The Essential Galileo

by Galilei, Galileo, Finocchiaro, Maurice A.

  SIMP. This was demonstrated first of all by Aristotle, in his Questions of Mechanics.9

  SALV. 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 the Equilibrium of Planes depends not only the explanation of the lever, but also those of most other mechanical devices.10

  SAGR. Now, since this principle is fundamental to all the demonstrations which you propose to set forth, would it not be advisable to give us a complete and thorough proof of this proposition, unless possibly it would take too much time?

  SALV. Yes, that would be quite proper. But it is better, I think, to approach our subject in a manner somewhat different from that employed by Archimedes. That is, I shall assume merely that equal weights placed in a balance of equal arms will produce equilibrium— a principle also assumed by Archimedes—and then prove two things: 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; and that it amounts to the same thing whether one places equal weights at equal distances or unequal weights at distances that bear to each other the inverse ratio of the weights.

  In order to make this matter clear, imagine a prism or solid cylinder, AB, suspended at each end to the rod HI, and supported by two threads HA and IB; it is evident that 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. Now suppose [153] 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; and likewise the other part DB would not change position if held by a thread FM placed at its middle point. Suppose now 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. Now let us consider that we have here two 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. It remains now only to show that 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 half of EH, and EF is 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 is to EF as FC is to 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 equidistance of ratios11 and by inversion, we have that the distance GC is to the distance CF as the weight BD is to the weight DA. This is what I desired to prove.

  [154] If what precedes is clear, you will not hesitate, I think, to admit that the two 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 anyone can doubt, if the two prisms AD and DB were transformed into cubes, spheres, or 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 quantity of matter does not vary. From this we may derive the general conclusion that any two heavy bodies are in equilibrium at distances that are inversely proportional to their weights.

  This principle established, I desire, before going further, 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 nonmaterial must be modified when we fill these figures with matter and so 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 balance the resistance offered by the heavy at Dthe same ratio as body D provided this force bears to the force at 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 that may be made either of wood or of iron—it is manifest that, when this weight has been added to the force at B, [155] 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 matter, we shall speak of taking it in an absolute sense; 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.

  SAGR. I must break my resolution about not leading you off into a digression, for I cannot concentrate my attention upon what is to follow until a certain doubt is removed from my mind. That is, you seem to compare the force at B with the total weight of the stone D, a part of which—possibly the greater part—rests upon the horizontal plane, so that …12

  SALV. I understand perfectly; you need go no further. However, please observe that I have not mentioned the total weight of the stone. I spoke only of its force at the point A, the extremity of the lever BA; this force is always less than the total weight of the stone and varies with its shape and elevation.

  SAGR. Good; but there occurs to me another question about which I am curious. For a complete understanding of this matter, I should like you to show me, if possible, how one can determine what part of the total weight is supported by the underlying plane and what part by the end A of the lever.

  SALV. The explanation will not delay us long, and I shall therefore be glad to grant your request. In the following figure, let us understand that the weight having its center of gravity at A rests with the end B upon the horizontal plane and with the other end upon the lever CG. Let N be the fulcrum of the lever to which a force is applied at G. Drop the perpendiculars, AO and CF, from the center A and the end C. Then, I say, the moment of the entire weight bears to the moment of the force 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 the ratio of FB to BO is the same as that of NC to X. But since the total weight A is supported by the two forces at B and at C, [156] it follows that the force at B is to that at C as the distance FO is to the distance BO. Hence, by addition, the sum of the forces at B and C, that is, the total weight 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 applied at C is to the force applied at G as the distance GN is to the distance NC. Hence it follows, by perturbed equidistance of ratios,13 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 supporting force at G a ratio compounded of that of GN to NC and of FB to BO. This is what had to be demonstrated.

  Let us now return to our original subject. If what has hitherto been said is clear, it will be easily understood why the following (Proposition 1) is true: A prism or solid cylinder of glass, steel, wood, or other breakable material, which is capable of sustaining a very heavy weight when applied longitudinally, is (as previously remarked) easily broken by the transverse application of a weight that may be much smaller in proportion as the length of the cylinder exceeds its thickness.

  Let us imagine a solid prism ABCD fastened into a wall at the end AB, and supporting a weight E at the other end; understand also that the wall is vertical and that the prism or cylinder is fastened at right angles to the wall. It is clear that if the prism breaks, fracture will occur at the point B where the edge of the slot in the wall acts as a fulcrum. The length BC acts as the part of the lever to which the force is applied. The thickness BA of the solid is the other arm of the lever in which is located the resistance. This resistance opposes the separation of the part of the solid BD lying outside the wall from the portion lying inside. From the preceding, it follows that the moment of the force applied at C bears to the moment of the resistance found in the thickness of the prism (i.e., in the attachment of the base BA to its contiguous parts) the same ratio which the length CB bears to half of BA. [157] Now if we call absolute resistance to fracture that offered to a longitudinal pull (in which case the stretching force moves by the same amount as the stretched body), then we can say that the absolute resistance of the prism BD is to the breaking load placed at the end of the lever BC in the same ratio as the length BC is to the half of AB in the case of a prism, or the radius in the case of a cylinder. This is our first proposition.

  Note that in what has here been said the weight of the solid BD itself has been left out of consideration, or rather, the prism has been assumed to be devoid of weight. But if the weight of the prism is to be taken into account in conjunction with the weight E, we must add to the weight E one half that of the prism BD. Thus, for example, if the latter weighs two pounds and the weight E is ten pounds, we must treat the weight E as if it were eleven pounds.

  SIMP. Why not twelve?

