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576. Conditions of Equilibrium. The conditions of equilibrium of three forces at right angles to one another have been already stated in § 470; and the conditions for three rectangular couples in § 567.

If a body be acted on by three forces and three couples simultaneously, all the conditions applicable when they act separately, must also be satisfied when they act conjointly, since a force cannot balance a couple. Six Equations of Equilibrium therefore are necessary and sufficient for a rigid body acted on by any number of forces. These

are

or

P, cos az + P, cos a, + &c.=0,
P, cos B. + P, cos 3, + &c.=0,
P, cos y + P, cos y + &c.=0,
G cos 6 + G, cos $2+ &c.=0,
G cos ni+G; cos 12+ &c.=0,

G, cos 0,+ G, cos 0, + &c.=0. 577. If the line of the resultant found by $ 575, is perpendicular to the plane of the couple, that is, if

1=l, u=m, v=n;
L M N

(21) x=y=z' the system cannot be reduced to another with a force and a smaller couple, and in this case the line found for the resultant force is the central axis of the system.

578. If, on the other hand, the plane of the couple is parallel to the line of the force, or the axis of the couple perpendicular to the line of the force, that is, if

in+mu+nv=0, LX+MY+NZ=0,

(22) the force and couple may ($ 568) be reduced to one force : and this

G force is parallel to the former, at a distance from it equal to in

R the plane of it and the couple. Thus, 00 will be perpendicular to the line of the resultant force, and to the axis of the resultant couple, and therefore its direction cosines are ($ 464, 6); mv-nj, nd - lv, lu- ma,

(23) each of which will be positive when Olies within the solid angle

G edged by OX, OY, OZ. Hence, remembering that 00=

and

R' using the expressions (19) and (20), we find for the co-ordinates of O YN-ZM ZL-XN XM-IL

(24) R2 R2

R2

or

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and we thus complete the specification of the single force to which the system is reduced when (22) holds.

579. If the line of the force is inclined at any angle to the plane of the couple, the resultant system can be further reduced by $ 569, to a smaller couple and a force in a determinate line, the central axis.' This couple is G cos 0, and according to the notation, may be thus expressed by $ 464, (7), if we substitute the values given in (19) and (20),

XL+YM+ ZN
G cos 0=

(25)

R The other component couple, G sin 0, lies in the same plane as R, and with it may be reduced by $ 568 to one force, which will be parallel to R, that is, in the direction (1, m, n), at a distance from it

G sin 0 equal to

Hence the direction cosines of 0 0 will be
R
mv— nu ni - lv lu- mi

(26) sin

sin o sin Substituting in each of these for 1, 1, &c., their respective values,

G sin 0 and multiplying each member by we have for the co-ordinates

R of the point 0), as in § 578, YN- ZM ZL-XN XM-PL

(27) R2 R2

R? A single force, R, through the point thus specified in the direction (1, m, n), with a couple in a plane perpendicular to it, and having

XL+YM+ZN

R for its moment, is consequently the system of force along central axis and minimum couple, to which the given set of forces is determinately reducible by Poinsot's beautiful method.

580. The position of the central axis may be determined otherwise; thus, instead of in the first place bringing the forces to 0, bring them to any point T, of which let (x, y, z) be the co-ordinates. Then instead of Y%+Y,%, + &c., which we had before ($ 574), we have now

Y (3,-2) +Y,(2,-2) + &c.,

Y 2,+Y222+ &c.—(Y+YA+&c.) 2, and so for the others. Then for the moments of the couples of transference we have

L=L-(Zy-Y2),

=M-(X2-Zx),

N=N-Yx-Xy). Now, let T be chosen, if possible, so as to make the resultant

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couple lie in a plane perpendicular to it. The condition to be fulfilled in this case is

L M

x=y=z'
which, when for L, &c., we substitute their values, becomes,

L—(Zy-Y%)_M—(X2–2x)_N-(Yx-Xy)
X
Y

Z which is the equation of the central axis of the system.

To show that 0, the point determined in š$ 578, 579, is in the central axis thus found; we have, substituting for x, y, z, the values given in (24),

Z(ZL-XN)+Y(XM-YL)
L
R2

=&c.

X Reducing, and remarking that

LR LY LZ=LX, we find that the first member becomes (LX+MY+NZX

=LX+MY+NZ,

X and is therefore equal to each of the two others. Thus is verified the comparison of the two methods.

