Page 36 - ebook fluid mechanics_finalize

FLUID MECHANICS

3.1.3 Problems Regarding Hydraulic Jack

Example 1

A force, P of 800 N is applied to the smaller cylinder of a hydraulic jack. The area, a of a
2
2
small piston is 20 cm while the area, A of a larger piston is 200 cm . Calculate the mass
can be lifted on the larger piston.

Solution :

2
-4
-4
Putting F = 800 N, a = 20x10 m , A = 200x10 m 2
p = p
1
2
A
F = W So that W = F
a A , a

200x 10 − 4
= 800  = 8000 N
20x 10 − 4

Mass lifted = W
g

= 80000 = 815 . 49 kg
. 9 81

If the two pistons are not at the same level, the pressure, p2 acting on the larger piston is

not same value to p1.

Figure 3.5 : Small piston high than large

piston
p = p + 
gh ….. 4
1
2

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