Here the cords a b, c d are loaded only with the natural weight, which is 2, one falling to each cord; and this happens because these cords have within them the central line of the weight, which extends from the centre of the weight to the point where such cords are attached.

All the angles of cords that sustain weights, which are greater than right angles, have their lever outside the sides of that angle; and acute angles have their lever in their sides; and the right angle has that lever in its own angle.

Here the natural weight is 2, and every other weight, like this one, is always distributed by half to its two supports a b and a c. Then to each of these supports is added the accidental weight, which must be received by the cord that draws the weight out of its perpendicular. And this accidental weight will be so much greater or smaller as the natural weight is drawn more or less out of its said perpendicular. Beyond this a third power is added, which is likewise of natural gravity: and this is the gravity of the opposite counterweight, which always comes to be equal to its opposite; and even if that counterweight were not actually present, but the cord were merely fixed at its ends with a strong tie to something, one must understand there to be a power of resistance equal to the weight.

Let the cord be b a c, which at the angle a sustains the weight 2; this 2 comes to be the natural weight. And this weight, being sustained by the two cords, is of necessity distributed by half to each of them, so that each cord feels one libra. Then to each is added the accidental weight, which arises from the weight 2 being drawn out of its perpendicular; in this arrangement, the weight being drawn below the middle of the line b c, it meets the point e. The cord b a is loaded with half of that unit which a c sustained when it stood perpendicular; whence the cord b a comes to sustain one libra and a half. The cord a c does the same. It remains that the cord b a on one side sustains the weight and on the other the counterweight, which joined together are 3 libre.

If 3 is 3/8 of 8, of what number will 4 be 3/8? Work by the rule of three and say: 3 is the denominator of 8, that is, it is 3/8 of it; of what will 4 be the denominator? Say: 8 times 4 makes 32. Divide that 32 by 3, and you get 10 and 2/3; and this 10 and 2/3 stands in the same proportion to 4 as 8 did to 3.

If you want the proof, say: if 3 gives me 8, what will 4 give me? And say, as above: 8 times 4 makes 32; divide it by 4, and you get 8 quarters. Then say: 3 times 4 makes 12; divide that 12 by 3, and you get 3 quarters. Therefore, just as 3 is 3/8 of 8, so 12 is 3/8 of 32.

Here the upright b e is 7, and the hypotenuse c e is 8. Therefore the 7, weight of the upright, is felt as eight by its two hypotenuses, four falling to each hypotenuse. The counter-proof follows.

a b, the counter-lever, is 4/7 of the lever a n. It follows that 4 of resistance at n can withstand 7 of descent at b. Behold that the second proof concludes the first to be true; that is, that each of these proofs shows that the hypotenuse e c feels 4 at c of the 7 that sustains the angle where the two hypotenuses a e and c e join.

In this arrangement of cords the 7 of weight, placed at e, generates one of accidental weight upon the two hypotenuses.

No heavy body will remain upon the support on which it is placed, if the central line of its weight does not touch the front edge of that support.

Let c b be the said support; let a be the heavy body placed upon that support; let a be the centre of its gravity; a c is the central line of that heavy body. I say that this heavy body cannot remain upon such a support if its central line does not touch the front of the support, because the greater part of its weight would be beyond its support, and so, falling, it would draw the lighter part down with it.