Sidebilder
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Hence

dx.d x NTS
dt =
2 yw {1

en -a;}
Let i
e tyre te

= eft; then 2

edu

X d x N TSW

1

dt =

216

dt =

Again, let e* = z; then

2

dz Х

1 d W TSW

1 and t =

x{1. +C}:
d N TSW

1+
Let x = a; then z = 0 and C = 0.

.. (2)

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

d

2

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TS

This being substituted in equation (1) gives

z = e = 1,

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. The resistance oc area x density x (vel.) :: R : R' :: A x D x V2 : 4A X 3D X 4V?

: 48.

737. Let y 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 wr*, that on the hemisphere is (Vince's Fluxions, p. 277,)

R

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(1 - )

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: the resistance on the segment cut off by my' is (Vince, p. 277,)

y? Х

R 2r2

28 and that upon the convex surface of the frustum is :: R

y? (1

Х 2

2r2

-)

y?

Hence the whole resistance upon the frustum is

y?
ý?

R

2y R

y' R – (1

R+

22

or

Х

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2

Hence by the question
R

+ 1) =

3

Х

R

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and this is sufficient to determine the required segment.

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

R Xa

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

"772 a R X

2ar

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2

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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)

F= M XS – M S = M X (S – S) which being constant gives by the question,

a’ = Fs = M X (S – S')s

:: S=S+ :

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 = B, KP = x, PH = y', PB=y, KB=s. Then since

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(1+ dy

a

dir ::a: M

?

dx? In the problem (Vince's Flur. prop. 130)

dy
2ax + x2
dx

2ax + x2
dy

a + x
1+
d.x2

aʻd.x
= [(a + x) dx - Satelit

ata (a + x)

al.

a + x

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dx

2010

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2

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

40r3

3

the weight of the whole apparatus is
4023

4713
S +

45r3

Ş= (S + S')
3
3

3

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

4753

(8 + S – 28).

3

Again, the resistance of the fluid upon each globe is (it is = that

upon the great O = wt. of a cylinder of the fluid of whose base is #ro, and altitude

29

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2

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.

3

39

1
х x 71XS

2g v being the velocity with which they are moving at the time. Hence, the balls descend with the force

4793
F= (S + $ – 2s)
3

2g
4713

ards and vdv = (S + S – 2s) dor

vodx

2g I being the space described from rest.

2v dv
8gr. (S + S – 2s) - 3svo

"
g

X {C - l. (8gr (S + S – 2s) — 3sv")}
r®STE
9

8gr.(S + S' - 28)
8gr.(S + S – 2s) 3sv?

B
A

B
by supposition.
Hence,

v=7B x (1 – e-Ar)
1

dr
Х

х.

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x l.

W(1-e-A

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