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Chapter 6c

The Effect Of Forces Acting On A Free Point

by Euler
March 20, 2025 25 minutes  • 5200 words

CONCERNING THE CURVILINEAR MOTION OF A FREE POINT IN A RESISTIVE MEDIUM

PROPOSITION 120. PROBLEM.

  1. A body is always attracted by some force in a medium with some kind of resistance towards the fixed point C (Fig.91); to determine the curve AM that the body describes projected in some manner.

SOLUTION

When the body is at M, the distance of this from the centre MC is equal to y, the element Mm is equal to ds; the speed at M corresponds to the height v. Also from M the normal Mr is drawn to mC, and mr = dy. Again with the tangent MT drawn, let the perpendicular CT = p be sent from C to that line and the radius of osculation at M is equal to r, ydy which is equal to dp . Now let the force which draws the body at M towards C be equal to P and the force of resistance at M is equal to R. Moreover by resolving the force P, the Pp Pdy normal force y and the tangential force − ds are produced, since the motion of the body is retarded. Moreover from the normal force this equation is had (552) :

…

Besides since the tangential force with a small resistance is given by − ds − R , we have (cit.). From these equations solved together, first the speed at particular points on the curve, and then the curve itself AM is known. Q.E.I.

[Recall in (910) we gave a sketch for determining the radius of curvature for rectilinear coordinates; here we repeat briefly the procedure for polar coordinates, illustrating Euler’s amazing grasp of elemental triangles :]

Corollary 1.

  1. On account of the similar triangles Mmr, CMT , we have

…

In which on substitution there is produced :

…

Therefore with v eliminated there is obtained an equation between y et p, which suffices to determine the curve.

Corollary 2.

  1. Since P = pdy , this value is substituted in the other equation. With this done there is produced From which equation, if R is a force of v, then the value of v itself can be found.

Corollary 3.

  1. Let the resistance be proportional to the square of the velocity and the medium 2vdp , which uniform, thus in order that R = vc . Hence we have therefore dv + p = − vds c equation on integrating gives vp 2 = bh 2e− c , where b is the height corresponding to the initial speed at A and h is the perpendicular from C sent to the tangent at A.

Corollary 4.

  1. Therefore according to this hypothesis of the resistance, since v = bh s , the force p 2e c is given by Therefore when P is given in terms of y, this equation is the equation sought for the curve [p.430] AM; moreover the speed at any place M varies inversely as the perpendicular to the tangent and also inversely to the number the logarithm of which is the path length divided by 2c shown.

Corollary 5.

  1. Therefore according to this hypothesis of resistance, the body describes the same path acted on by the centripetal force Vs , as it describes in a vacuum acted on by the ec force V. For in either case the equation sought for the curve is this : Vdy = 2bh 2 dp . p3 Whereby, as the body in this medium with resistance, describes the same curve as the body in a vacuum, the centripetal force must decrease in the ratio the logarithm of which is the distance described to the value of c applied.

Corollary 6.

  1. Let the resistance of the proportional speeds be as the 2mth power of the exponent m and with the medium uniform, thus in order that R = v m . Therefore we have c m 2vdp dv + p = − v mds . Which integrates to give c

Corollary 7. [p.431]

  1. Let the resistance be proportional to the speed, or m = 12 , this gives ∫ Moreover pds expresses twice the area ACM, which we take to be S. And taking away the constant, this gives with b and h having the same values as in coroll. 3. Corollary 8.
  2. Therefore in this hypothesis for the resistance, the speed of the body disappears when the sector of the body or the area completed is equal to h bc . Therefore this area is of such a size that the body is never able to complete this area. And the speed of the body at M varies directly as this absolute area of space now made very small, and inversely as the perpendicular to the tangent. Corollary 9.
  3. According to the same hypothesis of resistance Therefore the centripetal force is equal to Then truly the time, in which the arc AM is completed, is equal to

Hence an infinite time is needed before the body completes the area equal to h bc , or before all the motion is reduced to zero.

Scholium.

