PEESSURE INTENSITY 



17 



unaltered, it follows that the pressure intensity on the base must every- 

 where be the same. 



If ^Fis the weight of unit volume, or the intrinsic weight of the liqnid, 

 the weight of the column = W a h. 



If p is the pressure intensity at a depth h, the upward pressure on the 



base = p a. 



.. p a = W a h. 



p = W h. 



The pressure of still water against the sides or bottom of any vessel is 

 then simply due to the " head," or height of the level of the free surface 

 of the water above the point considered. Each square foot of the surface 

 at a depth h may be considered as supporting a column of water of 1 

 square foot cross sectional area, and of 

 height h, and therefore of weight 62'4 

 libs. 



/. Pressure per square 



foot, per foot of head = 62*4 Ibs. 

 Pressure per square 

 inch, per foot of head = '433 Ib. 

 /. Head equivalent to a pressure of 

 1 Ib. per square inch = 2'308 feet. 



The second of the above propositions 

 may be deduced by considering the 

 equilibrium of a triangular prism of the 

 liquid of unit length having its edges horizontal, and its ends perpen- 

 dicular to the sides. If a b c (Fig. 5) be a cross section of the prism, 

 a c b being a right angle and b c being horizontal, and pi, p%, p 3 be the 

 mean intensities of pressure on the sides a b, a c, c b, we have for 

 equilibrium, resolving parallel to the sides b c and c a, 

 p% ac = pi ab sin 6 



~ . TT _ ac cb 

 p 3 cb = pi ab cos 



Pi 



TT _ 

 W 

 & 



Putting ac = ab sin 6 ; be = ab cos 0, we get 

 = Pi 



sin cos 0. 



W 



= pi -\- 



A 



If now the sides of the prism be indefinitely diminished pi, p<i, and p 3 

 become, in the limit, the pressure intensities at the same point, but in 

 different directions. Also ab 2 , being of the second order of small quantities, 

 will vanish, so that in the limit 



Pi = P2 == 2/ 3 . 



H.A. c 



