Example: We wish to lift 8000 lb. with a compound lever as shown in Fig. 42, the first one being 10 ft. long with the weight 2 ft. from the end; the second 16 ft. long with the fulcrum 4 ft. from the end; what will be the necessary force, P2? P1X10-8000X2 16000 PXa=Wxb = 1600 lb. 1 10 W2-P1=1600 lb. 2 P2X12-1600X4 6400 P2 = 12 533 lb., Answer. Explanation: Taking the first, or lower lever, we find it to be an example of the second class. W has a weight arm b of 2 ft. The force has an arm equal to the whole length of the lever, or 10 ft. The necessary force on the end of this lever we find to be 1600 lb. The second lever must pull upward through the connection with a force of 1600 lb. In other words, the 8000 lb. weight on the first lever is equivalent to a 1600 lb. weight on the short end of the second lever. The first lever pulls downward the same amount that the second pulls up, or P1=W2. Having this 1600 as the weight, we find that a force of 533 lb. is needed on the end of the second lever. 94. Mechanical Advantage.-The ratio of the weight to the force is often called the Mechanical Advantage of the lever; this ratio is equal to the force arm the weight arm. In the compound lever of Fig. 42 the M. A. (mechanical advantage) of the first lever equals 10÷2 or 5; of the second, 12÷4 or 3; the M. A. of a compound lever is equal to the product of the M. A. of the separate single levers; hence, of the given compound lever the M. A. is 5×3 or 15. This means that a 1 lb. force will lift 15 lb.; 10 lb. will lift 150; or 100 lb. will lift 1500. The force multiplied by the M. A. gives the weight that can be lifted, or the weight divided by the M. A. gives the necessary force. In the case shown in Fig. 42, the mechanical advantage is 15 and consequently the necessary force is 8000÷15=5331⁄2 lb. If the mechanical advantage of a lever is 10, then 1 lb. will lift 10 lb., or 800 lb. will lift 8000 lb., etc.; but it must be remembered that the 1 lb. or the 800 lb. must travel 10 times as far as the 10 lb. or the 8000 lb. If a lever has a mechanical advantage of 10, the force must travel 10 times as far as it lifts the weight, and consequently a lever effects no saving in work. Work is the product of force, or weight, times the distance moved, and is the same for either end of the lever. It is similar to carrying a lot of castings to the top floor of a building. If I carry half of them at a time, I must make two trips; if I carry one-fourth of them at a time, I must make four trips. The lighter the load, the more trips I must make. The work done is the same whatever way I carry them and is equal to the product of the total weight times the height to which the load must be carried. 95. The Wheel and Axle.-This is a name given in mechanics to the modification of levers that enables them to be rotated continuously. Fig. 43 shows the principle of this: By wrapping a belt or rope around each of the two circular bodies, we find that the pulls in the cords are in inverse proportion to the radii of the circles. A little consideration shows that the wheel and axle may easily be studied as a force, P, with lever arm R, equal to the radius of the wheel; and a weight, W, with a lever arm r, equal to the radius of the axle. Two pulleys on a countershaft FIG. 43. The belt which The radius of this might be likened to a lever in the same way. drives the countershaft furnishes the force P. pulley is the force arm. The radius of the other countershaft pulley, which transmits the power to the machine, is the weight arm and the pull in this belt is the weight. Gears also are levers that can be rotated continuously. The simplest example of the use of the axle is probably the windlass, which we see used for hoisting, house-moving, etc. See Figs. 48 and 50. A geared windlass, such as shown in Fig. 50, is a case of compound levers. The crank and pinion form the first lever and the load on the gear teeth is transmitted to the teeth of the larger gear and becomes the force of the other lever, which consists of the large gear and the drum. Example: A geared windlass, such as shown in Fig. 50, has a crank 20 in. long; the small gear is 6 in. in diameter, the large gear is 30 in. in diameter, and the diameter of the drum is 6 in. What load could be raised by a man exerting a force of 25 lb. on the crank? 6÷2=3 in., radius of pinion 20÷3-6, M. A. of crank and pinion 6÷2-3 in., radius of drum 15÷3=5, M. A. of gear and drum 6×5=33 in., total mechanical advantage 25×33833+ lb., Answer. The solution of this problem might be shortened by writing all the work in a single equation: If we do not know the sizes of the gears in inches, but know the numbers of teeth, we can figure that the mechanical advantage of the pair of gears is the ratio of the numbers of teeth. In such a case with a hoist as shown in Fig. 50, we would first find the mechanical advantage of a simple windlass with the crank attached directly to the drum; then find the M. A. of the pair of gears, and by multiplying these two quantities together we would get the mechanical advantage of the entire hoist. 200 PROBLEMS 1800 12 FIG. 44. 161. The lever shown in Fig. 44 is 12 ft. long. Where should the fulcrum be placed so that a weight of 200 lb. will lift a weight of 1800 lb.? 162. In Fig. 45 where should the weight of 1800 lb. be placed so that it can be lifted by a force of 200 lb.? 163. Fig. 46 shows a safety valve V loaded with a 50 lb. weight at W. Find the total steam pressure on the bottom of V necessary to lift the valve. 4" 12" FIG. 46. 164. From the result of problem 163, find the steam pressure per square inch if V is 1 in. in diameter on the bottom where exposed to the steam. 165. Fig. 47 shows the clutch pedal for an automobile. What must be the length of the power arm a in order that a foot pressure of 15 lb. can open the clutch against a spring pressure of 60 lb. having an arm of 3 in.? 166. Fig. 48 shows an old fashioned windlass for raising water. If the crank is 15 in. long, and the drum is 5 in. in diameter, what pressure would be needed on the crank to raise a pail of water weighing 30 lb.? 167. Fig. 49 represents in an elementary way the levers of a pair of platform scales. How far from the fulcrum must the 5 lb. weight be placed to balance the 1000 lb. weight located as shown? 168. The hoist of Fig. 50 has an 18 in. crank; the drum is 10 in. in diameter; the diameter of the large gear is 30 in., and of the small gear 6 in. What weight can be raised by a force of 25 lb. on the crank? CHAPTER XIV TACKLE BLOCKS 96. Types of Blocks.-When a heavy weight is to be raised or moved through any considerable distance, either a windlass, such as described in Chapter XIII, or tackle blocks can be used. Referring to the figures in this chapter, the revolving part is called the Pulley or Sheave; the framework surrounding the pulleys is called the Block and, as generally used, includes both the frame and the sheaves contained in it. In Fig. 51 we have a single pulley which serves merely to give a change of direction. There is no mechanical advantage in a single fixed pulley such as this. The pull on the rope at P is transmitted around the pulley and supports W on the other side. We can look at the pulley in this case as a lever with equal arms. P on one end must equal W on the other end. Such a block would be used solely for the convenience it affords, since it is usually easier to pull down than up. |