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perpendicular to EF. Hence OM is the axis of the resultant couple.

566. Proposition IV. Any number of couples whatever are either in equilibrium with one another, or may be reduced to a single couple, under precisely the same conditions as those already investigated for forces acting on one point, the axes of the couples being now taken everywhere instead of the lines formerly used to represent the forces.

1°. Resolve each couple into three components having their axes along three rectangular lines of reference, ox, or, . Add all the components corresponding to each of these three lines. Then if the resultant of all the couples whose axes are along the line

OX, be denoted by L,
or.

M,

N, and if G be the resultant of these three, we have

G=N(L+Mo+N): and if [ n, , be the angles which the axis of this couple G, makes with the three axes Ox, OY, OZ, respectively, we have

L
M

N
cos

G 567. 2o. Conditions of equilibrium of any number of couples. For equilibrium the resultant couple must be equal to nothing : but as it is compounded of three subsidiary resultant couples in planes at right angles to one another, they also must each be equal to nothing. The remarks already made, and the equations already given in $$ 471, 472, apply with the necessary modification to couples also. Thus, for instance, the equations of equilibrium are

G cos + G, cos 6+G, cos (x + &c.,=0,
G, cos ni + G, cos ng + G, cos n; + &c.,=0,

G cos 0, + G, cos 0, +G, cos 6, + &c.,=0. 568. Before investigating the conditions of equilibrium of any number of forces acting on a rigid body, we shall establish some preliminary propositions.

1o. A force and a couple in the same or in parallel planes may be reduced to a single force. Let the plane of the couple be the plane of the diagram, and let its moment be denoted by G. Let R, acting in the A

G line OA in the same plane, be the force. Transfer the couple to an arm (which may

R

R be denoted by a) through the point 0, such

that each force shall be equal to R; and let its position be so chosen, that one of the YR forces shall act in the same straight line with R in OA, but in the opposite direction to it. A

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a=

R and G being known, the length of this arm can be found, for since the moment of the transposed couple is

Ra=G

G we have

R Through O then, draw a line 00' perpendicular to OA, making it equal to a. On this arm apply the couple, a force equal to R, acting on 0 in a line perpendicular to 00, and another in the opposite direction at the other extremity. There are now three forces, two of which, being equal and opposite to one another, in the line AA', may be removed. One, acting on the point Oʻ, remains, which is therefore equivalent to the given system.

569. 2°. A couple and a force in a given line inclined to its plane may be reduced to a smaller couple in a plane perpendicular to the force, and a force equal and parallel to the given force.

Let OA be the line of action of the force R, K

and let OK be the axis of the couple. Let A

the moment be denoted by G: and let AOK,

the inclination of its axis to the line of the B.

R

force, be 0. Draw OB perpendicular to OA. By Prop. IV ($ 566) resolve the couple into two components, one acting round OA as

axis, and one round OB. Thus the component round OA will be

G cos , and the component round OB,

G sin 8. Now as G sin 0 acts in the same plane as the given force R, this component together with R may be reduced by $ 568 to one force. This force which is equal to R, will act not at 0 in the line 0A, but in a parallel line through a point out of the plane of the diagram. Thus the given system is reduced to a smaller couple G cos 0, and to a force in a line which, by Poinsot, was denominated the central axis of the system.

570. 3o. Any number of forces may be reduced to a force and a couple. Let P, acting on My be one of a number of forces acting in

different directions on different points of a rigid A

body. Choose any point of reference 0, for the

12 different forces, and through it draw a line AA P.

M, parallel to the line of the first force P. Through O,

draw 00 perpendicular to AA' or the line of the 10

force P. In the line AA' introduce two equal opposite forces, each equal to P. There are now

three forces, producing the same effect as the given A

force, and they may be grouped differently: P, acting

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in 0 in the line OA, and a couple, P, acting at O', and P, at O in the line OA', on an arm 00. Reduce similarly all the other forces, each to a force acting on 0, and to a couple. But all the couples thus obtained are equivalent to a single couple, and all the forces are equivalent to one force. Hence, &c.

571. Reduction of any number of forces to their simplest equivalent system.

Suppose any number of forces acting in any directions on different points of a rigid body. Choose three rectangular planes of reference meeting in a point 0, the origin of co-ordinates. In order to effect the reduction it is necessary to bring in all the forces to the point O. This may be done in two different ways—either in two steps, or directly.

572. 1°. Let the magnitudes of the forces be P1, P2, &c., and the co-ordinates, with reference to the rectangular planes, of the points at which they act respectively, be (*1,91, 2), (x2, 92, 2.), &c. Let also the direction cosines be (1,, my, n), (12, m,, n.), &c.

