Proposition 128

PROPOSITION 129. PROBLEM

1087. If a body moves in a medium that resists as the square ratio of the speed, and if the force P is to the force Q as MP to MQ (Fig.94) or, which is the same, if the body is drawn to the centre C by some force, to determine the curve AMB described by the body.

SOLUTION

As before on placing CP = x, PM = y, Mm = ds and y = px, let the speed at M correspond to the height v and the exponent of the resistance is q; that is, R = qv . And since there is the ratio P : Q = y : x , then this gives (1080).

Which equation divided by vxdp becomes this equation : …

the integral of which is …

Truly this equation … and integrated gives (1079) multiplied by …

in which the value of v found substituted gives : …

This equation is differentiated with dx placed constant, and it becomes : Which is the equation for the curve sought. Q.E.I.

Corollary 1

1088. This equation for the curve found does not differ from the equation found in a … vacuum(1081), except that this has e ∫ q Qdx , when there − Adx .. only.

Corollary 2

1089. If the element dp is assumed for the constant, then this equation is produced : [p. 469] In which if we put x = 1z , there arises :

Corollary 3

1090. If the centripetal force attracting towards C is equal to : Whereby we have this equation for the curve given …

Which with the logarithms taken and differentiated , gives

Truly this is :

Corollary 4

1091. With everything the same in place : giving

Put z = e ∫ udp and this equation is produced :

Scholium

1092. I doubt that this second order differential equation can in any case be reduced to a differential equation of the first order [p. 470]; or yet that which we put in place above, where we considered the customary centripetal forces (1020). Therefore in a medium with resistance the method of working does not yet seem to be useful, as much as it brought in a vacuum, even for the cases in which n is either 1 or –2. On this account, since in this matter hardly anything more can be expected, I leave the motion made in a plane with a resisting medium, and I proceed to consider non–coplanar motion, connecting the body with the absolute forces and the force of resistance acting on it. Where in this business it is alright for a little easy understanding to lead to apparent knowledge, I am content to expound the rules, from which we are able to arrive at an equation for any proposed problem.

PROPOSITION 130. PROBLEM.

1093. In a medium with some resistance acting a body is acted on by three forces, of which one is along the tangent, and the remaining two are normal to the direction of the body, and in two planes with the normals between each normal in turn; to determine the motion of the body and the curve which it describes.

SOLUTION

From the element Mm (Fig. 95) which the body describes, from the ends M and m the perpendiculars MQ and mq are sent, and from the points Q and q the perpendiculars QP and qp are sent to the fixed axis AP in the fixed plane APQ [p. 471]. Then put AP = x, PQ = y and QM = z, with the height corresponding to the speed at M equal to v. Now let the tangential force be equal to T. The first of the normals , the direction of which lies in the plane Mq, is equal to N and the other, the direction of which is normal to the plane Mq, is equal to M. The force of the resistance is truly equal to V. Moreover since the force of the resistance V does not affect the normal forces, but only has a little effect on the tangential force, the effect of the normal forces N and M remains unchanged, but in effect the tangential force has to be defined by putting T – V in place of T. Whereby, when we determine the effect of these forces now above (809), the same equations prevail given there and here, if in this way T – V is put in place of T. On this account these equations are produced for the resisting medium : and (809).

From which with v eliminated there are two equations involving the three coordinates x, y, z, which express the nature of the curve sought. Moreover it is assumed that the element dx is assumed to be constant in these equations. Q.E.I. [p. 472]

Corollary 1

1094. The two latter equations can be solved together in order that v is eliminated and give this equation : Which are equally valid for some medium and for a vacuum (810).

Corollary 2

1095. From this equation, if N or M vanishes, the motion of the body is of such a kind. For on putting N = 0 it becomes : Truly dz is the tangent of the angle, by which the element Mm is inclined to Qq. ( dx 2 + dy 2 )

Whereby this angle is constant; since QM has a given ratio to the projection BQ of the described curve in the plane APQ.

Corollary 3

1096. If M = 0, then ddy = 0 and thus the projection BQ is a straight line. Therefore the total curve described by the body is put in a plane normal to the plane APQ and cutting the line BQ.

Corollary 4

1097. From the equation (1094) there arises

Whereby, since it becomes then

Corollary 5

1098. Whereby, if then the body also moves in a plane, since then ddz = 0 and dz = αdx. For the projection of the described curve is a straight line in the plane normal to the plane [p. 473] APQ with the normal AP.

Corollary 6

1099. Moreover the plane, in which the two elements Mm and mμ have been placed (Fig. 96) which the body describes, is determined in a like manner to that in the vacuum, since the determination of this plane only depends on the coordinates x, y and z. Truly if this plane is SMR, and it cuts the plane APQ in the line OR, then And the tangent of the angle that the plane RMS constitutes with the plane MQ APQ, or QV , is equal to (812).

Corollary 7

1100. Therefore the tangent of the angle, which the plane RMS makes with the plane ddz . APQ , is equal to the secant of the angle POR taken by ddy

Corollary 8

1101. Therefore in the case, in which the force N vanishes, since it is given by : dx or POR = RQS. Therefore it then follows that QV falls the tangent of the angle POR = dy on QS.

Truly the tangent of the angle, that RMS makes with RQS , is equal to Whereby this angle is constant, on account of (1095). Truly it is found that

Corollary 9

1102. As in corollary 1, the ratio is given between ddy and ddz for the normal forces M and N, and if the proportionals are substituted in their place then the position of the plane RMS is determined by a first order differential equation. But all these apply equally well for a vacuum and for a resisting medium. Whereby also this result agrees with these that were presented above in proposition 98 in a straightforward manner.