TRACT 3.8.

PROBLEMS.

345

A-

Let l = the whole length of the thread ;a — the weight of the same ;b = aw the dif. of lengths at first;d = w — w the dif. of the two weights;c — a weight applied to the circumference,such as to be equal to its whole wt. andfriction reduced to the circumference;s = w+ty+a+c the sum of the weights moved.

Then the weight of b is , and d — -y- is the moving forceat first. But if x denote any variable space descended by w,or ascended by w, the difference of the lengths of the thread

will be altered 2x ; so that the difference will then be b — 2x,b—%c

and its weight ——a; conseq. the motive force there will be

7 b — %x dl—ab + %ax , , ,, dl—ab + Qax r , i

a ~ —j—u = -j- ? and theref. ---= j the ac-

celerating force there. Hence then vv = 2 gfx — 2 gx x

---; the fluents of which give v l = 4 gx x ---,

or v = “2\/j l X V[ex J rx’ 1 ) the general expression for the

velocity, putting e = dl a ab . And when x — b, or w becomesas far below*® as it was above it at the beginning, it is barelyv — 2\/~ for the velocity at that time. Also, when a,

the weight of the thread, is nothing, the velocity is only

7T > tIle

d(r% . •.

2 v'—, as it ought.

Again, for the time, t or — = x

° ’ ’ » 1 v ag y /(^ex + a. 2 ) ‘

, si , aA + V( e + x ) .1

J— x log. —- -7 - the ge-

v c S ° aA b

fluents ,of which give t

neral expression for the time of descending any space x.

And if the radicals be expanded in a series, and the log.of it be taken, the same time will become

. ^ AX . dl x 32° o .

t = v't X V— -7 x (1 - — + —- &c).

dg dt—ab 6 e 40c 3 '

Wliich therefore becomes barely when a, the weight ofthe thread, is nothing, as it ought.