Resolve each force into three components, parallel to Ox, OY, OZ, respectively. Thus, if (X, Y, Z), &c., be the components of P, &c., we shall have X=P h; X=P,1,; &c.

(1) Y=P,m; Y,= P,m,; &c.

(2) Z=Pin; 2,=P,n,; &c.

(3) To transfer these components to the point 0. Let X,, in MK, be the component, parallel to OX, of the force P, acting on the point M. From M transmit it along its line to a

Z point N in the plane ZOY: the co-ordinates of this point will be 21

From Ndraw a perpendicular NB to OY, and

N: through B draw a line parallel to MK or OX. Introducing in this line a pair

„Y of balancing forces each equal to X, we have a couple acting on an arm %

1 B in a plane parallel to XOZ, and a

X

X single force X, parallel to OX in the plane XoY.

The moment of this couple is X,2,, and its axis is along Or. Next transfer the force X, from B to 0, by introducing a pair of balancing forces in X’OX, one of which, with the force X, in the line through B parallel to X'X and the direction similar to OxÝ, form a couple acting on an arm Yı. This couple, when Z and X are both positive, tends to turn in the plane XOY from OY to OX. Therefore by the rule, $ 201, its axis must be drawn from O in the direction OZ'. Hence its moment is to be reckoned as 2,x. Besides this couple there remains a single force equal to X,, in the direction OX, through the point 0. Similarly by successive steps transfer the forces Y, Z,

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to the origin of co-ordinates. In this way six couples of transference are got, three tending to turn in one direction round the axes respectively, and three in the opposite direction; and three single forces at right angles to one another, acting at the point 0. Thus for the force Pi, at the point (X1,Yı, 2.), we have as equivalent to it at the point O, three forces X, Y, Z, and three couples;

29-7%; moment of the couple round OX; (4) X,2,-2,x,; moment of the couple round OY; (5)

Y,x,-X,Yı; moment of the couple round OZ. (6) All the forces may be brought in to the origin of co-ordinates in a

similar way.

Y

573. 2°. Otherwise : Let P be one of the forces acting in the line

MT on a point M of a rigid body. Let T

O be the origin of co-ordinates; OX, Z

OY, OZ, three rectangular lines of res ference. Join OM and produce the line

to S. From O draw ON, cutting at right angles in the point N, the line MT produced through M. Let ON

be denoted by p, and the angle TMS N

by k. In a line through O parallel to X

MT (not shown in diagram) suppose introduced a pair of balancing forces each equal to P. We have thus a single force equal to P acting at 0, and a couple, whose moment is Pp, in the plane ONM. The direction cosines of this plane, or, which is the same thing, the direction cosines of a perpendicular to it, that is, the axis of the couple are ($ 464), if we denote them by 0, x, y, respectively,

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-- nx

Hence, if we substitute p for its value in the three preceding equations, the expression for the direction cosines are reduced to

ny- mz p=

(7) р Iz x=

(8) P mx-ly y

(9)

P To find the component couples round Ox, OY, oz, multiply these direction cosines respectively by Pp; whence we get

Pp.p=P (ny— m2), moment of couple round ox, (10)
Pp.X=P (12— nx), moment of couple round OY, (11)

Pp.v=P (mx-ly), moment of couple round OZ. (12) That this result is the same as that got by the other method will be evident, by considering that (equations 1, 2, 3),

Pl=X; Pm=Y; Pn=2. 574. When by either of the methods all the forces have been referred to 0, there is obtained a set of couples acting round OX, OY, OZ; and a set of forces acting along OX, OY, OZ. Find then the resultant moments of all the couples; and the sums of all the forces : if L, M, N be the resultant moments round Ox, OY, OZ respectively, we have L=(2,91-Y2,)+(Z29z-Y2%)+ &c.

(13) M=(X,24–27x,)+(X222-22x2)+&c.

(14) N=(Y,x,-X,71)+(,x,-X,72) + &c. (15) and if X, Y, Z be the resultant forces, X=X,+X,+Xg+&c.

(16) V=Y +Y,+Y2+ &c.

(17) Z=2, +2, +23+ &c.

(18) 575. Finally, find the resultant of the three forces by the formulae of Chap. VI, and the resultant of the three couples by Prop. IV ($ 566). Thus, if l, m, n be the direction cosines of the resultant force R, we have ($$ 463, 467)

Y

z
I=
m=Di

(19) R R and if , M, v be the direction cosines of the axis of the resultant couple, we have ($ 566)

L
M

N
a=
M= =;

(20) G

n ==

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