Proposition 98

PROPOSITION 98. THEOREM

802. There are three principal forces which can be put in place, and into which other forces must be resolved, in order that a body can move on a curve that does not exist in a plane; these individual forces are normal to each other; and of these one is the tangential force, and the remaining two are normal to that force, of which one lies in the given plane, and the direction of the other is normal to this plane, and nothing remains of the original forces to change the actions of these forces.

DEMONSTRATION

With the plane APQ assumed fixed [in space] (Fig. 75) and relative to that axis AP, an element Mm is described by a body. From the points M and m the perpendiculars MQ and mq are sent to the fixed plane and from the points Q and q, perpendiculars QP and qp are sent to the axis AP. Now, if the body is not acted on by any force, then it progresses along the line Mm produced with the speed that is has in Mm ; therefore in an equal small interval of time, equal to the time in which it traversed Mm, it arrives at n, with the element mn described equal to and in the same direction as that put for the element Mm. Whereby by also sending a perpendicular nr from n to the plane APQ, then the elements Qq and qr are also equal to each other and placed in the same direction ; because of this, the perpendicular rπ sent from r to the axis AP cuts off the element pπ = Pp. Let the speed in which the first element Mm is described, correspond to the height v, and in the first place the tangential force is considered, which has a direction along mn and the whole force is taken up with changing the speed. This tangential force T is put in place with the existing force of gravity equal to 1, and we have [p. 340] dv = T .Mm ,

and the element mn is completed with a speed corresponding to the height v + dv. Following this, in the plane Mr there is considered a force having a direction ms normal to the direction Mm of the body. This therefore has the effect that the body can deflect from mn and progress along the element mv placed in the same plane as Mr. Let this normal force be equal to N, and since the radius of osculation of the elements Mm and mν , with the perpendicular νε sent along fromν to mn is equal to mνεν [In triangle mεν , we can set ds = Rdθ , where ds = mν and dθ = εν / mν , giving the required result for the radius R], hence

…

2 v .νε = N mν 2

(561).

Truly νε is the sine of the angle nmν . On account of which, mν 2 vsin .nmν = N .mv = N .Mm , and thus sin .nmv = N .Mm .

[Note that Fig. 75 is drawn in perspective, so that the elements Mm = mn = mν , and the force normal to mn acts in this manner as previously shown for centripetal forces in a plane.]

The third force is normal to each of those set up on mn and ms, thus in order that its direction is along the normal mt to the plane Mr. This force neither impedes the actions of the preceding forces, nor is it allowed itself to be impeded by their actions. Therefore the whole effect of this force is to draw the body away from the plane Mr; the body is drawn from that plane from ν to μ , thus so that the plane νmμ is normal to the plane Mr, and the angleνmμ is the result of this force. Therefore from the same argument that we put in place for the preceding force, by evaluating the force in the same way, if this force is M, then it is given by

…

Therefore these three forces likewise have the effect that the body, after it has described the element Mm moves to the element mμ , with an increase in speed clearly corresponding to the height v + T .Mm. Moreover, any other forces acting on the body can also be resolved in this way into the forces, the directions of which lie along the directions mn, ms, mt. As we have determined the effect of these forces on the body, so [p. 341] likewise the effect of any forces can also become known. Q.E.D.

Corollary 1

803. By taking νμ in the plane nr π , and by sending a perpendicular μρ from μ to the plane APQ, μρ is parallel to rn itself. Therefore the three coordinates for the points M, m and μ are AP, PQ, QM; Ap, pq, qm, and Aπ , πρ , ρμ .

Corollary 2

804. Whereby if from μ the perpendicular μη is sent to mv, it is normal to the plane Mr; and likewise in a similar manner ρθ , which is perpendicular to qr, is normal to the same plane. On account of which, as ρ and μ on the line ρμ put parallel to this plane, then we have ρθ = μη and θη = ρμ .

Corollary 3

805. If the normal qT is drawn to Qq in the fixed plane APQ, this line qT is normal to the plane Mr. Therefore since mt is also normal to the same plane, then mt is parallel to qT; and between these the distance is the height mq.

Corollary 4

806. The three coordinates are called AP = x , PQ = y and QM = z . And we have :

Pp = pπ = dx , pq = y + dy , qm = z + dz and

πρ = y + 2dy + ddy and ρμ = z + 2dz + ddz = θη. But [from a binomial expansion on extracting the equivalent of qr] and hence:

Again we have πr = y + 2dy and rn = z + 2dz .

Then ..

and ..

[Thus, the length mμ is compounded from the lengths pπ or dx in the x-direction, πρ − pq = y + 2dy + ddy - y-dy = dy + ddy in the y-direction, and ρμ − qm = z + 2dz + ddz − z − dz = dz + ddz , in the z-direction; from which the length corresponding to mμ is extracted by a binomial expansion.]

Corollary 5

807. Since mq , θη and rν are parallel to each other, in the same plane and terminated by the lines qr and mν , then [The gradient of the line mη is the same as the gradient of the line mν in the plane mνrq , as are qθ and qr] For it is the case that : and Whereby and hence Thus, it is found that :EULER’S MECHANICA VOL. 1. Chapter Five (part e). Translated and annotated by Ian Bruce. page 492 [We apply the sine rule to the triangle nmν : it is easily found that the sine of the angle mnν in this triangle can be found from the right-angled triangle with hypotenuse Mn in the plane Mr with base MM’ parallel to Qr, to be given by sin .mnν = dx 2 + dy 2 dx 2 + dy 2 + dz 2 , while the length mν is equal to mn = dx 2 + dy 2 + dz 2 , where sums of powers of higher orders are ignored; from these on applying the sine rule, the result quoted emerges. The diagram here shows the coordinates of some of the points, and may be of some assistance to you if you want to establish the result for yourself.]

Corollary 6

808. Since rρ = −ddy and Qq : Pp = rρ : ρθ , it follows that On this account we have : [In triangle ημm , which projects normal to the plane Mr, we have from above, On using the sine rule, and noting that both the large angles are essentially right, we have, sin .νmμ / νμ = 1 / mμ , sin .νmμ = νμ / mμ , which gives the result quoted on neglecting the higher order terms in the denominator.]

Corollary 7

809. Therefore from the three given forces T, N and M that the body is acted upon, there arises the three following equations : and from which the speed of the body at individual points as well as the curve itself becomes known. [p. 343]

Corollary 8

810. The two latter equations joined together with v eliminated give this equation : For which, the nature of the surface in which the curve described by the body lies, can be assumed to be expressed by the equation,.

Scholium

811. Therefore from this proposition we can deduce the first rules, from which the motion of a body acted on can be deduced, in order that it does not move in the same plane. Indeed we have shown that all the forces can be resolved in terms of three, the effects of which we have determined ; and thus, whatever forces are proposed to be acting, whatever motion they produce on the body can become known. Indeed it is apparent, if the [second normal] force M is not present, then the motion of the body is entirely in its own plane, which does not concern us here. But if the tangential force T vanishes with the forces M and N left, then the body describes a non planar orbit, but yet is still carried around uniformly. From which the position of the orbit is generally known, and it is necessary to find the intersection of this inclined plane, in which the elements Mm and mμ are present, with the plane APQ. [The task of the next proposition.]