Proposition 28

PROPOSITION 28. Problem

253. A body is acted on by some force acting downwards ; to find the curve AM (Fig. 33), on which the body descends in a uniform motion downwards as it is carried forwards or equally as it recedes to the horizontal AB.

Solution

On placing AP = x, PM = y, AM = s and the force acting equal to P , let the initial speed of the body at A correspond to the height b; the speed at M corresponds to the height b + Pdx. Whereby the short interval of … time, in which the element Mm is traversed, is equal ds to .

Moreover since the motion along AM ( b + ∫ Pdx ) must correspond to a uniform motion along AP, it is understood that the motion of the body along AP is with the constant speed corresponding to the height b; the time to pass through Pp must be equal to the time to pass along Mm, hence we have the equation : dx = …

Moreover I put the initial speed agreeing with the speed of descent, in order that the curve A is a tangent to the vertical AP and the body in the beginning at first falls in a straight line. For since the motion on account of the acceleration by necessity is one of acceleration, in order that the curve continually becomes more inclined to the horizontal, the initial motion of this is most conveniently taken at A, where the curve is vertical.

Therefore the equation dy b = dx ∫ Pdx. is produced for this curve. Q.E.I.

[We can understand this second equation in terms of the squares of the speeds, which add according to the right angle rule : The speed downwards is u always, the speed horizontally is v at the point M, and the speed along ds is V, then V 2 = u 2 + v 2 . Now ..u 2 is proportional to b, v 2 is proportional to Pdx with the same constant of proportionality, from which the equality of the times gives the above equation.]

Corollary 1

254. Hence this curve has the property so that when the speed of the body is greater, in that place the more also is the inclination of the curve to the horizontal.

Corollary 2

255. Therefore in the highest place, where the speed of the body is the least, the inclination of the curve must be a minimum or the tangent to the curve at that place must be vertical.

Corollary 3

256. Therefore the initial speed b cannot be equal to zero, since the speed is equal to that respective speed, by which the body progresses downwards as it recedes to the horizontal AB.

Scholium 1

257. This curve is called the line of uniform descent, since a body descending on it is progressing with a constant motion downwards. The discovery of this line is set out in the Act. Erud. Lips. A 1689 for the hypothesis of gravity, or of a constant force acting. [G. W. Leibniz, Concerning isochronous lines, in which a weight falls without gravity, Acta erud. 1689, p. 234]

Moreover, there it is demonstrated that Neil’s cubic parabola satisfies this question, the same as that produced in the following example.

Example 1

258. Let a uniform force be acting or P = g; then ∫ Pdx = gx. [p. 114] Whereby this equation is obtained for the curve sought : dy b = dx gx , which on integration gives 9by 2 …

this equation : 3 y b = 2 x gx or 4 g = x3 , which is the equation for Neil’s parabola with the cusp at A tangent to the line AP , of which the parameter is 49bg . Therefore for which with another initial speed, another parabola is to be taken.

Example 2

259. Let the force acting P be proportional to some power of the abscissa increased by a given line, in order that P = … ; then there arises …

On account of which for the curve satisfying this equation, we have

If a = 0, thus so that the force P acting is proportional to a power n of the distances of the body from the horizontal AB, then : of which the integral is : or

But n + 1 must be a positive number ; otherwise ∫ Pdx is made infinite, since it must vanish on making x = 0. Hence n + 3 > 2; whereby the parabolas satisfy the tangents with the verticals at A tangent to AP. For, if n = 1 or P = xf , the parabola satisfies the parabola of Apollonius, of which the parameter is 2 2bf .

Scholium 2

260. From the solution of this proposition it is evident, how the inverse problem of this can be solved, in which the curve is given, [p. 115] which is a line of equal descent, and it is required to find the force acting. Since indeed it is given by : dy b = dx it becomes … then ..

From which there arises with dx put constant : …

Therefore it is evident that the force P depends on the initial speed b . Now the given curve thus has to be compared, so that it is a tangent with the vertical AP at A. If the radius of osculation at M is called r, then …

Whereby if for example the curve AM is a circle tangent to AP at A, the radius of this is equal to a, then

Therefore the circle, Now the speed at M corresponds to the height

Corollary 4

261. Likewise it is apparent from the equation, that we have found (253), the time, in which the arc AM is described, is equal to the time in which the body must move with the constant speed corresponding to the height b, traverses the vertical distance AP. In this equation clearly the nature of the line of equal descent arises.