244 Equilibrium between Living and Dead Forces. 



moving force, but is also exerting a force equal to its own weight 

 through the distance x. 



Since in practice there are many cases when the pile or other 

 matter driven is heavier than the ram or driving weight, it be- 

 comes necessary to introduce other relations. 



By the laws of momentum it is known that when an elastic 

 body w, moving with an uniform velocity v, strikes another 

 heavier body P at rest, the body w will return with a diminished 



' Fiff. 2. 



W 





[V — wv) 

 velocity = "^pIL 5 ^^^ ^^^' ^ velocity will be imparted to P 



2vw 

 equal to p^^ — Now, taking w and P as the weight of the ram 



and pile respectively, the problem resolves itself into the ques- 

 tion, What dead weight will hold in equilibrio a pile without 

 vis inertise, driven to the depth a; by a weight P (the pile itself) 



2vw 

 moving with the velocity p , ? Because, whatever velocity v 



the ram may have at the moment of impingement upon the pile, 



it will impart to the pile a velocity =pT~) ^^ which instant it 



will cease to act upon the pile, and the pile itself by its own 

 moving force will go down opposed only by friction, the pile 

 being thus, as it were, converted into the driving weight. 



The height H, then, through which the pile should fall to ac- 



quire the velocity p . , when v is considered in feet per se- 

 cond, (using 16 feet fall for the first second of time,) will be 



H = ^^/p ■ — r7, and the dead weight which it will bear, by 



equation (A), will be 



HP Vv^w^ 



Now if a be the height through which the ram w descends, 

 we have ?;2=a64, and by substitution, we have 

 64Paw;2 4P a^^" 



^=lij[V-\-wYx^{V-^wYx ••••(B) 



