In the same way it can be seen that the forces required to produce two or more types of 

 acceleration simultaneously are simply the vector sums of the forces required to produce each 

 type separately. 



In the last two chapters many expressions have been obtained for the kinetic energy, 

 kM'U^/2 or kl 'oj'^/2, of the fluid surrounding a moving or rotating body. To obtain k from these 

 expressions, it is only necessary to introduce the known value of M' or /'and to divide the 

 kinetic energy by M'U'^/2 or /'(u^/2. The values of the kinetic energy and of k are collected 

 for convenience of reference in a table following the next section. A self-explanatory picto- 

 rial representation of the body is appended in each case. 



147. NOTE ON UNITS. 



In the formulas a consistent set of dynamical units is understood to be employed, as 



was explained in detail in Section 18. The coefficient of inertia ^ is a pure numeric. 

 To illustrate the use of the units the following problem will be solved. 

 An ellipsoidal body, with semiaxes 6 feet, 3 feet, and 3 feet, weighing 25,000 pounds, 



is suspended in sea water with its major axis vertical. When released, what will be its initial 



acceleration? The density of sea water is 64 pounds per cubic foot- 

 Solution: The resultant force on the body acts vertically downward and is equal to the 



weight of the body minus the buoyant force. 



4 

 Force = 25,000 - — n-x6x3x3x64 = 10,500 pounds. 



. . , J ■ 25,000 lb-sec2 



The mass or the body is = 776 slues = 776 = 



32.2 ft 



The coefficient of inertia for prolate spheroids moving "end on," with a/b = 2.0, is k = 0.209. 

 Hence the effective mass of the fluid is 



4 64 



— 77 X 6 X 3 X 3 X X 0.209 = 94.0 slugs, 



3 32.2 



and the total effective mass of body and fluid is 776 + 94.0 = 870 slugs. From Newton's 

 second law of motion the acceleration is the resultant force divided by the mass or 



10,500 _ „ , , , 



— = 12.0 ft/sec2. 



870 



383 



