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each in space external to all : in other words, the outer boundary of every centrobaric body is a single closed surface.

Thus we see, by (a), that no symmetrical ring, or hollow cylinder with open ends, can have a centre of gravity; for its centre of gravity, if it had one, would be in its axis, and therefore external to its mass.

546. If any mass whatever, M, and any single surface, S, completely enclosing it be given, a distribution of any given amount, M', of matter on this surface may be found which shall make the whole centrobaric with its centre of gravity in any given position (G) within that surface.

The condition here to be fulfilled is to distribute M' over S, so as by it to produce the potential



EG any point, E, of S; V denoting the potential of M at this point. The possibility and singleness of the solution of this problem were stated above ($ 517). It is to be remarked, however, that if M'be not given in sufficient amount, an extra quantity must be taken, but neutralized by an equal quantity of negative matter, to constitute the required distribution on S.

The case in which there is no given body M to begin with is important; and yields the following :

547. A given quantity of matter may be distributed in one way, but in only one way, over any given closed surface, so as to constitute a centrobaric body with its centre of gravity at any given point within it.

Thus we have already seen that the condition is fulfilled by making the density inversely as the distance from the given point, if the surface be spherical. From what was proved in §§ 519, 524 above, it appears also that a centrobaric shell may be made of either half of the lemniscate in the diagram of $ 526, or of any of the ovals within it, by distributing matter with density proportional to the resultant force of m at I and m' at I'; and that the one of these points which is within it is its centre of gravity. And generally, by drawing the equipotential surfaces relatively to a mass m collected at a point I, and any other distribution of matter whatever not surrounding this point; and by taking one of these surfaces which encloses I but no other part of the mass, we learn, by Green's general theorem, and the special proposition of § 524, how to distribute matter over it so as to make it a centrobaric shell with I for centre of gravity.

548. Under hydrokinetics the same problem will be solved for a cube, or a rectangular parallelepiped in general, in terms of converging series; and under electricity (in a subsequent volume) it will be solved in finite algebraic terms for the surface of a lense bounded by two spherical surfaces cutting one another at any sub-multiple of two right angles, and for either part obtained by dividing this surface

in two by a third spherical surface cutting each of its sides at right angles.

549. Matter may be distributed in an infinite number of ways throughout a given closed space, to constitute a centrobaric body with its centre of gravity at any given point within it.

For by an infinite number of surfaces, each enclosing the given point, the whole space between this point and the given closed surface may be divided into infinitely thin shells; and matter may be distributed on each of these so as to make it centrobaric with its centre of gravity at the given point. Both the forms of these shells and the quantities of matter distributed on them, may be arbitrarily varied in an infinite variety of ways.

Thus, for example, if the given closed surface be the pointed oval constituted by either half of the lemniscate of the diagram of $ 526, and if the given point be the point I within it, a centrobaric solid may be built up of the interior ovals with matter distributed over them to make them centrobaric shells as above ($ 547). From what was proved in § 534, we see that a solid sphere with its density varying inversely as the fifth power of the distance from an external point, is centrobaric, and that its centre of gravity is the image ($ 530) of this point relatively to its surface.

550. The centre of gravity of a centrobaric body composed of true gravitating matter is its centre of inertia. For a centrobaric body, if attracted only by another infinitely distant body, or by matter so distributed round itself as to produce ($ 517) uniform force in parallel lines throughout the space occupied by it, experiences ($ 544) à resultant force always through its centre of gravity. But in this case this force is the resultant of parallel forces on all the particles of the body, which (see Properties of Matter, below) are rigorously proportional to their masses : and it is proved that the resultant of such a system of parallel forces passes through the point defined in $ 195, as the centre of inertia.

551. The moments of inertia of a centrobaric body are equal round all axes through its centre of inertia. In other words ($ 239), all these axes are principal axes, and the body is kinetically sym.metrical round its centre of inertia.

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552. FORCES whose lines meet. Let ABC be a rigid body acted on by two forces, P and l, applied to it at different points, D and E respectively, in

В. lines in the same plane.

Since the lines are not parallel, they will A meet if produced; let them be produced and

E meet in 0. Transmit the forces to act on that point; and the result is that we have simply the case of two forces acting on a material point, which has been already considered.

553. The preceding solution is applicable to every case of nonparallel forces in a plane, however far removed the point may be in which their lines of action meet, and the resultant will of course be found by the parallelogram of forces. The limiting case of parallel forces, or forces whose lines of action, however far produced, do not meet, was considered above, and the position and magnitude of the resultant were investigated. The following is an independent demonstration of the conclusion arrived at.

