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Then, in the triangles VCA VCB, since the angles VCA, CAV, are respectively equal to VCB, VBC, and VC is common to both, AC=CB, and the 2CVA= Z CVB. Again, in the triangles ACD, BCd, the angles DAC, CDA, are equal to the angles CBd, BdC, and AC=BC; therefore, DC=dC. In the same manner it may be shown that CF=Ce, and AE=Be; hence the sides AV, BV, of the triangle AVB, are cut proportionally in E and e; therefore Ee is parallel to AB (Euc. 2. 6), or perpendicular to CV; also, since CE=Ce,' and CF is common to the right-angled triangles CEF, CeF, we have EF=eF (Euc. 47. 1).

Now since DC and dC are equal, and in the directions of the forces upon the sides, they will represent them; resolve DC into two, DE, EC, of which DE produces no effect upon the wedge, and EC, which is effective (art. 149), does not wholly oppose the power, or force upon the back ; resolve EC therefore into two, EF, parallel to the back, and FC perpendicular to it, the latter of which is the only force which opposes the power. In the same manner it appears that eF, FC are the only effective parts of dC, of which FC opposes the power, and eF is counteracted by the equal and opposite force EF ; bence if 2CF represent the power, the wedge will be kept at rest ;* that is, when the force upon the back : the sum of the resistances upon the sides :: 2CF : DC+DC :: 2 CF : 2DC :: CF: DC; an.

CF : CE :: sin. CEF: rad. :: sin. CVE: rad.

CE: DC :: sin. CDE: rad.
Comp CF : DC :: sin CVE x sin. CDE : rad. 2

165. Cor. 1. The forces do not sotain cach other, because the parts DE, de, are not counteracted.

166. Cor. 2. If the resistances act perpendicularly upon the sides of the wedge, the angle CDE beri mes a righi angle, and P: the sum of the resistances :: sin. CVC x rad, : rad. 2 :: sin. CVE: rad. :: AC : AV.

167. Cor. 3. If the direction of the resistances be perpendicular to the back, the angle CDE=LCVF, and P; the sum of the resistances : sin. CVE : rad.:: AC : AV.

168. Cor. 4. When the resistances act parallel to the back, sin. CDA= sin. CAV, and P: the sun of the resistances :: sin. CVA X sin CAV : rad.)* :: CA XCV: AV2 :: CEXAVE: AV! :: CE: AV.

169. Cor. 5. In the demonstration of the proposition it has been supposed that the sides of the wedge are perfectly smooth ; if on account of the friction, or by any other means, the resistances are wholly effective, join Dd, fig. 44, which will cut CV at right angles in y, and resolve Dc, dC into Dy, yc, dy, yc, of which Dy and dy destroy each other, and 2yC sustains the power. Hence, the power : to the sum of the resistances :: 2yČ: 2DC :: yC: DC :: sin. CDy or DCA: rad,

170. Cor. 6. If Ee cnt DC and dC in a and 2, the force, xC, zl, when wholly effective, and the forces DC, dC, acting upon smooth surfaces, will sus tain the same power 2CF.

171. l'or. 7. If froin any point p in the side AV, PC be drawn, and the

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The directions of the three forces must ineet in a point, utherwise a rotatory motion will be given to he wedge.

# By similar triangles, CE: CA :: CV ; AV; therefore CEXAV CAX

resistance upon the side be represented by it, tlie effect upon the wedge will be the same as before ; the only difference will be in the part PE, which is ineffective.

172. Cor. 8. If DC be taken to represent the resistance on one side, and pC, greater or less than dC, represent the resistance on the other, the wedge cannot be kept at rest by a power acting upon the back ; because, on this sny position, the forces which are parallel to the back are unequal.

iliis proposition and its corollaries have been deduced from the actual resoInsion of the forces, for the purpose of showing what parts are lost, or de. su oyed by their opposition to each other; the same conclusions may, however, be very concisely and easily obtained from art. 175.

173. When three forces, acting perpendicularly upon the sides of a scalene wedge, keep each other in equilibrio, they are proportional to those sides.

