60. The weight of the bar has hitherto somewhat complicated our
calculations; the results would appear more simply if we could avoid
this weight; but since we want a strong bar, its weight is not so
small that we could afford to overlook it altogether. By means of the
arrangement of Fig. 21, we can _counterpoise_ the weight of the bar.
To the centre of A B a cord is attached, which, passing over a fixed
pulley D, carries a hook at the other end. The bar, being a pine rod, 4
feet long and 1 inch square, weighs about 12 ounces; consequently, if
a weight of twelve ounces be suspended from the hook, the bar will be
counterpoised, and will remain at whatever height it is placed.
[Illustration: FIG. 21.]
61. A B is divided by lines drawn along it at distances of 1" apart;
there are thus 48 of these divisions. The weights employed are
furnished with rings large enough to enable them to be slipped on the
bar and thus placed in any desired position.
62. Underneath the bar lies an important portion of the arrangement;
namely, the knife-edge C. This is a blunt edge of steel firmly
fastened to the support which carries it. This support can be moved
along underneath the bar so that the knife-edge can be placed under
any of the divisions required. The bar being counterpoised, though
still unloaded with weights, may be brought down till it just touches
the knife-edge; it will then remain horizontal, and will retain this
position whether the knife-edge be at either end of the bar or in any
intermediate position. I shall hang weights at the extremities of the
rod, and we shall find that there is for each pair of weights just one
position at which, if the knife-edge be placed, it will sustain the
rod horizontally. We shall then examine the relations between these
distances and the weights that have been attached, and we shall trace
the connection between the results of this method and those of the
arrangement that we last used.
63. Supposing that 6 lbs. be hung at each end of the rod, we might
easily foresee that the knife-edge should be placed in the middle, and
we find our anticipations verified. When the edge is exactly at the
middle, the rod remains horizontal; but if it be moved, even through
a very small distance, to either side, the rod instantly descends
on the other. The knife-edge is 24 inches distant from each end; and
if I multiply this number by the number of pounds in the weight, in
this case 6, I find 144 for the product, and this product is the same
for both ends of the bar. The importance of this remark will be seen
directly.
64. If I remove one of the 6 lb. weights and replace it by 2 lbs.,
leaving the other weight and the knife-edge unaltered, the bar
instantly descends on the side of the heavy weight; but, by slipping
the knife-edge along the bar, I find that when I have moved it to
within a distance of 12 inches from the 6 lbs., and therefore 36 inches
from the 2 lbs., the bar will remain horizontal. The edge must be put
carefully at the right place; a quarter of an inch to one side or the
other would upset the bar. The whole load borne by the knife-edge is
of course 8 lbs., being the sum of the weights. If we multiply 2, the
number of pounds at one end, by 36, the distance of that end from the
knife-edge, we obtain the product 72; and we find precisely the same
product by multiplying 6, the number of pounds in the other weight,
by 12, its distance from the knife-edge. To express this result
concisely we shall introduce the word _moment_, a term of frequent use
in mechanics. The 2 lb. weight produces a force tending to pull its
end of the bar downwards by making the bar turn round the knife-edge.
_The magnitude of this force, multiplied into its distance from the
knife-edge, is called the moment of the force._ We can express the
result at which we have arrived by saying that, when the knife-edge has
been so placed that the bar remains horizontal, _the moments of the
forces about the knife-edge are equal_.
65. We may further illustrate this law by suspending weights of 7 lbs.
and 5 lbs. respectively from the ends of the bar; it is found that
the knife-edge must then be placed 20 inches from the larger weight,
and, therefore, 28 inches from the smaller, but 5 × 28 = 140, and
7 × 20 = 140, thus again verifying the law of equality of the moments.
From the equality of the moments we can also deduce the law for the
distribution of the load given in Art. 53. Thus, taking the figures in
the last experiment, we have loads of 7 lbs. and 5 lbs. respectively.
These produce a pressure of 7 + 5 = 12 lbs. on the knife-edge. This
edge presses on the bar with an equal and opposite reaction. To
ascertain the distribution of this pressure on the ends of the beam,
we divide the whole beam into 12 equal parts of 4 inches each, and the
7 lb. weight is 5 of these parts, _i.e._, 20 inches distant from the
support. Hence the edge should be 20 inches from the greater weight,
which is the condition also implied by the equality of the moments.