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y==(A

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A+B(V—v),

Then their relative velocity in the direction AC is V-V; also the momenta before the stroke are AV and Bu, the sum of which is AV + Bo in the direction AC.

Again, put x for the velocity of A, and y for that of B, in the same direction AC after the stroke ; then their relative velocity is y-ty and the sum of their momenta, in the same direction, is Ar+ By.

But the momenta before and after the collision, estimated in the same direction, are equal, whatever be the nature of the bodies (by 79), as are also their relative velocities, in the case of elastic bodies, by the preceding article. Whence arise these two equations,

AV+Bv=Ax+By,

and V-u=y. The resolution of these equations gives (A-B)V+2Bv

for the velocity of A,
A+B
-(A-B)v+2AV

the velocity of B.

A+B From the above values of x and y, we find V-x, or the veiacus 2B

2A lost by A==

A+B(V—v) and y-v,or that gained by B= and these velocities are in the ratio of B to A, or, reciprocally, as the bodies themselves. 89. Cur. The velocity lost by A drawn into A, and the velocity gained by B

24B drawn into B, give each of them (V—v) for the momentum gained by

A+BY the one and lost by the other in consequence of the stroke; this increment and decrement being equal, they cancel one another, and leave the same momentum AV+Bv after the impact, as it was before it.

90. Hence, also, AV? + Bv:=Aro + By, or the sum of the vires viviarum, or living forces, are the same both before and after inpact.

For since AV+Bv=Ax+By, by transposition,

we find AV-Ar=By-Bv and (87) V+x=y+v, these two equations multiplied together give AV?--Ax'=Bye-Bv?, or

AV? + Bvo=B.x® + By2. 91. Cor. If u be negative, or if the body B move in the contrary directions before collision, or towards A, by changing the sign of o, the same theorems become,

_(A—13)V—2B0

the velocity of A,
A+B
(A-B)v+2AV
y=
A+B

the velocity of B in the direction AC.) i

92. If B were at rest before the impact, we have v=0; and, consequently, A-B

2A Ā+B*V, and y=A+B *V, the velocities in this case.

Х

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2A *V=V. There

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and y

Again, suppose B at rest, and the two bodies, A and B, equal to

2A

2A each other; then A-B=0, and XV= XVV

A+B fore, the body A will stand still, and the

body B will move forward with the whole velocity of the former. This we sometimes see happen in playing at billiards, and it would occur much more frequently, it the bal.s were perfectly elastic. 93. Cor. In the expressions

(A-BJV + 4B

A+B
2AV F(A-B)

A+B Let A=B, then I=+o, and y=V, and therefore the bodies interchange veloc. ties.

94. If the bodies be elastic only in a partial degree, the sum of the momenta will still be the same, both before and after collision, but the velocities after, will be less than in the case of perfect elas. ticity, in the ratio of the imperfection. Hence, employing the same notation as before, the two equations of article 88, will become

AV + Bu=Ax+By,

and V-v= = *(-); where m to n denotes the ratio of perfect to imperfect elasticity. The resolution of these two equations gives the following values of s and y, viz.

mtn B x=

m

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A+B(V)

for the velocities of the two bodies, after impact, in the case of imperfect elasticity : and these equations agree with the former when n=m.

From these expressions for I and y, we may easily solve the principal problems relating to the collision of imperfectly elastic bodies.

95. When a perfectly hard body impinges obliquely on a perfectly hard and immoveable plane AB, fig. 12, in the direction CD, after impact it will move along the plane, and the velocity before impact : the velocity after :: radius : cos. < CDA.

Take CD to represent the motion of the body before impact; draw CE parallel, and DE perpendicular, to AB. The motion CD may be resolved into the two CE, ED, art. 65, of which ED is wholly employed in carrying the body in a direction perpendicular to the plane, and since the plane is immoveable, this motion will be wholly de stroyed. The other motion CE, which is employed in carrying the body parallel to the plane, will not be effected by the impact ; and, consequently, there being no force to separate the body and the plane, the body will move along the plane ; and it will describe DB=CE,

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in the same time that it described CD before impact ; also these spaces are uniformly described ; consequently, the velocity before inpact: the velocity after :: CD:CE :: radius : sin. 2 CDE :: radius ; cos. < CDA.

96. Cor. If CD be taken to express the direction and absolute quaotity of oblique force with which the body impinges on the plane AB, it may, in the very same manner be shown, that ED will express the direct force, or energy of the stroke upon that plane, and we shall have oblique force : direct force :: CD : DE:: radius : sin. angle of incidence.

97. If a perfectly elastic body impinge upon an immoveable plane AB, fig. 13, in the direction CD, it will be reflected from it in the direction DF, which makes, with DB, the angle BDF, equal to the angle ADC.

Let CD represent the motion of the impinging body; draw CF parallel, and DE perpendicular, to AB; make EF=CE, and join DF. Then the whole motion may be resolved into the two CE, ED, of which CE is employed in carrying the body parallel to the plane, and must therefore remain after the impact ; and ED carries the body in the direction ED, perpendicular to the plane; and since the plane is immoveable, this motion will be destroyed during the compression, and an equal motion will be generated in the opposite direction by the force of elasticity. Hence, it appears that the body at the point D has two motions, one of which would carry it uniformly from D to E, and the other from E to F in the same time, viz. in the time in which it described CD before the impact; it will, therefore, describe DF in that time (art. 47). Again, in the triangles CDE, EDF, CE=EF, the side ED is common, and the Z CED=ZDEF; therefore, the 2CDE =ZFDE; hence, the ZCDA= Z FDB.

98. Cor. Since CD=DF, and these are spaces nniformly described in equal times, before and after the impact, the velocity of the body after the reflection is equal to its velocity before the incideuce.

