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or the ratio of force arm to weight arm.

W
P

If the M.A. equals 10,

a

we know that equals 10, and, therefore, equals 10.

b

In a compound lever the M.A. of the system of levers equals the product of the M.A.'s of all the separate single levers. The M.A. of the first lever of Fig. 48 equals

12

4

10
2'

or 5, and of the second,

or 3; hence, the M.A. of the compound lever of Fig. 48 is

5 X 3 15. The load W is 8000 lb., so that the force required at P2 equals the weight divided by the M.A., or 8000 ÷ 15 = 533 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.

89. The Wheel and Axle.-This is a name given in mechanics to the modification of levers that enables them to be rotated

b

W

FIG. 49. Simple lever pivoted at center.

continuously. That the wheel, pulley, or axle is a lever is not apparent at first glance, but a close study of Figs. 49 and 50 should make it clear. The simple lever of Fig. 49 is a straight piece pivoted on a shaft which furnishes the fulcrum, and since

a and b are equal, its M.A. is 1. In Fig. 50 is a wheel or pulley with a rope running over it, and a weight W at one end of the rope is balanced by a force P at the other end. By dotting in the lever of Fig. 49 on top of the wheel, for comparison, we see

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at once that the wheel is in reality a simple lever. The radius of the wheel is both the weight arm and the force arm and its

R
R'

M.A. is, therefore, or 1.

W

FIG. 51.-Wheel and axle.

By using two pulleys or wheels of different diameters attache to the same shaft, an M.A. greater than 1 can be obtained. This principle is illustrated by the wheel and axle of Fig. 51.

The force P exerted on the rope which is wound on the wheel, balances the weight W, hung from the rope which is wound on the axle. The force arm is thus equal to the radius, R, of the wheel, the weight arm equals the radius, r, of the axle, and the R

M.A. equals the ratio of force arm to weight arm, or M.A. =

r

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Two pulleys on a countershaft may be likened to a lever in the same way. The belt which drives the countershaft furnishes the force P. The radius of this 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

FIG. 52.-Geared hoist or windlass.

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, housemoving, etc.

A geared windlass, such as shown in Fig. 52, 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. 52, 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

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If we do not know the sizes of the gears in inches, but know the number of teeth on each, we can calculate the total M.A. in the same manner by using the number of teeth instead of the radius of each gear. This does not give the correct value for the M.A. of the separate parts, such as the crank and pinion, but the product of all the M.A.'s or the total M.A. will be the correct value.

90. Efficiency. In all classes of machines, the lever included, there is a certain amount of force, power, or energy that is lost within the machine itself due to friction. Previous to this, such losses have been neglected, and the forces were calculated on the assumption that the machines were frictionless.

As a

It is easily seen that if the windlass of Fig. 52 were unoiled and rusty it would require a greater force at the crank to pull a given load on the rope than if all the bearing surfaces were oiled and in good running condition. What then becomes of the additional force applied at the crank in the first condition, since the force delivered to the rope is the same in both cases? matter of fact, this extra force is simply used up in overcoming the additional friction in the machine due to the rusty bearings, etc. For any useful purpose it is lost beyond recovery. In spite of the best methods of oiling bearings, smoothing the surfaces, etc., friction cannot be eliminated entirely, and its presence is a very important factor in industry.

To draw an analogy, a machine might be compared to a leaky hose; the faster the hose leaks, the faster must the water be supplied at one end in order to deliver a constant stream at the other end. Thus, in a machine, friction is a leak, and a certain amount of the force or energy supplied to the machine is lost through this leak. The greater the friction, the greater is the loss of power.

Again considering the windlass of Fig. 52, it will be noted from the example of the previous paragraph that the windlass has a M.A. equal to 333. From the principles of levers, a force of 25 lb. at the crank will exert a pull of 25 × 33 = 833 lb. on

the rope. This is theoretically true (neglecting friction), but actually, however, in spite of well-oiled bearings, a certain amount of force will be lost due to friction, and the pull on the rope will be less than 833 lb., possibly no more than 750 lb. Supposing this to be the case, we then have a machine supplied with sufficient force to exert a pull of 833 lb. but actually only

delivering a pull of 750 lb.

750
833

= .90. In other words, the

force delivered is only .9 or 90% of the force supplied. This

750

fraction, 833' or 90%, is ca ed the efficiency of the machine.

Efficiency may be defined as the ratio of the power or energy delivered up by a machine to that supplied to it.

In calculating efficiencies such as the above, the force supplied must be calculated at the delivery end of the machine in order to compare it with the force delivered. In other words, the machine is supplied with a 25-lb. force at the crank which is equivalent to a force of 833 lb. at the rope, or delivery end. If the efficiency of the windlass had been 75%, the actual pull on the rope would be .75 X 833 = 625 lb. Expressed in words, 625 lb., the force delivered, is 75 % of 833 lb., the force supplied.

Example:

Assuming that the windlass of Fig. 52 has an efficiency of 85% and its M.A. is 33, what force will be required at the crank to deliver a pull of 1700 lb. at the rope?

Explanation: It is evident that 1700 lb. is the pull to be delivered. Since the efficiency is 85%, the pull delivered is 85% of the force supplied; therefore, 1700 is 85% of the force supplied.

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150. In the lever of Fig. 53, where should the fulcrum be placed so that the weight and the force will be balanced?

151. In the lever of Fig. 54 where should the weight of 1800 lb. be placed so that it may be lifted by a force of 200 lb.?

152. Figure 55 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.

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