Extend also CD to O and P and draw from the points E and F the lines EQ and FR perpendicular to that.It will be proved in the same way that the line EQ is greater than FR, and, since the weight placed at E will be farther from the line of direction OP than the weight at F, the weight at E will have greater heaviness than the weight at F.From this it follows that the balance EF moves downward on the side of E.
Thus Aristotle poses only two questions and leaves out the third; that is, the case in which the center of the balance is in the balance itself.But he left this out as a thing well known, as he usually did omit obvious things.Who can doubt that, if the weight is sustained at its center of gravity, it will remain at rest?But perhaps someone will take issue with the things that we have put forth in accordance with his opinion and will say that we have not brought in his entire thought.For in the second part of the second question he asks, "Why, when the support is below, the balance being carried downward and released, it does not rise again, but remains? "Here he affirms not that the balance moves downward, but that it remains, which he seems to have deduced in the last conclusion.But not only does this not bear against us; when properly understood, it greatly assists us.
33[Figure 33]For let there be the balance AB, parallel to the horizon, with its center E under the balance.And since Aristotle considers an actual balance, it is necessary to place the support or something else under the center E; let this be EF, and this will be the support that sustains the center E.Let ECD be the perpendicular .In order that the balance AB may move from this position, says Aristotle, let there be a weight at B which will move the balance downward on the side of B, say, to G, where the obstacle prevents its moving farther downward.Now Aristotle does not say that the balance moves down on the side of B as far as it can, and is left there, as we say; but he would have it that a weight is placed at B, which by its nature will always move downward until the balance rests against its support or something else.When B is at G, the balance will be at GH, in which position it will remain if the weight is taken away; for the side of the balance from the perpendicular toward G (that is, DG) is longer than DH; but the balance does not move farther downward because it will be on the support, or whatever else sustains the center of the balance.And if this did not support it, in his opinion the balance would move downward on the side of G, and DG, being greater, would necessarily be carried farther downward.
But someone might add to this that, if a very small weight were placed at B, it would indeed move the balance downward but not all the way to G, and in this position, according to Aristotle, it should remain if the weight were taken away.This is evident by experience, since the balance tilts more or less when at one end of the balance only there is placed a larger or smaller weight, and this is true enough so long as the center is placed above the balance, but not when it is below or in the balance, as we shall show by way of example.
34[Figure 34]Let there be the balance AB, parallel to the horizon, whose center C is above the balance; let the perpendicular CD be plumb to the horizon, and let this line be extended through D to H.Now, since we consider the weight of the balance, the point D will be the center of gravity of the balance.But if a small weight is placed with its center of gravity at the point B, the center of gravity D of the whole system composed of the balance AB and the weight placed at B will no longer be at D, but it will be in the line DB.Say it is at E, so that DE is to EB as the weight placed at B is to the weight of the balance AB.Now join C and E; and since the point C is fixed, when the balance moves, the point E will describe the circumference of the circle EFG with radius CE and center C.But since CD is plumb to the horizon, the line CE will not be so.Hence the weight composed of AB and the weight at B will not remain in this position but it will move downward along the circumference EFG according to its center of weight E, until CE becomes plumb to the horizon, that is, until CE gets to CDF.The balance AB will then be moved to KL, in which position the balance together with the weight will remain, nor will it move farther downward.If a heavier weight were placed at B, the center of gravity of the whole system will be closer to B, say at M, and then the balance will move downward until the line joining C and M comes to the line CDH.Hence when a greater or lesser weight is put at B, the balance will be tilted more or less.From this it follows that the weight B will always describe an angle less than a quadrant, since the angle FCE is always acute, nor will the point B ever go all the way to the line CH, because the center of gravity of the weight and the balance together will always be between B and D.The heavier the weight placed at B, the larger will be the arc described, beginning at E and approaching closer to the line CH.
35[Figure 35]But if the balance AB has its center C in the balance, C will also be the center of gravity of the balance; let the line FCG be drawn perpendicular to AB and to the horizon.Then put any weight you please at B, and let the center of gravity now be at E, so that CE is to EB as the weight placed at B is to the weight of the balance.And since CE is not perpendicular to the horizon, the balance AB and the weight at B will not remain in this position but will move downward on the side of B until CE becomes perpendicular to the horizon; that is, until the balance AB comes to FG.Whence it is clear that the weight placed at B always describes a full quadrant.
36[Figure 36]