[Illustration: FIG. 3.]
9. We now come to the important case where three forces act on a point:
this is to be studied by the apparatus represented in Fig. 3. It
consists essentially of two pulleys H, H, each about 2" diameter,[1]
which are capable of turning very freely on their axles; the distance
between these pulleys is about 5', and they are supported at a height
of 6' by a frame, which will easily be understood from the figure. Over
these pulleys passes a fine cord, 9' or 10' long, having a light hook
at each of the ends E, F. To the centre of this cord D a short piece
is attached, which at its free end G is also furnished with a hook. A
number of iron weights, 0·5 lb., 1 lb., 2 lbs., &c., with rings at the
top, are used; one or more of these can easily be suspended from the
hooks as occasion may require.
[1] We shall often, in these lectures, represent feet or inches in
the manner usual among practical men—1' is one foot, 1" is one inch.
Thus, for example, 3' 4" is to be read “three feet four inches.” When
it is necessary to use fractions we shall always employ decimals. For
example, 0"·5 is the mode of expressing a length of half an inch; 3'
1"·9 is to be read “three feet one inch and nine-tenths of an inch.”
10. We commence by placing one pound on each of the hooks. The cords
are first seen to make a few oscillations and then to settle into a
definite position. If we disturb the cords and try to move them into
some new position they will not remain there; when released they will
return to the places they originally occupied. We now concentrate our
attention on the central point D, at which the three forces act. Let
this be represented by O in Fig. 4, and the lines OP, OQ, and OS will
be the directions of the three cords.
On examining these positions we find that the three angles P O S, Q
O S, P O Q, are all equal. This may very easily be proved by holding
behind the cords a piece of cardboard on which three lines meeting at
a point and making equal angles have been drawn; it will then be seen
that the cords coincide with the three lines on the cardboard.
[Illustration: FIG. 4.]
11. A little reflection would have led us to anticipate this result.
For the three cords being each stretched by a tension of a pound, it
is obvious that the three forces pulling at O are all equal. As O is
at rest, it seems obvious that the three forces must make the angles
equal, for suppose that one of the angles, P O Q for instance, was
less than either of the others, experiment shows that the forces O P
and OQ would be too strong to be counteracted by O S. The three angles
must therefore be equal, and then the forces are arranged symmetrically.
12. The forces being each 1 lb., mark off along the three lines in Fig.
4 (which represent their directions) three equal parts O P, O Q, O S,
and place the arrowheads to show the direction in which each force is
acting; the forces are then completely represented both in position and
in magnitude.
Since these forces make equilibrium, each of them may be considered to
be counteracted by the other two. For example, O S is annulled by O Q
and O P. But O S could be balanced by a force O R equal and opposite
to it. Hence OR is capable of producing by itself the same effect as
the forces O P and OQ taken together. Therefore O R is equivalent to
O P and OQ. Here we learn the important truth that two forces not in
the same direction can be replaced by a single force. The process is
called the _composition of forces_, and the single force is called
the _resultant_ of the two forces. O R is only one pound, yet it is
equivalent to the forces O P and O Q together, each of which is also
one pound. This is because the forces O P and O Q partly counteract
each other.
13. Draw the lines P R and Q R; then the angles P O R and Q O R are
equal, because they are the supplements of the equal angles P O S and
Q O S; and since the angles P O R and Q O R together make up one-third
of four right angles, it follows that each of them is two-thirds of
one right angle, and therefore equal to the angle of an equilateral
triangle. Also O P being equal to O Q and O R common, the triangles O P
R and O Q R must be equilateral. Therefore the angle P R O is equal to
the angle R O Q; thus P R is parallel to O Q; similarly Q R is parallel
to O P; that is, O P R Q is a parallelogram. Here we first perceive
the great law that the resultant of two forces acting at a point is the
diagonal of a parallelogram, of which they are the two sides.
14. This remarkable geometrical figure is called the _parallelogram of
forces_. Stated in its general form, the property we have discovered
asserts that two forces acting at a point have a resultant, and that
this resultant is represented both in magnitude and in direction by
the diagonal of the parallelogram, of which two adjacent sides are the
lines which represent the forces.
[Illustration: FIG. 5.]
15. The parallelogram of forces may be illustrated in various ways by
means of the apparatus of Fig. 3. Attach, for example, to the middle
hook G 1·5 lb., and place 1 lb. on each of the remaining hooks E, F.
Here the three weights are not equal, and symmetry will not enable us,
as it did in the previous case, to foresee the condition which the
cords will assume; but they will be observed to settle in a definite
position, to which they will invariably return if withdrawn from it.
Let O P, O Q (Fig. 5) be the directions of the cords; O P and O Q being
each of the length which corresponds to 1 lb., while O S corresponds
to 1·5 lb. Here, as before, O P and O Q together may be considered
to counteract O S. But O S could have been counteracted by an equal
and opposite force O R. Hence O R may be regarded as the single force
equivalent to O P and O Q, that is, as their resultant; and thus it
is proved experimentally that these forces have a resultant. We can
further verify that the resultant is the diagonal of the parallelogram
of which the equal forces are the sides. Construct a parallelogram
on a piece of cardboard having its four sides equal, and one of the
diagonals half as long again as one of the sides. This may be done
very easily by first drawing one of the two triangles into which the
diagonal divides the parallelogram. The diagonal is to be produced
beyond the parallelogram in the direction O S. When the cardboard
is placed close against the cords, the two cords will lie in the
directions O P, O Q, while the produced diagonal will be in the
vertical O S. Thus the application of the parallelogram of force is
verified.
[Illustration: FIG. 6.]
16. The same experiment shows that two unequal forces may be compounded
into one resultant. For in Fig. 5 the two forces O P and O S may be
considered to be counterbalanced by the force O Q; in other words, O Q
must be equal and opposite to a force which is the resultant of O P and
O S.
17. Let us place on the central hook G a weight of 5 lbs., and weights
of 3 lbs. on the hook E and 4 lbs. on F. This is actually the case
shown in Fig. 3. The weights being unequal, we cannot immediately
infer anything with reference to the position of the cords, but still
we find, as before, that the cords assume a definite position, to
which they return when temporarily displaced. Let Fig. 6 represent
the positions of the cords. No two of the angles are in this case
equal. Still each of the forces is counterbalanced by the other two.
Each is therefore equal and opposite to the resultant of the other
two. Construct the parallelogram on cardboard, as can be easily done
by forming the triangle O P R, whose sides are 3, 4, and 5, and then
drawing O Q and R Q parallel to R P and O P. Produce the diagonal O R
to S. This parallelogram being placed behind the cords, you see that
the directions of the cords coincide with its sides and diagonal, thus
verifying the parallelogram of forces in a case where all the forces
are of different magnitudes.
18. It is easy, by the application of a set square, to prove that
in this case the cords attached to the 3 lb. and 4 lb. weights are
at right angles to each other. We could have inferred, from the
parallelogram of force, that this must be the case, for the sides of
the triangle O P R are 3, 4, and 5 respectively, and since the square
of 5 is 25, and the squares of 3 and of 4 are 9 and 16 respectively, it
follows that the square of one side of this triangle is equal to the
sum of the squares of the two opposite sides, and therefore this is a
right-angled triangle (Euclid, i. 48). Hence, since P R is parallel to
O Q, the angle P O Q must also be a right angle.