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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?

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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 mechanical advantage will be

h

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

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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.

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A comparison of these formulas with those for Case I shows that the mechanical advantage is greatest where the force P is exerted along the incline, as in Case I, because l, the length of the incline or the hypotenuse of a right triangle, is greater than b, the length of the base.

There may be other cases where the force acts in some other direction, but they are seldom seen in practice.

100. The Wedge.—The Wedge consists of two inclined planes placed base to base, the force acting parallel to the base, as shown in Fig. 62, where the horizontal center line of the wedge is

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the common base of the two inclined planes. Usually the wedge is moved instead of the object to be raised, but the effect is the same and the force relations are the same as if the object itself were being moved up a stationary incline. In Fig. 62 it will be seen that the weight W will be raised a distance h when the wedge is driven a distance l. The work expended in driving the wedge is PXl; the work accomplished in raising the weight is W×h; and, neglecting friction, these are equal, or

PXl=Wxh

From this we see that the mechanical advantage of the wedge is:

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The relation of P and W might, if desired, be written as a proportion, as follows:

Example:

P:W=h:l

Fig. 63 shows an adjustable pillow block for a Corliss engine, the bearing being raised or lowered by means of the wedge underneath. If the weight of the shaft and the fly-wheel upon this bearing is 6000 lb., and the wedge has a taper of 1 in. per foot of length, what pressure must be exerted on the wedge by the screw S in raising the shaft?

Mech. Adv. =12

6000 ÷ 12 =500 lb. Answer.

Explanation: If the taper of the wedge is 1 in. in 12 in., then a motion of 1 ft. would raise the bearing 1 in.; or a motion of 1 in. in the wedge would raise the bearing in. Hence, the Mech. Adv. of the wedge is 12 and

P=

W

12

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Note.-There would also be required, in addition to this 500 lb., a force sufficient to overcome the friction on the top and bottom of the wedge, which is neglected in this solution.

0000

FIG. 63.

101. The Jack Screw.-A screw is nothing but an inclined plane which, instead of being straight, is wrapped around or cut into a round rod or bar. Turning the screw gives the same effect as giving a straight push on an inclined plane or wedge.

When we raise an object with a jack screw, such as shown in Fig. 64, the weight presses down on the screw and, consequently, W

P

R

FIG. 64.

is borne by the threads (which are the inclined planes). The threads are advanced and the weight is raised by a pull (which we will call P) on the end of the rod whose length is marked R. The distance the weight W is moved for one revolution of the screw equals the lead of the screw expressed as a fraction of an inch. The lead of a screw is the distance it advances lengthwise in one turn or revolution. The force P moves through a distance

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