locity in the curve at F, is to the horizontal velocity there, as VD to ID, because VD is parallel to the curve or to the tangent at F, that is dr√(dx − x2) :: 8√√(x − a) : 8√√(x − a) × √(d — x) ̧ which is the horizontal velocity at F,

x

Nd

where the body is supposed to have that velocity a constant quantity; therefore also ( a) × √(dx), as well as (x − a) × (d - x) = ax + dx -ad-x is a constant quantity, and also ax + dx 22; but the fluxion of a constant quantity is equal to nothing, that is ax + dx – 2xx = 0 = a + d2x, and hence x = a + d = VI, the arithmetical mean between Vн and VC.

If the motion should commence at v, then x or vI would bed, and I would be the centre of the semicircle.

PROBLEM 22.

If a body begin to move from ▲, with a given velocity, along the quadrant of a circle AB; it is required to show at what point it will fly off from the curve.

Let D denote the point where the body quits the circle ADB, and then describes the parabola E. Draw the ordinate DF, and let GA be the height producing the velocity at 4. Put GA=α, AC or CD, AF = x; then the velocity in the curve at D will be the same as that acquired by falling through GF or a+x, which is, as before, 8(a+x); but the velocity in the curve is to the horizontal velocity as on to mn or as CD to CF by similar triangles, that is, as

BE

r: r− x : : 8 √ √ (x + a) ; 8 √ √ (x + a) x, which is to be a constant quantity where the body leaves the circle, therefore also (r−x)√(x+u) and (r−x)2 × (x + a) a constant quantity; the fluxion of which made to vanish, gives

Hence, if a = 0, or the body only commence motion at A, then x = r, or AF = AC when it quits the circle at D. But if a or ca were 7 or ac, then r 2a0, and the body would instantly quit the circle at the vertex A, and describe a parabola circumscribing it, and having the same vertex A.

PROBLEM

PROBLEM 23.

To determine the position of a bar or beam AB, being supported in equilibrio by two cords AC, BC, having their two ends fixed in the beam, at A and B.

By art. 210 vol. 2, the position will be such, that its centre of gravity G will be in the perpendicular or plumb line CG.

F

E

H

D

Corol. 1. Draw GD parallel to the cord AC. Then the triangle CGD, having its three sides in the directions of, or parallel to, the three forces, viz, the weight of the beam, and the tensions of the two cords AC, BC, these three forces will be proportional to the three sides CG, GD, CD, respectively, by art. 44; that is, CG is as the weight of the beam, GD as the tension or force of AC, and CD as the tension or force of BC.

Corol. 2. If two planes EAF, HBI, perpendicular to the two cords, be substituted instead of these, the beam will be still supported by the two planes, just the same as before by the cords, because the action of the planes is in the direction perpendicular to their surface; and the pressure on the planes will be just equal to the tension or force of the respective cords. So that it is the very same thing, whether the body is sustained by the two cords AC, BC, or by the two planes EF, HI; the directions and quantities of the forces acting at A and B being the same in both cases. Also, if the body be made to vibrate about the point c, the points A, B will describe circular arcs coinciding with the touching planes at ▲, B; and moving the body up and down the planes, will be just the same thing as making it vibrate by the cords; consequently the body can only rest, in either case, when the centre of gravity is in the perpendicular CG.

PROBLEM 24.

To determine the position of the beam AB, hanging by one cord ACB, having its ends fastened at ▲ and B, and sliding freely over a tack or pulley fixed at c.

G being the centre of gravity of the beam, CG will be perpendicular to the horizon, as in the last problem. Now as

the

A

G

the cord ACB moves freely about the point c, the tension of the cord is the same in every part, or the same both in AC and BC. Draw GD parallel to AC: then the sides of the triangle CGD are proportional to the three forces, the weight and the tensions of the string; that is, CD and DG are as the forces or tensions in CB and CA. But these tensions are equal; therefore CD DG, and conseq. the opposite angles DCG and DGC are also equal: but the angle DGC is the alternate angle ACG; theref. the angle ACG = BCG; and hence the line CG bisects the vertical angle ACB, and conseq. AC: CB:: AG: GB.

PROBLEM 25.

To determine the position of the beam AB, moveable about the end B, and sustained by a given weight g, hanging by a cord Acg, going over a pulley at c, and fixed to the other end A.