  1. Therefore these are the general laws obeyed by a body in a resisting medium acted on by some centripetal force. Moreover I have deduced these further in the cases where the resistance is proportional to the speed or to the square of the speed, while it is clear that this cannot be done in other hypothetical cases of resistance; consequently in the following we take to be our main concern these two resistances that we have been concerned with up to this stage. Moreover now we take the given centripetal forces as proportional to powers of the distances, and we investigate what differences resistance introduces to the curves described. Then just as in the preceding, we put in place the given curve that arises, together with enquiring about either the centripetal force, resistance, or speed in the remainder of the motion.

PROPOSITION 121. PROBLEM.

  1. If the centripetal force varies as some power of the distances from the centre, and the body moves in a medium with constant resistance, which resists in the square ratio of the speed, to determine the curve AM (Fig.91) that the body describes, and the motion of the body on this curve. SOLUTION. [p.433] By keeping as before: CM = y, CT = p, Mm = ds, with the speed at M corresponding to v there becomes : Hence this equation is found for the curve sought : However not a lot can be understood from that equation about the curve put in place, on s account of the complicating factor e c ; whereby with the logarithms taken, there is and on taking dy constant. Since indeed ds = ydy ( y2 − p2 ) , we have

…

Which is the equation between y and p sought for the curve. Q.E.I. Corollary 1.

