parullel to the axis : the circumference of the circle which the power describes.

Let BCD, fig.48, represent a section of the screw made by a plane per. pendicular toits axis, CE a part of the spiral thread upon which the weight is sustained ; then CE is a portion of an inclined plane, whose height is the distance between two threads, and base equal to the circum. ference BCD. Call F the power which acting at C in the plane BCD, and in the direction CI perpendicular to AC, will sustain the weight W, or prevent the motion of the screw round the axis; then, since the weight is sustained upon the inclined plane CE by a power F acting parallel to its base, F:W:: the height : the base (art. 153) :: the distance between two threads : the circumference BCD. Now, instead of supposing the power F to act at C, let a power P act perpendicularly at G, on the straight lever GCA, whose centre of motion is A, and let this power produce the same effect at C that F does; then, by the property of the lever, P:F:: CA : GA :: the circumference BCD: the circumference FGH. We have, therefore, these two proportions,

F: W:: the distance between two threads : BCD
P F ::


: FGH comp. P:W:: the distance between two threads : FGH. 178. Cor. 1. In the proof of this proposition the whole weight is supposed to be sustained at one point C of the spiral thread; if we suppose it to be dispersed over the wliole thread, then, by the proposition, the power at G necessary to sustain any part of the weight : that part :: the distance between two threads : the eircumference of the circle FGH ; therefore the sum of all these powers, or the whole power : the sum of all the corresponding weights, or the whole weight :: the distance between two threads : the circumference of the circle FGH.

179. Cor. 2. Since the power, necessary to sustain a given weight, depends opon the distance between two threads and the circumference FGH, if these remain inaltered, the power is the same, whether the weight is supposed to be sustained at C, or at a point upou the thread nearer to, or farther from, the axis of the cylinder.


180. In treating of the equilibrium of bodies on an inclined plane, we have shown, art. 158, that the absolute weight of the body, or the natural force of gravity, may be resolved into two forces, which, if we call W the absolute weight, may be represented by W sin. B, and W cos. B ; the latter denoting the resistance of the plane, the angle of its elevation being represented by B; and the former, W sin. B, the force or power which acting parallel to the plane, preserves the body in equilibrio, and which, therefore, necessarily also denotes the force with which the body has a tendency to descend. Hence W sin. B is the motive force; but W is the mass or body moved ; and

W sin. B therefore


= sin. B, is the accelerating force. Hence the accelerating force, or force with which the body descends down an inclined plane, is to the force of gravity as sin. B to radius.

We have, therefore, only to introduce into the equations of art. 41, on. Bxg, instead of g, and we shall obtain formulas which will exhibit all the circumstances of time, space, and velocity ; or since, if we denote the length of the plane by l, and its height by h, we have


T = sin. B, the equations of art. 41 become

hgt lve

1 4hg 2hgt 2s

hgs -=2


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sin. B=

tog 4sg From these equations we draw immediately the following remarkable properties relative to the descent of bodies down inclined planes.

181. If the time of descending through the whole plane were required, we should have

hl2 1 h s=l, whence t=

; hg

8 but the time of descending through the space h, by the free action of


h 1 gravity, = and


:: :1::1:h; therefore, 8

h the time which a lody employs in descending through an inclined plane, is to the time in falling freely through the height of the plane, us the length of the plane to the height.

182. Cor. If in different planes h is constant, toll. 11 or sine angle of inclination is constant, tOCV howl.

183. If the last acquired velocity on the plane were required, since

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in this case, also, l=s, we have v=2

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-2 hg, which is


cisely the expression for the velocity acquired by a heavy body falling freely through the space h.

Whence the velocity acquired by a body descending down an inclined plane, is equal to that acquired by a body falling freely through the height of the plane.

184. Cor. Since v= 2 whg, voch, and the velocities acqnired by descendo ing down any planes as the square roots of their beights.

185. Hence, also, the space descended down an inclined plane in any time t, is to the space descended freely in the same time, as the height of the plane to its length.

