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CHAPTER VII.

STATICS OF SOLIDS AND FLUIDS.

552. FORCES whose lines meet. Let ABC be a rigid body acted on by two forces, P and Q, applied to it at different points, D and E respectively, in lines in the same plane.

Since the lines are not parallel, they will meet if produced; let them be produced and meet in 0. Transmit the forces to act on that point; and the result is that we have simply the case of two forces acting on a material point, which has been already considered.

553. The preceding solution is applicable to every case of nonparallel forces in a plane, however far removed the point may be in which their lines of action meet, and the resultant will of course be found by the parallelogram of forces. The limiting case of parallel forces, or forces whose lines of action, however far produced, do not meet, was considered above, and the position and magnitude of the resultant were investigated. The following is an independent demonstration of the conclusion arrived at.

554. Parallel forces in a plane. The resultant of two parallel forces is equal to their sum, and is in the parallel line which divides any line drawn across their lines of action into parts inversely as their magnitudes.

1. Let P and Q be two parallel forces similar directions in lines AB and CD. their lines. In it introduce any pair of balancing forces, S in AĞ K-S and S in CH. These forces will not disturb the equilibrium of

acting on a rigid body in Draw any line AC across

the body. Suppose the forces GSA
P and S in AG, and Q and S in
CH to act respectively on the
points A and C of the rigid body.
The forces P and S, in AB and
AG, have a single resultant in
some line AM, within the angle

M

B

SL

E

SH

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GAB; and Q and S in CD and CH have a resultant in some line CN, within the angle DCH.

The angles MAC, NCA are together greater than two right angles, hence the lines MA, NC will meet if produced. Let them meet in 0. Now the two forces P and S may be transferred to parallel lines through O. Similarly the forces Q and S may be also transferred. Then there are four forces acting on O, two of which, S in OK and S in OL, are equal and directly opposed. They may, therefore, be removed, and there are left two forces equal to P and Q in one line on O, which are equivalent to a single force P+Q in the same line.

2o. If, for a moment, we suppose OE to represent the force P, then the force representing S must be equal and parallel to EA, since the resultant of the two is in the direction OA. That is to say,

SP EA: OE;

and in like manner, by considering the forces S in OL and Q in OE, we find that

Q: S: OE : EC.

Compounding these analogies, we get at once

Q: P:: EA: EC,

that is, the parts into which the line is divided by the resultant are inversely as the forces.

555. Forces in dissimilar directions. The resultant of two parallel forces in dissimilar directions 1, of which one is greater than the other, is found by the following rule: Draw any line across the lines of the forces and produce it across the line of the greater, until the whole line is to the part produced as the greater force is to the less; a force equal to the excess of the greater force above the less, applied at the extremity of this line in a parallel line and in the direction similar to that of the greater, is the resultant of the system.

M

Q-P

K L

A

B

E

P

YQ.

K

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M'

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Let P and Q in KK and LL', be the contrary forces. From any point A, in the line of P, draw a line AB across the line of Q cutting it in B, and produce the line to E, so (that AE: BE :: Q : P. Through E draw a line MM' parallel to KK' or LL'.

In MM' introduce a pair of balancing forces each equal to Q-P. Then P in AK' and Q-P in EM have a resultant equal to their

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1 In future the word 'contrary' will be employed instead of the phrase parallel and in dissimilar directions' to designate merely directional opposition, while the unqualified word opposite' will be understood to signify contrary and in one line.

sum, or Q. This resultant is in the line LL'; for, from the analogy,

we have

or

:

AE BEQ: P,
AE-BE: BE:: Q-P: P,

AB: BE: Q−P : P.

Hence P in AK', Q in BL', and Q-P in EM are in equilibrium and may be removed. There remains only Q-P in EM', which is therefore the resultant of the two given forces. This fails when the forces are equal.

A1 EA2 A3
P

556. Any number of parallel forces in a plane. Let P1, P2, P31 &c., be any number of parallel forces acting on a rigid body in one plane. To find their resultant in position and magnitude, draw any line across their lines of action, cutting them in points, denoted respectively by A1, A2, A3, &c., and in it choose a point of reference O. Let the distances of the lines of the forces from this point be denoted by a1, a2, A3, &c.; as OА1=a1, OA1=a,, &c. Also let R denote the resultant, and x its distance from O.