  1. From which an equation is produced, if the force is proportional to the distance or inversely proportional to the square of the distance, that is readily apparent from the equation found, if 1 or – 2 is substituted in place of n. Moreover all the substitutions of this kind do not help in bringing about the tractability of the general equation. Corollary 2.
  2. If the medium put in place is not uniform, but the exponent of this is the variable ds q, in place of e c there is put e ∫ q (873) [p.434] and this equation is found for the curve described : s Where, if q is made proportional to the distance y, the equation can be reduced to a differential equation of the first degree. Corollary 2. y
  3. Therefore let the exponent of the resistance q = α , and the curve described can be expressed by the following equation : In which, since the number of dimensions of the individual terms vanishes, reduction to a differential equation of the first degree can be put in place. Corollary 3.
  4. Moreover in this manner the differential equation of the first degree is found. Putting there is − sdt 2 . Again we have Whereby the equation becomes ddt = − dzdt z and From which we find :EULER’S MECHANICA VOL. 1. Chapter Six (part c). Translated and annotated by Ian Bruce. page 643 Corollary 5.
  5. If the centripetal force is put to vary inversely as the cube of the distance, then n = −3 . [p.435] Hence the curve described is contained in this equation : Corollary 6.
  6. If the centripetal force varies in proportion to the inverse square of the distance, then n = −2 . And the curve described is expressed by the following equation : In the same manner, if the centripetal force is proportional to the distances or n = 1 to be put in place, there is produced : Corollary 7.
  7. All these equations give curves in the vacuum if α = 0 is put in place. For in this case the exponent of the resistance is made infinitely great, and therefore the resistance infinitely small. Moreover, this equation is found : Scholium.
  8. Therefore when the exponent of the resisting medium, which is put to resist as the square of the ratio of the speeds, is proportional to the distances from the centre, then the equation for the curve described can be reduced to a differential equation of the first degree ; which is hardly possible to be done for the other hypotheses of the exponents of the resistance. Moreover I understand that such values of q, which only depend on the distances from the centre y, are clearly the only ones admitted to be put as a ratio. [p.436] Indeed it is not fitting to give q in terms of p, i. e. through the curve itself, which is still unknown. Yet meanwhile the differentio-differential equation can always be reduced to a first order equation [note that Euler considers differential equations as those written withEULER’S MECHANICA VOL. 1. Chapter Six (part c). Translated and annotated by Ian Bruce. page 644 single derivatives such as dx, dy, etc; while he considers the ratio of differentials such as e. g. dy/dx to be the ratio of a differential by a differential], as long as q is given as a function of a single dimension of y and p taken together. But with these equations, even if they are differentials of the first order, they are neither able to be separated nor integrated, and they serve no useful purpose. On this account, we consider the resistance proportional to the speed when it is joined with a centripetal force to any power of the distances. [Thus, there are now three kinds of variables to be considered : the power law governing the central attraction; the relation of the resistance to the speed; and the form of the function formed from the exponent of the resistance; only a small portion of these cases is soluble.] PROPOSITION 122. PROBLEM.
  9. In a uniform medium, which resists in the simple ratio of the speed, the body moves attracted to the centre C (Fig.91) by a force proportional to some power of the distance ; the determine the curve AM that the body describes. SOLUTION. By putting CM = y, CT = p, Mm = ds, with the speed at M corresponding to the height v and with the exponent of the resistance equal to q, the centripetal force is equal to yn fn and the ∫ area ACM = 12 pds = S . With these put in place we have (1012 and 1014), [p.437] where b is the height at A corresponding to the speed, and h is the perpendicular from C sent to the tangent at A. From which S can be eliminated, and the equation found put in this form : From which by differentiation with dp put constant there arises : orEULER’S MECHANICA VOL. 1. Chapter Six (part c). Translated and annotated by Ian Bruce. page 645 Which equation expresses the kind of the curve AM described. Truly by knowing this, at once the speed of the body can become known from the area of the curves and from the perpendicular p. Q.E.I. Corollary 1.
  10. If in place of dp the element dy is assumed constant, then this equation is produced: Moreover nothing can be concluded from this equation, since it cannot be reduced to a first order differential equation. Corollary 2.
  11. The above reduction can always be put in place (1020), if in the differential equation the indeterminates [p.438] p and y are agreed to have the same number of dimensions. Moreover, this happens if n = 1, i. e. if the centripetal force is proportional to the distance from the centre. Then indeed for the curve sought : with dy put constant. Corollary 3.
  12. Hence therefore for this hypothesis, put : and there is produced : With these substituted it is found that Or by putting tz = u there is produced:EULER’S MECHANICA VOL. 1. Chapter Six (part c). page 646 Translated and annotated by Ian Bruce. Corollary 4. 3
  13. This equation is possible to be integrated, if it is divided by tt( 1 + u ) 2 u ; from which there is produced : The integral of which is or Corollary 5. [p.439]
  14. Truly on the strength of the substitutions made: On account of which we have the equation for the curve sought: Corollary 6.
  15. Moreover when the differentials are made rational, there arises : by restoring p = yt. Therefore the following equation is found, in which the indeterminates y and t are in turn separated from each other, From which the equation for the curve can be constructed.EULER’S MECHANICA VOL. 1. Chapter Six (part c). Translated and annotated by Ian Bruce. page 647 Scholium.
