104. Projection of vectors in space on a plane. If the vectors are not coplanar, we may project them by parallel rays to a plane. A α B α' Fig. 71. It will be seen again that the sum of the projections of a number of vectors is the projection of the sum of the vectors and hence if the vector polygon be closed the sum of the projections = 0. 105. Theorem :-If a number of points with masses and their mass-centre be projected on a plane, then the mass-centre of the projections is the projection of the mass centre. The proof proceeds exactly as in § 102, the only difference being that a, B... are no longer in a plane. 106. The projection of a figure in space on a horizontal plane is called the Plan of the figure, and its projection on a vertical plane is called the Elevation. The theorem just proved can be then stated in slightly different terms, viz.: The mass-centre of the plan of a figure is the plan of the mass-centre of the figure. A similar theorem holds for the Elevation. Hence to find the mass-centre of a figure in space, we find the masscentres of the Plan and Elevation. This is most conveniently done by means of the Link and Vector polygons. Draw then a vector polygon for the masses, and two link polygons (at right angles) in the plan to determine M1, the mass-centre of the plan. Then draw one Link polygon in the elevation, and we get M the masscentre of the figure as shewn. (Only one Vector Polygon is necessary; to draw the link polygons whose links are perpendicular to the corresponding lines from the pole of the Vector Polygon, we have only to use the two perpendicular edges of a set-square.) EXERCISES VII. (1) Find by calculation and by the Link Polygon the masscentre of the lines shewn. (2) Find graphically and by calculation the position of the masscentre of the areas in the following figures: (3) A wire of length is bent into the form of a sector of a circle, the arc being twice the radius, calculate the position of the mass-centre, and determine it graphically. (4) A circular disc 1 ft. in diameter has a circular hole of 3" diameter cut out of it, the centre of the hole is 2 inches from the centre of the disc, find the mass-centre of the remaining part. (5) An area is formed of a right-angled triangle and the squares on its sides, find the distance of the mass-centre from the hypotenuse. (6) As in (5), the squares being replaced by semicircles. (7) Find the mass-centre of a semicircle and of a triangle by means of the equivalent figure and compare the results with those obtained by calculation. (8) Draw any irregular area, find the position of its mass-centre by the equivalent figure method and also by dividing it into strips, using the formula Emx=x2m. (9) The triangular faces of a wedge being equal and the other faces rectangles, shew that the mass-centre of the wedge is that of 6 equal mass-points placed at the corners. (10) A solid cylinder of height h and diameter d is surmounted by a solid hemisphere, find the position of the mass-centre (the volume of a sphere is r3, r being the radius). (11) From a right circular cone is cut off a part by a plane parallel to the base, find the mass-centre of the remainder (the volume of a cone is base x height). (12) The areas of the cross-sections of a tree trunk 40 ft. long are given by table as under: Distance from one end in ft. 04 285 12 16 3 20 4 25 2 | 301 | 34 3 | 381 | 40 61 48 40 3.2 2.8 2 1.5 end. Find approximately the distance of the mass-centre from one (13) Find the mass-centres of the surfaces of (iii) a rectangular box (without top), (iv) a cylindrical box with hemispherical lid. (14) Write out the proof for the theorem in § 105. (15) By aid of (14) establish the three scalar equations for the determination of the position of the mass-centre of a system of masses, viz. : (16) The masses and the rectangular coordinates of a number of mass-points being given by the accompanying table, determine the position of the mass-centre (a) by Vector and Link Polygons, (b) by calculation. |