PP'g

and a

x (W 1+A' + A) (P+P)(A® +1)* the actual tension,

N 2. PP'S

(Vi+ A' + A) (P+P')(A®+1) the pressure on the pulley.

If o be the inclination of the plane to the horizon, we have A = tan. 0, and +A? = sec. 0.

Hence

gPP

or o

(P + P') sec. 6

(sec. 0 + tan. 0)
gPP

(1 + sin. 6)
P + P
gPP / 2

x (1 + sin. 6)?

P + P which somewhat simplifies the expressions.

auda =

and c =

If 0 = 0. Then we get

gPP
P+P

.(a)
gPP' V2

P + P for the tension and pressure when the body P is drawn along an horizontal table by P' descending from its edge.

Again, if P draw P along an horizontal plane situated below the pulley, then we have

Y=0, A = 0 and B = 0 and equation (2) becomes

2gP(a - 1)

as P+P.

(X-a) Also

Hence (X - a) = 4* – nie

P.'(2-2).(18-m*)
::: d' = 29

(P+Pa' - Pm
Hence
v'du'

1

X
gda {(P + P')AS – P'm*}*
{P(P + Plat + m2 (PP – 2P2)

2P)* - 2PP'lm's + P"*} Also ds' =

da. Hence

(P + PP)* - 4moPP'+ 2PP'lm' o=P'g

{(P + P')a? – P'mo}

Imta a = 2

Xo which give the tension of the string and pressure on the pulley in this case.

2r=

Let m = 0. Then

gPP P + P

PP2 and

P + P which confirm the results marked (a).

601. Letr be the radius of the given circle, t the given time, & the altitude of the required diameter; then by the question

9

ť 2r

9. F =

2

602. Let zb be the length of the string to which W is attached, and x the space descended through in the time t; then the length of the pendulum being 26 - x, the time of one oscillation is

26

Hence N being the number of oscillations swung in the interval t, we have

be

Hence
1

dx
dN =

Х

✓ (26x - x*)

1
and N =

X vers.x.
b" N 2F
Let x = 20; then vers.-- 2b = 180o = b.
1

P+W
and N=

2(P-W) the number of vibrations required. It is evident that N is independent of the length of the string.

Let P = 00. Then

1 N= or the number of oscillations performed in the time

2

through 2b is

however great may be the power.

603. Let x denote the ratio required; then W and w being the weights of the wheel and axle R, r, their radii, we have

pRT qro

wr WR? pR+qr+ +

2 2

2pr - 2922
2p+ 2qxo + wxo+W
2px

29x2 2p+W+(29 + W)x? Now s being the given space described in the time t, we get

2(29 + w). (px — qxP) = (p – 29x) (2p + W + 2q + wx") :: (29 + w) x + 29(2p + W).c = (2p + W)p :. 28 + 29. 2p+W

P 29 + w 29 + w

2p+W

604. Let a be the height of the cone, r the radius of its base; then the distance of the centre of oscillation from the axis of rotation is (Venturoli,)

S FicdM
Mk

CdM

M

L =

MAM

= fx?dM

JedM x being the distance of the molecule dM from that axis. But dM = 2Tydz

2XT da s retao

Х

The centre of oscillation of a conical surface cannot possibly be in the base.