Proposition 12

PROPOSITION 12. Problem.

83. A body which is moving on the curve AM (Fig.12) is acted on everywhere by a force MF, the direction of which is parallel to the axis AP; to determine the speed of the body at individual points and the time taken for any part of the curve to be described by the body, with the force due to the curve acting on the body at individual points.

Solution.

The body describes the arc AM and the speed of the body at A corresponds to the height b while the speed at M corresponds to the height v. Now with AP = x, PM = y and with the arc AM = s the force MF which is called p is resolved into the sides, clearly the normal MN and the tangent MT; there is ds : dx = MF : MT and ds : dy = MF : MN .

Therefore there is hence produced the tangential force pdx pdy MT = ds and the normal force MN = ds . It is evident here that dv = − pdx and v = C − pdx (42). Moreover with the integral thus taken pdx ,∫ in order that it vanishes with x = 0, we have v = b − pdx ; [p. 40] from which equation the speed of the body is known at individual points. From the same equation it is found also the time in which the arc AM is completed; for with the time put as t, there results …pdy

The normal force MN = ds is completely devoted to pressing the body to the curve along MN (39), therefore it increases the pressing force arising from the centrifugal force, since MN falls in the region opposite to the radius of osculation MO. Whereby, with the radius of osculation put as MO = r , the centrifugal force is equal to 2rv (20), and the total

…

force pressing on the curve next to MN = ds + 2rv . Q.E.I.

Corollary 1

84. The speed at M is therefore of such a size as it would be at P, if the body rises with the same initial speed b along AP to the particular heights acted on by the same force.

Corollary 2.

85. Therefore the speed does not depend on the nature of the curve, but only on the height that the body traverses. Clearly if the height of the element were dx, then dv = – pdx or dv = pdx, as the body either rises or falls. ∫

Corollary 3.

86. Since we have v = b − pdx , if the abscissa x to be used is taken as far as AC, for which ∫ pdx = b , then the speed of the body corresponding to that height B is equal to zero. Therefore the body rises as far as B and there it is at rest ; now continuing by falling from B along BMA.

Corollary 4.

87. If the ascent along AMB is compared with the rectilinear ascent along APC, the time to pass along the element Mm to the time for Pp is as Mm to Pp, i. e. as ds to dx. Corollary 5.

88. Whereby if the line AMB is straight, as the ratio Mm to Pp is constant, the time to pass through AM to the time to pass along AP is in a constant ratio, surely that which the total sine has to the cosine of the angle A, or which the length AB has to AC.

Corollary 6.

− ds and thus the 89. With the element Pp placed constant, the radius of osculation r = dxddy 3 centrifugal force is equal to Whereby the total force pressing on the curve is equal to

Scholium 1.

90. As in this problem, from the given curve and force acting, the speed at individual points, the time to pass through any arc, and the pressing force on any point of the curve, are found : thus from these five things with any two, the remaining three can be found. From which ten problems arise, which all have a solution from the solution of this problem.

Scholium 2.

91. Similarly there can be ten questions of this kind, if the directions of the forces acting are not parallel, but either converge to a centre of force or have their directions determined in some other way. But if also the direction between these sought is put in place, then from the six quantities taken in the computation, from any three the other three can be found; and hence twenty problems are to be found. [One direction is fixed, and there are directions for the other four quantities, leaving 6 variables.]

Scholium 3.

92. Again indeterminate problems may arise, as if in place of the time through which some part of the curve is traversed can only be given by an integral along AMB ; for then an infinite number of solutions can be put in place. Besides if more ascents or descents are considered to be integrated upon various parts of the same curve, and the ratio of these is given, then the number of questions is increased much more. To this kind belongs the question of finding the curve, upon which all the ascents and descents to the same point are made in the same time, as these are the most difficult we will handle finally. Moreover now we take the first curve and force acting as given and we solve the problems pertaining to this. Next, we will show how from different given quantities, in what way the others are to be found.