PROPOSITION 94

PROPOSITION 94. Problem

851. To determine the effect of the normal deflecting force N on a body moving on any surface.

Solution

On placing as before AP = x, PQ = y and QM = z (Fig.93) the nature of the surface is expressed by this equation : dz = Pdx + Qdy and the body is moving with a speed corresponding to the height v through the element Mm; in traversing which, unless the deflecting force is present, it proceeds along the element mμ following the shortest line and it gives : and (835).

The force of the normal deflection N is added, which has the direction against increase. Therefore this force has the effect, that the body in describing the element Mm does not advance to mμ , but is deflected forwards from this direction. Therefore we place it to act along mν ; Mm and mν are two elements of the curve described by the body. Whereby with the perpendicular να sent from ν to the plane APQ then :

There is hence obtained : and

Now on placing for brevity: the radius of osculation corresponding to the angle between the elements μmν (72) is equal to : Therefore if we call the radius here equal to r, then N = 2rv or 2v = Nr , since here the angle is generated in the same way in which a body in a plane is deflected from a straight line by a normal force. Now it follows that : [p. 474] And in place of ddη and ddζ with the due values substituted, the radius becomes : But since through differentiation of the equation dz = Pdx + Qdy , there is dPdx + dQdy = ddz − Qddy , there is made on substituting this into the equation dz = Pdx + Qdy , on calling Then on this account, Q.E.I.

Scholium 1

852. This formula agrees with that which we have found previously (79) in determining the effect of this kind of force. For the difference is only in the sign of the letter N, as it was apparent that the force everywhere had to be taken as negative. And even here we cannot be certain of the sign, because we have extracted it from the root of a squared quantity, and it can be equally positive or negative. Now this doubt, if the calculus is adapted to this special case, is at once removed, because the formula must be of this kind, so that the point ν falls on this side of μ , if the force N is deflecting the body forwards, in order that the direction should be as we have put in place. From which with the help of an example the sign of the square root can also be determined and hence the formula itself found.

Corollary 1

853. If the deflecting force N vanishes, the body continues moving along its own shortest line ; which is indicated by the equation also. For on putting N = 0 there is obtained which is the equation for the shortest line :

Corollary 2.

854. Therefore whatever the pressing force and the tangential force and the force of the resistance acting on the body moving on the surface, only if nothing aids the deflecting force, then the body always moves along the shortest line [now called geodesic curves].

Scholium 2.

855. Moreover as concerning the position of this deflecting force N, that can be deduced as follows, since that force is placed on the surface in the tangent plane and likewise it is normal to the curve described, therefore let MG (Fig. 94) be the direction of this force and G is the point at which it crosses the plane APQ , thus so that the force N can be thought to pull along the line MG, while that force we have put before to be deflecting and pressing forwards. Therefore in the first place it is necessary to determine the intersection of the plane of the tangent of the surface at M with the plane APQ, which is the right line TVG ; now this can be found, if two tangents of the surface can be produced as far as the plane APQ, and the points in which they are incident on the plane APQ are joined by a right line. Therefore let MT be the tangent of the described line, which in addition is a tangent of the surface also ; then as we have now established,

Again the surface is understood to be cut by the plane PQM and let MV be the tangent of this cut ; then we have QV = Qz from the equation dz = Pdx + Qdy on putting dx = 0 [or, from the subtangent]. Therefore the point V is known, on account of which the line TV produced is the intersection of the tangent plane of the surface at M with the plane APQ. Therefore the point G, in which the line MG crosses the plane APQ, is placed on the line TV. Again there is taken on the line TQ : and MS is the normal described to the element Mm. And if the normal SG is drawn to QS, from this line SG all the right lines drawn to M are perpendicular to the element Mm. [Thus, the element Mm is normal to the plane MSG.] Whereby since MG is also normal to the element described, the point G is also placed on the line SG. Hence the point G is at the intersection of the lines TV and SG. Now it is the case that : [See annoted Fig. 94] and ang. ELG = ang. PQT. Putting GE = t; then tdy LE = dx and PE = ydy + tdy + zdz . dx Finally also on account of the similar triangles, we have FP : FT + PV = PE : GE − PV , that is : Hence there comes about : and thus the point G is determined. Therefore, if the line QG is drawn, then and

.. and

..