944 



THE IRKIGATION AGE. 



y instead of T where T' = T sin g and T" T cos g. So 

 instead of the 4 original forces we have here 8 forces acting 

 in rectangular axes; the four forces acting on the X X axis 

 can be combined into one, namely, P' + T' Q' S', which 

 will give one force, this we will call R' acting upward; like- 

 wise the 4 forces on the Y Y axis can be reduced to one = 

 p" -j- Q" _ 5" _ T", which we will call R' r ; then the final 

 combination is shown in Fig. 58, the two 

 forces R f and R" acting on point A under 

 right angle ; the resultant R is then the 

 diagonal of the rectangle and R 

 \/ Ri 2 -f- /?n 2 , also the directions of the 

 Resultant is readily found, for, let angle 



Fig. 58 



= n, then tg n = 



3. Parallel Forces Acting in the Same Direction. 

 If two parallel forces P and Q tending in the same di- 

 rection act upon a stiff bar A B, Fig. 59, then a resultant ./? 

 may be set for P and Q, which is equal to the sum P + Q, 

 also parallel to them and divides 

 the distance A B inversely as P and 

 Q, that means if R acts at the point 

 C then A C : B C = Q : P. This 

 is a very important principle; it is 

 seen if the point C be supported so 

 it could not move then the two 

 forces P and Q will be in equili- 

 brium ; the product of the force P 

 into its distance A C is called the 

 moment of the force P, likewise the 

 product of Q into the distance C B 



Fig. 69 



A C E CiMH F D S 



Fig. 60 



is the moment of the force Q with reference to point C. 

 This principle can be extended to any number of parallel 

 forces. The force of gravity acting on a body can be con- 

 sidered as a large number of 

 parallel and equal forces, each 

 acting upon a particle. Thus if 

 in Fig. 60 A B is a line com- 

 posed of a number of atoms A, 

 B, C, D, E, F. G, H, then on each 

 of them acts the force g down- 

 ward, so for Ag and Eg we can 

 set Mg = Ag + Eg, likewise 

 for Cg and Dg we can set gi gi 

 acting at M, also for Eg and Fg 

 we can set gi g 3 acting at M and 

 for Gg and Hg we can set g* g, 

 so that instead of all the parallel 

 forces we have but one force act- 

 ing in the middle point M. This 

 point is called the center of gravity; so if A B is a straight 

 line or bar of uniform thickness its center of gravity is in 

 its middle M, and if this point is supported then the whole 

 bar will be in equilibrio, i. e., will not move under the influ- 

 ence of gravity. It really acts as though the whole mass of 

 the bar was concentrated in the center of gravity. 



4. Center of Gravity for Various Shapes and Figures. 

 The center of gravity of a line is in its middle. 

 The center of gravity of any symmetrical figure is in 

 its geometrical center. Thus the centers of gravity of a 

 parallelogram, rectangle rhombus and square are in the inter- 

 section point of the diagonals. 



The center of gravity of a circle and the circle circum- 

 ference is in the center of the circle. 



The center of gravity of an ellipse lies midway between 

 the two foci. 



The center of gravity of a cylinder or right prism lies 

 in the middle of its axis. 



The center of gravity of a homo- 

 genous sphere lies in its center. 



The center of gravity of a tri- 

 angle lies in the intersection of its 

 medians, i. e., lines drawn from the 

 corners to the middle points of the 

 opposite sides : thus if in Fig. 61 

 the lines AM, BO and CN are medi- 

 ans intersecting in point P, then P 

 is the center of gravity of the tri- 

 angle and is located l /s of the length 

 Fig. ei of the median from the base. 



Fig. 63 



Every body or shape has its center of gravity, and when 

 it cannot be readily calculated it can be found experimentally 

 as follows: Suspend the figures (see Fig. 62) by one 

 corner A by sticking a pin through and having a thread A P 

 with a little weight P attached, let 

 the figure freely swing about the 

 point A, and when at rest draw a 

 light pencil line along A P over the 

 figure, then the center of gravity 

 must lie somewhere in this line A P. 

 Next hang up the figure from 

 some other corner B and obtain an- 

 other fine B P, which intersects the 

 line A P in O ; then O is the required 

 center of gravity. 



The center of gravity of an arc 

 of a circle is found as follows : let 

 the length of -arc A B = a and the radius C M = r and 

 the length of chord A B = c, and let g be the distance 



cr 



of the center of gravity, G from C then g = or in 



a 



words: multiply the radius with the chord and divide by 

 the length of the arc. 



The center of gravity of every symmetrical body lies 

 in its axis; thus the center of gravity of a sphere lies in 

 its center, of a cylinder or regular prism in the middle of 

 its axis, of a cone or regular pyramid upon the height }4 

 of the distance from the base. 



The center of gravity of the following circular figures 

 lies on the axis of symmetry in each case; in the sector 2 cr 

 H- 3 / from the center, of a semicircle .424 r from the center, 

 of a quadrant .6 r and a segment f -=- 12 a from the center, 

 r being length of a radius, c length of chord, /, center of arc 

 and a = area of segment. 



5. The Laws of Gravity. 



The force of gravity when acting upon a free falling 

 body produces an acceleration of 32.16 ft. per second, but 

 the distance that a body falls the first second is 16.08 ft. 

 This principle is illustrated in Fig. 63 in which the horizontal 



line indicates the time in 

 fi 



seconds, and the ordi- 



nates V 



i, 



and 



the velocities reached at 

 the ends of the respect- 

 ive seconds, thus Vi = 

 32.16 Vt = 64.32, v, = 

 96.48, v t = 128.64, etc., 

 the areas of the triangles 

 represent the spaces 

 fallen through ; thus 

 triangle OlA has an area 



1 X 32.16 

 of = 16.08, 



hence a body falls 16.08 

 ft. during the first second; the space fallen through in 2 



2 X 64.32 

 seconds is represented by triangle 02B =: -- = 64.32 



2 

 ft. ; the space fallen through in 3 seconds is represented by 



3 X 96.48 

 the area of triangle 03C - = 144.72 ft. ; the space 



2 

 fallen through in 4 seconds is represented by the area of 



4 X 128.64 

 triangle 04D = - = 257.28 ft. From these con- 



2 



siderations the general law may be derived ; if g is set for 

 32.16, then Vi = 2g, v* = 3g, v* 4g., etc., thus the velocity 

 at the end of t seconds will be tg ; now to determine the total 

 space fallen through : the base of the triangle represents the 

 time t and the height represents the velocity tg hence the 



t'g 



area = - , so if h = space fallen through in * seconds h 

 2 



fy 



=-- - and if V = velocity at the end of the time t then V 

 2 