Let GI, HI, DI, the directions of the forces, meet in I, fig. 45; then, sir ce the forces keep each other at rest, they are proportional to the th-ee sides of a triangle, which are respectively perpendicular to those directions (art. 58) ; that is, to the three sides of the wedge.

174. Cor. If the lines of direction, passing through the points of impact, do not meet in a point, the wedge will have a rotatory motion communicated to it; and this motion will be round the centre of gravity of the wedge as will be shown hereafter.

175. Cor. 2. When the directions of the forces are not perpendicular to the sieks:, the effective parts must be found, and there will be an equilibrium when to: e parts are to each other as the sides of the wedge.

ON THE SCREW.. 176. Def. The Screw is a mechanical power, which may be conceived to be generated in the following manner :

I et a solid and a hollow cyliuder of equal diameters be taken, and let ABC, fig. 46, be a rigbi-angled plane triangle, whose base BC is equal to tie circumference of the solid cylinder ; apply the triangle to the convex surface of this cylinder, in such a manner that the base BC may coincide with the circunference of the base of the cylinder, and BA will form a spiral thread on its surface. By applying to the cyTinder, triangles in succession, similar and equal to ABC, in such a Ipanner, that their bases may be parallel to BC, the spiral thread may be continued : and, supposing this thread to have thickness, or the cylinder to be protuberant where it falls, the external screw will be forned, in which the distance between two contiguous threads, measured in a direction parallel to the axis of the cylinder, is AC. Again, let the triangles be applied in the same manner to the concave surface of the hollow cylinder, and where the thread falls let a groove be made, and the internal screw will be formed. The two screws being thus exactly adapted to each other, the solid or hollow cylinder, fig. 47, as the case requires, may be moved round the common axis, by a lever perpendicular to that axis ; and a motion will be produced in the direcului (f the axis, by means of the spiral thread.

177. When there is an equilibrium upon the screw, P:W:: the dislunce between two contiguous threads, measured in a direction

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

Let BCD, fig.48, represent a section of the screw made by a plane perpendicular to its axis, CE a part of the spiral thread upon which the weight is sustained ; then CE is a portion of au inclined plane, whose height is the distance between two threads, and base equal to the circumference 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

a 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 ::

BCD

: 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 whole 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 circumterence 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 npon the distance between two threads and the circumference FGH, if these remain unaltered, the power is the same, whether the weight is supposed to be sustained at C, or at a point upon the thread nearer to, or farther from, the axis of the cylinder.

OF THE MOTIONS OF BODIES ON INCLINED PLANES.

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

W

= 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.

S

thg Atv

ev

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2hg

V

V

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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

h denote the length of the plane by 1, and its height by h, we have

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

bglve
hgt

1
2hgt

2s thgs

=2
1
lv
2s

Is

hg h

v2 sin. B=

T

2tg 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

hl 7 h s=l, whence t=

; hg

h

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

h
1

h
:1::1:h; therefore,

:
h

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, as the length of the plane to the height. 182. Cur. If in different planes h is constant, tal. If

, or sine angle of joclination is constant, tov hoc ml.

183. If the last acquired velocity on the plane were required, since in this case, also, l=s, we have v=24

, 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=? whg, voe wh, and the velocities acqnired by descend 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 time !

gravity, =V, and VVT

2Vhod=2 vig, which is pre

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:: h:1.

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hgt down an inclined plane

T

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

hgt

h

1

T 186. Therefore, if ABC be taken to denote any plane, and the perpendicular

D 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 : BD.

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.

:

B

a

Examples on the descent of Bodies on inclined Planes. Ex. 1. Let the length of an inclined plane be twice its height, what pace will a body descend on this plane in 3 seconds ?

hgto_1x1672x32 Here s=

=723 feet, the space required. 1

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

2hg! 2x1x161 X 3 Here v=

=48£ feet

per

second 1

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

2 x 12 Here (=

1x161

=1:49=1'22 nearly.

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OF THE MOTION OF PROJECTILES IN A NON

RESISTING MEDIUM. 188. 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 parallel

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