Sidebilder
PDF
ePub

605. Generally let R,R', &c., be the radii of the wheels, T, ', &c., those of the axles; then x denoting the weight to be raised each time, and P the power, the moving force of Pis

rr. P-XX

RR' ....

Pt

F =

P - x.

and the mass moved reckoning the inertia is
x + wd + w'd'R' + w"d"?R?R" + &c.

R’R'R"&c. wherein w, w', w" &c. are the weights of the several pairs of wheels and axles, and d, d', d" &c., the distances of their centres of gyration from the respective axes of rotation. Hence the accelerating force is

rr'....
R.R'..

X(RR'R"....)
P + x + 'wd" + wód”R” + &c.
P.RR'....

x RR'R" &c.
P.R?R'x +wd? + wd?R+&c.
and the force which accelerates the ascent of x is ..
F

PM - xm
F =

RR'R" &c. PM +x+wd + w'd?R? + &c.
where

M = RR'&c, and m = rr' &c. Now n being the number of ascents,

Q the whole quantity raised through the space s in the whole time T, we have

Q nx = Q and .. n =

.....-X.rr.....

[merged small][merged small][merged small][ocr errors][merged small][ocr errors][merged small][merged small][ocr errors][merged small][merged small][merged small]
[ocr errors]

D being substituted for wds + w'd'R + &c.

2PM:- 3m.ro PM.x2 23 m

PM +D+ (PMR + D+x) and we finally obtain

3m PM+3mD-PM 2PM(PMR +D) 22+

2m which being resolved will give x, and therefore the quantity required, when for M, m and D we substitute their values, viz.,

RR', rr', wd + w'd"

2m

606. Let either weight be denoted by P and the additional one by p, and put P + p = Q, also let l be the length of the whole string, a, b the lengths of the parts of it to which P, p are attached previously to the motion, and suppose the weight Q to have descended x feet. Then since the weight of the strings may be represented by their lengths the moving force is

Q-P + b + x - (a - x) = Q - P + b -a + 2x and the mass moved is P + Q +7

:. the accelerating force is
Q-P+b-a+2x M + 2x

by hypothesis
P+Q+b-a

N
Hence
vdv = Fdx gives)

2gM 29
v=

x + X N N

2

[ocr errors]

6a

29
and v=

x (Mx + x2)
N
29

x Q

♡ (Q-P+b-a)x + x2 P+Q+b+a Let b = a = 0. Then the velocity resulting from the actions of P, Q alone is

29 d':

v (Q-P)x + r?

PER and therefore that which is due to the weight of the string is

v – v', which being put = 0 according to the question gives

(Q-P+b-a).x+x® Q-Px+xs
P+Q+6+a

P+Q
Q-P+b-a Q Р
Q+P+6+a Q + P

Q

:: x= 2(P6- Qa)

a + b

which gives the part of the descent required.

d =

607. The distance required (d) is that of the centre of oscillation from the axis of suspension, that is

S fr’dM
Mk

SædM
M being the mass (see Venturoli.)

0 But M = x x x tan.

2

[blocks in formation]

Let x = a, the altitude of the triangle; then

3

a.

608. Let r be the radius of the base of the paraboloid, W its weight, and I the length of the chain, (whose weight is represented by its length), a that part of it which is unwound at the commencement of the motion, and a + x that part unwound after the body has moved t seconds; then since the accelerating force is

(a +x)

W

F =

[merged small][ocr errors][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][subsumed][ocr errors][merged small][ocr errors][merged small]

which being integrated between t = 0 and t = t", will give t" in terms of x, and therefore x in terms of t", the value required.

[ocr errors]

609.

? тр

denote the ratios of the num-
ber of teeth in the pairs of wheels and pinions, and N, N,
N, their respective numbers of revolutions in any time t; then
since

N = r,N
N, = N,= r,r,N
&c. = &c.

[ocr errors]
[ocr errors][merged small]

610.

Let W be the weight of the given cylinder, r the

radius of its base, 8 the required inclination of the plane to the horizon, and I its length, then the moving force is

W sin. 0 and the mass is

W + W

[merged small][merged small][merged small][ocr errors][merged small][merged small][ocr errors][ocr errors][merged small][merged small]
[ocr errors]

611. Letr be the radius of the circle; then D being the distance of the required centre from the axis, we have

D=

M M being the mass, and a the distance of the particle dM from the axis of gyration. But .

M = 2yd.x = 2dr

s 2x*dx / ma—
:: D' =

M
Let P = x (7* – 2)}. Then
dP = rdx vp - 3* - 4x'dx ?? - ?
:: S xdx pe – 20 AM – (78 – zegt
M – 2

2x (ne – zey? and D =

M
Let x = r. Then

M
M

[ocr errors]

2

612.

Generally if S denote the moment of inertia, M the

2 G

VOL. II.

« ForrigeFortsett »