q and R. PQ > PR > PP

iv) Name the principle used. Explain the principle.Bernoulli’s Principle. When a fluid moves through an area with high velocity, the pressure in the area will decrease.

b) A passenger is standing by the side of a railway track. He seems to be attracted to the railway track when a fast moving train travels in front of him. Explain the observation. The fast moving train causes the surrounding air to be fast moving too. This causes a region of low pressure. The higher pressure behind him pushes him forward.

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Direction of flow of air

Text Book : Mastery Practice 3.6 (Page 102) Practise Your Skill – Question 3 (Page 106)

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Application of Bernoulli’s Principle

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