  16. I will not delay any more over this equation, although I suspect that it is possible to be integrated anew. This indeed is certain, if C 2c = − f ; in which case as the integral is made composite, as here I did not wish to make the change. From which it is understood that the integral obtained is very complicated, so thus hardly anything could be deduced about understanding the motion. On which account with these in place I go on to the inverse problems. PROPOSITION 123. [p.440] PROBLEM.
  17. If the curve AM is given (Fig.91) that the body describes, and the resistance is given at the individual points M, to determine the centripetal force always directed towards the centre C, and the speed of the body at individual points. SOLUTION. As before there are put in place CM = y, Mm = ds, CT = p, with the height corresponding to the speed at M equal to v, with the resistance equal to R, and the centripetal force equal to P. With these in place this equation is found : (1007), from which, since the curve AM and the resistance R are given, v is found from the integral p 2v = − Rp 2ds , truly v = − ∫ ∫ Rp 2 ds . Moreover with v found, again it is found that p2 (1005). Q.E.I. Corollary 1.
  18. If the initial speed at A is put equal to b and the perpendicular to the tangent at A sent from C is equal to h, then in the case of zero resistance or in a vacuum, these equations are produced:EULER’S MECHANICA VOL. 1. Chapter Six (part c). Translated and annotated by Ian Bruce. page 648 Corollary 2.
  19. Therefore in a medium with resistance, if the arc AM, this becomes ∫ Rp ds is thus taken, as it vanishes with 2 Corollary 3. [p.441]
  20. If the body in a vacuum is moving along the curve AM with the same initial speed at A, and if the speed that the body has at M is said to be u , and the centripetal force at M is equal to V, then there arises : Whereby we have : Corollary 4.
  21. Therefore since in this problem, in which the curve AM and the initial speed at A is given, the centripetal force is truly sought, now it is as the solution in the preceding chapter, and from the same solution likewise this problem is solved. For with ∫ Rp ds found, at once the difference of the centripetal forces in a vacuum and in the 2 resisting medium becomes known, and thus the centripetal force in the resisting medium itself. Example 1.
  22. If the curve AM is a circle of radius a having its centre at C and the same resistance everywhere R equal to a constant λ, then y = p = a and h = a . Whereby there is found : Therefore the speed is always decreasing and Cleary vanishes with the arc described equal to λb , in which place also the centripetal force goes to zero. Moreover this centripetal force is everywhere as the square of the speed. [p.442]Besides the time, in which the arc AM is traversed, is equal toEULER’S MECHANICA VOL. 1. Chapter Six (part c). page 649 Translated and annotated by Ian Bruce. 2 b and the time in which the body is returned to rest is equal to λ . Example 2.
  23. Let the curve AMC (Fig. 92) be a logarithmic spiral, the centre of which is C, and the resistance is as some power of the distance CM, truly R= yn . Hence we have p = αy and with β = fn (1 − α 2 ) dy ds = − β . Therefore with AC = a, we have h = αa . Therefore the equation [in this prop. above] becomes : and hence And In which case, when n = −3 , for which the integral which depends on logarithms, is And Corollary 5.
  24. If the initial speed impressed on the body at A is such that then also everywhere Therefore in this case the centripetal force P is to the resistance R as 2 to ( n + 3 )β , i. e. in the given ratio.EULER’S MECHANICA VOL. 1. Chapter Six (part c). page 650 Translated and annotated by Ian Bruce. Corollary 6. [p.443]
  25. In the same case we have : and Therefore the resistance is in the n2+n1 − multiple ratio of the speeds with the medium f present being uniform, clearly the exponent of this is ( n +3 )β . Corollary 7.
  26. If n = 1, then the resistance is in the ratio of the speeds and the exponent of the f medium is 4 β . Therefore the body can describe a logarithmic spiral in this medium, if y the centripetal force is in proportion to the distance, truly equal to 2 βf , and if initially it is projected from A with the speed a . Besides in any medium with a uniform 2 βf resistance, a spiral can be described by the body, except in the case in which the resistance is proportional to the square of the speed. Scholium. [p.444]
  27. Centripetal force and resistance of such a kind may be required in order that the body moves in a logarithmic spiral, now to be cited more often by Newton and Bernoulli set out in the Princip. Phil. and in the Act. Lips. 1713. In the following therefore, we offer more examples concerning this.EULER’S MECHANICA VOL. 1. Chapter Six (part c). page 651 Translated and annotated by Ian Bruce. PROPOSITION 124. PROBLEM.
  28. If the resistance is proportional to some power of the speed and the exponent at particular places is given, the centripetal force arising is to be found in order that the body moves on the given curve AM (Fig.91). SOLUTION. With CM = y, CT = p, Mm = ds, remaining as before, with the speed at M equal to m v and with the exponent of the resistance equal to q , let the resistance R = v m and the q centripetal force equal to P. With these put in place (1005) : (1007). On integrating, this equation gives : Truly in the case when m = 1 , it is given by [p.445] 2vdp Moreover from finding v likewise P becomes known from the equation P = pdy . Q.E.I. Corollary 1.
  29. If the speed, by which the body is projected at A, corresponds to the altitude b, and the perpendicular from C sent to the tangent at A is equal to h. And is thus taken, in order that it vanishes by making s = 0 or M is incident on A ; and thus with the integral put equal to S. Therefore with the constant added, it becomes : Now make S = 0, then p = h and v = b and thus C = b1− m h 2( 1− m ) . Hence with the constant C determined, we have :EULER’S MECHANICA VOL. 1. Chapter Six (part c). page 652 Translated and annotated by Ian Bruce. Corollary 2.
  30. In the case m = 1, which requires a special integration, if ∫ dsq is thus taken in order that it vanishes on making s = 0, then it becomes : In a vacuum it produces [p.445] Hence the centripetal force in a vacuum to the centripetal force in this medium − ds with resistance is as 1 to e ∫ q . Corollary 3.
  31. Moreover with m denoting some number other than one, From which is produced In a vacuum the centripetal force equal to 2bh 2 dp is produced; which if it is said to be p 3 dy equal to V, gives the ratio And hence

