Cor. 2. If the case should arise in which where one at least of A and B has some determinate finite value, the system reduces to a couple; and in this case there is no centre of parallel forces in finite space. If the forces are the weights of parts of a body they act in the same direction, and therefore their sum can never be zero, so that every body and system of bodies must have one, and only one centre of gravity, which can be determined by the above equations. N.B.-For examples on this Proposition see Art. 69. CHAPTER V. OF THE CENTRE OF GRAVITY. 63. Definition of the Centre of Gravity. It has been already remarked that the weight of a body is an instance of a distributed force, and that it can be treated as a single force by supposing it to be collected at a certain point, called its centre of gravity. The centre of gravity of any system of heavy points is the centre of the system of parallel forces composed of the weights of those points. If the points form a solid body, it is plain that, if the centre of gravity be supported, the body will rest in any position under the action of gravity only, since the resultant of the applied forces will in all cases pass through the fixed point. It is also plain that no point but the centre of gravity has this property. That, as a matter of fact, every body has a centre of gravity, is shown in the corollary to Proposition 16. In determining the centre of gravity of any figure, it is assumed that a heavy line is made up of heavy points, a heavy plane of heavy parallel lines, and a solid of heavy parallel planes. It is also assumed that every figure is of uniform density, unless the contrary is specified. Ex. 271. Determine the centre of gravity of a uniform straight line AB. The line AB may be conceived to be made up of a number of equally heavy points distributed uniformly along it (like beads on a wire); now if we take the two extreme points, the resultant of their weights will pass through the middle point of AB, and in like manner that of each successive pair; consequently the weight of the whole will act through the middle point of AB, which is therefore the centre of gravity of the whole, or of the heavy line AB. 64. Method of determining the Centre of Gravity of ... FIG. 69. B D E F This method FIG. 70. A a Plane Area. Let ABCD be the plane area; we may conceive it to be made up of a set of parallel heavy lines, such as BD, EF drawn in any direction. If we can find a set of parallel lines all bisected by a single line AC, the centre of gravity of each line must be in AC, and therefore that of the whole figure must be in AC. If, moreover, we can determine a second line bisecting another set of parallel lines, we know that the centre of gravity must also be in this second line, and must therefore be at its point of intersection with Ac. enables us to determine the centre of gravity of many simple figures: it also suggests a practical means of determining the centre of gravity of any plane area whatever. Suppose the figure to be cut out carefully to the required shape in cardboard or tin; suppose it to be suspended by a fine thread from any point B; now the forces in equilibrium are the tension of the string and the weight of the body; they must therefore act along the same line, so that the required centre of gravity must be in the prolongation BC Of AB; this prolongation can easily be marked by suspending a plumb-line from A. Again, suspend the body by a fine thread DE fastened to any other point E, and draw the prolongation of this line, viz. EF; the centre B C B F D E C G of gravity must be in EF, and therefore at G, the point of intersection of EF and вс. Ex. 272. Show that the centre of gravity of the area of a circle is at its centre. Since any diameter bisects all lines in the circle drawn perpendicularly to it, the centre of gravity must be in any diameter, and therefore at the centre of the circle. Ex. 273. Show that the centre of gravity of an ellipse must be at its centre. Ex. 274. Determine the centre of gravity of a triangle. B D C Let ABC be any triangle, bisect BC in D and join A D; draw any line KL parallel to BC cutting AD in H; then by similar triangles we have KH: HA::BD: DA .. (ex æquali) кKH: HL::BD:DC But BD is equal to DC, therefore KH is equal to HL, or KL is bisected by AD; and the same being true of any line drawn parallel to BC, the centre of gravity of the triangle must be in AD. Again, if Ac is bisected in E and BE is drawn, the centre of gravity will be in BE, and therefore must be at G, the point of intersection of AD and BE. It can be easily proved that GD=AD. For join ED, then because AE=EC, and BD=DC we have AE:EC::BD: DC, and therefore ED is parallel to AB; hence the triangle DEG is similar to ABG and EDC to ABC; therefore and therefore (ex æquali) But DG: DE::GA: AB DE:DC:AB: BC DG:DC::GA:BC DC=BC..DG=GA=DA. Ex. 275. Show that the centre of gravity of a parallelogram is at the intersection of the diagonals. 65. Centre of Gravity of Solids. - The above method can easily be extended to the case of solids; we may suppose them to be made up of heavy parallel planes: if we can show that the centres of gravity of these all lie along a line, we know that the centre of gravity of the solid must be in that line, and if two such lines can be found, the centre of gravity of the solid must be at their point of intersection. Ex. 276. Show that the centre of gravity of a sphere is at its centre. Ex. 277. Show that the centre of gravity of a cylinder is at the middle point of its axis. [It may be regarded as evident that the same rule will hold good of any prism.] |