such coplanar system can in general be reduced to a single
resullant acting in a definite line. As exceptional cases the
system may reduce to a couple, or it may be in equilibrium.
The moment of a force about a point O is the product of the
force into the perpendicular drawn to its line of action from
° O, this perpendicular being reckoned
positive or negative according as O
P lies on one side or other of the line
of action. If we mark off a segment
AB along the line of action so as to
represent the force completely, the
moment is represented as to magnitude
by twice the area of the triangle
OAB, and the usual convention as
to be reckoned positive or negative
according as the letters 0, A, B, occur in “ counterclockwise ”
or “ clockwise ” order.
The sum of the moments of two forces about any point O
is equal to the moment of their resultant (P. Varignon, 1687).
Let AB, AC (fig. 16) represent the two forces, AD their resultant;
we have to prove that the sum of the triangles OAB, OAC is
equal to the triangle OAD,
regard being had to signs. Since
the side OA is common, we have
~." to prove that the sum of the
perpendiculars from B and C on
OA is equal to the perpendicular
from D on OA, these perpendiculars
being reckoned positive
or negative according as they lie
to the right or left of AO.
Regarded as a statement concerning the orthogonal projections
9 9 5 9
/A
FIG. 15.
to sign is that the area is
C D
o
A:
FIG. 16.
of the vectors AB and AC (or BD), and of their sum AD, on a
line perpendicular to AO, this is obvious.
It is now evident that in the process of reduction of a coplanar
system no change is made at any stage either in the sum of the
projections of the forces on any line or in the sum of their
moments about any point. It follows that the single resultant
to which the system in general reduces is uniquely determinate,
i.e. it acts in a definite line and has a definite magnitude and
sense. Again it is necessary and sufficient for equilibrium
that the sum of the projections of the forces on each of two
perpendicular directions should vanish, and (moreover) that
the sum of the moments about some one point should be zero.
T he fact that three independent conditions must hold for equilibrium
is important. The conditions may of course be expressed
in different (but equivalent) forms; e.g.the sum of the moments
of the forces about each of the three points which are not collinear
must be zero.
The particular case of three forces is of interest. If they
are not all parallel they must be concurrent, and their vector sum
must be zero. Thus three forces acting perpendicular
Flo. 17.
to the sides of a triangle at the middle points will be in equilibrium
provided they are proportional to the respective sides,
and act all inwards or all outwards. This result is easily
extended to the case of a polygon of any number of sides; it
has an important application in hydrostatics.
Again, suppose we have a bar AB resting with its ends on two
smooth inclined planes which face each other. Let G be the centre
of gravity (§ II), and let AG=a, GB=b. Let o., 13 be the inclinations
of the planes, and 0 the angle which the bar makes with the
vertical. The position of equilibrium is determined by the consideration
that the reactions at A and B, which are by hypothesis normal to
the planes, must meet at a point I on the vertical through G. Hence
IG/a =sin (0a.)/sin a., IG/b =sin (H+/3)/sind,
whence cot 6 7  (6)
If the bar is uniform we have a =Ig, and
C0t0=% (cot acotH). (7)
The problem of a rod suspended by strings attached to two points
of it is virtually identical, the tensions of the strings taking the place
of the reactions of the planes.
1
rn
/Y~.' X
/ g, ,, .»
IB .1
C

1
/
I/
I: o.
I 1
I
», gif *' .
!:
A
a'.' ':H
" C
FIG. 18.
just as a system of forces is in general equivalent to a single
force, so a given force can conversely be replaced by combinations
of other forces, in various ways. For instance, 'a given
force (and consequently a system of forces) can be replaced
in one and only one way by three forces acting in three assigned
straight lines, provided these lin.es be not concurrent or parallel.
Thus if the three lines form a triangle ABC, and if the given force
F meet BC in H, then F can be resolved into two components
acting in HA, BC, respectively. And the force in HA can
be resolved into two components acting in BC, CA, respectively.
A simple graphical construction is indicated in fig. 19, where
A P
R ~" !
B 'H R
FIG. 19.
the dotted lines are parallel. As an example, any system of
forces acting on the lamina in fig. 9 is balanced by three
determinate tensions (or thrusts) in the three links, provided
the directions of the latter are not concurrent.
If P, Q, R, be any three forces acting along BC, CA, AB, respectively,
the line of action of the resultant is determined by the consideration
that the sum of the moments about any point on it must
vanish. Hence in “tr1hnear" coordinates, with ABC as fundamental
triangle, its equation is Po.lQ/3+Ry=0. If P: Q: R=
a: b: c, where a, b, c are the lengths of the sides, this becomes the
“ line at infinity, ” and the forces reduce to a couple.
The sum of the moments of the two forces of a couple is the
same about any point in the plane. Thus in the figure the sum
of the moments about O is P . OAP OB or P AB, which is
independent of the position of 0
O. This sum is called the '
moment of the couple; it must
of course have the proper sign
attributed to it. It easily '
follows that any two couples
of the same moment are
P A
ii, P equivalent, and that any number of couples can be replaced by a single couple whose moment is the sum of their moments. Since a couple is for our purposes sufficiently represented by its moment, it has been proposed to substitute the name torque (or twisting effort), as free from the suggestion of any special pair of forces. A system of forces represented completely by the sides of a plane polygon taken in order is equivalent to a couple whose FIG. 20.