Corollary 4.

  1. Whereby if the centripetal force is found, which produces the given curve AM in a vacuum, from this with the help of this ratio the centripetal force can be found for any resistance present, if the value of this integral is determined ∫ ( m −1 ) p 2( 1−m ) ds . qmEULER’S MECHANICA VOL. 1. Chapter Six (part c). Translated and annotated by Ian Bruce. page 653 Example 1.

  2. If the given curve is a circle having the centre C, of which the radius MC = a, y = p = a and h = a. [p.447] Besides let the medium be uniform or q = c; then the integral becomes : Therefore we have : or Thus there arises : If the resistance is in the simple ratio to the speed, then m = 12 and then the speed itself is given by 2 bc − s and the time, in which the body completes the arc 2 c AM , is given by : Hence an infinite time is needed , before the body completes the arc equal to 2 bc ; in which when it arrives, all the motion has gone and likewise the centripetal force has vanished. If the resistance is in proportion to the square of the speed, then Moreover the motion of the body in the periphery of the circle clearly agrees with the rectilinear motion, in which the motion of the body has been lost by the impressed resistance. For the centripetal force, which is always normal, clearly does not affect the speed, but yet turns the body in a circle.EULER’S MECHANICA VOL. 1. Chapter Six (part c). Translated and annotated by Ian Bruce. page 654 Example 2. [p.448]

  3. The body descends from A towards the centre C in a logarithmic spiral AM (Fig.

  1. and the exponent of the resistance q is put as any power of the distance MC = y, thus in order that q = y n+1 . From the nature of the logarithmic spiral it follows that p = αy and fn with AC = a then h = αa and by putting β = ( 1 − α 2 ) there becomes Hence there arises Besides let then Therefore the centripetal force varies inversely as the power of the distance, the exponent . of which is 1mn −m Corollary 5.
  1. If the centripetal force is equal to the constant the body is able to move in a logarithmic spiral, of which the angles of intersection of the radii with the cosine on the curve is β, with the exponent of the resistance being equal to y [p.449] and with the initial speed corresponding to the heightEULER’S MECHANICA VOL. 1. Chapter Six (part c). Translated and annotated by Ian Bruce. page 655 Corollary 6.
  2. If the centripetal force is as the distance y raised to the power k, then hence Therefore the exponent of the resistance must become Corollary 7.
  3. Besides the time in which the arc AM is completed is equal to : Therefore the time, in which the body descends as far as the centre C, is finite if k < 1 or k > 1. Corollary 8.
  4. Let the resistance be proportional to the square of the speed, and the exponent of y the resistance be equal to δ ; then [p.450] Hence it is found that Therefore in a medium with this resistance a body can describe some kind of logarithmic spiral, if the centripetal force is equal to y βi . 2by i −2 and the exponent of resistance is equal to a i −2EULER’S MECHANICA VOL. 1. Chapter Six (part c). Translated and annotated by Ian Bruce. page 656

Scholion.

