HYDROSTATICS. 



131 



How can we 

 calculate the 

 pressure upon 

 the bottom of 

 a Tessel con- 

 taining water ? 



the pressure on A B a portion of the base, E F, is measured by the column 

 A B C D as before ; but the other portions of the Hquid not resting on the 

 sides also press upon the bottom, E F ; and as the pressure of the column A 

 B C D is transmitted equally, every portion of the base, E F, sustains an 

 equal pressure as that portion of the base, A B, which is directly beneath the 

 column, A B C D ; therefore the whole pressure on the base, E F, is the 

 same as if the vessel had been cjdindrical, and filled throughout to the height 

 indicated by the dotted lines, G H. 



296. Hence, to' find the pressure of water upon the bottom of any vessel, 

 we have the following rule : 



297. Multiply the area of the base by the 

 perpendicular dejDth of the water, and this 

 product by' the weight of a cubic foot of 

 water. '••■ 



Thus, suppose the area of the base of a vessel to be 2 square feet, and the 

 perpendicular depth of the water to be 3 feet; required the pressure on 

 the bottom of the vessel, the weight of a cubic foot of water being assumed 

 to be 1,000 ounces (see § 82). 

 2X3=6 cubic feet. 

 6X1,000=6,000 oz.=pressure on the base of the vessel 



• " The actual pressure of water may also be calculated from the following data. It is 

 ascertained that the weight of a cubic inch of water of the common temperature of C'2' 

 Fahrenheit, is a portion of a pound expressed by the decimal 0-0360G5. The pressure, 

 therefore, of a column of water one foot high, having a square inch for its base, will be 

 found by multiplying this by 12, and consequently will be 0-4"2S lb. 



" The pressure produced upon a square foot by a column one foot high, will be found 

 by multiplying this last number by 144, the number of square inches forming a square 

 foot; it wUl therefore be 623232 lbs. 



Table showing the pressure in lbs. per square inch and square foot, produced by water 

 at various deptlis. 



" By the aid of the above t.ible, the actual pressure of water on each part oi the surface 

 «f a vessel containing it can always be determined, the depth of such part being given. 

 Thus, for example, if it be required to know the pressure upon a square foot of the bot- 

 tom of a vessel where the depth of the water is 25 feet, we find, from the above table, that 

 the pressure upon a square foot at the depth of 2 feet is 124-6464 lbs. ; and, consequently, 

 the pressure at the depth of 20 feet is 1246-464 lbs. ; to this, let the pressure at the depth 

 of 5-feet, as given in the table, be added: 1246-464-f311-61G = 155S-0S0 lbs., which is, there- 

 fore, the required pressure. 



" If the liquid contained in the vessel be not water, but any other whose relative weight 

 compared with water is known, the calculation is made first for water, and the result being 

 multiplii^d by (he number expressing the proportion of the weight of the given liquid t» 

 that of Wikter, the result will be the required pressure."— ittJ'dner. 



