{1

es}2

X

× { 1.

then

dx. d x √ πs

2 √ w√ {1 − eTM«-«)}'

(x a)

2

(x-a)

1

4w

2

d

#sd2v°) + C }

(x-a)

#d2sw

4

- dz

22

e"du

216

47

d ↓ xsw

Let aa; then z = 0 and C = 0.

-)

TS

+ C}.

Now the greatest velocity that the man can acquire in his descent is that which takes place when the retarding force the accelerating force. Hence the velocity is

4w

το

บ

πd's

......

This being substituted in equation (1) gives

z=e" = 1,

(2)

whence by equation (2) the time of acquiring this velocity

is oo.

In this solution the resistance of the air upon the man's body is not considered; so that it holds good but for the descent of the solid semicircle or cylinder whose weight is w.

736.

.. R

737. Let be the radius of the sphere, y that of the smaller end of the required segment; then if R denote the resistance on the circle r2, that on the hemisphere is (Vince's Fluxions, p. 277,)

The resistance

area x density x (vel.) R :: A x D x V2: 4A × 3D × 4V2

:: 1 : 48.

and the resistance on the end y3 is

2

TY3 x R = 2 R

y2

772

r2

or

R

.. the resistance on the segment cut off by zy is (Vinee, p. 277,)

:).

(1

×

X
2.2

and that upon the convex surface of the frustum is ..

R

y2 − (1

y2

272

2.2

R

y2

2r2

Hence the whole resistance upon the frustum is

2

R(1) R+R

2

R

( + 1).

Hence by the question

R

2

(4 + 1) = 2 R

4

which gives

VA AS

and this is sufficient to determine the required segment.

and

738.

The sections of a cylinder made by planes parallel to the base are circles. Hence, if R denote the resistance upon the diameter (2r) of one of these sections, and a the axis of the cylinder, the resistance upon the semi-surface of the cylinder is (Vince's Flux. p. 277).

2/ Rx a

y = rx

Also the resistance upon the end of the cylinder when it moves parallel to the axis is

739. Let M be the magnitude of the body, and S and S' the specific gravities of the fluid and the body, respectively; then the descending force is (Vince's Hyd. prop. XVIII)

FM XS-MX SM x (S - S')

which being constant gives by the question,
a2=Fs Mx (S-S')s

a2

:.S=S'+

Ms

740. For the general construction see the note 192, Vol. II, of Newton's Prin. edit. of PP. Le Seur and Jaquier. From this it appears that the resistance (r) upon the curve KA : resist(R) upon the base KC:: area of KQHEC: area of the rectangle KI.

Now let CI=m, CK = a, CA = ß, KP = x, PH = y', PB=y, KB = s. Then since

4xr3

dy*

dx2

2ax + x2

2

Let x = a; then we get

a2 l.

- a2 l.

√(1+

a'dx

a+x

mdx2

ds

a + x

α

dx

a + x

α

dy2

dx2

2ax + x2

741.

Let S, S' and s denote the specific gravities of the wood, iron, and sea-water respectively. Also, let 2r be the diameter of either ball; then since the volume of each ball is

4πr3
3

the weight of the whole apparatus is

4r3

4#r3

S+

S'=

dx

(S+ S')

HYDRODYNAMICS IN GENERAL.

and the force which accelerates the descent of the balls is (Vince's

Hyd.)

=

4πrs (S+S' - 28).

3

Again, the resistance of the fluid upon each globe is (it is that upon the great 4 wt. of a cylinder of the fluid of

whose base is r2, and altitude

and vdv =

r being the

v2

2

2g

v being the velocity with which they are moving at the time. Hence, the balls descend with the force

F

4773

(S+S' - 2s)

3

space

3g

=

4πrs

3

3

by supposition.

t =

(S+S' - 28) dx

described from rest.

2v dv

Sgr. (S+S'- 2s) - 3sv

x l.
B

2g

1

A/B

v = √ B × √ (1 — e −12)

× {C − 1. (8gr (S + S′ — 2s) — 3sv*)}

B

Let 1 eu2; then

1.

8gr.(SS'28) Sgr.(SS-2s)

v2

1

2du

1 = √√√3 × √ 121

t

X

AB

u2

mrs?

2g

-

1 + u

1

- 26

-

3sv2