Proposition 23

PROPOSITION 23. Problem.

214. If a body is acted on by any two forces, of which the direction of one is along the vertical MQ (Fig. 29), and the other MP is horizontal ,to define the motion of the body from these forces acting on a given curve AMB.

Solution

Let the speed at B be zero, and at M it corresponds to the height v. The force acting along MQ is equal to P and that along MP is equal to Q. Put BR = t, RM = z, the arc BM = w, which letters we use for the descent of the body from rest at B. But for the ascent from A with any initial speed, which motion is referring to oscillations, let AP = x = QM, PM = AQ =y and the arc AM = s; now the speed of the body at A corresponds to the height b; hence t + x = const., likewise z + y = const. and w + s = const., thus dt + dx = 0 [, dz + dy = 0] and dw + ds = 0. With the forces P and Q resolved into normal and the and tangential components, the tangential force that arises from P is equal to Pdt dw normal force that arises from P is equal to Pdz pulling along MN. Then the tangential .. , which is force arising from Q is equal to dw and the normal force from Q is equal to Pdt dw .. contrary to that normal force. And beyond that the tangential force along BM accelerates the motion and thus dv = Pdt + Qdz and v = Pdt + Qdz … with these integrations thus made, in order that they vanish with t and z = 0.

And for the ascent from A there is : v = b − Pdx − Qdy , with these integrations thus made, in order that they vanish with x .. and y put equal to 0. Therefore with t = BD and z = AD put in this equation : v = Pdt + Qdz , we find that v = b. Whereby the time to traverse BM is equal to : ..and the time for AM is equal to .. dw and thus …

With the element dt or dx taken as constant, the radius of osculation at M = dtddz the centrifugal force, the direction of which is along MN , is equal to : Hence the total force, by which the curve is pressed upon at M along MN, is equal to : Q.E.I.

Corollary 1

215. But if P is some function of x or t and Q some function of y or z, so that Pdx as well as Qdy can be integrate; and thus the speed v can be exhibited and with the help of the equation for the curve the time too.

Corollary 2

216. Because whatever and however many forces are acting, but if the directions of these are in that plane as the curve AMB, these forces can be resolved into two forces of this kind, and this proposition extends widely and embraces all the cases in which the directions of the forces and the curve are in the same plane. Scholium.

217. Also it is apparent that this proposition is of wider applicability if a few cases are added on and examined, in which not all the directions of the forces are in the plane of the curve.

For then these forces are to be resolved into two components, of which the one are in the plane of the curve itself, and the other normal to this plane. Therefore these situated in the plane of the curve, so that in the proposition we have used, the analysis gives the acceleration of the body and the compression force along MN ; the other forces, because they are normal to the curve, are only devoted to pressing on the curve. Whereby hence a twofold compression arises, which the curve sustains, the one directed along MN , and the other normal to the plane of the curve. Therefore of these two compressions, if the direction of the mean is taken, there is produced, and there is produced the direction of the equivalent force by which the curve is pressed.

On this account there is no need for us to explain cases of this kind, but we will mention briefly a few in which the motion of bodies are on a curve which is not itself placed in the plane in which the forces act, which we take as constant and in the downwards direction.

PROPOSITION 24. Problem

218. With a uniform force acting in a downwards direction, to determine the motion of a body on some curve AM (Fig. 30)not set up in the same plane. Solution.

Let the projection of the curve AM be the curve AQ in the horizontal plane, and with the perpendiculars MQ and mq sent from some nearby points M and m to this plane, and there are drawn to the axis AP taken as you please, the normals QP and qp and put AP = x, PQ = y and QM = z. Let the speed of the body at A correspond to the height b, and the speed at M correspond to the height v. Now the force is equal to g, by which the body at M is acted on along MQ. With the tangent MT drawn, and in that from Q to the perpendicular QT the force g in is resolved into tangential and normal [components]. Since the tangential force is equal to :

Since the normal force is equal to : Moreover since the tangential force slows the motion, then we have dv = − gdz and v = b − gz , hence the time in which the arc AM is completed, is made equal to :

The normal force brings about a compression of the curve by the body at M with so much force along a direction normal to Mm and situated in the plane QMmq. Now the curve is acted on in addition by the centrifugal force along the opposite direction of the position of the radius of osculation by a force equal to 2( b − gz ) , with r designating the r radius of osculation at M. Moreover we found above (71) the position of the radius of osculation, from which the direction of the centrifugal force can hence become known.

Now the magnitude of the centrifugal force is given from the radius of osculation, which has been found (72) ; clearly it is given by : Q.E.I.

Corollary 1

219. Therefore the speed of the body in this case also depends only on the height. And the speed at M is of such as magnitude as the body has ascending through QM , when it has a speed corresponding to the height b at Q.

Corollary 2

220. Hence the body is unable to ascend to a greater height than to bg . For if we set b − gz = 0 , the body has lost all its speed at that height and begins to descend again.

Corollary 3

221. Also it is understood, if the force cannot be taken as constant, but is the variable P, then the speed at M can be found corresponding to the height b − Pdz .

Scholium 1

222. If the curve AM is considered to be in the vertical plane (Fig. 31) and with the curve related to the horizontal axis AQ, and AQ is equal to the curve AQ in the previous figure, and QM = QM in the preced. fig., then the curve AM is also equal to the preceding curve AM. If now the body on the curve AM ascends with an initial speed at A corresponding to the height b and acted on by the same force g, then it also has the speed at M corresponding to the height b − gz .

Thus the time of ascent along AM also agrees with the times of ascent along AM in the preced. fig. Therefore by this reason the motion of the body not in the same plane can be reduced to motion on a curve placed in the same plane.

For it is not possible to distinguish between the motions ; but the forces acting on these two curves are different. On account of which this compression can be varies as it pleases, with the motion on the curve remaining the same.

Scholium 2

223. Up to the present we have put the curve in place upon which the body is moving, and the force acting in one given direction, and from these we have deduced the motion of the body and the compression of the curve.

Therefore now, since these should suffice, we progress to other questions, in which other quantities are taken as given, and the remaining quantities are to be found. First indeed the compression is given at individual points on the curve and the force acting; from which the curve itself and the motion on the curve must be found. Then from other combinations made from these things, which are come upon in the computation, we will form other questions.