30 OF RECTANGULAR PARALLELOGRAMS DIVIDED INTO 



but by the Trigonometrical Tables, the natural sine of 59 38' is 

 .86281 ; therefore, we have 



p'=i 30 2 X 20 x .86281 X i= 1941.3225, 



and again, by a similar process we have 

 p= 30 2 X 20 X .86281 x 1 = 5823.9675. 



Now, these results are obviously expressed in cubic feet of water, 

 for they are respectively equal to the solidity of a fluid column, whose 

 base is equal to one half the given parallelogram, and whose altitude, 

 in the one case, is expressed by \l sin. $ rz 7.5 x .86281, and in the 

 other by |J sin. <j>=: 22.5 X .86281 ; but the weight of one cubic foot 

 of water is .equal to 62J Ibs. ; consequently, the pressures expressed 

 in Ibs. avoirdupois, are 



p f = 1941.3225 X 62.5 = 121332.65625 Ibs. 

 and;? =5823.9675 X 62.5 = 363997.96875 Ibs. 

 When the plane is perpendicular to the surface of the fluid, the 

 pressure is a maximum, and in that case, the respective pressures on 

 the two portions of the parallelogram, are 



p' = 30 2 x 20 X 62.5 x I = 140625 Ibs. 

 and jo rz30 2 X 20 x 62.5 x |zr 421875 Ibs. 



and the sum of these, is obviously equal to the whole pressure on the 

 plane ; hence we get 



P = 140625 + 421875 == 562500 Ibs. 



COROL. If the plane, instead of being immersed in the fluid, as we 

 have hitherto supposed it to be, should only be in contact with it, as 

 we may conceive the surface of a vessel to be in contact with the fluid 

 which it contains ; then, the pressure will be the same ; for the quan- 

 tity of pressure at any given depth upon a given surface, is always the 

 same, whether the surface pressed be immersed in the fluid or just in 

 contact with it, and whether it be parallel to the horizon, or placed in 

 a position perpendicular or oblique to it. 



5. OF RECTANGULAR PARALLELOGRAMS IMMERSED IN NON -ELASTIC 

 FLUIDS, AND DIVIDED INTO TWO PARTS SUCH THAT THE PRESSURES 

 OF THE FLUID UPON THEM SHALL BE EQUAL BETWEEN THEMSELVES. 



PROBLEM VI. 



46. A rectangular parallelogram is obliquely immersed in an 

 incompressible and non-elastic fluid, in such a manner, that one 

 side is just coincident with the surface : 



It is required to divide the parallelogram into two parts by 

 a line drawn parallel to the horizon, so that the pressures on 

 the two parts shall be equal to one another. 



