90 OF FLUID PRESSURE ON CIRCULAR PLANES AND ON SPHERES. 



RULE. Multiply one third of the diameter of the given 

 plane by the square root of 6, and the product will be the 

 length of the chord required. Or thus : 



Multiply the given diameter by the constant number 

 .81647, and the product will be the length of the chord 

 required. 



100. EXAMPLE 20. A circular plane whose diameter is equal to 

 36 feet, is perpendicularly immersed in a fluid, so that the upper 

 extremity of the vertical diameter is in contact with the horizontal 

 surface ; what is the length of a chord, which being inflected from 

 the lower extremity of the diameter, sustains a greater pressure than 

 any other chord which can be drawn from the same point ? 



Here by the rule, we have 

 f = i X 36 x VQ = 29.3929 feet, 

 and by the second part of the rule, we have 

 1=36 X .81647 29.3929 feet. 



PROBLEM XV. 



101. If two spheres or globes of different diameters, be 

 immersed in a fluid, in such a manner, that the uppermost point 

 on their surface is just in contact with the horizontal surface 

 of the fluid : 



It is required to determine the pressure on each of the 

 spheres, and to compare the pressures with one another. 



Let ABD and abd be the two spheres, whose diameters AB and ab, 

 have their upper extremities A and a in 

 contact with EF, the horizontal surface of 

 the fluid. 



Bisect the diameters AB and a b, respec- 

 tively in the points c and c ; then are the 

 points c and c, the centres of magnitude of 



the respective spheres ; but in a sphere, the centre of magnitude and 

 the centre of gravity occur in the same point ; therefore, c and c are 

 the centres of gravity of the spheres ABD and abd, and AC and 

 ac are the perpendicular depths below the surface of the fluid. 



