Proposition 13

PROPOSITION 13. Problem.

93. If the force acting is uniform and acting downwards everywhere, to determine the descent of the body on a given curve AM (Fig.13), beginning from A at rest, and to find the force pressing on the curve at individual points M.

Solution.

With the vertical AP or with the parallel in the direction of the force drawn MF and with the connected line MP at right angles, let AP = x, PM = y, with the curve AM = s.

The force MF is put equal to g with the force of gravity present equal to 1 and the speed at M corresponding to the gdy height v. With these in place, the normal force = ds and the tangential force = ds (83). Because in this case the tangential force accelerates, we have dv = gdx and v = gx as the speed at A = 0.

Hence since the radius of osculation … + ds with dx placed along the direction MO is equal to dxddy …3 constant, the centrifugal force = + dvdxddy , the direction of which is MN.

Now the ds 3 gdy … normal force ds acts along the same direction. whereby the total force pressing on the curve along MN at M is equal to as v = gx. Truly the time in which the body traverses the arc AM is equal to

…

Q.E.I.

Corollary 1

94. Therefore the speed at M only depends on the height AP through which the body descends, and the body acquires the same amount as falling from A to P and acted on by the same force g.

Corollary 2

95. Therefore, whatever the curve the body falls along from rest, acted on by the constant force g, the speeds are as the square roots from the squares [of the speeds] proportional to the heights traversed; that is as v , i. e. as gx .

Corollary 3

96. The time, in which the first element Aa is traversed, is … ∫ dsgx with x evanescent.

If therefore the angle PAa is less than a right angle, or s = nx, then the time to pass through Aa is infinitely small and thus the time to pass along AM is finite, unless the curve either rises between A and M beyond A or it progresses to infinity.

But if the angle PAa is right, for the point A, s n = ax with a number n present greater than one and thus Whereby if n is less than two, the time to pass along Aa is infinitely small and the time to traverse AM is finite.

But if n is greater than or equal to 2, the time to pass through the first element Aa is infinitely great or the body never escapes from A. [p. 45]

Corollary 4.

97. But whenever n < 2, to the radius of osculation at A is indefinitely small. Whereby in this case, in which the tangent to the curve at A is normal to AP, the body does not descend in this case, unless the radius of osculation at A is infinitely small.

Scholium 1.

98. From which, as the first element is traversed in an infinitely short time, it is correctly concluded that the time to traverse the arc AM is finite; for since the body descends along AM with an accelerated motion, the following elements are described more quickly and on this account the corresponding time must be finite. Moreover everything is illustrated in the following.

Example 1.

99. Let the line AM (Fig. 14) be straight and inclined at some angle to the vertical AP and the cosine of the angle A = n; then x = ns. Therefore the time, in which the body descends along AM, is equal to

…

or the time to travel along the line at any inclination varies directly as the root of the length of the line and inversely as the root of the cosine of the angle of inclination MAP.

Moreover the centrifugal force is equal to 0 [i.e. infinite radius of curvature], whereby the line AM is only acted on by the normal force, which is equal to

Corollary 5.

100. Therefore the time to pass along AM is to the time to pass along AK as AM to AK . But the time to pass along AM is to the time along AP as AM to AP (88).

Whereby if it is the case that AM : AP = AM : AK or AM : AP = AP : AK , which comes about if PK is perpendicular to AM , then the descent time along AK is equal to the descent time along AP.

Corollary 6.

101. It is also apparent that the descent time along the perpendicular PK is equal to the …

Whereby, since the time descent time along AP. For the cosine of the angle APK = PK .. along AP to the time along KP is as AP to

KP : PK

, this is an equal ratio. AP

Corollary 7

102. From this it is evident in the circle APPB (Fig. 15), that all the descents along the chords AP drawn from the uppermost point A and all the descents along the chords drawn to the lowest point B are to be made in equal times, clearly each in that time in which the body falls freely perpendicularly along the diameter AB.

Example 2.

103. If the curve AMB (Fig. 16) is a circle and the radius BC = a and AP is a tangent to the circle, then ..

Hence we have ds = adx and hence v = gx ( a2 − x2 )

and r = a , and the centrifugal force is equal to

… 2 gx

and on this account dy =

…

, and the … ( a2 − x2 )

total force sustained by the circle at M is equal to

…

Therefore the total force pressing on the curve is three times as great as the normal force.

The time then, in which the arc AM is traversed, is equal to

… ∫ g( aadxx− x ) , [p. 47]

the integration of which neither depends on circular nor hyperbolic quadrature, but with the help of rectification of the elastic curve it is able to be constructed. Meanwhile the time along the quadrant AB is equal to

Corollary 8

104. When the body reaches the lowest point B, there it has the speed corresponding to the height ga. Therefore this ascends in the other quadrant BD and reaches D, where its speed vanishes, and thus again it descends to B and then it re-ascends to A along BA. Now in a similar way is the ascent and descent along the sides of a square, since the body either in the ascent or in the descent, has the same speed at the same points.

Scholium 2.

105. We will not offer other examples now, as in what follows, where we consider more descents on a given line, we are to report on more examples.

We disclose in the first place these questions which pertain to the motion on a given line starting from rest at a given point, and the following problem is of this kind.