PROP. LXIV. (Fig. 3. Pl. IX.)

IF A HEAVY BEAM, OR ONE BEARING A WEIGHT, BE SUSTAINED AT C, AND MOVEABLE ABOUT A POINT C; WHILST THE OTHER END B LIES UPON THE WALL BE; AND IF HGF BE DRAWN GRAVITY G, PERPENDICULAR TO

THROUGH THE CENTRE OF

THE HORIZON, AND BF, CH PERPENDICULAR TO BC, AND CF BE DRAWN; THEN

THE WHOLE WEIGHT
PRESSURE AT B

FORCE ACTING at C,

ARE, RESPECTIVELY, AS

HF
HC

CF,

AND IN THESE DIRECTIONS.

For the end B is sustained by the plane CB, and (by Cor. 2. Prop. LIII.) the end C may be supposed to be sustained by a plane perpendicular to FC, or by a cord in direction CF. Then since HC is parallel to BF, the weight, force at C, pressure at B, are, respectively, as HF, CF, HC; by Cor. 1. Prop. LII., or Cor. 1. Prop. LIII.

Cor. But if, instead of lying upon the inclined plane at B, the end B laid upon the horizontal plane AB, then the weight, the pressure at B and C, are, respectively, as BC, GC, and BG; and in this case there is no lateral pressure.

For BF will be perpendicular to BA, and parallel to HF, and, consequently, CF is, also, parallel to HF; therefore (by Cor. 5. Prop XIX.) the forces at C, G, B, are as BG, BC, and CG.

PROP. LXV. (Fig. 4. Pl. IX.)

IF A HEAVY BEAM BC, WHOSE CENTRE OF GRAVITY IS G, BE SUPPORTED UPON TWO POSTS BA, CD; AND BE MOVEABLE ABOUT THE POINTS, A, B, C, D; AND IF AB, DC, PRODUCED, MEET IN ANY POINT H, OF THE LINE GF, DRAWN PERPENDICULAR TO THE HORIZON; AND IF, FROM ANY POINT F, IN THE LINE GF, FE BE DRAWN PARALLEL TO AB; I SAY,

For the points A, B, C, D, being in a plane perpendicular to the horizon, the body may be supposed to be supported by two planes at B, C, perpendicular to AB, DC; or by two ropes BH, CH. And, in either case, the weight in direction HG, the pressure at B, C, in directions HB, HC, are as HF, EF, and HE.

Cor. Hence, whether a body be sustained by two ropes BH, CH, or by two posts AB, CD, or by two planes perpendicular to BA, CD; the body then can only be at rest, when the plumb line HGF passes through G, the centre of gravity of the whole weight sustained. Or, which is the same thing, when AB, DC intersect in the plumb line HGF passing through the centre of gravity.

SCHOLIUM. By the construction of these four last propositions, there is formed the triangle of pressure, representing the several forces. In which, the line of gravity (or plumb line passing through the centre of gravity) always represents the absolute weight, and the other sides the corresponding pres

sures.

PROP. LXVI. (Fig. 5. Pl. IX.)

IF SEVERAL BEAMS AB, BC, CD, &c. BE JOINED TOGETHER AT
B, C, D, &c., AND MOVEABLE ABOUT THE POINTS A, B, C, &c.
BE PLACED IN A VERTICAL PLANE, THE POINTS A, F, BEING
FIXED, AND THROUGH B, C, D, DRAWING ri, sm, tp, PERPENDI-
CULAR TO THE HORIZON ; AND, IF SEVERAL WEIGHTS BE
LAID ON THE ANGLES B, C, D, &c., SO THAT THE WEIGHT ON
S.BCD
ANY ANGLE C MAY BE AS
THEN ALL
S.mCB x S.mCD'

THE BEAMS WILL BE
WEIGHTS.

KEPT IN EQUILIBRIO BY THESE

Produce DC to r. Then (by Cor. 2. Prop. VIII.) S. LABC:

S.LABr weight B: force in direction BC=

BXS.ABr

S.ABC

S.BCD S.DCs:: weight: force in direction CB=

; and

CXS.DCs.