99. To determine the motions after the impact, in two bodies which strike one another obliquely.

Let the two bodies A and B, fig. 14, move in the oblique directions AK, BK, and strike each other at K, with velocities which are in proportion to the lines AK, CK; to find their motions after the impact. Let CKH represent the plane in which the bodies touch in the point of concourse ; to this plane draw the perpendiculars AC, BD, and complete the rectangles CE, DF. Then the motion in AK is resolved into the two AC, CK; and the motion in BK is resolved into the two BD, DK; of which the antecedents AC, BD, are the velccities with which they directly meet, and the consequents CK, DK, are parallel ; therefore, by these the bodies do not impinge on each other, and consequently the motions, according to these directions, will not be changed by the impulse ; so that the velocities with which the bodies meet are AC and BD, or their equals EK, FK. The motions of the bodies A, B, directly striking each other with the velocities EK, FK, will be determined by articles 76, 87, &c. according as the bodies are elastic or non-elastic ; which being done, let KG be the velocity so determined, of one of them, as A, and since there "emains also in the body a force of moving in the direction parallel to AE, with a velocity as AE, make KH=AE, and complete the rectangle GH. Then the two motions in KH, and KG, or HI, are compounded into the diagonal KI, which, therefore, will be the path and velocity of the body A after the stroke. And after the same manner may the motion of the other body B be determined after the impact

If the elasticity of the bodies be imperfect in any given degree, then the quantity of the corresponding lines must be diminished in the same proportion.

ON THE MECHANICAL POWERS. 100. The mechanical powers are the most simple instruments used for the purpose of supporting weights, or communicating motion to bodies, and by the combination of which, all machines,

owever complicated, are constructed.

These powers are six in number, viz. the lever, the wheel and axle, the pulley, the inclined plane, the wedge, and the screw.

Before we enter upon a particular description of these instruments and the calculation of their effects, it is necessary to premise, that when any forces are applied to them, they are themselves supposed to be at rest, and consequently that they are either without weight, or that the parts are so adjusted as to sustain each other. They are also supposed to be perfectly smooth; no allowance being made for the effects of adhesion.

When two forces act upon each other by means of any machine, one of them is, for the sake of distinction, called the power, and the other the weight.

On the Lever. 101. Def. The Lever is an inflexible rod, moveable upon a point which is called the fulcrum, or centre of motion.

The power and weight are supposed to act in the plane in which the lever is moveable round the fulcrum, and tend to turn it in opposite directions.

Ar. 1. If two weights balance each other apon a straight lever, the prese sure upon the fulcrum is equal to the sum of the weights, whatever be the length of the lever.

Ax. 2. If a weight be supported npon a lever which rests on two fulcrums, the pressure npon the fulcrums is equal to the whole weight.

Ar. 3. Equal forces, acting perpendicularly at the extremities of equal arras of a lever, exert the same effort to turn the lever round.

102. If two equal weights act perpendicularly upon a straight lever,

• The effect produced by the gravity of the lever is not taken into consideration, unless it be expressly mentionedo,

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the effort to put it in motion, rouna any fulcrum, will be the same as if they acted together at the middle point between them.

Let A and B be two equal weights, acting perpendicularly upon the lever FB, whose fulcrum is F, fig. 15. Bisect AB in C, make CE=CF, and at E suppose another fulcrum to be placed.

Then, since the two weights A and B are supported by E and F, and these fulcrums are similarly situated with respect to the weights, each sustains an equal pressure ; and, therefore, the weight sustained by E is equal to half the sum of the weights. Now let the weights A and B be placed at C, the middle point between A and B, and consequently the middle point between E and F; then, since É and F support the whole weight C, and are similarly situated with respect to it, the fulcrum E supports half the weight; that is, the pressure upon E is the same, whether the weights are placed at A and B, or collected in C, the middle point between them; and, therefore, the effort to put the lever in motion round F, is the same on either supposition.

103. Cor. If a weight be formed into a cylinder AB ( fig. 16) which is every where of the same density, and placed parallel to the horizon, the effort of any part AD), to put the whole in motion round C, is the same as if ti.is part were collected at E, the middle point of AD.

For the weight AD may be supposed to consist of pairs of equal weights, equally distant from the middle point.

What is here affirmed of weights, is true of any forces which are propor. tional to the weights, and act in the same directions.

104. Two weihts, or two forces, acting perpendicularly upon a straight lever, will lalance each other, when they are reciprocally proportional to their distances from the fulcrum.

Case 1. When the weights act on contrary sides of the fulcrum.
Let r and

у be the two weights, and let them be formed into the cylinder AB, fig. 16, which is every where of the same density. Bisect AB in C, then this cylinder will balance itself upon the fulcrum C (art. 103). Divide AB into two parts in D, so that AD: DB :: X : y, and the weights of AD and DB will be respectively x and y; bisect AD in E and DB in F; then, since AD and DB keep the lever at rest, they will keep it at rest when they are collected at E and F (art. 103); that is, x, when placed at E, will balance y, when placed

AD BD AB-BD AB-AD at F; and x:y :: AD : BD ::

2

2 C3-BF : AC-AE :: CF : CE.

Case 2. When the two forces act on the same side of the centre of motion.

Let AB, fig. 17, be a lever whose fulcrum is C, A and B two weights acting perpendicularly upon it; and let A:B :: BC : AC, then these weights will balance each other, as appears by the former case. Now suppose a power sufficient to sustain a weight equal to the sum of the weights A and B, to be applied at C, in a direction opposite to that in which the weights act; then will this power supply the place of the fulcrum (art. 101, al. 1); and the centre of motion

:

2

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