But the triangle OA'B' is similar to the triangle OA,B1, a is the altitude of OA'B', and if y is the altitude of OA,B1, we have, since AB = A'B' To find the distance of the mass-centre of the area from XX we have to perform the summation Σmy, i.e. multiply each mass by its distance from XX and add; the sum is equal to the whole mass of the figure multiplied by the distance of the mass-centre from XX. Calling the whole mass M and y the distance of the mass-centre, we have Emy= Eam = Mỹ. or But in Zam1, a is the same for each term of the sum Now my is the mass of one strip of the inner figure and Em, is the mass of the whole inner figure, calling this M1, we have If A is the area of the given figure, and A, the area of the constructed figure-called the Equivalent Figure y does not determine the mass-centre but only a line on which it must lie, viz. a line parallel to XX and It is therefore necessary to find another line on which the mass-centre must lie. If the area has an axis of symmetry, the mass-centre lies on that, if not, then the above process must be repeated with another base line XX inclined at some convenient angle to the first. We may verify the above by considering the case of a parallelogram. Take for XX, and X'X' two of the parallel sides, and the pole O at one corner, and let the parallels from the ends of the strips AB be the other two sides. Then obviously all the strips will give the same two points A'B' on X'X', and the equivalent figure will be the triangle OA'B', which is half the parallelogram. Hence the mass-centre lies on a line parallel to XX halfway up OA'. 92. We may also find the mass-centre of any irregular area by aid of the Link Polygon. For divide up the area into a number of equal width strips, bisect each of these strips by a line parallel to the lengths of the strips, then we may (as an approximation) suppose the mass of each strip concentrated at the mid-point of each of these lines, and find the mass-centre of the whole by the link polygon method. Another approximate method (a mixture of drawing and calculation) is to divide up the area as in the previous method and then use the formula Mx=Σmx. In this formula m is the length of one of the strips (i.e. the length of its mid-line), a the distance of the midline from some fixed parallel line and M the sum of the lengths of the strips. The general method for the determination of masscentres of areas by the Link Polygon is as follows:— Divide the area into such parts that the area and mass-centre of each part can easily be found, either graphically or by calculation. Give the M.C. of each part a mass proportional to the area and determine by a Link Polygon the M.C. of this system of mass-points. It will be the M.C. of the given area. In thus dividing an area it is often convenient to add one or more pieces to the given area, which thus has the sum of positive and negative parts (§ 89). This does not alter the Link Polygon construction provided that in the Vector Polygon negative masses are set off in a negative sense. 93. Mass-centres of Volumes. If we suppose a distribution of points with equal mass throughout a volume, such that the mass in any part of the volume, however small the part, is proportional to that volume, then we have a uniform distribution of mass throughout the volume. The mass-centres of such volumes can often be found by simple methods. If the volume has a plane of symmetry, then the mass-centre must lie in that plane. By § 77, if z denote the distance of any point from a plane (measured parallel to some fixed direction, which may be to the plane), then the distance of the mass-centre from the plane is given by z, where = Σmz If the plane is a plane of symmetry, then to every mass-point at a distance z corresponds an equal mass-point at the distance - z, hence Σmz=0, and the mass-centre lies in the plane of symmetry. Thus the mass-centre of this book when shut lies somewhere in its middle page or leaf. If the volume has two such planes, the mass-centre must be in their line of intersection, the axis of symmetry. If there are three such planes, the mass-centre is at their point of intersection. Thus the mass-centre of a circular cylinder with parallel ends, is half-way up the axis of the cylinder. That of a parallelopiped is at the intersection of planes half-way between opposite faces, and therefore at the point of intersection of the diagonals. That of a sphere is at its centre. 94. Prism of any cross-section, with parallel ends. If we divide the prism up into a great number of equally thin slices, parallel to Dbase, then the whole may be looked on as a number of evenly loaded equal areas. Since the areas are all equal, the masscentres of all the areas will lie in the line joining the mass-centres of the base and top. Hence we may replace the prism by the evenly loaded line M1M, whose mass-centre is at its mid-point M. When the number of sides increases indefinitely while their size diminishes indefinitely, then the base polygon ABCDEF becomes a closed curve and the prism becomes a cylinder. Hence the mass-centre of any cylinder with parallel ends lies half-way up the line joining the mass-centres of top and base. 95. Mass-centre of a tetrahedron. As in the last article, suppose the volume divided up into a great number of equally thin slices parallel to the base, then we have a number of evenly loaded areas and the mass of each may be supposed concentrated at its mass-centre. The areas being triangles, the mass-centre of each is at the intersection of the medians, and is the same as for three equal particles placed at the vertices of the triangle. If we consider the base, the mass-centre M, is given by 1 a+B+Y 3 where a, ẞ, y are the three edges OA, OB, and OC. If we consider any parallel section DEF, where OD=aa, OE=aß, OF=ay, then its mass-centre is given by Hence the mass-centre of all the sections lies in the straight line joining the mass-centre of the base to the vertex. |