root of the product. Thus, if the thing weigh i 16 ounces in the one scale, and only 14 in the other : then the product of 16 multiplied by 14 is 224, the square root of which gives 143% for the true weight, or nearly 15 ounces. Or indeed the just weight is found nearly by barely adding the two numbers together, and dividing the sums by 2. Thus 16 and 14 make 30, the half of which, or 15; is the true weight very nearly.

PROBLEM IV.

To find the centre of gravity of several weights.

As the solution of various problems in mechanics depends on a knowledge of the nature and place of the centre of gravity, we shall here explain the principles of its theory.

The centre of gravity of a body, is that point around which all its parts are balanced, in such a manner, that if it were suspended by that point, the body would remain at rest in every position, in which it might be placed around that point.

It may be readily seen that, in regular and homogeneous bodies, this point can be no other than the centre of magnitude of the figure. Thus, the centre of gravity in the globe and spheroid, is the centre of these bodies; in the cylinder it is in the middle of the axis.

The centre of gravity between two weights, or bodies of different gravities, is found by dividing the distance between their points of suspension into two parts, which shall be inversely proportional to their weights; so that the shorter part shall be next to the heavier body, and the longer part

towards the lighter. This is the principle of balances with unequal arms, by means of which any bodies of different weights may be weighed with the same weight, as in the steel yard.

When there are several bodies, the centre of gravity of two of them must be found by the above rule; these two are then supposed to be united in that point, and the common centre of gravity between them and the third is to be found in the same manner, and so of the rest.

Let the weights A, B and C, for example, be suspended from three points of the line or balance DF (pl. 1 fig. 1), which we shall suppose to have no weight. Let the body A, weigh 18 pounds; B 144, and C 180; and let the distance DE be 11 inches, and E F 9.

First find the common centre of gravity of the bodies B and C, by dividing the distance EF, or 9 inches, into two parts, which are to each other as 144 and 180, or as 5 to 4. These two parts will be 5 and 4 inches; the greater of which must be placed towards the smaller weight: the body B being here the smaller, we shall have E G equal to 5 inches, and F G to 4; consequently D G will be 16.

If we now suppose the two weights B and C, united into one in the point G, and consequently equal in that point to 324 pounds; the distance DG, or 16 inches, must be divided in the ratio of 108 to 324, or of 1 to 3. One of these parts will be 12 and the other 4; and as A is the less weight, DH must be made equal to 12 inches, and the point H will be the common centre of gravity of all the three bodies, as required.

The result would have been the same, had the bodies A and B been first united,

In short, the rule is the same whatever be the number of the bodies, and whatever be their position in the same straight line, or in the same plane,

This may suffice here in regard to the centre of gravity. For many curious truths, deduced from this consideration, recourse may be had to books which treat on mechanics; we shall however mention one beautiful principle in this science deduced from it, which is as follows:

If several bodies or weights be so disposed, that by com municating motion to each other their common centre of gravity remains at rest, or does not deviate from the horizontal line, that is to say neither rises nor falls, there will then be an equilibrium.

The demonstration of this principle is almost evident from its enunciation; and it may be employed to demonstrate all the properties of machines. But we shall leave the application of it to the reader.

REMARK.

As this is the proper place, we shall here discharge a promise made in the preceding volume, p. 418, viz, to resolve a geometrical problem, the solution of which, as we said, seems to be only deducible from the property of the centre of gravity.

Let the proposed irregular polygon then be ABCDE (pl. 1 fig. 2 N° 1); the sides of which are each divided into two equal parts, in a, b, c, d and e, from which results a new polygon abcdea; let the sides of the latter be each divided also into

two equal parts, by the points a', b', c', d', e', which when joined will give a third polygon a' b' d' d'e' a'; and so on. In what point will this division ter minate?

To solve this problem, if we suppose equal weights placed at a, b, c, d, e, their common centre of gravity will be the point required.

But, to find this centre of gravity, we must proceed in the following manner, which is exceedingly simple, First draw a b (fig. 2 N° 2), and let the middle of it be the point f; then draw fe, and divide it in g, in such a manner that fg shall be one third of it; draw also gd, and let g h be the fourth of it; in the last place, draw he and let hi be the fifth of it: the weight e being the last, the point i, as may be demonstrated from what has been before said, will be the centre of gravity of the five equal weights placed at a, b, c, d and e; and will solve the proposed problem.

PROBLEM V.

When two persons carry a burthen, by means of a lever or pole, which they support at the extremities, to find how much of the weight is borne by each person.

IT may be readily seen that, if the weight C were exactly in the middle of the lever A B (pl I fig. 3), the two persons would each bear one half.. But if the weight is not in the middle, it can be easily demonstrated, that the parts of the weight borne by the two persons, are in the reciprocal ratio of their distance from the weight. Nothing then is necessary but to divide the weight according to this ratio; and the greater portion will be that supported by

the person nearest the weight, and the least that supported by the person farthest distant. The calculation may be made by the following propor

tion.

As the whole length of the lever A B, is to the length A E, so is the whole weight to the weight supported by the power or person at the other extremity B; or as A B is to BE, so is the whole weight, to the part supported by the power or person placed at A.

If AB, for example, be 6 feet, the weight C. 150 pounds, AE 4 feet, and BE 2, we shall have this proportion: as 6 is to 4, so is 150 to a fourth term, which will be 100. The person placed at the extremity B, will therefore support 100 pounds, and consequently the one placed at A will have to support only 50.

REMARK.

The solution of this problem affords the means of dividing a burthen or weight proportionally to the strength of the agents employed to raise it. Thus, for example, if the one has only half the strength of the other, nothing is necessary but to place him at a distance from the weight double to that of the other.

PROBLEM VI.

How 4, 8, 16 or 32 men may be distributed in such a manner, as to carry a considerable burthen with eafe.

IF the burthen can be carried by four men, after