Cor. It is evident that if the prism or cylinder is cut by another plane inclined at any angle to the base, the volume contained between the cutting planes equals the area of the perpendicular section multiplied into the part contained between the planes of a line drawn through the centre of gravity of the perpendicular section at right angles to its plane.

Ex. 319. Show that Propositions 17 and 18 are true in the case when the curve is a closed curve and revolves round an axis wholly without it. Ex. 320.-In Proposition 19 show that Q is the centre of gravity

of DCF.

Ex. 321. An equilateral triangle revolves round its base, whose length is a; find the area and volume of the figure described.

Ex. 322. An equilateral triangle revolves round an axis parallel to the base, the vertex of the triangle being between the axis and the base; the base is 6 in. long and the distance from the vertex to the axis is 9 in.; determine the volume of the ring described. Ans. 1220-7 cub. in.

Ex. 323. Determine the volume of a ring formed like that in the last Example having given that each side of the triangle is 6 in. and the external diameter of the ring 3 ft. Ans. 1593-4 cub. in.

Ex. 324. The section of a ring is a trapezoid, its height is 3 in. and its parallel sides are respectively 7 in. and 3 in. long, they are parallel to the axis, the shorter being the nearer to the axis and at a distance of 11 in.; find the volume of the ring. Ans. 1196.9 cub. in.

Ex. 325. In the last Example if the longer side of the trapezoid had been the nearer to the axis, the external diameter of the ring being the same in both cases, what would have been the volume?

Ans. 1159.2 cub. in.

Ex. 326. Determine the volume and surface of a ring with a circular section whose internal diameter is 12 in. and thickness 3 in.

Ans. (1) 333-1 cub. in. (2) 444.1 sq. in.

Ex. 327. Determine the volume and surface of a ring whose section is a regular hexagon, whose circumscribing circle has a radius a, and whose centre is at a distance b from the axis of revolution.

Ans. (1) 3πδα2 √3. (2) 12παδ. Ex. 328. Find the centre of gravity of the arc of a semicircle. Ans. Distance from centre ==

diam.

Ex. 329. Find the centre of gravity of the area of a semicircle.

Ex. 330. A cylindrical shaft is cut off obliquely at an angle of 45° to the axis, its radius is 6 in. and its extreme height is 2 ft. 6 in.; find its solid contents. Ans. 1.5708 cub. ft.

Ex. 331. A cylindrical shaft is cut obliquely at an angle of 60° to the axis, the radius of the base is 10 in., the extreme height of the shaft 3 ft.; find its volume. Ans. 9497 cub. in.

Ex. 332. A right prism stands on a triangular base the angles of which are A, B, C, the angles of the other end being D, E, F; the sides AB, AC, are each 15 ft. long, Bc is 18 ft. long; the edges AD, BE, CF are each 30 ft. long; through the edge вс passes a plane making an angle of 60° with the base; determine the volumes of the parts into which the prism is divided. Also if the prism were cut by a plane parallel to the former and cutting an at a distance of 24 ft. above a, find the volumes of the two parts.

Ans. (1) 748-3 and 2491.7 cub. ft. (2) 1095.6 and 2144.4 cub. ft. Ex. 333. Show that if any triangular prism be cut by a plane so that the edges perpendicular to the base are respectively a, b, c, and the area of the base a, then the volume of the frustum will be a (a + b + c).

FIG. 85.

d

C

D

a

Ex. 334. Let abcd represent the plan and ABCD the section of a portion of a ditch; AD=20 ft.; depth of ditch 8 ft.; slope of AB is 2 in 1, and that of DC is 1 in 1; ab and cd are respectively 20 and 40 ft. long. Find the volume; and determine the error that would be committed 6 if we had found the volume by multiplying the area of the section by half the sum of ab and dc.

Ans. (1) 3264 cub. ft. (2) Error 96 cub. ft. [Compare Ex. 311.]

Ex. 335.- Let ABCD be the plan of square redoubt, each A side of which is 150 ft., the corners of the ditch are quadrants of circle whose centres are respectively A, B, C, D. So that the ditch has a uniform width of 24 ft., its depth is 9 ft., the inside slope is 3 in 1 and the outside 1 in 1. Find the volume of the ditch. Ans. 108,057 cub. ft.

