These pulleys, A and B, are fastened solidly together and rotate as one about a fixed axis; the pulley C is in the movable block. An endless chain passes over the pulleys as shown, the rims of the pulleys being grooved and fitted with lugs to prevent the chain from slipping. The loop np hangs free and is the pulling loop.

From the figure it is easily seen that, if we pull down on p until pulley A is turned once around, the branch m will be shortened a length equal to the circumference of A. Since B is attached to A, it also will turn once around and the branch o will be lengthened a distance equal to the circumference of B.

Hence, the loop mo will be shortened by an amount equal to the difference of the circumferences of A and B; and the pulley C will rise one-half this amount.

We can express the difference in the distances moved by m and o as

where D and d represent the diameters of large and small pulleys, A and B. Hence C will move up one-half of this or of TX (D-d). To cause this motion of C upward, the chain p was moved a distance of XD.

The mechanical advantage of the hoist is obtained by dividing the motion of P by the motion of W.

This can be simplified by cancelling 7 out of both numerator and denominator of the fraction, leaving

This formula might be written as a rule in the following words: "The mechanical advantage of a differential hoist is obtained by dividing the diameter of the larger pulley in the upper block by half the difference between the diameters of the larger and smaller pulleys."

A differential hoist can actually lift about 30% of the theoretical load with a given pull; that is, the efficiency is about 30%. Likewise, to lift a given weight will require about 100 or 31

times the theoretical force. In other words, the actual force must be such that the 30% that is really effective will equal the theoretical force.

Example:

Calculate the actual pull required to lift 600 lb. with a differential hoist having 10 and 8 in. pulleys and an efficiency of 30%.

Explanation: In this case the letters D and d of the formula are 10 in. and 8 in., and we find the M. A. to be 10. It should, therefore, only require a force of 60 lb. to raise the 600 lb. weight. But we find that this type of hoist has only an efficiency of 30%, that is, it only does 30% of what we might expect it to do from our theories. Then to lift 600 lb. will require a force such that 30% of it will be 60 lb. This necessary force is 200 16.

PROBLEMS

169. A weight of 2000 lb. is to be lifted with a four sheave and three sheave pair of blocks, as shown in Fig. 57. The four sheave is used as the movable block. Neglecting friction and assuming each man to be capable of pulling 125 lb., how many men are necessary?

170. A windlass and tackle blocks, as shown in Fig. 58, are used for moving a house. If the team can exert a steady pull of 200 lb. at the end of the sweep, find the theoretical pull on the house. Also find the actual pull on the house if the efficiency of the whole mechanism is 65%.

171. Draw a sketch of a pair of blocks, each having 3 pulleys, and indicate which should be the movable block in order to secure the greatest mechanical advantage. What would the mechanical advantage be?

172. When the geared windlass of the dimensions shown in Fig. 59 is used with the pair of pulley blocks, find the weight at W that can be lifted by a force of 25 lb. on the crank.

We

173. A differential hoist has pulleys 7 in. and 6 in. in diameter. attach a weight of 200 lb. to the hoist and find that a pull of 58 lb. is required to raise the weight.

(a) Find the theoretical force required to raise 200 lb. with

this hoist.

(b) From this and the force actually required, calculate the

efficiency of the hoist.

174. Three men pull 70 lb. apiece on a pair of pulley blocks, two sheaves above and one below. The single block is movable. Find the weight that can be lifted:

(a) Neglecting friction;

(b) Assuming that 40% of the work is lost in friction. 175. A load of 2 tons is to be lifted with a differential hoist. The pulleys are 12 in. and 10 in. in diameter.

(a) What is the theoretical pull required to lift the load?
(b) What is the actual pull required, if the efficiency of the
hoist is 30%?

CHAPTER XV

THE INCLINED PLANE AND SCREW

98. The Use of Inclined Planes.-An Inclined Plane is a surface which slopes or is inclined from the horizontal. Any one who has had experience in raising heavy bodies from one level to another knows that inclined planes are very useful for such work. The Wedge is a form of an inclined plane, the powerful effect of which in splitting wood, quarrying stone, aligning machinery, and performing many other heavy duties is well known. The inclined plane, like the lever and the tackle block, enables us to lift a heavy weight with a smaller force.

99. Theory of the Inclined Plane.-The work done in moving a body up an inclined plane is merely the work of raising the body vertically. If we skid an engine base from the shop floor onto a flat car, the work accomplished is the raising of the base from the floor level to the car level, and is the same as if it was raised straight up by a crane, or by tackle blocks. The effect of the long incline is similar to that of a long force arm on a lever. It enables the force doing the work to use a greater distance, and hence the force will be smaller than the weight raised.

Neglecting friction or, in other words, supposing bodies to be perfectly smooth and hard, no work is done in moving the bodies

in a horizontal direction; hence the work done upon a body when it is moved equals the weight of the body times the vertical height to which it is raised. In studying the theory of the inclined plane, we find that the force generally acts in one of two directions in raising the body: either parallel to the incline or parallel to the horizontal base.

In Case I (Fig. 60) the force P is exerted along the incline, and, in raising the weight to the top, will act through a distance l. Meanwhile, it will raise W a distance h. Consequently, the

If we remember that the work put in equals the work got out of a machine (neglecting what is lost in friction) we see that we have the formula

PXl=WXh

To sum up, when the force is exerted parallel to the surface of the inclined plane, the force times the length of the inclined plane equals the weight times the vertical height through which

the weight is raised by the plane; or the mechanical advantage equals the length of the inclined surface divided by the height. If the weight to be raised is great as compared with the force available, a comparatively long incline must be used to give the necessary mechanical advantage.

In Case II (Fig. 61) the force acts parallel to the base of the inclined plane; that is, along the horizontal. This case is not often found in this elementary form, but is seen in jack screws, in worm gearing, in wedges, and in cams, all of which are modifications of inclined planes. When the force acts parallel to the base, the work expended by it is the product of the force times the length of the base; the work accomplished is, as before, the product of the weight times the height.