Experimental Mechanics / A Course of Lectures Delivered at the Royal College of Science for Ireland

by Robert S. Ball

THE LEVER OF THE SECOND ORDER.

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.

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.

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].).