CHAMBERS'S INFORMATION FOR THE PEOPLE. 



appears that the pressure on a square inch of tho 

 side next to the bottom is less than that on a square 

 inch of the bottom itself, by the weight of half a ; 

 cubic inch of water. 



By similar reasoning, the pressure at any point 

 on the side is represented by a horizontal line 

 equal to the depth of the point from the surface, 

 or the head of water of the point, as it is called. 

 Thus, at 3, the pressure is 3^, equal to 3A. Now, 

 by the property of similar triangles, the extremities 

 of all such lines, e, I, h, are in the same straight 

 line with A ; and by joining them, we have a tri- 

 angle, CeA, representing the amount and mode 

 of distribution of the lateral pressure on the 

 side CA. 



What is true of one vertical row of square inches 

 in the side, is true of all the vertical rows of which 

 it is composed ; and therefore the pressure on the 

 whole rectangular side is equal to the weight of a | 

 wedge-shaped mass of liquid, AC?, whose base is 

 equal in area to that side. But the content of 

 AO is evidently equal to that of a uniform rect- ' 

 angular column of mean depth, 3^. If, then, we 

 suppose the length of the side to be 10 inches, its 

 depth being 6 inches, the area will be 60 square 

 inches ; and this multiplied by 3, the mean depth, 



Fig. 9. 



gives 1 80 cubic inches of water, the weight of which 

 is equivalent to the lateral pressure. The truth thus 

 arrived at may be expressed in the following gene- 

 ral proposition : The lateral pressure of a liquid, 

 Perpendicular to the side of a vessel, is equal to the 

 weight of a column of the liquid whose base is the 

 side and height equal to the depth of the middle 

 point, or centre of gravity of the side. 



In the case of a vessel with a sloping side, AC 

 (fig. 9), the pressure perpendicular to AC is still 

 represented by a wedge-shaped mass, AC<?; but 

 here Ce is not equal to CA, but to Ca, the per- 

 pendicular depth of the water ; and 3^ to 30', the 

 mean perpendicular depth. In all cases, when 

 speaking of depth or head of water, it is perpen- 

 dicular depth that is meant 



We have as yet 

 considered only 

 rectangular sides, 

 in which the 

 middle point, or 

 point of mean 

 pressure, lies mid- 

 way between the 

 surface and the 

 bottom, or at half 

 the depth. But 

 in the case of a vessel with a triangular end or 



22S 



side, like ABC (fig. 10 or 11), it is evident that 

 the middle point of the figure is not at m, the 

 middle of the vertical line Brf. For in fig. 10 the 

 number of points or units of surface that lie below 

 m, -and are therefore pressed with greater force, is 

 far less than of those that lie above m, and 

 sustain less pressure. The real middle point ot 

 any surface is its centre of gravity ; and the centre 

 of gravity of a triangular surface is found by 

 drawing a line from any angle, B, to the middle 

 of the opposite side, d, and taking the point g at 

 one-third of the distance from d to B. The mean 

 depth, then, or mean head of water, in fig. 10, is 

 dg ; and in fig. II, B^. 



The pressure of a liquid on a surface is the 

 same whether the surface forms part of the 

 bottom or sides of the vessel containing the 

 liquid, or belongs to a body immersed in the 

 liquid. If an empty box, having for one of its 

 sides a triangle like ABC, fig. 10, is sunk in water 

 to any depth, the water will exert a pressure tend- 

 ing to force in the side, whether it be uppermost 

 or undermost, vertical or oblique ; and the amount 

 of the pressure could be shewn, as in the cases 

 already examined, to depend on the area of the 

 surface and the depth of its centre of gravity from 

 the surface of the water. 



A cubic foot of pure water weighs very nearly 

 1000 oz. avoirdupois, or 62*5 Ibs. More exactly, 

 a cubic inch of distilled water, at the temperature 

 of 62 F. and barometric pressure of 30 inches, 

 weighs 252'46 grains ; and a cubic foot weighs 

 62*32 Ibs. or 997 oz. The cubic foot of sea-water 

 may be taken on an average at 1026 oz. By 

 means of these data, we can solve such a question 

 as the following : 



Required the pressure on a lock-gate 12 feet 

 wide, the depth of the water being 10 feet ? The 

 area of the surface pressed is 12 X 10 = 120 

 square feet ; 120 multiplied by 5 (the depth of the 

 middle point of the surface) gives 600 cubic feet of 

 water as the pressing column ; and 600 X 62^5 

 = 37,500 Ibs. is the pressure in weight. 



The circumstance of pressure increasing in pro- 

 portion to depth, makes it necessary to increase 

 the breadth of embankments for dams and canals 

 from the top downwards ; also to increase the 

 strength of the lower hoops of large vats, to pre- 

 vent their bursting. It likewise teaches the pro- 

 priety of making dams, ponds, canals, and vessels 

 for liquids generally, as shallow as is consistent 

 with convenience or their required purpose. In 

 every case, it is important to recollect that the 

 degree of pressure on the sides is irrespective of 

 shape or size of the contents, and depends exclu- 

 sively on the perpendicular depth of the liquid. 



When the poet speaks of ' the broad ocean 

 leaning against the land,' we are apt to think of 

 the broadness as adding to the weight that the 

 land has to sustain. But the pressure against a 

 square foot of the shore of the broad Atlantic is 

 no greater than against a foot of the shore oppo- 

 site Dover, or a foot of the side of a canal, at the 

 same depth. The pressure against the gates of a 

 canal is the same whether the next lock is a mile 

 or whether it is 50 yards off. It might be brought 

 within a foot, or even a fraction of an inch, and 

 the weight on the gate would remain unchanged 

 so long as the depth was the same. This being 

 the case, it is really no more difficult to embank 

 the calm ocean than a small lake of the same 



