228 



Mr. HOPKINS, ON THE TRANSPORT OF ERRATIC BLOCKS. 



R' = — pi be sin' 0. 



Therefore, the vertical pressure on the base 



= {p - pi) gU + R cos 9 



= 1 (|0 - pi) gabc sm 6 + — p\bc sin' 0cos 



= 1 fee sin 6 \{p - p\) ga + v^p^ sin QcosG}. 



If we suppose the force (/") opposing the body's sliding to follow the ordinary law of friction, 

 we shall have 



/' = /i . vertical pressure on the bottom, 



= It- \{p - pi) g U + Rcos 0}, 

 (where /x = coefficient of friction) ; and the condition of the prism being on the point of moving 

 will be 



R' = F. 



Hence we obtain 



— p, sin^S = li {(p - pi) a ■{ p, sin cos 0} , 



or, 



Ui-')' 



2g sin (sin - /j. cos 0) 



This shews that a triangular prism with its axis perpendicular to the current cannot be moved 

 by sliding unless tan he > p., whatever be the velocity of the current. 



If the section ABC be equilateral, = 6o", and we shall have 



— = Vi ' 



2g v/3(v/F-^) 



If we take — = 2, ."5, which may be assumed as a mean value of that ratio, we shall have 



V v'sl ^g 



14. Let us now take the rectangular parallelopiped, of which ABCD is the transverse 

 section. Let AB = a, AD = c, and the length = 6. Then 



„2 D c 



,6c; 



R'='- 



pi- 



and in order that the body may be on the point of rolling 

 round the edge perpendicular to the plane of the paper 

 through B, we must have 



V c a 



-pibc.-= (p - p,)gabc-, 