  SALV. The weight E, my dear Simplicio, hanging at the extreme end C acts upon the lever BC with its full moment of ten pounds. If suspended at the same point, the solid BD would also exert its full moment of two pounds. But as you know, this solid is uniformly distributed throughout its entire length, BC, so that the parts which lie near the end B are less effective than those more remote. Accordingly, if we strike a balance between the two, the weight of the entire prism may be considered as concentrated at its center of gravity, which lies at the midpoint of the lever BC. But a weight hung at the extremity C exerts a moment twice as great as it would if suspended from the middle. Therefore, [158] if we consider the moments of both as located at the end C, we must add to the weight E one-half that of the prism.

  SIMP. I understand perfectly. Moreover, if I am not mistaken, the force of the two weights BD and E, thus disposed, would exert the same moment as would the entire weight BD together with twice the weight E suspended at the middle of the lever BC.

  SALV. Precisely so, and a fact worth remembering. Now we can readily understand Proposition 2: How and in what proportion a rod, or rather a prism, whose width is greater than its thickness offers more resistance to fracture when the force is applied in the direction of its width than in the direction of its thickness.

  For the sake of clarity, take a ruler ad whose width is ac and whose thickness, cb, is much less than its width. The question now is why will the ruler, if stood on edge, as in the first figure, withstand a great weight T, while, when laid flat, as in the second figure, it will not support the weight X, which is less than T. The answer is evident when we remember that in the one case the fulcrum is at the line bc, and in the other case at ca, while the distance at which the force is applied is the same in both cases, namely, the length bd. But in the first case the distance of the resistance from the fulcrum—half the line ca—is greater than in the other case where it is only half of bc. Therefore, the weight T is greater than X in the same ratio as half the width ca is greater than half the thickness bc, since the former acts as a lever arm for ca, and the latter for cb, against the same resistance, namely, the strength of all the fibers in the cross section ab. We conclude, therefore, that any given ruler, or prism, whose width exceeds its thickness, will offer greater resistance to fracture when standing on edge than when lying flat, and this in the ratio of the width to the thickness.

  Proposition 3: Consider now the case of a prism or cylinder lying horizontal and growing longer in a horizontal direction.We must find out in what ratio the moment of its own weight increases in comparison [159] with its resistance to fracture.This moment I find increases in proportion to the square of the length.

  In order to prove this, let AD be a prism or cylinder lying horizontal with its end A firmly fixed in a wall. Let the length of the prism be increased by the addition of the portion BE. It is clear that merely changing the length of the lever from AB to AC will, if we disregard its weight, increase the moment of the force tending to produce fracture at A in the ratio of CA to BA. But, besides this, the weight of the solid portion BE, added to the weight of the solid AB, increases the moment of the total weight in the ratio of the weight of the prism AE to that of the prism AB, which is the same as the ratio of the length AC to AB. It follows, therefore, that, when the length and weight are simultaneously increased in any given proportion, the moment, which is the product of these two, is increased in a ratio that is the square of the preceding proportion. The conclusion is then that the bending moments due to the weight of prisms and cylinders that have the same thickness but different lengths bear to each other a ratio that is the square of the ratio of their lengths, or, what is the same thing, the ratio of the squares of their lengths.

  We shall next show in what ratio the resistance to fracture in prisms and cylinders increases with increasing thickness while [160] the length remains unchanged. Here I say that (Proposition 4): In prisms and cylinders of equal length but of unequal thicknesses, the resistance to fracture increases in the same ratio as the cube of the diameter of the thickness, i.e., of the base.

  Let A and B be two cylinders of equal lengths DG, FH; let their bases be unequal, namely, the circles with the diameters CD, EF. Then I say that the resistance to fracture offered by the cylinder B is to that offered by A as the cube of the diameter EF is to the cube of the diameter CD. For if we consider the resistance to fracture by longitudinal pull as dependent upon the bases, i.e., upon the circles EF and CD, no one can doubt that the resistance of the cylinder B is greater than that of A in the same proportion in which the area of the circle EF exceeds that of CD; this is so because it is precisely in this ratio that the number of fibers binding the parts of the solid together in the one cylinder exceeds that in the other cylinder. But in the case of a force acting transversely, it must be remembered that we are employing two levers in which the forces are applied at distances DG and FH, and the fulcrums are located at the points D and F; and the resistances act at distances that are equal to the radii of the circles CD and EF, since the fibers distributed over these entire cross sections act as if concentrated at the centers. Remembering this and remembering also that the arms, DG and FH, through which the forces G and H act are equal, we can understand that the resistance located at the center of the base EF and acting against the force H is greater than the resistance at the center of the base CD opposing the force G in the ratio of the radius EF to the radius CD. Accordingly, the resistance to fracture offered by the cylinder B is greater than that of the cylinder A in a ratio which is compounded of that of the area of the circles EF and CD and that of their radii, or of their diameters. But the areas of circles are as the squares of their diameters. Therefore, the ratio
of the resistances, being the product of the two preceding ratios, is the same as that of the cubes [161] of the diameters. This is what I set out to prove.

  Moreover, since the volume of a cube varies as the third power of its edge, we may say that the resistance of a cylinder whose length remains constant varies as the third power of its diameter. So from the preceding we are also able to derive a Corollary: The resistance of a prism or cylinder of constant length varies as the three-halves power of its volume or weight. This is evident as follows. The volume of a prism or cylinder of constant altitude varies directly as the area of its base, i.e., as the square of a side or diameter of this base. But as just demonstrated, the resistance varies as the cube of this same side or diameter. Hence, the resistance varies as the three-halves power of the volume—and consequently also of the weight—of the solid itself.