B

D E

g

Let the weight of the beam, and G denote the place of its centre of gravity. Produce the direction of the cord CA to meet the horizontal line BE in D; also let fall AE perp. to BB: then AE is the direction of the weight of the beam, and DA the direction of the weight g, the former acting at G by the lever BG, and the latter at A by the lever BA; theref. the intensity of the former is wx BG, and that of the latter g x BA; but these are also proportional to the sines of their angles of direction with AB, that is, of the angles BAE, and BAD; therefore the whole intensity of the former is X BG X sin. BAE, and of the latter it is g× BA X sin. BAD. But, since these two forces balance each other, they are equal, viz, w × BG X sin. BAE= g × BA X sin. BAD, and therefore w: g:: BA X sin. BAD BG sin. BAE, or w × BG g X BA :: sin. BAD: sin. BAE.

1

PROBLEM 26.

To determine the position of the beam AB, sustained by the given weights m, n, by means of the cords ACM, BDn, going over the fixed pulleys C, D.

Let

the sine of the angle of direction B; but these two effects are equal, because they balance each other; that is, mɩ × AG × sin. A = 1 × BG X sin. B; theref. m × AG : N × BG :: sin. B sin. A.

PROBLEM 27.

To determine the position of the two posts AD and BE, supporting the beam AB, so that the beam may rest in equilibrio.

Through the centre of gravity G of the beam, draw CG perp. to the horizon; from any point c in which draw CAD, CBE through the extremities of the beam; then AD and BE will be the positions of the two posts or props required, so as AB may be sustained in equilibrio; because the three

G

D

F

forces sustaining any body in such a state, must be all directed to the same point c.

Corol. If GF be drawn parallel to CD; then the quantities of the three forces balancing the beam, will be proportional to the three sides of the triangle CGF, viz, CG as the weight of the beam, CF as the thrust or pressure in BE, and FG as the thrust or pressure in AD.

Scholium. The equilibrium may be equally maintained by the two posts or props AD, BE, as by the two cords AC, BC, or by two planes at A and B perp. to those cords.-It does not always happen that the centre of gravity is at the lowest place to which it can get, to make an equilibrium; for here when the beam AB is supported by the posts DA, EB, the centre of gravity is at the highest it can get; and being in that position, it is not disposed to move one way more than another, and therefore it as truly in equilibrio, as if the centre was at the lowest point. It is true this is only a tottering equilibrium, and any the least force will destroy it; and then, if the beam and posts be moveable about the angles A, B, D, E,

which

which is all along supposed, the beam will descend till it is below the points D, E, and gain such a position as is described in prob. 26, supposing the cords fixed at C and D, in the fig. to that prob. and then G will be at the lowest point, coming there to an equilibrium again. In planes, the centre of gravity G may be either at its highest or lowest point. And there are cases, when that centre is neither at its highest nor lowest point, as may happen in the case of prob. 24.

PROBLEM 28.

Supposing the beam AB hanging by a pin at в, and lying on the wall ac; it is required to determine the forces or pressures at the points A and B, and their directions.

Draw AD perp. to AB, and through G, the centre of gravity of the beam, draw GD perp. to the horizon; and join BD. Then the weight of the beam, and the two forces or pressures at A and B, will be in the directions of the three sides of the triangle ADG; or in the directions of, and proportional to, the three sides of the triangle GDH, having

A

G

E

drawn GH parallel to BD; viz, the weight of the beam as GD, the pressure at A as HD, and the pressure в as GH, and in these directions.

For, the action of the beam is in the direction GD; and the action of the wall at A, is in the perp. AD; conseq. the stress on the pin at в must be in the direction BD, because all the three forces sustaining a body in equilibrio, must tend to the same point, as D.

Corol. 1. If the beam were supported by a pin at A, and laid upon the wall at B; the like construction must be made at B, as has been done at ▲, and then the forces and their . directions will be obtained.

Corol. 2. It is all the same thing, whether the beam is sustained by the pin в and the wall AC, or by two cords BE, AF, acting in the directions DB, DA, and with the forces HG, HD.

PROBLEM 29.

To determine the Quantities and Directions of the Forces, exerted by a heavy beam AB, at its two Extremities and its Centre of Gravity, bearing against a perp. wall at its upper end B

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