And SA represents the resultant of these two pressures. similarly RA, the diagonal of the parallelogram RSAP,, represents in magnitude and direction the resultant of SA and P1, that is, of P1, P, and P,, since SA is the resultant of P, and P.. It is evident that the fourth pressure necessary to produce an equilibrium with P., P., P, being equal and posite to their resultant, is represented in magnitude and direction by AR. 13. Three pressures, P., P2, P, being in equilibrium, it is required to determine the third P, in terms of the other two, and their inclination to one another. Let AP, and AP, represent the pressures P, and P, in magnitude and airection, and let the inclination P, AP, of P, to P, be represented by 0. Complete the parallelogram AP, RP,, and draw its diagonal AR. Then does AR represent the resultant of P, and P, in magnitude and direction. But this resultant is in equilibrium with P, since P, and P, are in equilibrium with P. It acts, therefore, in the same straight line with P,, but in an opposite direction, and is equal to it. Since then AR represents this resultant in magnitude and direction, therefore RA represents P, in mag nitude and direction. Now, AR=AP,'—2AP, . P‚R . cos. AP,R+P,R2; also, AP,R —P,AP,=¬—,02, P,R=AP,, and AP, AP, AR, represent P1, P,, P,, in magnitude. 1 .. P ̧2=P,'—2P,P2 cos. (7~,02)+P,2. Now cos. (―0)=—cos. ‚o1⁄2, .. P‚2=P ̧2+2P,P2 cos. ‚§2+P2', 1 29 14. If three pressures, P., P2, P, be in equilibrium, any two of them are to one another inversely as the sines of their inclinations to the third. Let the inclination of P, to P, be represented by 9, and that of P, to P, by 203. 1 3 1 Now P,AR=7-P,AP,=7-0 PRA=P,AR=¬—P,AP,=¬— -0. 2 39 .. sin. P,AR=sin.,, ; .. sin. PRA=sin. ,0,. That is, P, is to P, inversely, as the sine of the inclina tion of P, to P, is to the sine of the inclination of P, to P. Therefore, &c. &c. OF PARALLEL PRESSURES. [Q. E. D.] 15. The principle of the equality of moments obtains in respect to pressures in the same plane whatever may be their inclinations to one another, and therefore if their inclinations be infinitely small, or if they be parallel. In this case of parallel pressures, the same line AB, which is drawn from a given point A, perpendicular C to one of these pressures, is also perpendicular to all the rest, so that the perpendiculars are here the parts of this line AM,, AM, &c. intercepted between the point A and the directions of the pressures respectively. The principle is not however in this case true only in respect to the intercepted parts of this perpendicular line AB, but in respect to the intercepted parts of any line AC, drawn through the point A across the directions of the pressures, since the intercepted parts Am, Am, Am, &c. of this second line are proportional to those, AM,, AM,, &c. of the first. Thus taking the case represented in the figure, since by the principle of the equality of moments we have, 2 3 AM, . P1+AM, . P1=AM, . P2+AM ̧P ̧+AM ̧P ̧; dividing both sides by AM,, AM, Am, AM, But by similar triangles, AM,Am, AM, Am 2, &c.=&c Am, Am, 3 Therefore multiplying by Am, Am,. P1+ Am ̧ . P1 =Am, . P, +Am, . P,+Am, . P ̧. Therefore, &c. [Q.E.D.] 16. To find the resultant of any number of parallel pressures in the same plane. It is evident that if a pressure equal and opposite to the resultant were added to the system, the whole would be in equilibrium. And being in equilibrium it has been shown (Art. 8.), that if the pressures were all moved from their present points of application, so as to remain parallel to their existing directions, and applied to the same point, they are such as would be in equilibrium about that point. But being thus moved, these parallel pressures would all have their directions in the same straight line. Acting therefore all in the same straight line, and being in equilibrium, the sum of those pressures whose tendency is in one direction along that line must equal the sum of those whose tendency is in the opposite direction. Now one of these sums includes the resultant R. It is evident then that before R was introduced the two sums must have been unequal, and that R equals the excess of the greater sum over the less; and generally that if IP