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CHAPTER III.

ON THE THEORY OF COUPLES.

WE saw in the last chapter that two equal and parallel forces acting in opposite directions and at different points of a body had no single resultant, but constituted a statical couple, tending to produce rotatory motion. This tendency can be balanced only by an equal and opposite tendency produced by an opposite couple. Statical couples have peculiar properties, which we will now discuss, and chiefly by employing the super-position of equilibrium. See page 6.

18. PROP. A couple may be turned round in any manner in its own plane without altering its statical effect.

P4

E

Pi

P5

P2

b

D

B

P6

a

Let PABP be the original couple; take ab=AB, and turned round any point C; apply equal and opposite forces, P3, P4, perpendicular to ab at a; and similarly, P5 and P6 at b; these will not affect the system, being in equilibrium amongst themselves: let each of them equal P1 or P2. Then P1 at A, and P4 at a, are equivalent to a force bisecting the angle P1EP between them, or a force in CE; and P are equivalent to an equal force in CD. being equal and opposite may be removed; that is, we may remove from the system the forces P1, P2, P4, P6, and we have remaining the forces P3 and P, at a and b, forming the couple PabP, which is the same as if we had turned the original couple round the point C until its arm came to the position ab.

5

P3

similarly, P2 These forces

19. PROP. A couple may be removed to any other part of its own plane, its arm remaining parallel to the original direction; and it may be removed to any other plane, in the body on which it acts, parallel to its own plane, its arm being parallel to the original arm.

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rallel to the original direction; similarly, P2 at B, and P4 at a, are equivalent to a force = 2P2 at C, opposite to the former; these will consequently balance each other, and may be removed, or the forces P1, P2, P4, P5, may be removed, and we have remaining the couple Pab P6, equivalent to the original couple removed parallel to itself in its own plane.

1

D

P2

F

F6

A

B

P4

C

Secondly. Let the arm AB of the original couple be removed from its own plane DE, parallel to itself, to a b in the parallel plane FG; let equal forces, each equal to P1 or P2, be applied at a and b, as in the former case. Join Ab and Ba; these lines will bisect each other in C. The forces P1 and P will be equivalent to a force =2P1 at C, parallel to the original di

a

Pi

E

P3

P5

G

rection, and P2 and P4 will be equivalent to an equal and opposite force at the same point; these equal and opposite forces at C may be removed, and there remains the couple PзabP6, equivalent to the original couple removed to the plane FG.

20. PROP. All statical couples are equivalent to each other whose planes are parallel and moments equal.

DEFINITION. The moment of a couple is the product of one of the forces into the arm, or P.AB in the foregoing propositions.

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and forces Pat A, and Q at C, have a resultant parallel to their directions and equal to their sum at B. This force, P+Q=P1, will balance P2, and therefore P at A, Q1 at C, and P, at B, may be removed; and there remains the couple Q.AC, which is therefore equivalent to the original couple.

By the previous propositions this couple, Q.AC, may be removed into any plane parallel to its own, and turned round in any manner in that plane.

DEFINITION. The axis of a couple is a line perpendicular to the plane of the couple; and its length being taken proportional to the moment of the couple, represents it in magnitude. The tendency to rotatory motion being round the axis, and the length of the axis representing this tendency as measured by the moment, the axis represents completely the couple. The effect of the previous propositions is consequently this: that the

axis, of fixed length, may be removed any where within the body acted on, parallel to itself.

When any number of couples act upon the body, they can be compounded into one resultant couple.

21. PROP. When any number of couples act upon a body in parallel planes, the moment of the resultant couple equals the sum of the moments of the component couples.

Let P, Q, R, &c. be the forces; a, b, c, &c. their arms respectively; the couples can be removed all into one plane, turned round, and moved in that plane, and their arms changed to a common arm, whilst their moments remain unchanged. Let m be the common arm AB; P', Q', R', &c., the PV forces; so that P.m=P. a, Q.m=Q.b, q' R'.m=R.c, &c.; but the forces P', Q', R R', &c. at A are equivalent to a force

Ρ

A

P.a Q.b R.c

P+Q'+R'+ &c. = + + + &c.

m

m

And similarly, the forces at B are equal to the same sum; and the moment of the resultant couple (P' + Q'+R'+ &c.)AB

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=(P' + Q'+R'+ &c.) m

= Pa+Qb+Re+ &c.

m

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B

R

'Q'

K L M

Р

This is the same thing as taking the algebraic sum of the axes, as OK, KL, LM, &c. for the resultant axis, when the component axes are parallel. If any of the couples tend to cause rotation the contrary way round, we must take them with contrary signs, or their axes must have been measured in the opposite direction from 0. An axis is therefore balanced by an equal and opposite axis, or a couple by an equal and opposite couple. If PABP, QDEQ, were couples whose moments were equal and

Q A

B

E

opposite, or P.AB=-Q.DE, they would evidently make equilibrium with each other.

22. PROP. If two sides of a parallelogram represent the axes of two component couples, the diagonal represents the axis of the resultant couple.

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each other as the axes themselves are. The couples can be moved and turned round, each in their own plane, until their arms are in the intersection of their planes; and their moments being kept the same, they can be brought to have the same arm. Let AB be this common arm in the intersection of the planes; PABP, QABQ, the couples. Completing the parallelograms on the lines representing P and Q respectively, the diagonal represents their resultants R, at A and B, and the two couples are equivalent to a couple RABR.

If be the angle between P and Q, or between OK and OL, R2=P2+Q2+2PQ cos. 0... from the triangle of forces.

.. R.AB=AB. √ P2+Q2+2PQ cos. Ø

= √ P2.AB2 + Q2.AB2+2P.AB × Q.AB cos. Ꮎ

= √OK2+OL2+20K. OL cos. Ꮎ

= OM

And OK, OL being respectively perpendicular to the planes of the couples PABP, QABQ, we have OM perpendicular to the plane of the resultant couple RABR; therefore OM represents the axis of the resultant couple. Let L and M be the

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