Proposition 54

PROPOSITION 54. Problem

475. If the body is always acted on by a uniform force g downwards in a medium with some resistance, to determine the motion of the ascending body on the given curve AM (Fig.58) and the force pressing on the curve sustained at individual points M.

Solution

In the vertical AP place the abscissa AP = x, PM = y and AM = s, and let the height corresponding to the speed of the body at A be equal to b and at M the corresponding height is v, and the resistance at M is equal to R. Therefore it is the case, while the body rises, so the force acting g as well as the resistance R to be contrary to the motion. On this account likewise as in the previous proposition, dv = − gdx − Rds.

From which equation v is thus to be determined, so that on putting x = 0 makes v = b. Then with the resistance not present in the pressing force experienced by the curve, as above the total pressing force [which we would call the normal reaction now], that the curve sustains at M along the direction of the normal MN, …

with dx put constant; where ds the normal force and − both placed along the direction MN. Q.E.I. 2vdxddy the centrifugal force, ds 3

Corollary 1

476. Hence in the ascent of the body on any curve, the speed of the body is continually made less and the point D is reached, at which the speed of the ascent of the body vanishes, if in the equation : dv = − gdx − Rds after integration, putting v = 0.

Corollary 2

477. If the body descending on the curve DMA should have this [hypothetical] equation dv = − gdx + Rds (470), from which it is understood that the ascent is not similar to the descent, as in a vacuum. But if [also] the resistance were to become negative or accelerating, then the ascent would be similar to the descent. Whereby the descent in the medium with resistance agrees just as much in the resisting medium with the ascent, and in turn with the acceleration.

Corollary 3.

478. Since the equation for this ascent yet differs from the equation for the descent, since the value of the resistance R is put negative, it is understood from the same cases, [p. 241] in which the equation for the descent are to be separated or integrable, from which the equation for the ascent too can be treated in the same way.

Corollary 4.

479. If we set R = VK , then this equation is found for the ascent on the curve AM : dv = − gdx − Vds . K But for the descent there is had : dv = − gdx + Vds . K Whereby if the other equation is to be integrated, likewise also the integral of this equation is had only on putting –K in place of K. [On rising, both dx and ds are considered as positive, and thus the height corresponding to the speed diminishes; however, on descending, dx is negative and thus –dx gives a positive contribution to the height and speed; however, energy dissipation means that the resistance term is negative. Thus, Euler’s musings are here more connected with solving an equation than correctly handling the physical situation.] Scholium.

480. Therefore following the three cases mentioned above, in which the equation found can be either separated or integrated, so that we can handle the descent as well as the ascent, clearly if the curve is given upon which the motion can be performed. Moreover then, from the given force acting, we can investigate the curve for resistance and the force acting on it. Thirdly if the motion should have a certain proposed property, we determine the curve which it satisfies according to the hypothesis of resistance it satisfies. Besides other problems follow, in which of these four quantities – resistance, motion, force pressing, and the curve – two are given, and the remaining two are required. Then we also have indeterminate problems, for which all the curves are required, upon which the descending body either acquires the same speed or completes these in the same time. [p. 242] Then the doctrine of brachistochrone lines follows, and finally the chapter ends with a treatment of oscillatory motion.