represent the sum of any number of parallel pressures, those whose tendency is in one direction being taken with the positive sign, and those whose tendency is in the opposite direction, with the negative sign; then the sign of R indicating whether it act in the direction of those pressures which are taken positively, or those which are taken negatively. Moreover since these pressures, including R, are in equi 'librium, therefore the sum of the moments about any point, of those whose tendency is to communicate motion in one direction, must equal the sum of the moments of the restthese moments being measured on any line, as AC; but one of these sums includes the moment of R; these two sums must therefore, before the introduction of R, have been unequal, and the moment of R must be equal to the excess of the greater sum over the less, so that, representing the sum of the moments of the pressures (R not being included) by 2 m P, those whose tendency is to communicate motion in one direction, having the positive sign, and the rest the negative; and representing by a the distance from A, measured along the line AC, at which R intersects that line, we have, since R is the moment of R, ≈R = 2 m2 P, where the sign of R indicates the direction in which R tends to turn the system about A, but R = ΣP, Equations (15) and (16) determine completely the magnitude and the direction of the resultant of a system of parallel pressures in the same plane. 17. To determine the resultant of any number of parallel pressures not in the same plane. Let P, and P, be the points of application of any two of these pressures, and let the pressures themselves be represented by P, and P,. Also let their resultant R, intersect the line joining the points P, and P, in the point R,; produce the line P, P., to intersect any plane given in position. in the point L. Through the points P, P., and R,, draw P,M,, P,M,, and R, N, perpendicularly to this plane: these lines will be in the same plane with one another and with P,L; let the intersection of this last mentioned plane with the first be LM,, then will P, M,, P, M., and R,N, be perpendiculars to LM,; moreover by the last proposition, y Let now the resultant, R,, of R, and P, intersect the line joining the points R, and P, in the point R, and similarly let the resultant, R,,. of R, and P, intersect the ine joining the points R, and P, in the point R,, and so on: then by the last equa tion. 4 Similarly, 2 P1. P‚M,+P,. P,M1 = R, R‚N, . 29 R. RN,+P,. P,M1 = R, R ̧Ñ„, &c. + &c. = &c. R„−2 . Rn—‚Ñn—, +Pn. PnMn=Rn— . Rn—, Nn—,. Adding these equations, and striking out terms common té both sides, P ̧ . P ̧M ̧+P‚P‚M2+ . . . +Pn.P„M=R2-, . Rn, Nn~, (17) 2 2 Now, RP,+P„, R=R,+P,=P1+P2+P„ 2 + ..... + P.; 3 .'. R2-1 Në−1 . P1+P2+P,+&c.+P2=P, . P ̧M,+P ̧. P,M,+ + P2. P2Mn; P, P,M,+P, P,M, + +Pr. P2Mn ... P1+P2+P2+ ... +P (18); in which expression those of the parallel pressures P1, P2, &c. which tend in one direction, are to be taken positively, whilst those which tend in the opposite direction are to be taken negatively. The line R-1 N-1 represents the perpendicular distance from the given plane of a point through which the resultant of all the pressures P1, P. . . . Pn, passes. In the same manner may be determined the distance of this point from any other plane. Let this distance be thus determined in respect to three given planes at right angles to one another. Its actual position in space will then be known. Thus then we shall know a point through which the resultant of all the pressures passes, also the direction of that resultant, for it is parallel to the common direction of all the pressures, and we shall know its amount, for it is equal to the sum of all the pressures with their proper signs. Thus then the resultant. pressure will be completely known. The point R-1 is called the CENTRE OF PARALLEL PRESSURES. 18. The product of any pressure by its perpendicular distance from a plane (or rather the product of the number of units in the pressure by the number of units in the perpendicular), is called the moment of the pressure, in respect to that plane. Whence it follows from equation (17) that the sum of the moments of any number of parallel pressures in |