OF THE CENTRE OF PRESSURE. 411 



upper extremity being 2 feet below the surface, and the lower extremity 

 6 feet ? 



Here we have d 3 3 :z: 6 s 2 3 208, 



and (F c~ 6 2 2'zz: 32 ; 

 consequently, by division, we obtain 



208 



-55= 6.5 feet, 



and by taking two thirds of this, we get 



fm i= -| of 6.5 rz 44 ft. 

 and finally, by subtracting the depth of the upper end, we obtain 



511. If the upper extremity of the line had been in contact with 

 the surface of the fluid, then would 



imzzfof 4z=2tft. 



This is manifest from equation (304), and if a rectangle be described 

 upon the vertical line, the distance of its centre of pressure below the 

 surface of the fluid will be expressed by the equation (303 or 304), 

 according as the upper extremity is situated below, or in contact with 

 the surface, and this distance will obviously be measured in the line 

 by which the rectangle is bisected. 



512. If the upper side of the rectangle coincides with the surface 

 of the fluid, as in the annexed diagram, where 



adfe is the rectangular parallelogram, having T _ _ 

 the upper side ad in contact with IK ; then, 

 according to equation (304), bm the distance 

 of the centre of pressure, is equal to two 

 thirds of be, the whole length of the paral- 

 lelogram, and consequently, by subtraction, 

 we have mc^bc; therefore, the tendency 

 of the plane to turn about the base ef, is 



equal to the pressure which it sustains, drawn into the length of the 

 lever mc \l, where I denotes the whole length of the plane. 



Put b nr ad or ef, the horizontal breadth of the plane, 

 p =: the entire pressure which it sustains, and 

 s iz: the specific gravity of the fluid in which it is immersed. 



Then, by equation (8), Problem III. Chapter II. the whole pressure 

 sustained by the immersed plane, is expressed by 



