1. The document discusses Pascal's law and how hydraulic systems work by transmitting pressure through an incompressible liquid. It provides equations to calculate pressure, force, and displacement in hydraulic systems. Examples are given to demonstrate calculating pressure, load lifted, and displacement of pistons in hydraulic presses and jacks. 2. The document explains that Pascal's law states that pressure is transmitted equally throughout a liquid. It then provides equations showing that the pressure, force and displacement are proportional to the ratio of the piston areas in a hydraulic system. Examples are worked out to calculate pressure, load lifted, and piston displacement. 3. Hydraulic systems are described as using an incompressible liquid to transmit pressure between

1. 1.4 Pascal’s Law and The Hydraulic System
Pascal’s Law
Pascal’s Law states that when a liquid completely fills a vessel and
pressure and pressure is applied to it, the pressure is transmitted equally in
all directions throughout the liquid.
Figure 1.20 shows a vessel with two cylinder and pistons. The vessel is
completely filled with an incompressible liquid. The small piston has a
cross-sectional area A1, and that of the other piston is A2,
Figure 1.20 Illustrating Pascal’s Law
From : Loo Wan Yong and Others. Elective Physics. (2012:10)
A force F1 is applied on the small piston to produce a pressure.
1
1
1
A
F
P = …………………….(Eq. 1.12)
The pressure at the other piston is given by
2
2
2
A
F
P = ……………………(Eq. 1.13)
According to Pascal’s Law, the pressure produced by F1 is transmitted to
the other piston through the liquid. Hence the pressure at both piston are
the same.
21 PP = …………………….(Eq. 1.14)
..…………………...(Eq. 1.15)
2
2
2
1
A
F
A
F
=

2. 23
Hydraulic System
Hydraulic Systems work by using liquids under pressure. They make use of
two properties of liquid.
: Liquid are incompressible,
: If pressure is applied to an enclosed liquid, the pressure is transmitted to
all parts of the liquid.
Figure 1.21 show how a simple hydraulic press works. A small force
applied to the small piston can lift a greater load on the large piston.
The input pressure P1, supplied by an input force F1 on a small piston area
A1 is transmitted to the large piston, which has a area A2.
Since 21 PP = , where P2 is the output
Pressure,
2
2
1
1
A
F
A
F
=






=
1
2
12
A
A
FF …………….(Eq. 1.16)
Since A2 > A1, it follows that F2 > F1.
The snag in hydraulic press is that we need to push the smaller piston
through a long distance d1 in order for the bigger piston to move through a
small distance d2, This is due to the fact that the volume of liquid pushed
by the smaller piston V1 is equal to the volume of liquid that goes below
the bigger piston V2.
Figure 1.21 Hydraulic press.
From : Loo Wan Yong and
Others. Elective Physics.
(2012:11).

3. 24
12 VV = ……………..……...(Eq. 1.17)
1122 dAdA = …………………….(Eq. 1.18)






=
2
1
12
A
A
dd …………………….(Eq. 1.19)
Since A2 > A1, it follows that d2 < d1.
Example 1.10
Figure 1.21 shows a simplified form
of a hydraulic press. A force of 50N
is applied to the smaller piston of
0.01 m2
.
(a) The large piston has an area
of 0.20 m2
. Calculated
(i) the pressure exerted on
the liquid by the smaller
piston, Figure 1.22 Example 1.10
the pressure exerted on From: Loo Wan Yong and Others.
(ii) the large piston, Elective Physics. (2012:11).
(iii) the load that can be lifted.
(b) If the smaller piston is pushed down through a distance of 10 cm,
how much will the large piston move up?
Solution :
(a) (i) Pressure
A
F
P =
2
01.0
50
m
N
=
Pa5000=
(ii) This pressure is transmitted in the liquid and is the same
everywhere within the liquid. Thus the pressure under the
bigger piston must also be 5000Pa
(iii) The bigger piston has an area of 0.20 m2
. The load that it
can support is given by
PAF =
2
02.05000 mPax=
N1000=

4. 25
(b) The volume of liquid that moves out from the smaller cylinder
equals the volume of liquid that moves into the bigger cylinder.
12 VV =
cmxmxdm 1001.020.0 2
2
2
=






= 2
2
2
20.0
1001.0
m
cmxm
d
cm50.0=
Example 1.11
Figure 1.23 the car has 1,000 kg. on the right of hydraulic, an diameter
14 cm. Area of the small hydraulic has diameter 7 cm. Calculate the
force to move up the car.
Figure 1.23 Example 1.11
From: Loo Wan Yong and Others. Elective Physics. (2012:11).
Solution :
Find area smaller piston
21
1 )
2
(
d
A π=
2
2
1 )
2
107
(
−
=
x
A π
)
4
107
(
4
1
−
=
x
A π
Find area bigger piston
22
2 )
2
(
d
A π=
2
2
2 )
2
1014
(
−
=
x
A π
)
4
10196
(
4
2
−
=
x
A π

5. 26
From
2
2
1
1
A
F
A
F
= ; Force to lift mgF =2
21
1
A
mg
A
F
=
2
1
1
A
mgA
F =
π
π
)
4
10196
(
)
4
107
)(/101000(
4
4
2
1 −
−
=
x
x
smkgx
F
N14.357=

6. 27
Problem 1.4 Hydraulic System
1. The diagram on above shows a simple hydraulic jack. Assuming that
the jack is frictionless:
a) What is the pressure at A?
………………………………………………………………..
……………………………………………………………………
……………………………………………………………………
……………………………………………………………………
b) What is the output force?
………………………………….…………………………………
………………………………….…………………………………
………………………………….…………………………………
………………………………….…………………………………
c) What is the pressure at B?
…………………….…………………………………
………………………………….…………………………………
………………………………….…………………………………
………………………………….…………………………………
2. In the jack on the above, what would be the effect of:
a) increase the area of the output piston?
…………………………………………………………..
…………………………………………………………..
b) decrease the area of the input piston?
…………………………………………………………..
………… ………………………………………………..
…………………………………………………………..