554. Parallel forces in a plane. The resultant of two parallel forces is equal to their sum, and is in the parallel line which divides any line drawn across their lines of action into parts inversely as their magnitudes.

1. Let P and be two parallel forces acting on a rigid body in similar directions in lines AB and CD. Draw any line AC across their lines. In it introduce any pair of balancing forces, S in AĠ K-S. and S in CH. These forces will not disturb the equilibrium of the body. Suppose the forces G-SA

CSH P and S in AG, and Q and S in CH to act respectively on the

YP points A and C of the rigid body. The forces P and S, in AB and


N AG, have single resultant in

B some line AM, within the angle


GAB; and sand S in CD and CH have a resultant in some line

CD CN, within the angle DCH.

The angles MAC, NCA are together greater than two right angles, hence the lines MA, NC will meet if produced. Let them meet in 0. Now the two forces P and S may be transferred to parallel lines through 0. Similarly the forces Q and S may be also transferred. Then there are four forces acting on 0, two of which, S in OK and S in OL, are equal and directly opposed. They may, therefore, be removed, and there are left two forces equal to P and Q in one line on 0, which are equivalent to a single force P+Q in the same line.

2°. If, for a moment, we suppose OE to represent the force P, then the force representing S must be equal and parallel to EA, since the resultant of the two is in the direction OA.

That is to say, S:P :: EA : OE; and in like manner, by considering the forces S in OL and Q in OE, we find that

Q:S :: OE : EC.

Compounding these analogies, we get at once

Q:P :: EA : EC,

that is, the parts into which the line is divided by the resultant are inversely the forces.

555. Forces in dissimilar directions. The resultant of two parallel forces in dissimilar directions', of which one is greater than the other, is found by the following rule: Draw any line across the lines of the forces and produce it across the line of the greater, until the whole line is to the part produced as the greater force is to the less; a force equal to the excess of the greater force above the less, applied at the extremity of this line in a parallel line and in the direction similar to that of the greater, is the resultant of the system.

Let P and Q in KK” and LL', be the K L


contrary forces. From any point A, in

1Q-P the line of P, draw a line AB across the A B

E line of cutting it in B, and produce the YQ

line to E, so that AE:BE :: Q:P.

Through E draw a line MM' parallel to K L


KKor LL'. In MM introduce a pair of balancing forces each equal to Q-P. Then P in AK” and (-P in EM have a resultant equal to their



1 In future the word ' contrary' will be employed instead of the phrase ' parallel and in dissimilar directions' to designate merely directional opposition, while the unqualified word ' opposite' will be understood to signify contrary and in one line.


ho le


sum, or Q. This resultant is in the line LL'; for, from the analogy,

AE: BE :: Q:P, we have

AE-BE : BE :: Q-P:P,

AB: BE :: Q-P:P. Hence Pin AK, Q in BL', and Q-P in EM are in equilibrium and may be removed. There remains only Q-P in EM', which is therefore the resultant of the two given forces. This fails when the forces are equal.

556. Any number of parallel forces in a plane. Let P1, P2, P3, &c., be any number of parallel forces acting on a rigid body in one plane. To find their resultant in position and

A1 E'A

magnitude, draw any line across their
lines of action, cutting them in points,


TRU denoted respectively by A, A2, A3, &c., and in it choose a point of reference 0. Let the distances of the lines of the forces from this point be denoted by an, , 23, &c.; as 0A,=@,, 0A,=2,, &c. Also let R denote the resultant, and x its distance from O.

Find the resultant of any two of the forces, as P, and P2, by $ 554. Then if we denote this resultant by R', we have

Divide A, A, in E' into parts inversely as the forces, so that

PAXA,E=P, E' Az.
Hence if we denote OE' by x' we have

P,x(x' – a)=P,x(az - x')

(P,+P2) x'=P, 2, +P, 22, that is

R'x'=P, a,+P,2,.
Similarly we shall find the resultant of R' and P, to be

R"=R' + P =P,+P,+P3; and

R"x"=R'x' + Pzaz=P, Q,+P, 2, + Pgag.
Hence, finally we have
R=P,+P+P +


(1) and Rx=P, 2, +P,a,+P, az + ............ +Prane (2)

In this method negative forces or negative values of any of the quantities a,,a,, ..., may be included, provided the generalized rules of multiplication and addition in algebra are followed.

557. Any number of parallel forces not in one plane. To find the resultant, let a plane cut the lines of all the forces, and let the points


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