This expression gives the distance of the centre of gravity of the frustum of a cone from the greater end; hence, if R, r represent the radii of the greater and less ends of the frustum, and h its altitude, we have the distance h R2+2Rr + 31a and when r = 0, we have 4R+ Rr +

of the centre of gravity

=

the distance of the centre of gravity of a cone from its base =

fourth of the altitude as found above.

= one

Ex. 3. Four bodies, whose weights are w1, w, w1, w, pounds, are placed at the successive angles of a square whose side is 2a inches; required the position of their common centre of gravity, the square being considered without weight.

Take O, the centre of the square, as the origin of co-ordinate reference, and the two rectangular axes parallel and perpendicular to the sides; then we have (w2 + w3 — w1 — w1) a, and y = w1 + w2 + w3 + W1

X =
w1 + w2 + w3 + w1
Thus if w=3, w2=4, w ̧=5, w1=6; and 2a= 12 inches;

4

a.

.. X=0 and y =

6= -14= OG = distance of

18

centre of gravity below O on the axis of y.

PROBLEMS FOR EXERCISE.

69. Ex. 1. Find the centres of gravity of

(1.) The common parabola and the paraboloid.
(2.) A semicircle, and the segment of a circle.

(3.) A hemispheroid, and a hemisphere.

(4.) The sector of a circle, and a spheric sector.

(5.) The surface of a spheric segment, and that of a cone.

Ex. 2. Two cones are placed with their equal bases in contact, and the altitude of the one is three times that of the other; find the position of their common centre of gravity.

Ex. 3. The surface generated by a plane line or curve revolving about an axis in the plane of the figure, is equal to the product of the generating line or curve, and the path described by its centre of gravity.

Ex. 4. The volume of the solid generated by the revolution of a plane figure about an axis in the plane of the figure, is equal to the product of the generating surface, and the path described by its centre of gravity.

Ex. 5. From a given rectangle ABCD of uniform thickness, to cut off a triangle CDO, so that the remainder, ABCO when suspended at O, shall hang with AB in a vertical position.

of the prismatic mass of earth which lies above the surface of a bank that would be itself supported. But this prismatic mass is partly supported by friction, and we must therefore ascertain how much of the horizontal thrust is counteracted by friction.

Suppose a weight W to be placed on a plane, inclined to the vertical at an angle i; and let H be the horizontal force, which, with the friction, just sustains the weight W. Resolve each of the forces W, H into two others, the one parallel and the other perpendicular to the plane; and those parallel to the plane act in opposite directions, while those perpendicular to the plane concur in direction; hence we have

force parallel to the plane

W

H

force perpendicular to the plane = W sin i + H cos i. And the first of these forces must be precisely equal to the friction; that is, equal to a force that will just support the weight upon the plane; hence W cos iH sin i = ƒ W sin i +ƒ H cos i

.. H =
cos if sin i
W =
sin if cos i

1-ftan i w
tan i +ƒ

Now to apply this to the investigation of the horizontal thrust of the prism BCH, we shall put BC= a, a variable part Cb= x, bb' = dx, and s the specific gravity of the earth. Then the area of bb'hh' xdx tan i, and its weight= =sxdx tan i; hence the horizontal thrust against bb' will be

1-ftan i sxdx tan i = -ftan i

tani + f

where M =

fos Mxdx

.

f tan i

1+fcot i

1+fcot i

sxdxsMxdx

; hence, integrating, we have

fa's M = whole horizontal thrust of triangle BCH.

=

Again, the length of the lever Bb ax, and the moment* of the thrust of the element bb'hh'

sMx (a — x) dx = as Mxdx

• If lines be expressed numerically, the product of a force acting on a lever, and the perpendicu➡ lar from the axis of motion on its direction is called the moment of that force,

fas Mxdx-f`sMx2dx = 4a3sM — 4a3s M = ja3s M = moment of the whin horizonal thrust.

-ftan i

The expression

1 + f cot i

will vanish when tan i = 0, or tan i = ''; and

between these limits there is a value which gives both the horizontal thrust and its moment a maximum.

Let then

1-ftan i
= maximum, and differentiating we have
1 + f cot i

Hence M =

=

cosec

= tan2 i = (−ƒ+ √T +ƒ3ï3

.. horizontal thrust

jas(-f+ √[ +ƒ2)2 = fa2s tan2 i.

and moment of thrust = fa3s( —ƒ + √1 +ƒ ̃32)2° = fa3s tan2 i.