  1. Therefore all the cases are contained in this example and adjoining corollaries, in which the body is able to describe a logarithmic spiral in a medium with some kind of resistance, acted upon by a centripetal force in proportion to some power of the distance. Where the case arises, in which the resistance is proportional to the square of the speed and the exponent varies as the distances from the centre, then in this special case it occurs that immediately the centripetal force can be given proportional to some power of the distance, which in other hypotheses of resistance are finally obtained after the determination of the initial speed in a certain way. Moreover in that hypothesis of the y resistance, with the exponent of the resistance being δ , if the body is projected at A with some speed b along a direction at an angle of inclination to AC, of which the cosine is β, and the centripetal force at A is equal to 2ab , then the body always moves in a logarithmic spiral, [p.451] if in addition the centripetal force varies as y i −3 ; moreover i is given, since i = δβ . Therefore for these, in which the motion observed is a logarithmic spiral, the cases have been explained well enough.

PROPOSITION 125. PROBLEM.

  1. If the curve AM is given (Fig.91) which the curve describes, and the centripetal force acts towards the centre C, to find the requisite resistance at the individual points M and the speed of the body.

SOLUTION.

By putting CM = y, CT = p, Mm = ds, let the centripetal force at M be equal to P. Then the resistance at M is put equal to R and the height corresponding to the speed at M is equal to v. With these in place we have (1005) (1007). On account of the given curve and centripetal force, from that equation, it is Ppdy found that v = 2dp and by differentiating with dy placed as constant, it becomes In which place with the values v and dv substituted, there resultsEULER’S MECHANICA VOL. 1. Chapter Six (part c). Translated and annotated by Ian Bruce. page 657 If the resistance is in proportion to the square of the speed, the exponent of this is taken as q, then R = qv et q = Rv . Whereby it is found that : [p.452] Hence from the given curve or the equation between y and p, we have found the centripetal force, as well as the resistance R and the speed, at individual points. Q.E.I.

Corollary 1.

  1. The resistance can be expressed in another way : From which it is evident, if P varies as dp , that the resistance vanishes. For indeed in p 3 dy this case the centripetal force is sufficient to produce the given curve.

Corollary 2.

  1. With the initial speed at A put equal to b and with the perpendicular sent from C to the tangent at A equal to h let the centripetal force be equal to V, which can be produced in the vacuum in order that the body moves along this curve; then it is given by (591). Hence on this account

Corollary 3.

  1. If the body moves in a vacuum in this curve acted on by the force V , the speed of 2 this at M corresponds to the height u and it becomes u = bh2 . Thus we have this ratio p u : v = V : P. And in general this theorem is obtained : the speeds of the body at the same place M are in the square root ratio of the centripetal forces [in the vacuum and with resistance]. [p.453] Corollary 4.
  2. If the centripetal force is constant or P = g, then we haveEULER’S MECHANICA VOL. 1. Chapter Six (part c). Translated and annotated by Ian Bruce. page 658 Example.
  3. The body descends in an hyperbolic spiral AM, the nature of this is expressed by this equation : and if the centripetal force is raised by some power of the distance from the centre of the yn fn spiral C, evidently P = . There is the equation : Thus there is produced : and from these we find : Therefore if medium resists in the square ratio of the speed, then the exponent of the resistance is equal to ( a2 + y2 ) . But if the medium resists in the simple ratio of the n +3 speed, and the exponent of the resistance is q, then we have Therefore in this hypothesis of the resistance the medium is uniform if n = −1; that is, if the centripetal force is in the inverse ratio of the distances. For indeed it becomes 2 q = af . But if the centripetal force is inversely as the square of the distances, then the exponent of the resistance becomes equal to [p.454] 2a 2 y , or it is in proportion to the f2 distances themselves from the centre.

Scholium

  1. Just as here the problem ought to follow our custom, in which the centripetal force as well as the resistance are sought from the given curve and the speeds at individual places; but since the solution of this is much easier and from the rules given above (1007), it immediately becomes soluble and besides from that, nothing noteworthy can be deduced, and so I pass over this; moreover it is found that which formulae solve the problem. Truly I add in place of this problem another related problem, in which besides the curve the angular motion around the centre of force is given and so the centripetal force as well as the resistance is sought.
The Effect Of Forces Acting On A Free Point
The Rectilinear Motion Of A Free Point Acted On By Absolute Forces

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