POSITION, CENTRE OF — Suppl. 3rd (1801)

POSITION, CENTRE OF, is a point of any body, or system of bodies, so selected, that we can estimate with propriety the situation and motion of the body or system by the situation and motion of this point. It is very plain that, in all our attempts to accurate discussion of mechanical questions, especially in the present extended sense of the word mechanism, such a selection is necessary. Even in common conversation, we frequently find it necessary to ascertain the distance of objects with a certain precision, and we then perceive that we must make some such selection. We conceive the distance to be mentioned, neither with respect to the nearest nor the remotest point of the object, but as a sort of average distance; and we conceive the point so ascertained to be somewhere about the middle of the object. The more we reflect on this, we find it the more necessary to attend to many circumstances which we had overlooked. Were it the question, to decide in what precise part of a country parish the church should be placed, we find that the geometrical middle is not always the most proper. We must consider the populousness of the different quarters of the parish, and select a point such, that the distances of the inhabitants on each side, in every direction, shall be as equally balanced as possible.

In mechanical discussions, the point by whose position and distance we estimate the position and distance of the whole, must be so selected, that its position and distance, estimated in any direction whatever, shall be the average of the positions and distances of every particle of the assemblage, estimated in that direction.

This will be the case, if the point be so selected that, when a plane is made to pass through it in any direction

Position whatever, and perpendiculars are drawn to this plane from every particle in the body or system, the sum of all the perpendiculars on one side of this plane is equal to the sum of all the perpendiculars on the other side. If there be such a point in a body, the position and motion of this point is the average of the positions and motions of all the particles.

Plate XL. For if $P$ (fig. 1.) be a point so situated, and if $QR$ be a plane (perpendicular to the paper) at any distance from it, the distance $P\rho$ of the point from this plane is the average of the distances of all the particles from it. For let the plane $APB$ be passed through $P$ , parallel to $QR$ . The distance $CS$ of any particle $C$ from the plane $QR$ is equal to $DS - DC$ , or to $P\rho - DC$ . And the distance $GH$ of any particle $G$ , lying on the other side of $APB$ , is equal to $HP + GH$ , or to $P\rho + GH$ . Let $n$ be the number of particles on that side of $AB$ which is nearest to $QR$ , and let $o$ be the number of those on the remote side of $AB$ , and let $m$ be the number of particles in the whole body, and therefore equal to $n + o$ . It is evident that the sum of the distances of all the particles, such as $C$ , is $n$ times $P\rho$ , after deducting all the distances, such as $DC$ . Also the sum of all the distances of the particles, such as $G$ , is $o$ times $P\rho$ , together with the sum of all the distances, such as $GH$ . Therefore the sum of both sets is $n + o \times P\rho + \text{sum of } GH - \text{sum of } DC$ , or $m \times P\rho + \text{sum of } GH - \text{sum of } DC$ . But the sum of $GH$ , wanting the sum of $DC$ , is nothing, by the supposed property of the point $P$ . Therefore $m \times P\rho$ is the sum of all the distances, and $P\rho$ is the $m$ th part of this sum, or the average distance.

Now suppose that the body has changed both its place and its position with respect to the plane $QR$ , and that $P$ (fig. 2.) is still the same point of the body, and $P\rho$ a plane parallel to $QR$ . Make $\rho$ equal to $P\rho$ of fig. 1. It is plain that $P\rho$ is still the average distance, and that $m \times P\rho$ is the sum of all the present distances of the particles from $QR$ , and that $m \times P\rho$ is the sum of all the former distances. Therefore $m \times P\rho$ is the sum of all the changes of distance, or the whole quantity of motion estimated in the direction $P\rho$ . $P\rho$ is the $m$ th part of this sum, and is therefore the average motion in this direction. The point $P$ has therefore been properly selected; and its position, and distance, and motion, in respect of any plane, is a proper representation of the situation and motion of the whole.

It follows from the preceding discussion, that if any particle $C$ (fig. 1.) moves from $C$ to $N$ , in the line $CS$ , the centre of the whole will be transferred from $P$ to $Q$ , so that $PQ$ is the $m$ th part of $CN$ ; for the sum of all the distances has been diminished by the quantity $CN$ , and therefore the average distance must be diminished by the $m$ th part of $CN$ , or $PQ$ is $= \frac{CN}{m}$ .

But it may be doubted whether there is in every body a point, and but one point, such that if a plane pass through it, in any direction whatever, the sum of all the distances of the particles on one side of this plane is equal to the sum of all the distances on the other.

It is easy to show that such a point may be found, with respect to a plane parallel to $QR$ . For if the sum of all the distances $DC$ exceed the sum of all the distances $GH$ , we have only to pass the plane $AB$ a little nearer to $QR$ , but still parallel to it. This will dimin-

nish the sum of the lines $DC$ , and increase the sum of the lines $GH$ . We may do this till the sums are equal.

In like manner we can do this with respect to a plane $LM$ (also perpendicular to the paper), perpendicular to the plane $AB$ . The point wanted is somewhere in the plane $AB$ , and somewhere in the plane $LM$ . Therefore it is somewhere in the line in which these two planes intersect each other. This line passes through the point $P$ of the paper where the two lines $AB$ and $LM$ cut each other. These two lines represent planes, but are, in fact, only the intersection of those planes with the plane of the paper. Part of the body must be conceived as being above the paper, and part of it behind or below the paper. The plane of the paper therefore divides the body into two parts. It may be so situated, therefore, that the sum of all the distances from it to the particles lying above it shall be equal to the sum of all the distances of those which are below it. Therefore the situation of the point $P$ is now determined, namely, at the common intersection of three planes perpendicular to each other. It is evident that this point alone can have the condition required in respect of these three planes.

