Table of Contents

PROPOSITION 55. Problem.

In a medium with some kind of uniform resistance and under the hypothesis of uniform gravity g acting, to determine the motion of the body descending on the given straight line AMB (Fig.59) inclined in some manner to the horizontal.

Solution

With AP = x then AM = s = nx; PM = y and since the resistance of the medium is uniform, then let the resistance be R = VK . Therefore with the height corresponding to the speed at M equal to v, then dv = − gdx − nVdx. K

Hence this equation becomes : Kdv = dx , gK − nV in which equation the indeterminates are separable from each other in turn; hence the equation is : x = gKKdv , − nV ∫

in which the integration is effected so that putting x = 0 makes v = 0, if indeed with the descent starts from A at rest. But if now it has an initial speed, this has to be introduced through the integration. The time to traverse the interval AM is equal to ndx . ∫ v

Therefore with the value in terms of v put in place of dx , the time to traverse AM is equal to nKdv , ∫ ( gK −nV ) v which integral is thus to be taken, so that it vanishes on putting v equal to the initial speed at A . Now the pressing force, that the line sustains at any point M is constant, surely equal to the normal force : g ( n 2 −1) gdy

, ds n since the centrifugal force vanishes on account of ddy = 0. Q.E.I. [p. 243]

Corollary 1.

Hence the speed of the body for some time is accelerating, as during that time gK > nV. But if once gK = nV, then the body is neither accelerated nor retarded. Now the speed of the body is diminished if at the start at A it should be the case that nV > gK.

Corollary 2

Therefore if the body starts to descend from A at rest, then the motion is always increasing thus yet as gK > nV always, clearly the final speed is that acquired at last from an infinite descent distance

Corollary 3

Since the greater the angle BAC, the less is the terminal speed that the body is able to acquire. Now the maximum terminal speed by which the body can progress uniformly is acquired by descending along a vertical straight line AC.

Corollary 4

If the resistance were as the 2m power of the index of the speed, then V = v m and K = k m , hence this equation is obtained : [p. 244] x= and the time to pass along AM = ∫ k m dv gk m − nv m ∫ ( gk Nk−nvdv) v . m m m Example 1.
The medium resists in the simple ratio of the speeds ; hence 2m = 1 and dv = gdx − ndx v . k Hence this gives : or by the series : if indeed the descent starts from A at rest. Moreover the time to complete the interval AM is equal to : Whereby if the time for AM is put equal to t, then this gives : and in the series expansion: Hence if the body in the descent along AB acquires the speed corresponding to the height b, from this there is found the height

Example 2. 486. The medium resists in the square ratio of the speeds; hence m = 1 and ∫ gk x = gkkdv = kl , − nv n gk − nv if indeed the body starts its descent from A at rest. Whereby if e is the number of which the logarithm is one, then [p. 245] On this account, if the body has a speed at B corresponding to the height b, then gk AC = kn l gk − nb . And if the body descends through an infinite distance, it has a speed corresponding to the gk height n . Now the time to traverse the distance AM is equal to : So the distance x as the time can conveniently be expressed by a series, that we show generally for any value of the letter m in the following example.

Example 3.

Let the resistance be expressed by the 2m th power of the speed; then dx = k m dv ; gk m − nv m [recall that always Euler takes the speed as proportional to the square root of a height v.] which gives on conversion to a series : From which there is found : But if the time to traverse AM = t, which is dt = ndx , v then we have :

Proposition 55