581. In one respect, this reduction of a system of forces to a couple, and a force perpendicular to its plane, is the best and simplest, especially in having the advantage of being determinate, and it gives very clear and useful conceptions regarding the effect of force on a rigid body. The system may, however, be farther reduced to two equal forces acting symmetrically on the rigid body, but whose position is indeterminate. Thus, supposing the central axis of the system has been found, draw a line AA', at right angles through any point C in it, so that CA may be equal to CA For R, acting along the central axis, substitute į R at each end of AA'. Thus, choosing this line AA' as the arm of the couple, and calling it a, we have at

G each extremity of it two forces, perpendicular to the central axis, and IR parallel to the central axis. Compounding these, we get

ao

through A and A' respectively, perpendicular to A A, and equally inclined at the angle

2G tan-1

on the two sides of the plane through AA' and the central aR axis.

582. It is obvious, from the formulae of $ 195, that if masses proportional to the forces be placed at the several points of application of these forces, the centre of inertia of these masses will be the same

a

point in the body as the centre of parallel forces. Hence the reactions of the different parts of a rigid body against acceleration in parallel lines are rigorously reducible to one force, acting at the centre of inertia. The same is true approximately of the action of gravity on a rigid body of small dimensions relatively to the earth, and hence the centre of inertia is sometimes ($ 195) called the Centre of Gravity. But, except on a centrobaric body ($ 543), gravity is not in general reducible to a single force: and when it is so, this force does not pass through a point fixed relatively to the body in all positions.

583. The resultant of a system of parallel forces is not a single force when the algebraic sum of the given forces vanishes. In this case the resultant is a couple whose plane is parallel to the common direction of the forces. A good example of this is furnished by a magnetized mass of steel, of moderate dimensions, subject to the influence of the earth's magnetism only. As will be shown later, the amounts of the so-called north and south magnetisms in each element of the mass are equal, and are therefore subject to equal and opposite forces, all parallel to the line of dip. Thus a compass-needle experiences from the earth's magnetism merely a couple or directive action, and is not attracted or repelled as a whole.

584. If three forces, acting on a rigid body, produce equilibrium, their directions must lie in one plane; and must all meet in one point, or be parallel. For the proof, we may introduce a consideration which will be very useful to us in investigations connected with the statics of flexible bodies and fluids.

If any forces, acting on a solid or fluid body, produce equilibrium, we may suppose any portions of the body to become fixed, or rigid, or rigid and fixed, without destroying the equilibrium.

Applying this principle to the case above, suppose any two points of the body, respectively in the lines of action of two of the forces, to be fixed—the third force must have no moment about the line joining these points; that is, its direction must pass through the line joining them. As any two points in the lines of action may be taken, it follows that the three forces are coplanar. And three forces in one plane cannot equilibrate, unless their directions are parallel or pass through a point.

585. It is easy and useful to consider various cases of equilibrium when no forces act on a rigid body but gravity and the pressures, normal or tangential, between it and fixed supports. Thus, if one given point only of the body be fixed, it is evident that the centre of gravity must be in the vertical line through this point-else the weight and the reaction of the support would form an unbalanced couple. Also for stable equilibrium the centre of gravity must be below the point of suspension. Thus a body of any form may be made to stand in stable equilibrium on the point of a needle if we rigidly attach to it such a mass as to cause the joint centre of gravity to be below the point of the needle.

P

586. An interesting case of equilibrium is suggested by what are called Rocking Stones, where, whether by natural or by artificial processes, the lower surface of a loose mass of rock is worn into a convex form which may be approximately spherical, while the bed of rock on which rests in equilibrium is, wheth convex or concave, approximately spherical, if not plane. A loaded sphere resting on a spherical surface is therefore a type of such cases. Let O, O be the centres of curvature of the fixed and rocking O bodies respectively, when in the position of equilibrium.

Take any two infinitely small equal arcs PQ, Pp; and R at Q make the angle O QR equal to POp. When, by

displacement, Q and p become the points in contact, QR will evidently be vertical ; and, if the centre of gravity G, which must be OPO when the movable body is in its position of equilibrium, be to the left of QR, the equilibrium will obviously be stable. Hence, if it be below R, the equilibrium is stable, and not unless.

Now if p and o be the radii of curvature OP, OP of the two surfaces, and the angle Pop, the angle

po QOR will be equal to ; and we have in the triangle

o

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po

::0:0+p (approximately).

02 Hence

PR=0

otp pto and therefore, for stable equilibrium,

po PG<

pto If the lower surface be plane, p is infinite, and the condition becomes (as in § 256)

PG<o. If the lower surface be concave, the sign of p must be changed, and the condition becomes

po

PG< which cannot be negative, since p must be numerically greater than o in this case.

587. If two points be fixed, the only motion of which the system is capable is one of rotation about a fixed axis. The centre of gravity must then be in the vertical plane passing through those points, and below the line adjoining them.

588. If a rigid body rest on a fixed surface, there will in general be only three points of contact, $ 380; and the body will be in stable

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