1

Find the resultant of any two of the forces, as P, and P2, by § 554. Then if we denote this resultant by R', we have

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Divide A, A, in E' into parts inversely as the forces, so that

2

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and

Similarly we shall find the resultant of R' and P, to be

R"=R'+P=P1+P2+P3;

R"x"=R'x' +P ̧a ̧=P1α1+Р2α2+P3ɑ3•

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In this method negative forces or negative values of any of the quantities a,, a, ..., may be included, provided the generalized rules of multiplication and addition in algebra are followed.

557. Any number of parallel forces not in one plane. To find the resultant, let a plane cut the lines of all the forces, and let the points

in which they are cut be specified by reference to two rectangular axes in the plane. Let the plane be YOX: OX, OF, the axes of reference, O the origin of co-ordinates, and A1, A2, A,, &c., the points in which the plane cuts the lines of the forces, P1, P2, P3, &c. Thus each of these points will be specified by perpendiculars drawn from it to the axis. Let the co-ordinates of the point A, be denoted by x11; of A2, by x2 y2; and so on; that is, ON1=x1, N1 A1=y1; ON2=x2, N2 A, =y, &c.; let also the final resultant be denoted by R, and its co-ordinates by x and y.

Y

E

K

-X

the plane of reference. Then

2

Find the resultant of P1 and P2 by joining A1 A2, and dividing the line inversely as the forces. Suppose E' the point in which this resultant cuts

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To find the co-ordinates, which may be denoted by x'y', of the point E' with reference to OX and OF; draw E'N' perpendicular to OX and cutting it in N, and from A, draw A, K parallel to OX, or perpendicular to A, N, and cutting it in K and E'N' in M. Then (Euclid VI. 2)

Hence

or

whence we get

and since

we have

and similarly,

We may find the

Iwith all the forces.

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resultant of R' and P, in like manner, and so Hence we have for the final resultant,

R=P1+P2+P2+ + Pn.

1

2

.....

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(3)

(4)

(5)

These equations may include negative forces, or negative coordinates.

558. Conditions of equilibrium of any number of parallel forces. In order that any given parallel forces may be in equilibrium, it is not sufficient alone, that their algebraic sum be equal to zero.

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From this equation it follows that if the forces be divided into two groups, one consisting of the forces reckoned positive, the other of those reckoned negative, the sum, or resultant (§ 556) of the former is equal to the resultant of the latter; that is, if ‚R and 'R denote the resultants of the positive and negative groups respectively,

R='R.

But unless these resultants are directly opposed they do not balance one another; wherefore, if (x,y) and (xy) be the co-ordinates of ‚R and 'R respectively, we must have for equilibrium

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But,R,x is equal to the sum of those of the terms P1x1, P2X2, &c., which are positive, and 'R'x is equal to the sum of the others each with its sign changed: and so for R,y and 'R'y. Hence the preceding equations are equivalent to

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We conclude that, for equilibrium, it is necessary and sufficient that each of the following three equations be satisfied :

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559. If equation (6) do not hold, but equations (7) and (8) do, the forces have a single resultant through the origin of co-ordinates. If equation (6) and either of the other two do not hold, there will be a single resultant in a line through the corresponding axis of reference, the co-ordinates of the other vanishing. If equation (6) and either of the other two do hold, the system is reducible to a single couple in a plane through that line of reference for which the sum of the products is not equal to nothing. If the plane of reference is perpendicular to the lines of the forces, the moment of this couple is equal to the sum of the products not equal to nothing.

560. In finding the resultant of two contrary forces in any case in which the forces are unequal-the smaller the difference of magnitude between them, the farther removed is the point of application of the resultant. When the difference is nothing, the point is removed to an infinite distance, and the construction (§ 555) is thus rendered · nugatory. The general solution gives in this case R=0; yet the forces are not in equilibrium, since they are not directly opposed. Hence two equal contrary forces neither balance, nor have a single resultant. It is clear that they have a tendency to turn the body to

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