328

MISCELLANEOUS

TRACT 3S.

in fluxions —^— — -d— = 0, and hence x = ia, that is,Ac = yAB, gives the point c fallen from.

PROBLEM VIII.

A cylinder of oak is depressed in water till its top is justlevel with the surface, and then is suffered to ascend; it is re-quired to determine the greatest altitude to which it will rise,and the time of its a.-cent.

Let a — the length, and b the area or base of the cylinder,m the specific gravity of oak, that of water being 1, also xany variable height through which the cylinder has ascended.Then, a — x being the part still immersed in the water,(a — x) x b x 1 = (a — x)b is the force of the water upwardsto raise the cylinder ; and a x b x m — abm is the weightof the cylinder opposing its ascent; therefore the efficaciousforce to raise the cylinder is (a — x)b — abm ; and, the massbeing abm, the accelerating force is

(a——abm a — x—am an —.r -

abm am am

putting n — l — m the difference between the specific gra-vities of water and oak.

Now if v denote the velocity of ascent at the same timewhen x space is ascended, then by the theorems for variableforces, vv = 32/* = — x ( anx — xx), therefore

if r= x (2 anx — x *), and v — 8*/———but when thecylinder has acquired its greatest ascent, v and v z —Q, there-fore then 2anx — x z = 0, and hence x = Ian the part of thecylinder that rises out of the water, being = T 5a or ofits length.

To find when the velocity is the greatest, the factor 2 anx

— x l in the velocity must be a maximum, then 2 anx — 2xx

— 0, and jp =an, being the height above the water when thevelocity is the greatest ; and which it appears is just equal tothe half of 2 an above found for the greatest rise, when the