8494483 pascals-principle

GOOD
MORNING
EVERYONE!

HYE THERE!! HOW
ARE YOU TODAY?
SEEN THAT YOU
LOOK NOT
INTERESTED…WHY
NOT WE DO AN
EXERCISE FOR A FEW
MINUTES?

Do you believe or
not that I can raise
an Elephant?

DO YOU KNOW
WHAT TOPIC
WE WANT TO
LEARN TODAY?

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.

Transmission of Pressure in a
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

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 cm 2
= 20 N cm -2

(b) F = P x A
= 20 N cm -2 x 8 cm 2
= 160 N

Problem Solving
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 cm 2 = X N / 4.5 cm 2
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/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

8494483 pascals-principle

8494483 pascals-principle

More Related Content

What's hot

Viewers also liked

Similar to 8494483 pascals-principle

8494483 pascals-principle