262. COR. 1. Suppose the resultant K of the forces XYZ, acting at the point (xyz), is in the normal to the sur

face at that point, so that

X = KVdu, Y = KVdu, Z = KVdzu,

the equation (A) then takes the form

or substituting A, B, C for the coefficients of du, dyu, dzu, the equation is

Adu + Bd1u + Cdu = 0.

Now the equation to the osculating plane at the point

(xyz) is

A (x – x) + B (y' - y) + C (x' – x) = 0

and the equations to the normal are

xy'x' being the current co-ordinates of a point in the plane or normal: and when the plane (9) contains the normal (10) we have the condition

Adu + Bd1u + Cdzu = 0.

Hence equation (8) expresses that the osculating plane contains the normal : now this is the property of the shortest line between two points on the surface.

263. PROP. To find the pressure on the surface at any

point.

(1). Cos a + (2). cos β + (3). cos y = 0, gives - R = X cos a + Y cos ẞ + Z cos y + T {der cos a

S

2

+ day. cos ẞ + d2z cos y}. Now if p be the radius of absolute curvature at the point (xyz), and λ, μ, v the angles it makes with the axes, we have

2

2

cos x = pd2, cos u = pdy, cos v = pd.2x;

Rds = Xds cos a + Yds cos β + Zds cos y

+ T. - (cos a cosx + cos ẞ cos u + cos y cos v).

Let e be the angle between the radius of absolute curvature, in the direction of which the resultant of the tensions on the extremities of ds acts, and the normal to the surface, then

cos θ

=

cos a cosx + cos β COS μ + cos y

COS v.

Substituting in the above equation, we have then

pressure on a portion ds of the surface

264. COR.

so that X = 0,

When there are no forces acting on the string, Y = 0, Z = 0, we have

265. PROP. To find the form of equilibrium when the string is not attached to a surface.

The equations (1), (2), (3), will give the equations of equilibrium, by putting R = 0: and eliminating T between (1), (2), (4), and also between (1), (3), (4), we have the two following equations to the form of the string.

v(dxdy - dyd2x) = Ydx

Xday...(B).

v(dxdz - dxdx) = Zdx - Xdx

266. COR. In the case of gravity, supposing the axis

of & vertical and measured upwards, we have

X = 0, Y = 0, Z = g;

..dxdy - dyd2x = 0.

This is the differential equation to a straight line in the plane ay, so that the chain hangs in a vertical plane. Take this plane for the plane of xx, and the lowest point as origin. We have, since

d. T = gd, and ... T = g(x + c),

or x + c = cds; since when x = 0, dx = 1.

2

.. x + C = c loge {x + c + √(x + c) – c°},

and when x = 0, x = 0;

267. COR. 2. The Catenary also possesses the property of being the curve of total minimum tension, supposing gravity alone to act.

Thus tension = g (x + c).

Hence to find the curve having the above property, we have

f(x + c) ds = a minimum,

2

or when f(x + c). 1 + (dx)2 = a minimum.

by the principles of the Calculus of Variations; or when x + c = cds, and this has been just shewn to be the differential equation to the Catenary.

N

CHAPTER XI.

PROBLEMS.

1. GIVEN the magnitudes of two forces which act on a point, and the angle between the lines in which they act; to find the magnitude of their resultant.

Let P, Q be the two forces acting upon the point O (fig 81) in the directions OA, OB. Take OA, OB to represent them, and complete the parallelogram OBCA; the diagonal represents their resultant R.

Let a = AOB the given angle. Then from the triangle OAC we have

OC® = OA2 - 2OA. AC cos OAC + AC2

=

OA2 + 2OA . OB. cos POQ + OB2;

.. R2 = P2 + 2 PQ cos a + Q2.

2. Three forces acting on a point are found to balance each other when their directions make angles 105°, 120°, 135° with each other. Find the relation of the forces to each other.

Let F1, F2, F3 be the forces respectively opposite to the angles 105°, 120°, 135°. Then by Art. 28 we have

F1:F2: F3 :: sin 105° : sin 120° : sin 135°

:: cos 15°: cos 30° : cos 45°

:: cos (45° - 30°): cos 30° : cos 45°