71. The angle whose tangent is ƒ + √1 + ƒ2 is just half of that whose

For since tan 2i =

=

1 is

is the angle of the slope which the earth would naturally assume

if unsustained by any wall.

For if i be the inclination of a plane to the vertical, and g the accelerating force of gravity; then the force g resolved into two, parallel and perpendicu lar to the plane, gives g cos i and g sin i; hence the friction =ƒ g sin i, and being counteracted by the force g cos i, we must have

g cos i = fg sin i, or tan i = or f = cot i.

Hence if BK be the natural slope of loose earth, and BH bisect the angle KBC; then the prismatic mass CBH will exert the greatest force against the vertical wall BC.

=

72. In loose earth the natural slope is about 60° from the vertical, and in tenacious earth this angle is about 54°; hence in the former case i= 30; tan i = tan 30° = √}=}√3, and in the latter i=27°, tan i = tan 27°, nearly. Therefore, for loose earth, the horizontal thrust = - as, and its momenta3s, and for tenacious earth, the horizontal thrust is 4a2s, and its momenta3s. Now put AB the breadth of the wallx, BC= a, and the specific gravity=S; then the moment of the resistance of the wall is = ax2S, which, in the case of equilibrium, must be equal to the moment of the horizontal thrust; hence, for tenacious earth we have

Ex. 2. Let the wall be triangular, as in the annexed figure, and let z = its breadth; then the moment of the resistance will be = 3x x + axS = {ax2S; nence we must have

73. To determine the thickness of a pier necessary to support a given arch

LA

KL, LA, KA. So that, if A denote the weight or area of the arch; then KL A

will be its force at A in the direction LA; and

LA
KL

.

the lever GA to overset the pier, or to turn it about the point F.

Again, the weight or area of the pier, is as EF. FG; and therefore EF. FG. FG, or EF. FG', is its effect on the lever FG, to prevent the pier from being overset; supposing the length of the pier, from point to point, to be no more than the thickness of the arch.

But that the pier and arch be in equilibrio, these two effects must be equal.

LA
KL

Therefore we have EF. FG2 = GA. A, and consequently the thickness

2GA. AL

of the pier is FG = ✓ EF. KL X A.

=

=

Example 1. Suppose the arc ABN to be a semicircle; and that DC or AO 45, BC 6, and GA 18 feet. Then KL will be found 40, AL 15 nearly, and EF = 69; also, the area ABCD or A= 704. Therefore FG = 2GA. AL

A is

36.15
69.40

70411 nearly, which is the thickness

of the pier.

Example 2. Suppose, in the segment ABN, AN = 100, OB = 41, BC = 6, 58, KL = 35, AL = 15 nearly, and ABCD or GA.AL. A = /20.15

and AG 10. Then EF

A = 842. Therefore FG =

nearly, the thickness of the pier in

EF.KL

this case.

58.35

842 = 11

GGG

DYNAMICS.

DEFINITIONS AND PRINCIPles.

1. A body is said to be in motion when it is continually changing its position in space.

2. Motion is said to be uniform when the spaces described in equal successive intervals of time are equal, and variable when these spaces are unequal.

3. The velocity of a body is the space it would describe in a unit of time, were the motion to become uniform at the commencement of that unit.

4. Motion is said to be accelerated when the velocity continually increases, and retarded when it continually decreases; and an accelerating or retarding force is said to be uniform or variable, according as the increments or decrements of velocity in equal times are equal or unequal.

5. The momentum or quantity of motion of a body is the sum of the motions of all its particles; and, as the motion of a particle is measured by its velocity, and the number of particles in a body constitutes its mass; hence the momentum will be equal to the product of the mass and velocity, when all the particles move, with the same velocity.

6. Inertia is the opposition offered by a body to a change of state, either of rest or of motion, by the action of a force impressed upon it.

7. If a system of particles, m, m1, m2, ... revolve round an axis, and r, r', r'"',... be their respective distances from that axis; then mr2 + m ̧r22 + m2r? + (mr2) is called the moment of inertia of the system.

... or

ON THE COLLISION OF SPHERICAL BODIES.

PROP. I.

8. If a spherical body strike or act obliquely on a plane surface, the force or energy of the stroke or action, is as the sine of the angle of incidence.

Or, the force on the surface is to the same if it had acted perpendicularly, as the sine of incidence is to radius.

Let AB express the direction and the absolute quantity of the oblique force on the plane DE; or let a given body A, moving with a certain velocity, impinge on the plane at B; then its force will be to the action on the plane, as radius to the sine of the angle ABD, or as AB to BC, drawing BC perpendicular, and AC parallel to DE.

A

B