And if the area is symmetrical on each side of the axis of x, we need only one equation, viz. X = Syxdx III. For a surface of revolution round the axis of x. Here one equation only is necessary; and since du differential of the surface 2 yds, we have X = xyds IV. For a solid of revolution round the axis of x. Here dv differential of the volume= Ty2dx, and hence X = Syxdx We shall now apply these formulæ to a few examples. EXAMPLES. 68. Ex. 1. To find the centre of gravity of a circular arc. Here the curve is symmetrical on each side of the axis of x, and the equation is y2 = 2ax - x2; hence we have that is, the distance of the centre of gravity of a circular arc from the vertex is = a centre = ay, and therefore the distance of the centre of gravity from the Ex. 2. To find the centre of gravity of a cone. Let x1, x2 represent any two parts of the axis of the cone, measuring from the vertex, and y1, y2 the radii of the circular sections of the cone corresponding to the altitudes x1, x2; then, if x denote any variable part of the axis, and y the radius of the corresponding circular section, we have y2. .. y2 = x2. ¥¦2; ye føde x = y 3 24 3 = 4 x3 4 where x, y, are constants; hence x2dx x= distance from vertex. But integrating between the limits x, and x2, we have 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 + 3r2 of the centre of gravity= and when = 0, we have 4 R2 + Rr + go?' the distance of the centre of gravity of a cone from its base = = onefourth of the altitude as found above. Ex. 3. Four bodies, whose weights are w1, w2, wз, w, pounds, are placed at the successive angles of a square whose side is 2a inches; required the posi tion 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 w1 + W2 — W3 — W4 X = (w2 + w3 — w1 — w1) w1 + w2+w3 + W4 and 2a= 12 inches; a, and y = w1 + w2+ W3 + W4 Thus if w=3, w2=4, w3=5, w1 = 6; 4 a. .. X=0 and y = — .6=- -1}= OG = distance of 18 centre of gravity below O on the axis of y. y WI PROBLEMS FOR EXERCISE. 69. Ex. 1. Find the centres of gravity of (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 H force perpendicular to the plane = W sin i + H cos i. W 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 i H sin if W sin i +fH cos i .. H = cos i -f sin i If then the weight W were sustained the weight W against the wall would be. 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 fsMxdx=4a2s M whole horizontal thrust of triangle BCH. Again, the length of the lever Bb = a — x, and the moment* of the thrust of the element bb'hh' = sMx (a — x) dx = as Mxdx sMx dx; hence * If lines be expressed numerically, the product of a force acting on a lever, and the perpendicular from the axis of motion on its direction is called the moment of that force. fas Mxdx-f°sMx2dx = 4a3sM — a3s M = ¿a3sM = moment of the whol horizonal thrust. The expression 1 between these limits there is a value which gives both the horizontal thrust and its moment a maximum. Let then ... Hence M = tan i 1 + f cot i .. horizontal thrust = - ftan i) sec2 i = coseci tani = ( − ƒ + √T + ƒ2)2 fa2s(-ƒ+ √] +ƒ ̃2)2 = 4a2s tan2 i. and moment of thrust = fa3s( —ƒ + √1 +ƒ ̃31)2 = 4a3s tan2 i. 71. The angle whose tangent is - ƒ + √1 + ƒ2 is just half of that whose and tan -1 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 perpendicular to the plane, gives g cos i and g sin i; hence the friction being counteracted by the force g cos i, we must have g cos i = fg sin i, or tan i = شر or f= cot i. f g sin i, and 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°: = √4, nearly=4. Therefore, for loose earth, the horizontal thrust = ja2s, and its moment = Ta's, and for tenacious earth, the horizontal thrust is 4a3s, and its moment= a's. Now put AB the breadth of the wall = x, 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 x = its breadth; then the moment of the resistance will be ax$ = ax2S; 73. To determine the thickness of a pier necessary to support a given arch KL, LA, KA. So that, if A denote the weight or area of the arch; then will be its force at A in the direction LA; and LA 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. 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 = 7043. Therefore FG = 2GA. AL EF. KL of the pier. 36.15 A is 69.40 704 114 nearly, which is the thickness Example 2. Suppose, in the segment ABN, AN = 100, OB = 41, BC = 6, and AG 10. Then EF 58, KL = 35, AL = 15 nearly, and ABCD or /2GA. AL A = nearly, the thickness of the pier in this case. /20.15 842 = 11 58.35 GGG |