Page 24 - Aerodynamics for Engineering Students

Basic concepts and definitions 7

Fig. 1.2 The prism for Pascal's Law

Resolving forces horizontally,

p~(Sxtana)Sz-pz(Sxseca)Szsina = 0
Dividing by Sx Sz tan a, this becomes

PI -P2 = 0
i.e.
PI = Pz
Resolving forces vertically,
p3SxSz -pz(Sxseca)Szcosa - W = 0
Now

w = pg(sxl2 tan a 62-12

therefore, substituting this in Eqn (1.2) and dividing by Sx 62,
1
p3 -p2 --pg tanabz = 0
2
If now the prism is imagined to become infinitely small, so that Sx 4 0 and Sz + 0,
then the third term tends to zero leaving

P3-p2=0
Thus, finally,
P1 = Pz = p3 (1.3)
Having become infinitely small, the prism is in effect a point and thus the above
analysis shows that, at a point, the three pressures considered are equal. In addition,
the angle a is purely arbitrary and can take any value, while the whole prism could be
rotated through a complete circle about a vertical axis without affecting the result.
Consequently, it may be concluded that the pressure acting at a point in a fluid at rest
is the same in all directions.