tion that F = 0, when x=c, and putting 2.(c+r)x−c.(c+2r=X,

the sum of the direct attractions, of every particle in the portion cut off by the section which is distant from P, by the interval x. Again, when c0, P is in the surface, and then we have 2r = x + xl. 2. . . . . . . (1')

for

F'

clcl.cl.0° 110, &c.

Hence it follows that the sum of the attractions of that part of the sphere, whose radius is, which is between P and the

same section is

2

Again, for the section which passes through the intersection of the spherical surfaces, it is easily shewn that

Also, when x = c + r, equation (2) becomes

f' = « (c + r)

and hence the attraction of the part scooped out of the given sphere is

But by equation (1), when x = c + 2r we get, for the whole attraction of the sphere

which being twice as great as that of the part scooped out, resolves the problem.

636.

Let a be the distance of the two centres of force, the length of the bar; also let x be the distance of any point in the bar from one centre; then the attraction of this centre at the distance a being A, we have

for the attraction of the particle: and if y, y + 1, be the distances of the extremities of the rod from this centre of force, when in the required position, the attraction upon the whole rod is the above integral taken between x = y + 7 and x = y; it is

Aa {l. (y + 1) − l.y}

or Aa. 1. y + 1

y

Hence, and by the question, we easily get, for the whole attraction of the other force

2 Aa l. a-y
a-y-l

But to effect the equilibrium, these attractions must be equal; :: 1. y + 1

y

= 21.

a-y
a = y + l

whence y is easily found, and with it the required position of the

rod.

637.

axis. Now

The spheroid revolves, of course, round its minor

the terms involving et, e, &c. being neglected because of their smallness.

Again, since the solid content of the spheroid, (see Vince p. 95) is

Απα
3

to find the sphere of the same volume with the spheroid, suppose its radius a, and we have

Again, if PG be the normal at P, meeting the axis b in G, and

the attraction of the spheroid is (see Simpson, Vol. II. p. 387).

Hence since the eccentricity is small, and at P, x2 = ** = 1/2 PC2

3

= a(1 − 2 ‹") × (1 + — — ) (1 + — ) nearly.

3

= √(15) = * nearly.

a2(1 e3)

.. the whole attraction of the spheroid as

Απα

or

3

), which differs so little from the attraction of

the sphere whose radius is a (1

a (1— —2) viz.

Απα
3

(1 - 0)

see Simpson's Fluxions, Art. 381, that they may be considered as equal. Q. E. D.

638. Since the accelerating forces are a, b, c and the masses A, B, C the moving forces are Aa, Bb, Cc. Let A, B, C be joined and in the ABC assume P for the point required; then through P let a line be drawn parallel to the direction of the forces, and let x, y, z be the distances of A, B, C from this line. Then when there is an equilibrium, by the property of the lever,

Aa x + Bby + Ccz = 0.

The remainder of the question is merely geometrical and consists in finding from the given sides and angles of the A▲ ABC, and the given inclination of the line passing through P, the quantities x,y,z.

639. The body will evidently move along the diameter of the sphere. Moreover it will be accelerated through the first half of the diameter and retarded through the remaining half. Suppose the body has descended through a; then the attraction of any circle whose plane, x, is distant from the body by the interval u, and whose radius is w, is measured by (Vince, Art. 69.)