127. The single movable pulley. Now let the force of the earth's attraction upon the mass R be overcome by a single movable pulley, as shown in Fig. 115. Since the weight of R (R representing in this case the weight of both the pulley and the suspended mass) is now supported half by the strand C and half by the strand B, the force E which must act at A to hold the weight in place, or to move it slowly upward if there is no friction, should be only one half of R. A reading of the balance will show that this is indeed the case. Experiment thus shows that in the case of the single movable pulley the effort E is just one half as great as the resistance R. R B FIG. 115. The single movable pulley But when we again consider the work which the force E must do to lift the weight R a distances, we see that A must move upward 2 inches in order to raise R 1 inch; for when R moves up 1 inch, both of the strands B and C must be shortened 1 inch. As before, therefore, since R= 2 E and 8 = √ 8, EX 8=RX 8'; that is, in the case of the single movable pulley, as in the case of the fixed pulley, the work put into the machine by the effort E is equal to the work accomplished by the machine against the resistance R. 128. Combinations of pulleys. Let a weight R be lifted by means of such a system of pulleys as is shown in Fig. 116, either (1) or (2). Here, since R is supported by 6 strands of the cord, it is clear that the force which must be applied at A in order to hold R in place, or to make it move slowly upward if there is no friction, should be but 1⁄2 of R. The experiment will show this to be the case if the effects of friction, which are often very considerable, are eliminated by taking the mean of the forces which must be applied at E to cause it to move first slowly upward and then slowly downward. The law of any combination of movable pulleys may then be stated thus: If n represents the number of strands between which the weight is divided, E=R/n. (2) (3) But when we again consider the work which the force E must do in order to lift the weight R through a distance s', we see that, in order that the weight R may be moved up through 1 inch, each of the strands must be shortened 1 inch, and hence the point A must move through n inches; that is, s' s/n. Hence, ignoring friction, = 8 FIG. 116. Combinations of pulleys 129. Mechanical advantage. The above experiments show that it is sometimes possible by applying a small force E to overcome a much larger resisting force R. The ratio of the resistance R to the effort E (ignoring friction) is called the mechanical advantage of the machine. Thus, the mechanical advantage of the single fixed pulley is 1, that of the single movable pulley is 2, that of the system of pulleys shown in Fig. 116 is 6, etc. If the acting force is applied at R instead of at E the mechanical advantage of the systems of pulleys of Fig. 116 is for it requires an application of 6 pounds at R to lift 1 pound at E. But it will be observed that the resisting force at E now moves six times as fast and six times as far as the acting force at R. We can thus either sacrifice speed to gain force, or sacrifice force to gain speed; but in every case, whatever we gain in the one we lose in the other. Thus in the hydraulic elevator shown in Fig. 13, p. 18, the cage moves only as fast as the piston; but in that shown in Fig. 14 it moves four times as fast. Hence the force applied to the piston in the latter case must be four times as great as in the former if the same load is to be lifted. This means that the diameter of the latter cylinder must be twice as great. QUESTIONS AND PROBLEMS 1. Although the mechanical advantage of the fixed pulley is only 1, it is extensively used in connection with clothes lines, awnings, open wells, and flags. Explain. 2. If the hydraulic elevator of Fig. 14, p. 18, is to carry a total load of 20,000 lb., what force must be applied to the piston? If the water pressure is 70 lb. per square inch, what must be the area of the piston? 3. Draw a diagram of a set of pulleys by which a force of 50 lb. can support a load of 200 lb. 4. Draw a diagram of a set of pulleys by which a force of 50 lb. can support 250 lb. What would be the mechanical advantage of this arrangement? 5. Two men, pulling 50 lb. each, lifted 300 lb. by a system of pulleys. Assuming no friction, how many feet of rope did they pull down in raising the weight 20 ft.? WORK AND THE LEVER 130. The law of the lever. The lever is a rigid rod free to turn about some point P called the fulcrum (Fig. 117). First let a meter stick be balanced as in the figure, and then let a mass of, say, 300 g. be hung by a thread from a point 15 cm. from the fulcrum. Then P FIG. 117. The simple lever let a point be found on the other side of the fulcrum at which a weight of 100 g. will just support the 300 g. This point will be found to be 45 cm. from the fulcrum. It will be seen at once that the product of 300 × 15 is equal to the product of 100 x 45. Next let the point be found at which 150 g. just balance the 300 g. This will be found to be 30 cm. from the fulcrum. Again, the products 300 × 15 and 150 × 30 are equal. No matter where the weights are placed, or what weights are used on either side of the fulcrum, the product of the effort E by its distance from the fulcrum (Fig. 118) will be found to be equal to the product of the resistance R by its distance l' A B from the fulcrum. Now the FIG. 118. Illustrating the law of moments, namely, El = Rl' perpendicular distances l and l' from the fulcrum to the line of action of the forces are called the lever arms of the forces E and R, and the product of a force by its lever arm is called the moment of that force. The above experiments on the lever may then be generalized in the following law: The moment of the effort is equal to the moment of the resistance. Algebraically stated, it is It will be seen that the mechanical advantage of the lever, namely R/E, is equal to /'; that is, to the lever arm of the effort divided by the lever arm of the resistance. 131. General laws of the lever. If parallel forces are applied at several points on a lever, as in Figs. 119 and 120, it will be found, in the particular cases illustrated, that for equilibrium and 300 x 20+50 x 40 = 100 x 15+ 200 × 32.5; that is, the sum of all the moments which are tending to turn the beam in one direction is equal to the sum of all the moments tending to turn it in the opposite direction. If, further, we support the levers of Figs. 119 and 120 by spring balances attached at P, we shall find, after allowing for the weight of the stick, that the two forces indicated by the balances are respectively 200 + 100+ 100 = 400 - = and 300+100+200 50 550; that is, the sum of all the forces acting in one direction on the lever is equal to the sum of all the forces act ing in the opposite direction. 50 200 40 32.5 B1 Condition of equilibrium of a bar acted upon by several forces These two laws may be combined as follows: If we think of the force exerted by the spring balance as the equilibrant of all the other forces acting on the lever, then we find that the resultant of any number of parallel forces is their algebraic sum, and its point of application is the point about which the algebraic sum of the moments is zero. a 0 132. The couple. There is one case, however, in which parallel forces can have no single force as their resultant, namely, the case represented in Fig. 121. Such a pair of equal and opposite forces acting at different points on a lever is called a couple and can be neutralized only by another couple tending to produce rotation in the opposite direction. The moment of such a couple is evidently F, xoa + F x ob = F, x ab; that is, it is one of the forces times the total distance between them. The forces applied to the steering wheel of an automobile illustrate the couple. 1 2 F 1 FIG. 121. The couple |