58 PARALLELOGRAM DIVIDED TO SUSTAIN EQUAL FLUID PRESSURES. 



root of the product by the length of the parallelogram, and 

 divide by the whole number of sections, and the quotient will 

 express the distance of any particular point from the upper 

 extremity of the divided length. 

 If the length of any particular section, or the distance between any 



two contiguous points should be required, which is the condition 



expressed by each of the above equations ; then, 



Calculate for each of the points according to the preceding 

 rule, and the difference of the results will give the length of 

 the required section. 



72. EXAMPLE 12. A rectangular parallelogram whose length is 25 

 feet, is perpendicularly immersed in a fluid, in such a manner, that 

 its breadth or upper side is just in contact with the surface ; now, if 

 it be proposed to divide the parallelogram by lines drawn parallel to 

 the horizon, into five parts sustaining equal pressures ; it is required 

 to determine the distance of each point of section from the surface of 

 the fluid, and the respective distances between the several points ? 



Here then, we have given I 25 feet, and n =r 5 an abstract 

 number ; therefore, by proceeding according to the rule, we shall have, 

 for the distance of the first point, 



25V5--- 5 = 11. 18034 feet. 

 For the distance of the second point, we get 



25V2 x 5 + 5 = 15.81139 feet. 

 For the distance of the third point, we obtain 



25 /3~x^5 H- 5 = 19.36492 feet, 

 and for the distance of the fourth point, it is 



25 V 4~X~5 ~ 5 = 22.36068 feet. 



The preceding is all that is necessary to be calculated, for the 

 distance of the fifth point is manifestly equal to the whole length of 

 the parallelogram, and consequently, by the question, it is a given 

 quantity. 



Having thus determined the distances of the several points of 

 section below the upper surface of the fluid, the respective distances 

 between them, or the breadths of the several sections can easily be 

 ascertained, since they are merely the consecutive differences of the 

 quantities above calculated ; hence, we have 



Distances 11.18034, 15.81139, 19.36492, 22.36068, 25. 



Differences 4.63105, 3.55353, 2.99576, 2.63932 ; 



