3.4 PASCAL PRINCIPLE

LEARNING OUTCOMES

• State Pascal’s principle,

• Explain hydraulic systems

• Describe applications of Pascal’s

principle

• Solve problem involving Pascal’s

principle

• A car at a workshop can be raised up for service by pushing only one switch. This is one application of Pascal’s Principle in our daily life.

Activity 1

• Squeeze a toothpaste tube. How does the pressure exerted by the thumb transfer to the mouth of the tube?

The pressure is transferred from the place we press the thumb to the mouth of the tube in every part of the liquid

• When the piston is push, why does water spurt out in all direction with the same speed?

The pressure is transmitted in every part of the water and equally.

State Pascal’s Principle

• Pascal’s principle states that pressure applied to an enclosed fluid is transmitted equally to every part of the liquid.

Experiment to verify the Pascal’s Principle

• The water shoots out with same speed means the pressure are the same.

• The pressure acting

on the water is transferred uniformly throughout the water.

HYDRAULIC SYSTEM

• A hydraulic system is a device in which a small applied force can give rise to a larger force.

• A small input force, F1 is applied to the small piston resulting in a large output force, F2.

HYDRAULIC SYSTEM

• Use liquid – liquid cannot be compressed

• Oil – less density, high boiling point (cannot evaporate easily / change state to gas easily)

• Pressure at small cylinder = Pressure at large cylinder

• Based on the Pascal’s Principle, the pressure is transmitted uniformly in all directions,

• F1 = F2

A1 A2

HYDRAULIC SYSTEM

2. The figure shows a simple hydraulic system. Assume there is no friction in the system.

(a)what is the pressure at A?

(b)what is the pressure at B?

P = 20/0.1 = 200 Pa

200 Pa

(c) what is the upward force produced?

(d) What is the effect on the hydraulic system if the cross-sectional area of the large piston is increase?

F = PA = 200 x 0.5 = 100 N

Force will increases. F directly proportional to area

3. A basic hydraulic system has small and large pistons with cross-sectional areas of 2 cm2

and 50 cm2 respectively. When a force of 20 N is applied to the small piston, it pushes down the piston by 20 cm. Calculate

(a) the pressure transmitted in the hydraulic fluid,

P = 20/2 = 10 N/cm2

20 N

Area2 cm2 Area

50 cm2

(b) the force acting on the large piston,

(c) the distance moved by the large piston

(d) the magnification of the force.

F = 10 x 50 = 500 N

volume down = volume up(2)(20) = (50) x x = 0.8 cm

F2 = (A2/A1) F1 F2 = (50/2)F1 = 25 F1

Hydraulic Jack

• When the handle is pushed, valve A is closed and valve B opens so the pressure can be sent to the larger piston. The large piston will rise.

• While the small piston is pulled out, valve B closes and valve P opens so that the oil in the tank enter into the hydraulic cylinder as a result of atmospheric pressure.

• By moving the push-pull handle a number of times the large piston can be raised carrying a heavy load.

Hydraulic Brake

Hydraulic Brake

• When the brake pedal is pushed, the piston in the master cylinder exerts a force on the brake fluid

• The pressure is transmitted equally to all the other small pistons.

• The pressure on the pistons forces the brake shoes or pad .

• The frictional forces between the brake drum and the brake pedal cause the vehicle to slow down or stop.

Explain how does the brake system work?When the driver’s foot presses the brake pedal, the piston in the master cylinder exerts _____________ on the brake fluid. This pressure is ____________ to the wheel cylinder of all wheels.

pressure

transmitted

On the figure above indicate the direction of motions of the front brake cylinder and the rear brake cylinder.

State the relationship between the liquid pressure in the front brake cylinder and the rear brake cylinder.

Equal

Explain why the piston of the wheel cylinder is larger than that of the master cylinder.

larger surface area, large output force.

What is the function of the return spring?

return the brake shoe to original position

Why is the brake system in the car is less effective air bubbles are present in the brake fluid?

air can be compressed. So part of the force is used to compress air.

Why is the cross-sectional area of the rear brake cylinder of the right wheel is the same as the cross-sectional area of the rear brake cylinder of the left wheel?

3.A boy invented a machine to compress old newspaper. The machine consists of a hydraulic system with two pistons. Piston A has a smaller cross-sectional area compared to piston B.

(a) the function of this machine is based on a principle in physics. Name the principle,

Pascal’s principle

(b) Explain why a small force exerted by the boy can result in a much bigger force acting on the old newspaper.

pressure is transmitted through the oil from piston A to piston B equally.Pressure at A = pressure at BLarge surface area of piston B Produce large force on piston B

2.Figure shows a hydraulic lift commonly used in a workshop. Highly compressed air is used to push the piston upwards. The force produces is enough to lift a car.

• By referring to the diagram, explain how a car can be lifted by the hydraulic lift.

Open inlet valveAir enter to the tube and press down the oil.Pressure is transmitted to the larger piston through the oil.Produce larger force and lift the car.

• Explain how the car can be lowered after repair works are done.

Open the release valve The air will be release.The piston will be lowered.

4. Figure shows dentist treating a patient. The dentist presses on the input piston to raise the patient. The cross-sectional areas of the input piston and the output piston are 400 cm2 and 600 cm2 respectively.

(a) Describe how the patient is raised to a certain height.

Press the input piston.Pressure is transmitted to the larger piston through the oil.Produce larger force on the output piston and lift the patient.

(b) How much force is exerted by the dentist in order to raise a patient of 54 kg?

(c) If the patient is raised to a height of 8 cm, how far should input piston be pushed down?

F1/A1 = F2/A2F/400 = 540/600

F = (400/600) x 540 F = 360 N

(x)(400) = (8)(600)x = 12 cm