Let a the length and P= the weight of the rod, and W the weight of the beetle. = Then the original distance of the centre of gravity of the rod and beetle from A is equal to aP P+W' and its distance from 4, when the beetle arrives at B, is equal to aP+aW hence the centre of gravity has moved through a space equal to Wa (4) A circular piece is cut out of a rectangular board ABCD, fig. (62), the two sides AB, AD, touching the circumference of the circle to find the position of the centre of gravity of the remaining portion. Produce AB, AD, indefinitely to x, y, Ax, Ay, being taken as axes of co-ordinates. Let AB=a, AD=b, r = the radius of the circle. Let x, y, be the co-ordinates of the centre of gravity of the remaining portion of the board. Then (5) A weight P is supported on a smooth inclined plane by a string, parallel to the plane, which passes over a fixed pully, and is attached to a weight Q; to prove that, when Q is moved vertically, the centre of gravity of P and Q will neither rise nor fall. Let x be the distance of P and y the distance of Q from the pully then the depth of the centre of gravity of P and Q below the pully is equal to Px sin a + Qy But, since P is at rest, we have, resolving the forces, which act upon it, parallel to the plane, Q=P sin a. Hence, c being the length of the string, the depth of the centre of gravity is equal to which is independent of the depth of Q below the pully. (6) The inscribed circle being cut out of a right-angled triangle, the sides of which are 3, 4, 5; to find the centre of gravity of the remainder. Let OA 3, OB 4, and therefore AB= 5. = = OBy, fig. (63), be taken as the co-ordinate axes. Let OAx, Since the centre of gravity of the circular area and the remainder of the triangle must coincide with the centre of gravity of the triangle, we have, x, y, being the co-ordinates of the required centre of gravity, OA × area of triangle = rad. of circle × its area and +xx area of remainder of triangle, OB area of triangle = rad. of circle x its area The value of a shews that the required centre of gravity lies in a line through the centre of the circle parallel to OB. (7) Three weights, placed at the corners of a triangle, are proportional to the opposite sides; to shew that their centre of gravity coincides with the centre of the circle inscribed in the triangle. Let a, b, c, be the sides of the triangle, and Xa, λb, λc, the weights at the opposite angles: let x, y, z, denote the distances of the centre of gravity of the weights from the sides a, b, c, respectively: then, p denoting the perpendicular upon the side a from the opposite angle, where A is the area of the triangle. Similarly These results shew that the distances of the centre of gravity from the three sides are all equal to the radius of the inscribed circle. The centre of gravity of the weights must therefore coincide with the centre of the circle. (8) To find the centre of gravity of the periphery of a triangle formed by a piece of uniform wire. Let ABC, fig. (64), be the triangle. Bisect the sides BC, CA, AB, in a, B, y, respectively, and join By, ya, aß. The centre of gravity of the triangle ABC will be the same as that of three weights in the proportion of BC, CA, AB, placed respectively at a, B, y: but BC, CA, AB, are proportional to By, ya, aß; hence the centre of gravity of ABC coincides with that of three weights, proportional to ẞy, ya, aß, placed at the angular points a, B, y, of the triangle aßy. Hence, by the preceding theorem, the centre of gravity of ABC coincides with the centre of the circle inscribed in the triangle aßy. 1 n (9) One corner of a triangle, equal to th part of its area, is cut off by a line parallel to its base: to find the centre of gravity of the remainder. Let B'C', fig. (65), be a line parallel to the base BC of the triangle ABC, the area AB'C' being one nth of that of ABC. Bisect BC in P and join AP, which will bisect B'C' in P'. Let u denote the area of ABC, and the distance of G, the centre of gravity of BCC'B', from A. Then But, the triangles AB'C', ABC, being similar, and therefore Hence 、 = 2 3 n + n + 1 (10) A piece of uniform wire is formed into a triangle; to find the distance of the centre of gravity of the periphery of the triangle from each of the sides; and to shew that, if x, y, z, be the three distances, and r the radius of the inscribed circle, then Let a, b, c, be the sides of the triangle. Then, p denoting the distance between the side a and the opposite angle, but hence similarly and Hence {(b + c) (c + a) (a + b) − bc (b + c) — ca (c + a) − ab (a+b)} (11) AOB, fig. (66), is a bent lever of uniform thickness; to find the distance of its centre of gravity from 0. = Let OA=2a, OB 2b, AOB= w. Let G be the centre ▲ draw GH parallel to yBO, and join OG: The weights of OA, OB, supposed to of gravity of the lever: let OH=x, GH=7. be collected at their centres of gravity, may be represented by λα, λό. Hence |