241. In the lever of the second order the power is at one end, the
fulcrum at the other end, and the load lies between the two: this lever
therefore differs from the lever of the first order, in which the
fulcrum lies between the two forces. The relation between the power
and the load in the lever of the second order may be studied by the
arrangement in Fig. 39.
242. The bar A C is the same rod of iron 72" × 1" × 1" which was used
in the former experiment. The fulcrum A is a steel edge on which the
bar rests; the power consists of a spring balance H, in the hook of
which the end C of the bar rests; the spring balance is sustained by
a wire-strainer, by turning the nut of which the bar may be adjusted
horizontally. The part of the bar between the fulcrum A and the power
C is divided into five portions, each 1' long, and the points A and C
are each 6" distant from the extremities of the bar. The load employed
is 56 lbs.; through the ring of this weight the bar passes, and thus
the bar supports the load. The bar is counterpoised by the weight of 19
lbs. at G, in the manner already explained (Art. 231).
243. The mode of experimenting is as follows:—Let the weight B be
placed 1' from the fulcrum; the strain shown by the spring balance
is about 11 lbs. If we calculate the value of the power by the rule
already given, we should have found the same result. The product of
the load by its distance from the fulcrum is 56, the distance of the
power from the fulcrum is 5; hence the value of the power should be
56 ÷ 5 = 11·2.
[Illustration: FIG. 39.]
244. If the weight be placed 2' from the fulcrum the strain is about
22·5 lbs. and it is easy to ascertain that this is the same amount as
would have been found by the application of the rule. A similar result
would have been obtained if the 56 lb. weight had been placed upon any
other part of the bar; and hence we may regard the rule proved for the
lever of the second order as well as for the lever of the first order:
that, the power multiplied by its distance from the fulcrum is equal to
the load multiplied by its distance from the fulcrum. In the present
case the load is uniformly 56 lbs., while the power by which it is
sustained is always less than 56 lbs.
[Illustration: FIG. 40.]
245. The lever of the second order is frequently applied to practical
purposes; one of the most instructive of these applications is
illustrated in the shears shown in Fig. 40.
The shears consist of two levers of the second order, which by their
united action enable a man to exert a greatly increased force,
sufficient, for example, to cut with ease a rod of iron 0"·25 square.
The mode of action is simple. The first lever A F has a handle at one
end F, which is 22" distant from the other end A, where the fulcrum
is placed. At a point B on this lever, 1"·8 distant from the fulcrum
A, a short link B C is attached; the end of the link C is jointed to a
second lever C D; this second lever is 8" long; it forms one edge of
the cutting shears, the other edge being fixed to the framework.
246. I place a rod of iron 0"·25 square between the jaws of the shears
in the position E, the distance D E being 3"·5, and proceed to cut the
iron by applying pressure to the handle. Let us calculate the amount by
which the levers increase the power exerted upon F. Suppose for example
that I press downwards on the handle with a force of 10 lbs., what is
the magnitude of the pressure upon the piece of iron? The effect of
each lever is to be calculated separately. We may ascertain the power
exerted at B by the rule of moments already explained; the product of
the power and its arm is 22 × 10 = 220; this divided by the number of
inches, 1·8 in the line A B, gives a quotient 122, and this quotient
is the number of pounds pressure which is exerted by means of the
link upon the second lever. We proceed in the same manner to find the
magnitude of the pressure upon the iron at E. The product of 122 and 8
is 976. This is divided by 3·5, and the quotient found is 279. Hence
the exertion of a pressure of 10 lbs. at F produces a pressure of 279
lbs. at E. In round numbers, we may say that the pressure is magnified
28-fold by means of this combination of levers of the second order.
247. A pressure of 10 lbs. is not sufficient to shear across the bar
of iron, even though it be magnified to 279 lbs. I therefore suspend
weights from F, and gradually increase the load until the bar is cut. I
find at the first trial that 112 lbs. is sufficient, and a second trial
with the same bar gives 114 lbs.; 113 lbs., the mean between these
results, may be considered an adequate force. This is the load on F;
the real pressure on the bar is 113 × 27·9 = 3153 lbs.: thus the actual
pressure which was necessary to cut the bar amounted to more than a ton.
248. We can calculate from this experiment the amount of force
necessary to shear across a bar one square inch in section. We may
reasonably suppose that the necessary power is proportional to the
section, and therefore the power will bear to 3153 lbs. the proportion
which a square of one inch bears to the square of a quarter inch; but
this ratio is 16: hence the force is 16 × 3153 lbs., equal to about
22·5 tons.
249. It is noticeable that 22·5 tons is nearly the force which
would suffice to tear the bar in sunder by actual tension. We shall
subsequently return to the subject of shearing iron in the lecture upon
Inertia (Lecture XVI.).