The resistance of solids

This resistance of solids offer to large tractive forces [violenta attrazzione] there can at least be no doubt of its existence; and though this resistance is very great in the case of a direct pull, it is found, as a rule, to be less in the case of bending forces [nel violentargli per traverso].

Thus, for example, a rod of steel or of glass will sustain a longitudinal pull of a thousand pounds while a weight of fifty pounds would be quite sufficient to break it if the rod were fastened at right angles into a vertical wall. It is this second type of resistance which we must consider, seeking to discover in what proportion it is found in prisms and cylinders of the same material, whether alike or unlike in shape, length, and thickness.

In this discussion I shall take for granted the well-known mechanical principle which has been shown to govern the behavior of a bar, which we call a lever, namely, that the force bears to the resistance the inverse ratio of the distances which separate the fulcrum from the force and resistance respectively.

Yes, I am willing to concede him priority in point of time; but as regards rigor of demonstration the first place must be given to Archimedes, since upon a single proposition proved in his book on Equilibrium depends not only the law of the lever but also those of most other mechanical devices.

Yes. But we should approach our subject in a way different from that employed by Archimedes.

We assume merely that equal weights placed in a balance of equal arms will produce equilibrium. This principle is also assumed by Archimedes.

Then we prove that it is no less true that unequal weights produce equilibrium when the arms of the steelyard have lengths inversely proportional to the weights suspended from them.

In other words, it amounts to the same thing whether one places equal weights at equal distances or unequal weights at distances which bear to each other the inverse ratio of the weights.

Imagine a prism or solid cylinder AB suspended at each end to the rod [linea] HI and supported by 2 threads HA and IB.

If I attach a thread C at the middle point of the balance beam HI the entire prism AB will according to the principle assumed, hang in equilibrium since one-half its weight lies on one side, and the other half on the other side, of the point of suspension C.

Suppose the prism to be divided into unequal parts by a plane through the line D, and let the part DA be the larger and DB the smaller: this division having been made, imagine a thread ED attached at the point E and supporting the parts AD and DB, in order that these parts may remain in the same position relative to line HI and since the relative position of the prism and the beam HI remains unchanged, there can be no doubt but that the prism will maintain its former state of equilibrium.

But circumstances would remain the same if that part of the prism which is now held up, at the ends, by the threads AH and DE were supported at the middle by a single thread GL.

Likewise, the other part DB would not change position if held by a thread FM placed at its middle point.

Suppose the threads HA, ED, and IB to be removed, leaving only the two GL and FM, then the same equilibrium will be maintained so long as the suspension is at C.

We have 2 heavy bodies AD and DB hung at the ends G and F, of a balance beam GF in equilibrium about the point C, so that the line CG is the distance from C to the point of suspension of the heavy body AD, while CF is the distance at which the other heavy body, DB, is supported.

These distances bear to each other the inverse ratio of the weights themselves, that is, the distance GC is to the distance CF as the prism DB is to the prism DA—a proposition which we shall prove as follows:

Since the line GE is the half of EH, and since EF is the half of EI, the whole length GF will be half of the entire line HI, and therefore equal to CI: if now we subtract the common part CF the remainder GC will be equal to the remainder FI, that is, to FE, and if to each of these we add CE we shall have GE equal to CF: hence GE:EF=FC:CG.

But GE and EF bear the same ratio to each other as do their doubles HE and EI, that is, the same ratio as the prism AD to DB. Therefore, by equating ratios we have, convertendo, the distance GC is to the distance CF as the weight BD is to the weight DA, which is what I desired to prove.

The 2 prisms AD and DB are in equilibrium about the point C since one-half of the whole body AB lies on the right of the suspension C and the other half on the left.

In other words, this arrangement is equivalent to two equal weights disposed at equal distances.

I do not see how any one can doubt, if the two prisms AD and DB were transformed into cubes, spheres, or into any other figure whatever and if G and F were retained as points of suspension, that they would remain in equilibrium about the point C, for it is only too evident that change of figure does not produce change of weight so long as the mass [quantità di materia] does not vary. From this we may derive the general conclusion that any two heavy bodies are in equilibrium at distances which are inversely proportional to their weights.

This principle established, I desire, before passing to any other subject, to call your attention to the fact that these forces, resistances, moments, figures, etc., may be considered either in the abstract, dissociated from matter, or in the concrete, associated with matter. Hence the properties which belong to figures that are merely geometrical and non-material must be modified when we fill these figures with matter and therefore give them weight.

Take, for example, the lever BA which, resting upon the support E, is used to lift a heavy stone D.

The principle just demonstrated makes it clear that a force applied at the extremity B will just suffice to equilibrate the resistance offered by the heavy body D provided this force [momento] bears to the force [momento] at D the same ratio as the distance AC bears to the distance CB; and this is true so long as we consider only the moments of the single force at B and of the resistance at D, treating the lever as an immaterial body devoid of weight.

But if we take into account the weight of the lever itself—an instrument which may be made either of wood or of iron—it is manifest that, when this weight has been added to the force at B, the ratio will be changed and must therefore be expressed in different terms.

Hence before going further let us agree to distinguish between these two points of view; when we consider an instrument in the abstract, i. e., apart from the weight of its own material, we shall speak of “taking it in an absolute sense” [prendere assolutamente];

But if we fill one of these simple and absolute figures with matter and thus give it weight, we shall refer to such a material figure as a “moment” or “compound force” [momento o forza composta].

I have not mentioned the total weight of the stone. I spoke only of its force [momento] at the point A, the extremity of the lever BA, which force is always less than the total weight of the stone, and varies with its shape and elevation.

In Fig 15, the weight has its center of gravity at A resting with the end B upon the horizontal plane and with the other end upon the lever CG.

Let N be the fulcrum of a lever to which the force [potenza] is applied at G. Let fall the perpendiculars, AO and CF, from the center A and the end C.

Then I say, the magnitude [momento] of the entire weight bears to the magnitude of the force [momento della potenza] at G a ratio compounded of the ratio between the two distances GN and NC and the ratio between FB and BO.

Lay off a distance X such that its ratio to NC is the same as that of BO to FB; then, since the total weight A is counterbalanced by the two forces at B and at C, it follows that the force at B is to that at C as the distance FO is to the distance OB. Hence, componendo, the sum of the forces at B and C, that is, the total weight A [momento di tutto ’l peso A], is to the force at C as the line FB is to the line BO, that is, as NC is to X: but the force [momento della potenza] applied at C is to the force applied at G as the distance GN is to the distance NC; hence it follows, ex æquali in proportione perturbata,

that the entire weight A is to the force applied at G as the distance GN is to X. But the ratio of GN to X is compounded of the ratio of GN to NC and of NC to X, that is, of FB to BO; hence the weight A bears to the equilibrating force at G a ratio compounded of that of GN to NC and of FB to BO: which was to be proved.