If a body be supported at its centre of gravity, and be turned about that point, it will remain in equilibrium in all positions. The resultant of the parallel forces of gravity acting upon a body is obviously equal to the weight of the body, and if an equal force be applied, acting in a line passing through the centre of gravity of the body, the body will be in equilibrium. Examples of Centres of Gravity. - Centre of Gravity of Lines. Straight Lines. By a line is here meant a material line whose transverse section is very small, such as a very fine wire. The centre of gravity of a uniform straight line is at its middle point. This proposition is too evident to require demonstration. The centre of gravity of the perimeter of a triangle is at the centre of the circle inscribed in the lines joining the centres of the sides of the given triangle. Thus, let ABC (Fig. 10) be the given triangle. To find the centre of gravity of its perimeter, find the middle points, D, E, and F, and connect them by straight lines. The centre of the circle inscribed in the triangle formed by these lines will B be the centre of gravity sought. Symmetrical Lines. The centre of Fig.10 gravity of lines which are symmetrical with reference to a point will be at that point. Thus the centre of gravity of the circumference of a circle or an ellipse is at the geometrical centre of those figures. The centre of gravity of the perimeter of an equilateral triangle, or of a regular polygon, is at the centre of the inscribed circle. The centre of gravity of the perimeter of a square, rectangle, or parallelogram, is at the intersection of the diagonals of those figures. Centre of Gravity of Surfaces. Definition. A surface here means a very thin plate or shell. Symmetrical Surfaces. If a surface can be divided into two symmetrical halves by a line, the centre of gravity will be on that line: if it can be divided by two lines, the centre of gravity will be at their intersection. The centre of gravity of the surface of a circle or an ellipse is at the geometrical centre of the figure; of an equilateral triangle or a regular polygon, it is at the centre of the inscribed circle; of a parallelogram, at the intersection of the diagonals; of the surface of a sphere, or an ellipsoid of revolution, at the geometrical centre of the body; of the convex surface of a right cylinder at the middle point of the axis of the cylinder. Irregular Figures. — Any figure may be divided into rectangles and triangles, and, the centre of gravity of each being found, the centre of gravity of the whole may be determined by treating the centres of gravity of the separate parts as particles whose weights are proportional to the areas of the parts they represent. - Triangle. To find the centre of gravity of a triangle, draw a line from each of two angles to the middle of the side opposite: the intersection of the two lines will give the centre of gravity. Quadrilateral. To find the centre of gravity of any quadrilateral, draw diagonals, and, from the end of each farthest from their intersection, lay off, toward the intersection, its shorter segment: the two points thus formed with the point of intersection will form a triangle whose centre of gravity is that of the quadrilateral. Fig. II D B Thus, let Fig. 11 be a quadrilateral whose centre of gravity is sought. Draw the diagonals AD and BC, and from A lay off AF= ED, and from B lay off BH = EC. From E draw a line to the middle of FH, and from Fa line to the middle of EH. The point of intersection of these two lines is the centre of gravity of the quadrilateral. This is a method commonly used for finding the centre of gravity of the voussoirs of an arch. Table of Centres of Gravity. Let a denote a line drawn from the vertex of a figure to the middle point of the base, and D the distance from the vertex to the cen In a frustum of a cone or pyramid, let h = height of complete cone or pyramid, h' = height of frustum, and the vertex be at apex 3(h-h's) of complete cone or pyramid; then D = 4(h3-h'a) The common centre of gravity of two figures or bodies external to each other is found by the following rule: Multiply the smaller area or weight by the distance between centres of gravity, and divide the product by the sum of the areas or weights: the quotient will be the distance of the common centre of gravity from the centre of gravity of the larger area. EXAMPLE. rule and tables, let us find the common centre of gravity of the semicircle and triangle shown in Fig. 12. -As an example under the above We must first find the centres of gravity of the two parts. Fig.12 The centre of gravity of the semicircle is 0.425 R from A, or 2.975. The centre of gravity of the triangle is of 8", or 2.666" from A; and hence the distance between the centre of gravity is 2.975" + 2.666", or 5.641". 3 × 49 The area of the semicircle is approximately or 77 square inches. The area of the triangle is 7 × 8, or 56 square inches. The sum of the areas is 133 square inches. Then, by the above rule, the distance of the common centre of gravity from the centre 56 × 5.641 133 of gravity of the semicircle is or = = 2.37", 2.9752.37 = 0.605 inches from A. Centre of Gravity of Heavy Particles. - Centre of Gravity of Two Particles. Let P be the weight of a particle at A (Fig. 13), and W that at C. PW B P Fig. 13 W The centre of gravity will be at some point, B, on the line joining A and C. (A. The point B must be so situated, that if the two particles were held together by a stiff wire, and were supported at B by a force equal to the sum of P and W, the two particles would be in equilibrium. The problem then comes under the principle of the lever, and hence we must have the proportion, or P+W:P :: AC: BC, PX AC BC= P+W If WP, then BC= AB, or the centre of gravity will be half way between the two particles. This problem is of great importance, for it presents itself in many practical examples. Centre of Gravity of Several Heavy Particles. · Let W1, W2, W 3, W4 and W. (Fig. 14) be the weights of the particles. W1 A B W3 W2 Fig.14 W1 Join W, and W2 by a straight line, and find their centre of gravity A, as in the preceding Ws problem. Join A with W3, and find the cen tre of gravity B, which will be the centre of gravity of the three weights W1, W2, and W3. Proceed in the same way with each weight, and the last centre of gravity found will be the centre of gravity of all the particles. In both of these cases the lines joining the particles are supposed to be horizontal lines, or else the horizontal projection of the real straight line which would join the points. CHAPTER V. RETAINING WALLS. A Retaining Wall is a wall for sustaining a pressure of earth, sand, or other filling or backing deposited behind it after it is built, in distinction to a brest or face wall, which is a similar structure for preventing the fall of earth which is in its undisturbed natural position, but in which a vertical or inclined face has been excavated. Fig. 1 gives an illustration of the two kinds of wall. Retaining Walls. — A great deal has been written upon the theory of retaining walls, and many theories have been given for computing the thrust which a bank of earth exerts against a retaining wall, and for determining the form of wall which affords the greatest resistance with the least amount of material. There are so many conditions, however, upon which the thrust exerted by the backing depends, such as the cohesion of the earth, the dryness of the material, the mode of backing up the wall, etc., that in practice it is impossible to determine the exact thrust which will be exerted against a wall of a given height. It is therefore necessary, in designing retaining walls, to be guided by experience rather than by theory. As the theory of retaining walls is so vague and unsatisfactory, we shall not offer any in this article, but rather give such rules and cautions as have been established by practice and experience. In designing a retaining wall there are two things to be considered, the backing and the wall. The tendency of the backing to slip is very much less when it is |