C

B

Ex. 336. If the ditch in the last Example were surrounded with a glacis 3 ft. high whose outside slope is 1 in 10 and inside slope 1 in 1; find its volume. Ans. 40,897 cub. ft.

CHAPTER VI.

FRICTION OF

PLANE SURFACES INCLINED PLANE, WEDGE, SCREW.

SECTION I.

72. Reaction of Surfaces. It nearly always happens that amongst the forces which keep a body at rest is the reaction of one or more sur

FIG. 86.

weigh 1000 lbs.; that weight must be supported by the table, which must therefore exert upwards a force of 1000 lbs. in a direction opposite to the direction of the weight. If we consider the case particularly we shall see that this reaction is an instance of a distributed force, for the under surface of CD will be in contact with the table at many points, and at each point there will be a reaction; what are the magnitudes of the reactions respectively at the points we do not commonly know; they must, however, be such that their resultant shall act vertically upward through the centre of gravity of M and shall equal 1000 lbs. And, in general, if a body is at rest when pressed against a surface the various points of that surface must supply reactions whose resultant is equal and opposite to the resultant of the forces by which that body is urged

K

against the surface; this resultant reaction is called the reaction of the surface.

73. The Limiting angle of Resistance. The question now arises-Under what circumstances is the plane capable of supplying the reaction necessary to produce equilibrium? There will be equilibrium if the plane do not break, if the body do not turn over, and if the reaction keep the body from sliding; it is with the last condition we are here

P

A

K

C

FIG. 87.

M

L

E

H

P

w

R

D

B

concerned. Let us revert to the example discussed in the last article, and let us suppose a rope to be fastened to the point k by means of which the body is pulled horizontally by a force P; we know that if P have a certain magnitude it will just make the body slide, but if it be less than that certain magnitude the body will

continue at rest; suppose that a force of 190 lbs. will just not make the body slide; produce PK to meet the vertical through the centre of gravity in L, let LE represent P (190) and LF represent w (1000), complete the parallelogram and draw the diagonal LH, this must be the direction of the resultant R, and its direction makes with a perpendicular to AB an angle of 10° 45'; now, if the force p is less than 190 lbs. the direction of the resultant will fall within the angle RLW; but if P is greater than 190 lbs. the direction of the resultant will fall without the angle RLW; in the former case the surface A B can supply a reaction which prevents motion, in the latter it cannot; and thus in the case we have supposed the surface A B can supply a reaction in any required direction which makes an angle less than 10° 45′ with the normal, i. e. the perpendicular, to the surface; and when the body is in the state bordering on motion, the direction of the reaction will make an angle equal to 10° 45′ with the normal.

Now it appears from experiment that if the surface AB were of cast iron, and the mass M of wrought iron, a force of 190 lbs. would be required just not to produce motion in the case above discussed; and it also appears from experiment that within very considerable limits, the same proportions are preserved, irrespectively of the extent of the surface pressed and the amount of the force; so that we may state as a fact of experience, that when wrought iron rests on cast iron the former will exert a reaction in any direction required to produce equilibrium that does not make with the normal an angle greater than 10° 45′; and when motion is about to ensue, the direction of the reaction will make an angle with the normal equal to 10° 45'; this angle is therefore called the limiting angle of resistance in the case of cast iron upon wrought. It further appears from experiment, that in the case of any two surfaces whatever, there is a limiting angle of resistance proper to those surfaces, and depending on their physical character; for instance, in the case of wrought iron on oak, the angle is 31° 50′, and similarly in other cases. Values of this angle in several cases are given in Table XI.

Hence if a body is urged against a fixed surface by any force or forces, the direction of the reaction of that surface can never make with the normal an angle greater than a certain angle. That angle is called the limiting angle of resistance; its magnitude is fixed by the physical nature of the surfaces in contact.

If the resultant of the forces which urge the body against the fixed plane be found, the body will continue at rest, provided the direction of the resultant makes with the normal an angle less than the limiting angle of resist