168 OF THE PRESSURE OF FLUIDS OF VARIABLE DENSITY. 



draw the chord EGF parallel to AB the diameter of the semicircle; 

 then is EGF the chord, on which the pressure is a maximum. 



That the line DG corresponds with x in the equation marked (168), 

 may be thus demonstrated. 



By reason of the parallel lines AB and KC, the angles ABII and 

 KCH are equal to one another ; but the angle ABH is manifestly equal 

 to half a right angle or 45 degrees, therefore, the angles KCH and 

 CHI, are each of them equal to half a right angle, and the lines ci 

 and HI are equal, being respectively the sine and cosine of 45 degrees 

 to the radius CH or CD. 



Now, according to the principles of Plane Trigonometry, the sine 

 and cosine of 45 degrees to the radius unity, are respectively expressed 

 by \ V% > hence we have 



ci^JvC 



and by the property of the rightangled triangle, it is 



Dizr- v /ci*-4-DC 2 =:r- v /|-, 

 and by similar triangles, we have 



r V i : r ' : ' r ' D G - = - r V $ 



The length of the chord line EF is very easily found, for by reason 

 of the right angled triangle EDG, of which the two sides DE and DG 

 are known, it is 



E G 2 ~ D E ? D G 2 ; 



but by the elements of geometry, the square of a line is equal to four 

 times the square of its half, therefore, we have 



EF 2 H=4(DE 2 DG 2 ); 



hence, by extracting the square root, we get 



E F Zr 2 ^ D E 2 D G 2 J 



now D L 2 1=1 r*, and DG*I=: f r 9 ; therefore it is 



EF=z|rV~3. (131). 



Wherefore, if we take the radius of the semi-circle equal to 27 

 inches, as in the preceding example, the whole length of the chord 

 will be 18X1. 7321=31. 176inches. 



PROBLEM XXVII. 



168. If a given conical vessel be filled with fluid, and sup- 

 ported with its axis inclined to the horizon at a given angle : 



It is required to determine, on vjhat section parallel to the 

 base the pressure is a maximum. 



