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CHAPTER X.

Rules for Moments and Centres of Plane Areas-Rules for Moments and Centres of Volumes.

RULES FOR MOMENTS AND CENTRES OF PLANE AREAS.

LET the plane figure in question be such as that represented in Fig. 8, Chapter VIII., its curved boundary being defined by means of a series of equidistant ordinates; and let it be required to find the moment of that figure relatively to a transverse axis, AB, traversing the origin; its moment relatively to the base-line, AD; and the abscissa and ordinate of its centre. The rules for those purposes are as follows:

I. To find the Moment relatively to the transverse Axis: Multiply each ordinate by the corresponding abscissa; treat the products as if they were the ordinates of a curve, of the same length with the given figure; the area of that curve, found by the proper rule, will be the moment required.

II. To find the Abscissa of the Centre: Divide that moment by the area of the figure.

In applying Rule I. it is often convenient to multiply each ordinate first by its proper multiplier, according to Simpson's rule, and then, not by the abscissa itself, but by the number of intervals contained in the abscissa, whether integral or fractional. The sum of the products is finally to be multiplied by one-third of the square of a whole interval, to obtain the moment required.

III. To find the Moment relatively to the Base-line or Axis of Abscissa: Take the half-square of each ordinate; treat those halfsquares as if they were the ordinates of a curve, of the same length with the given figure; the area of that curve, found by the proper rule, will be the moment required.

IV. To find the Ordinate of the Centre: Divide the last-mentioned moment by the area of the figure.

RULES FOR MOMENTS AND CENTRES OF VOLUMES.

I. To find the Moment relatively to the Transverse Plane traversing the Origin: Multiply each sectional area by the corresponding abscissa; treat the products as if they were the ordinates

of a curve of the same length with the given figure; the area of that curve, found by the proper rule, will be the moment required.

II. To find the Abscissa of the Centre: Divide that moment by the volume of the figure.

In applying Rule I. it is often convenient to multiply each sectional area, first by its proper multiplier according to Simpson's rule, and then by the number of intervals in the abscissa. The sum of the products is finally to be multiplied by one-third of the square of a whole interval.

The usual method of finding the moment of a solid figure relatively to a plane different from that first chosen, and the distance of the centre from that plane, is to conceive the solid to be divided into layers by transverse sections parallel to the new plane of moments, and proceed as in Rules I. and II. above stated. Example.-Quarter-Spheroid as in Fig. 10.

B

FIG. 10.

C

A

Calculation of the moment relatively to the transverse plane traversing A, and of the distance of the centre of mean distances from the plane. Length 120 feet, in six intervals of 20 feet radius of greatest section 12 feet.

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Moment, 814315

Moment relatively to end plane at A...

Volume, 18096-45, abscissa of centre of mean distances, being its distance

from the end plane.

To find the moment relatively to the longitudinal plane through A and B, and the ordinate, or distance of the centre, from that plane, conceive the solid to be divided into six horizontal layers by horizontal plane sections at intervals of 2 feet, and proceed as above. It is unnecessary to give the details of the calculation. The results are

Moment, 81431.5;

Ordinate of centre, 4.5.

CHAPTER XI.

Centre of Gravity-Centres of Gravity and Moments of Bodies.

THE Centre of Gravity of a heavy body, or of a system of heavy bodies, is a point which is always traversed by the resultant of the weight of the body or system. Supposing a set of weights to act at detached points, their common centre of gravity is found by the following:

RULE I.-Having chosen a fixed plane to which to refer the positions of the weights, take their moments relatively to that plane by multiplying each weight by its perpendicular distance from the plane; find the resultant of those moments, and divide it by the sum of the weights; the quotient will be the distance of their common centre of gravity from the fixed plane. By a similar process, find the distances of the same point from two other fixed planes at right angles to the first plane and to each other; the position of the centre of gravity will then be completely known.

The three planes are called, as in other cases, co-ordinate planes, and the distances of the weights and of their centre of gravity from those planes, co-ordinates.

The weights and their resultant (or sum) are to be regarded as all positive; but if some of the weights lie at one side, and some at the opposite side of a co-ordinate plane, their co-ordinates and moments relatively to that plane must be distinguished into positive and negative.

The sign of the resultant moment will show at which side of the plane the centre of gravity lies. When the resultant moment relatively to a plane is nothing, the centre of gravity is in that plane; and when each of the resultant moments is nothing, the centre of gravity is at the point of intersection of the three planes, or origin of co-ordinates. In other words, the moment of a set of weights relatively to their common centre of gravity is nothing.

The two following particular cases are useful:

II.—The common centre of gravity of two weights divides the

straight line joining them into parts inversely proportional to the weights respectively furthest from them. For example, in Fig. 11, let A and B be the two weights; join AB; the centre of gravity, G, divides that line in the following proportion :

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III.— The common centre of gravity of three weights is in the same plane with them; and if from the centre of gravity three straight lines be drawn to the three weights, those lines will divide the triangle formed by the weights into three triangles, each of an area proportional to the weight furthest from it. In Fig. 12, let A, B, C, be the weights, G their common centre of gravity; then,

A+B+C : A : B B :

C

:: Triangle ABC : GBC : GCA : GAB.

[In algebraical symbols, the general problem of finding the common centre of gravity of a set of weights is expressed as follows:-In Fig. 13, let - XO + X, YO + Y, - ZO+Z, be − the axes of co-ordinates, being the three straight lines in which the three co-ordinate planes cut each other. The distance or

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and regarded as positive or negative, according as the point lies to one side of the plane or to the opposite side.

Let the weights be denoted by W1, W2, W3, &c.; their co

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