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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.

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

Pressure = Force (compression) Surface area of liquid

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

Transmission of Pressure in a Liquid

Applications of Pascal’s Principle

APPLICATION

• A hydraulic lift for automobiles is an example of a force multiplied by hydraulic press, based on Pascal's principle. The fluid in the small cylinder must be moved much further than the distance the car is lifted.

For example, if the lift cylinder were 25 cm in diameter and the small cylinder were 1.25 cm in diameter, then the ratio of the areas is 400, so the hydraulic press arrangement gives a multiplication of 400 times the force. To lift a 6000 newton car, you would have to exert only 6000 N/400 = 15 N on the fluid in the small cylinder to lift the car. However, to lift the car 10 cm, you would have to move the oil 400 x 10cm = 40 meters. This is practical by pumping oil into this small cylinder with a small compressor.

Problem Solving

Example 1:In hydraulic brake, a force of 80 N is applied to a piston with area of 4 cm2.

1. What is the pressure transmitted throughout the liquid?

2. 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

Problem SolvingExample 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 = 10 / 1.5 x 4.5 = 30 N

Problem Solving

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.

Problem Solving

(a)Calculate the force which acts on piston A

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

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

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

Solutions

(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

Solutions

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

F1/A1 = F2/A22 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

PASCAL’S PRINCIPLE