COROLLARY 3.—h hence all the properties that have been demonstrated of three forces acting upon a body supposed void of weight, will equally flew from a heavy body sup ported by two strings, by substituting the weight of the body for the middle force; and hence, it' the direction of any force supporting a heavy body be given, the other may easily be tinind.
PnoPosmos V1.—Giveu the position in which a body shoold be placed, and the position of a plane supporting the body at one end ; to find the position of another plane, to support it at another given point ; and to find the pressure on the planes, the weight of the body being given.
Through the centre of gravity of the body draw a vertical line, anl through the point on which the body rests on the given plane draw a line perpendicular to that plane, meeting the crtieal line; from their intersection draw a line to the other point to be supported ; from that point draw a plane at right angles to this line, which will be the direction of the plane required.
To find the intensity of the forces, take any distance on line to represent the weight of the beam from the intersection ; on that line. as a diagonal, complete a parallelogram. whose sides are in the directions of the lines perpendicular to the supporting planes; and the side of the pandlebigram perpendicular to either plane will represent the force on that plane.
'through the centre of gravity, 0, of the body, draw the vertical line a F; draw C F perpendicular to c B; join F B; and draw B I perpendicular to F B; and B I is the direction of the plane required. On the vertical line. F, make F M to represent the weight of the body, and complete the parallelo gram L N F.; then F N represents the tiirce on the wall-head in the likeetilen C r, and F L the foree acting, perpendicular to the plane in B F. But it' the vertical and hori
zontal thrusts on the wall at c lie required, draw s r perpen dicular to F a, meeting it in r; then the force F N is resolved into two forces, F. r and rx;rx representing the horizontal part of the force, viz., that which pushes the wall in a direc tion parallel to the horizon, and F r the other part which tends to press it downwards in a direction perpendicular to the horizon.
Examide '2.—Figure S.—Let the body, A B C D. be supported by a wall at its lower end, u, which coincides with the surtitee of the body; and let G he the centre of gravity ; it is required to cut a notch out of the body at the upper end. c, so that it may rest upon the top of a all which is made to fit the noteh, and to lied the pressure on the walls.
Draw the vertical ; from D draw u E perpendicular to u c E C. and draw c at right. angles to it ; then the notch. it I' F. being cut. the body. A n c D, will be at rest.
lied the pno.ssuer on the walls, complete the parallelo gram E 1 K r,, hat ing a given angle, D F c. and its diagonal on the given line E G: now if E represent the weight of the body, t E will represent the pressure in the n the wall at O, and L E the pressure in the direction c E.
The horizontal and perpendicular pressures upon each wall !nay lie tiamd, as in the first example, by resolving each of the forces, t E and L E, into two, one of which is perpen dicular to the horizon, and the other parallel to it.
Sellout :,t,—It must be observed in this example, that the notch cut mit at c will remove the centre of gravity nearer the Inver end, il, and conAiquently alter the slope. C F; but as this can only be in a small degr the equilibrimn will hardly he affected by it when the notch is minute.
Example 8. Figure 9.—Let one of the corners of a sloping body. A ti c n. rest upon the level top of a wall at D; it is required to find the position of a notch cut out of the upper end, c, so that the body may rest upon a wall made to fit the notch.