seen that the wheel X goes just so much slower than W, Y so much slower than X, and so on. 195. From what is above said, it will be perceived how, by means of toothed wheels, the velocity may be augmented in any Fig.107. given ratio. Let there be, for example, the toothed wheel W, acting upon the pinion w; it is clear, that during one revolution of W, the pinion w will turn as many times as the number of leaves in the pinion is contained in the number of teeth of the wheel; N that is, during one revolution of the wheel, the pinion will turn บ times, N denoting the number of teeth in the wheel, and the number of leaves in the pinion. If therefore the axis of the pinion w carries a wheel, which acts also on a pinion x, we shall see that during one revolution of the wheel X, or of the pinion w, the pinion x will turn N V times, N' denoting the number of teeth in the wheel X, and the number of leaves in the pinion x. Therefore while the wheel X N makes a number of turns expressed by that is, during one ν revolution of the wheel W, the pinion a revolves a number of this manner for a greater number of wheels and pinions, it will be perceived that the number of times that the last pinion turns, during one revolution of the first wheel, is expressed by a fraction having for its numerator the product of the number of teeth in the several wheels, and for a denominator the product of the number of leaves in the several pinions. When it is asked, therefore, what must be the number of teeth and leaves for a proposed number of wheels and pinions, in order that the velocity of the last piece shall be to that of the first in a given ratio, the question is indeterminate, that is, one which admits of several answers. Two examples will suffice to show how we ought to proceed in questions of this kind. We will suppose that it is required to find how many teeth must be given to the two wheels W and X, and how many leaves to the pinions w and x, in order that the pinion x may make 50 revolutions while the wheel W makes one. We shall have NN' We know in this case only the quotient obtained by dividing NN by vv; we do not know either the dividend or the divisor. Let us take, therefore, arbitrarily for the divisor vv a number composed of two factors which shall be neither too small nor too great for the number of leaves to be allowed to the pinions. Suppose, for example, v v = 7 x 8 = 56, v being 7, and 8. We shall then have = 50, or NN' = 50 × 56. Now 50 and 56 not exceeding the number of teeth that can be given to the wheels W and X, I will suppose N to have 50; and consequently those of N' will be 56. If these two factors, or one of them, should happen to be too great, I should decompose them into their prime factors, and see if from the combination of these factors there would not result two smaller factors; or another number might be taken for vv. Suppose, for a second example, that it is proposed to find the number of teeth and leaves to be given to three wheels and their pinions, in order that while the last pinion turns once in twelve hours, the first wheel shall require a year to make one revolution. The common year consisting of 365,25 × 24 × 60 or 525949 minutes, and 12 hours being equal to 12 × 60 or 720 minutes, it is evident that during one revolution of the first wheel the last pinion will make a number of revolutions expressed by 196. If a body p of any figure whatever, touching a plane Fig.108. XZ in any point C, is urged by a single force, it can remain at rest on this plane only when the direction of this force is perpendicular to the plane, and is such at the same time as to pass 38. 137. 110. through the point C. The necessity of the first condition is evident. As to the second, it will be seen, with a moment's attention, that this is not the less necessary; since, if the direction AD of the body p', for example, although perpendicular to the plane, does not pass through the point of contact C', the resistance of the plane, which cannot be exerted except according to the perpendicular at C, would not be directly opposed to the force AD, and consequently would not destroy it, even when it is supposed equal to this force. 197. If the body, instead of touching the plane only in one Fig. 109, point, touches it in several points, it is not indispensable that the single force AD, which acts upon it, should pass through any one of these points; but it is necessary that it should be perpendicular to the plane, and that it should be capable of being decomposed into as many forces perpendicular to the plane, as there arc points which rest upon it, and that they should be such as to pass through these points. Thus if the body p, for example, were Fig.109. in contact with the plane at the points C, C', and the force AD were not in the plane which passes through the two perpendiculars raised at the points C, C', an equilibrium would not take place, because the force AD could not be decomposed into forces passing through C and C, without a third arising which would not be counterbalanced. 193. Hence, if a body which touches a plane in one or in several points, be urged by several forces directed at pleasure, it is necessary, (1.) That these forces should admit of being reduced to a single one perpendicular to the plane; (2.) That this, in the case where it does not pass through one of the points of contact, should be capable of being decomposed into as many forces parallel to it, as there are points of contact, and that these should pass each through one of the points of contact. 199. If the single force which urges a body be gravity, it is necessary that the plane should be horizontal; and if the vertical plane, drawn through the centre of gravity of the body, do not pass through one of the points of contact, it is necessary, at least, that it should not leave all the touching points on the same side. 200. If therefore, the body be urged only by two forces, it is necessary; (1.) That the two forces should be in the same plane; (2.) That this plane should be perpendicular to that on which the body rests; (3.) That the resultant (which must be always perpendicular to this last plane), should not leave all the points of contact on the same side; and if one of these forces be gravity, it is necessary, moreover, that this plane should be vertical and pass through the centre of gravity of the body. 201. Let us now see what ratio must exist between two forces which hold a body in equilibrium upon a plane. Let Fq, Fp, be the directions of these two forces, and AB the intersection of Fig.111. the plane of these forces with that upon which the body rests; having drawn the perpendicular FH upon AB, let us suppose that on this line, as a diagonal, and upon Fq, Fp, as sides, that the parallelogram FEDC is constructed. In order that the resultant of the two forces q and p may be directed according to FD or FH, it is necessary that the two forces q and p should be to each other as FC to FE; and then the two forces p and q, and the pressure which they exert upon the plane, and which I shall represent by g, will be such as to give the proportion q: P ୧ :: FC: FE : FD. 202. According to article 48, we have likewise :P : ୧ :: sin EFD: sin CFD: sin EFC. 203. From the two points A, B, taken arbitrarily in AB, we let fall upon the directions of the two forces q, p, the perpendicu lars AG, BG. The triangle ABG having its sides perpendicular respectively to those of the triangle FDE, the two triangles will be similar; hence AG BG AB :: sin ABG sin BAG sin AGB, : 38. Trig. 32. therefore qpp sin ABG sin BAG: sin AGB; : that is, when two forces only act upon a body to retain it in equilibrium upon a plane; if we imagine two other planes to which the forces are perpendicular, these two forces and the pressure upon the given plane, are represented each by the sine of the angle comprehended between the planes to which the two other forces are perpendicular. 204. Since the ratios which we have established, take place Fig.112. whatever be the nature of the two forces p and q, they will hold true when one of the forces p for example is gravity; in this case the plane BG is horizontal, and the intersection BG is called the base, and AL, perpendicular to BG, the height of the plane. if, therefore, knowing the weight p, the power q, and the angle HF p, which the direction of the weight p makes with the perpendicular to the plane, we would determine the angle which the direction of the power q must make with the same perpendicular, we shall obtain it by the above proportion, which gives But, when an angle is determined by its sine, there is no reason for taking as the value of this angle, the angle itself found in the Trig. 13. tables, rather than its supplement. Accordingly, the same weight may be supported upon the same plane, by the same power, directed in two different ways. These two directions must therefore be such that the two angles HF q, HF q, which they form with the perpendicular FH, may be supplements to each other. Now if we produce the perpendicular HF, toward I, the greater of these two angles HFq is the supplement of q FI; therefore, since it must also be the supplement of the smaller angle HFq, it follows that q FI is equal to the smaller angle HF q. Hence |