For, from the above formulas, the space descended in the tine

Tgl ::


hgl down an inclined plane


and the space descended freely in the same time (41) is =gt, and


h T

I :: h:1. 186. Therefore, if ABC be taken to denote any plane, and the perpendicular DB be drawn, the body will descend from C to D along the plane, in the same time as it would fall freely through the height CB, because AC : BC :: BC :


A 187. As this result is wholly independent of the inclination of the plane, it follows that, in any rightangled triangle BDC, having its hypothenuse BC perpendicular to the horizon, a body will descend D down any of its three sides BD, BC, DC, in the same time ; or if on the diameter BC a semicircle be de

E scribed, the time of descending down any of the chords BD, BE, BF, or DC, EC, FC, will be the same as

F the time in falling down the diameter BC.

Here s=

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Examples on the descent of Bodies on inclined Planes. Er. 1. Let the length of an inclined plane be twice its height, what space

will a body descend on this plane in 3 seconds ?
hgro _1x1676 x 3?

=723 feet, the space required.

2 Er. 2. On the same suppositions as in the preceding example, wliat will be the velocity acquired in the 3 seconds ?

2hgl_2x1 X 16 17 x3 Here v=

=481 feet 2


per Er. 3. In what time will a body fall through 12 feet down the same plane?

2 x 12 Here i

=149=1'22 nearly. I X161

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RESISTING MEDIUM. 189. Def. When a heavy body is impelled by an instantaneous force in any direction, either vertical, parallel, or oblique to the hori. zon, it is said to be projected; the body itself is called a projectile, and the curve or line is denominated the path of the projectile. These terms, however, are more generally applied to those cases in which the original direction of the projectile is a line either oblique or paralel

to the horizon ; as in the projection of balls and shells from mortars, or other pieces of ordnance.

189. Let a body, be projected from A, in any direction, not vertical as

D AD; and let AC, AD, be the spaces that the body would describe in the times i and t', from the uniform velucity of projection, and CE and DB the spaces through which it would de

C scend by the action of gravity in the

B. same times; then it is obvious, from the composition of the two motions that the body will be found at the ends of those times in the points E and B.

Now, by the laws of uniform motions (art. 26) AC=tv, and AD=thy (v being the velocity of projection), therefore,

E AC: AD :: t and, by the laws of falling bodies(41), sat”, and therefore CE : DB :: 2 : t's

F whence CE : DB :: AC' : AD , which is a known property of the parabola ; and, as the same las place for every point in the path of the projectile, it follows, that the curte described ly the projectile is a parabola.

190. In order now to investigate the laws of the bodies' motion,

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let A be the point of projection AB, or AB', the plane on which the body is projected, passing through A, and which also denotes the range.

Let AC be drawn parallel, and BCD perpendicular to the horizon;

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cos. B


let the angle of elevation CAD-A, and the angle of inclination of the piane CAB=B, the velocity of projection =v, the time of flight = t, and the range AB =r, also let164 feet =

= 6 Then we know from the laws of uniform motion, that the body at the end of the time t, if gravity did not act, would be found in the point D; while by the laws of falling bodies, it would in the same time pass through the perpendicular DB ; consequently

AD=Iv; and DB=ge Whence we have sin. LABD : sin. <BAD AD: DB

tv sin. (A + B)
or cos. B : sin. (A + B) :: tv :
and therefore

gl_sin. (A+B)
cos. B

(Equation I.) Again sin. ZADB : sin. <BAD :: AB : DB

sin. (A+B)
cos. A : sin. (A + B) ::
and therefore gle_sin.


(Equation II.) And equating the value t in equations 1 and 2, we obtain sin. (A + B) cos. A

(Equation 111.) From these three equations, all the relations between the time, velocity, range, and angle of elevation, are readily determined. For example :

If the time and elevation be given to find the velocity and range ;



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8 cos. B

ja this case,

gt cos. B

Equation 1 gives va

sin. (A+B) Equation 2 gives =

gle cos. A

sin. (A+B) If the range and elevation be given to find the time and velocity we have from

sin. (A+B) Equation 2 ...


=V (SIN. CA. #B)

(sin. (A + B) cos. A)

tg cos.* B

Equation 3 If the velocity and elevation be given, to find the time and range, we obtain from

v sin. (A+B) By Equation 1...,

8 cos. B

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