But it still remains to be determined whether the same condition will hold true for the point thus found, in respect to any other plane passing through it; that is, whether the sum of all the perpendiculars on one side of this fourth plane is equal to the sum of all the perpendiculars on the other side. Therefore

Let $AGHB$ (fig. 3.), $AXYB$ , and $CDFE$ , be three planes intersecting each other perpendicularly in the point $C$ ; and let $CIKL$ be any other plane, intersecting the first in the line $CI$ , and the second in the line $CL$ . Let $P$ be any particle of matter in the body or system. Draw $PM$ , $PO$ , $PR$ , perpendicular to the first three planes respectively, and let $PR$ , when produced, meet the oblique plane in $V$ ; draw $MN$ , $ON$ , perpendicular to $CB$ . They will meet in one point $N$ . Then $PMNO$ is a rectangular parallelogram. Also draw $MQ$ perpendicular to $CE$ , and therefore parallel to $AB$ , and meeting $CI$ in $S$ . Draw $SV$ ; also draw $ST$ perpendicular to $VP$ . It is evident that $SV$ is parallel to $CL$ , and that $STRQ$ and $STPM$ are rectangles.

All the perpendiculars, such as $PR$ , on one side of the plane $CDFE$ , being equal to all those on the other side, they may be considered as compensating each other; the one being considered as positive or additive quantities, the other as negative or subtractive. There is no difference between their sums, and the sum of both sets may be called $o$ or nothing. The same must be affirmed of all the perpendiculars $PM$ , and of all the perpendiculars $PO$ .

Every line, such as $RT$ , or its equal $QS$ , is in a certain invariable ratio to its corresponding $QC$ , or its equal $PO$ . Therefore the positive lines $RT$ are compensated by the negative, and the sum total is nothing.

Every line, such as $TV$ , is in a certain invariable ratio to its corresponding $ST$ , or its equal $PM$ , and therefore their sum total is nothing.

Therefore the sum of all the lines $PV$ is nothing; but each is in an invariable ratio to a corresponding perpendicular from $P$ on the oblique plane $CIKL$ . Therefore the sum of all the positive perpendiculars on this plane is equal to the sum of all the negative perpendiculars, and the proposition is demonstrated, viz. that

in every body, or system of bodies, there is a point such, that if a plane be passed through it in any direction whatever, the sum of all the perpendiculars on one side of the plane is equal to the sum of all the perpendiculars on the other side.

The point P, thus selected, may, with great propriety, be called the CENTRE OF POSITION of the body or system.

If A and B (fig. 4.) be the centres of position of two bodies, whose quantities of matter (or numbers of equal particles) are $a$ and $b$ , the centre C lies in the straight line joining A and B, and $AC : CB = b : a$ , or its distance from the centres are inversely as their quantities of matter. For let $a C \beta$ be any plane passing through C. Draw $A \alpha$ , $B \beta$ , perpendicular to this plane. Then we have $a \times A \alpha = b \times B \beta$ , and $A \alpha : B \beta = b : a$ , and, by similarity of triangles, $CA : CB = b : a$ .

If a third body D, whose quantity of matter is $d$ , be added, the common centre of position E of the three bodies is in the straight line DC, joining the centre D of the third body with the centre C of the other two, and $DE : EC = a + b : d$ . For, passing the plane $\alpha E \beta$ through E, and drawing the perpendiculars $D \delta$ , $C \gamma$ , the sum of the perpendiculars from D is $d \times D \delta$ ; and the sum of the perpendiculars from A and B is $a + b \times C \gamma$ , and we have $d \times D \delta = a + b \times C \gamma$ ; and therefore $DE : EC = a + b : d$ .

In like manner, if a fourth body be added, the common centre is in the line joining the fourth with the centre of the other three, and its distance from this centre and from the fourth is inversely as the quantities of matter; and so on for any number of bodies.

If all the particles of any system be moving uniformly, in straight lines, in any directions, and with any velocities whatever, the centre of the system is either moving uniformly in a straight line, or is at rest.

For, let $m$ be the number of particles in the system. Suppose any particle to move uniformly in any direction. It is evident from the reasoning in a former paragraph, that the motion of the common centre is the $m$ th part of this motion, and is in the same direction. The same must be said of every particle. Therefore the motion of the centre is the motion which is compounded of the $m$ th part of the motion of each particle. And because each of these was supposed to be uniform and rectilinear, the motion compounded of them all is also uniform and rectilinear; or it may happen that they will so compensate each other that there will be no diagonal, and the common centre will remain at rest.

Cor. 1. If the centres of any number of bodies move uniformly in straight lines, whatever may have been the motions of each particle of each body, by rotation or otherwise, the motion of the common centre will be uniform and rectilinear.

Cor. 2. The quantity of motion of such a system is the sum of the quantities of motion of each body, reduced to the direction of the centre's motion. And it is had by multiplying the quantity of matter in the system by the velocity of the centre.

The velocity of the centre is had by reducing the motion of each particle to the direction of the centre's motion, and then dividing the sum of those reduced motions by the quantity of matter in the system.

By the selection of this point, we render the investi-

gation of the motions and actions of bodies incomparably more simple and easy, freeing our discussions from numberless intricate complications of motion, which would frequently make our progress almost impossible.