S.BCD

which, to preserve the equilibrium, must be equal to the force in

direction BC, that is,

BXS.ABr CX S.DCs
=
2; whence B: C
S.ABC S.BCD

..S.ABC S.BCD And, by the same way of reasoning, C: D::

S.ABr S.DCs

S.BCD, S.CDE

S.BCs S.ED

SABC

Therefore, ex equo, weight B: weight D::

S.ABrXS.BCs S.DCs XS.ED S.ABix S.CBi S.CDpx S.EDp

Cor. 1. Produce CD, so that Dw may be equal to Cr, and draw

wx parallel to Dp, cutting DE in x. Then the weight C, the forces in directions CB and CD, are as rB, CB and Cr respectively." And weight C is to the weight D, as Br to wx.

Cor. 2. The force or thrust at C, in direction CB, or at B, in direction BC, is as the secant of the elevation of the line BC above the horizon.

For, force in direction CB: force in direction CD :: CB Cr: S.CrB or rCm or sCD; S. rBC:: cosine elevation of CD: cosine elevation of CB:: secant elevation of CB secant elevation CD; because the secants are, reciprocally, as the

cosines.

Cor. 3. Draw Cp, Dm parallel to DE, CB; then the weights on C and D to preserve the equilibrium, will be as Cm to Dp. And, therefore, if all the weights are given, and the position of two lines CD, DE; then the positions of all the rest CB, BA, &c. will be successively found.

For, let the force in direction CD or DC be CD, then Cp is the force in direction DE, and Dm, in Direction CB. And Dp or the weight D, is the force compounded of DC, Cp; and Cm or the weight C is the force compounded of CD, Dm, by Cor. 2. Prop. VII.

Cor. 4. If the weights lie not on the angles B, C, D, &c. let the places of their centres of gravity be at g, h, k, l. And let g, h, k, l, also express their weights. And take the weight B= Ag hC Bh kD 58 + h, C= h+ k, D= k+ 1, &c.; then B, ΑΒ BC BC CD

IE

Ck
CD DE

C, D, &c. will be the weights lying upon the respective angles.
This is evident by Cor. 5. Prop. XIX.

Cor. 5 If the weights were to act upwards, in the directions_mC, PD, &c. or, which is the same thing, if the figure A, B, C, D, E, F, was turned upside down, and the weights remain the same, and the points A, F be fixed as before; all the angles at B, C, D, &c. and, consequently, the whole figure will remain the same as before; and that, whether the lines AB, BC, CD, &c. be flexible or inflexible cords or timbers.

This will easily appear by the demonstration of the Prop. For the ratio of the forces at any angle C, will be the same, whether they act towards the point C, or from it; by Prop. VIII. that is, it will be the same thing whether the weight at any angle C, acts in direction Cm or Cs. And as the forces were

supposed before to thrust against C, the same forces now do pull from it.

SCHOLIUM.-(Fig. 6. Pl. IX.) If DABF be a semi-circle, whose diameter is DF; draw ÁG perpendicular to DF; then the force or weight at any place A, to preserve the equilibrium, will be, reciprocally, as AG3, or directly as the cube of the secant of the arch BA.

Likewise it follows from Cor. 5. that if any cords of equal lengths be stretched to the same degree of curvature, the stretching forces will be as the weights of the cords.

PROP. LXVII. (Fig. 7. Pl. IX.)

THE LATERAL STRENGTH OF ANY PIECE OF TIMBER IN ANY PLACE, WHOSE SECTION IS A RECTANGLE, IS DIRECTLY AS THE BREADTH AND SQUARE OF THE DEPTH.

Let BD be any beam, placed horizontally, and fixed at the end BC. And let AFG be the perpendicular section. Divide the depth AF into an infinite number of equal parts at a, b, c, d, &c. whose number is AF or n; through which, suppose lines drawn parallel to FG. And let any force be applied at P in direction DP, to break the beam at AF; then, since the strength of the timber is nothing but the force by which the parts of the timber at a, b, c, &c. cohere together; the breaking the timber is nothing but overcoming this force, and separating the parts at a, b, c. Suppose 1 force of cohesion of any of the parts