equal to the circumference of a circle whose radius is the length of the handle; if we represent the length of the handle or lever by R, then the distance traversed by P in one revolution is π×2×R. If we let the letter L represent the lead of the screw then we will have the work accomplished in one revolution of the screw= WXL. Meanwhile, the work expended in doing it Assuming that there is no friction in the screw, we have If stated in words, these formulas would read: "The force. multiplied by the circumference of the circle through which it moves equals the weight multiplied by the lead of the screw." "The mechanical advantage of a jack screw equals the circumference of the circle through which the force moves divided by the lead of the screw" (the amount the screw advances in one turn). Example: With a 1 in. jack screw having 3 threads per inch and a pull of 50 lb. at a radius of 18 in., calculate: (a) The theoretical load that can be lifted by the screw; (a) Mech. Adv. Explanation: If there are 3 threads per inch, the lead is in., and if the radius is 18 in., we have the Mech. Adv. =339.3. In theory then we should be able to lift 50 lb. X339 = 16950 lb. with this screw. But a screw has considerable friction and for this reason only 18% of the energy expended in this case is effective, the remaining 82% being all lost in friction. actual weight lifted is, therefore, only 18% of 16950 lb. or 3051 lb. The 102. Efficiencies.-In explaining the machines of this chapter and of Chapters XIII and XIV, it was assumed that no work is lost in friction within the machines. In a properly mounted lever there is little energy lost. In a tackle block the loss depends on the size of the pulleys as compared with the size of the rope and on the nature of the pulley bearings. The efficiency may vary from 60 to 95%. The more pulleys there are, the lower will be the efficiency, because each bend in the rope and each pulley means a loss in friction. With inclined planes, the efficiency may vary all the way from 0 to nearly 100%. It will be lowest if the weight is merely slid on the plane and will be much higher if wheels or rollers are used. In any machine, if the weight will start back of its own accord when the force is removed, the friction is less than 50% and the efficiency is greater than 50%. If the weight will not start back, the efficiency is less than 50%. This can be shown as follows: Of the force applied to a machine, part of it is absorbed in overcoming the friction within the machine. The balance goes through the machine and is effective in accomplishing the work to be done. Of the whole force applied, the per cent which this effective force represents is called the Efficiency. If the efficiency is less than 50%, it shows that the friction absorbs more than half of the total force and, therefore, that the friction is greater than the effective force. Now, suppose we had a simple machine such as a jack-screw, being used to raise a weight. If the applied force is removed, the friction will remain the same, but will now act to hold the weight from going back. If the friction is sufficient to hold the weight, it must at least equal the effective or theoretical force required to raise the weight. Therefore, if a machine does not run backward when the force is removed, the friction must be more than one-half of the total force required to raise the weight, and the effective force must be less than onehalf of this total force. Hence, the efficiency in such a case is less than 50%. A jack-screw will not go down of its own accord when the force is removed and therefore its efficiency is less than 50%. In reality, for the usual dimensions of screws, it has been found to be only from 15 to 20%. Mr. Wilfred Lewis has derived, from experiment, a simple formula which gives the average efficiency for a jack-screw under ordinary conditions. Example: Find the probable efficiency of the screw given in the example under Article 101. One can get an approximate idea of the efficiency of any machine by observing, as before explained, whether or not it will run backward of its own accord when the force is removed. This will tell whether the efficiency is above or below 50%. If it is above 50% and a considerable force is required to keep the weight from going back, then the efficiency is high. If, however, a very slight pull will hold it from going back, then the efficiency is not very much above 50%. If we find the efficiency to be under 50% but find that only a very small pull will start the weight down, then the efficiency is not far under 50%. On the other hand, if it seems as if almost as great a force is required to lower the weight as to raise it, this signifies that the efficiency of the machine is extremely low. PROBLEMS 176. An engine weighing 5 tons is to be loaded onto a car, the floor of which is 6 ft. from the ground. If 16 ft. timbers are used for the runway, find the pull necessary to draw the engine up the slope, neglecting friction. 177. How many pounds must a locomotive exert to pull a train of 50 cars, each weighing 50 tons, up a grade of 3 in. in 100 ft.? 178. A building is to be raised by means of 4 jack-screws; the screws are 2 in. in diameter, with 4 threads to the inch. The lever is 20 in. long and a 30 lb. force is required on each handle. Calculate the theoretical weight which the four screws should lift under these conditions. 179. Calculate the probable efficiency of these jack-screws from Lewis' formula and estimate the probable weight of the building. 180. A windlass with an axle 8 in. in diameter and crank 18 in. long is used in connection with an inclined plane 20 ft. long and 5 ft. high, as shown in Fig. 65. Neglecting friction, what weight can be pulled up the slope with a force of 150 lb. on the crank? CHAPTER XVI WORK, POWER, AND ENERGY; HORSE-POWER OF BELTING 103. Work.-Whenever a force causes a body to move, work is done. Unless the body is moved, no work is accomplished. A man may push against a heavy casting for hours and, unless he moves it, he does no work, no matter how tired he may feel at the end of the time. It is evident that there are two factors to be considered in measuring work-force and distance. In the study of levers, tackle blocks, and inclined planes we dealt with the problem of work. In any of these machines the work accomplished in lifting a weight is measured by the product of the weight and the distance it is moved. The work expended or put into the machine to accomplish this is the product of the force exerted times the distance through which this force must act. We found that, if we neglect the work lost in friction, the work put into a machine is equal to the work accomplished by it. The actual difference between the work put in and the work accomplished is the amount that is lost in friction. The following expressions may make these relations clearer: Work lost in Friction = Work put in- Work got out. 104. Unit of Work.-The unit by which work is measured is called the Foot-pound. This is the work done in overcoming a resistance of one pound through a distance of 1 ft.; that is, if a weight of 1 lb. is lifted 1 ft., the work done is equal to 1 footpound. All work is measured by this standard. The work in foot-pounds is the product of the force in pounds and the distance in feet through which it acts. In lifting a weight vertically, the resistance, and hence, the force that must be exerted, is equal to the weight itself in pounds. The work done is the product of the weight times the vertical distance that it is raised. If a weight of 80 lb. is lifted a distance of 4 ft., the work done is 80X4 or 320 foot-pounds. It would require this same amount of work to lift 40 lb. 8 ft., or to lift 20 lb. 16 ft. When a body s moved horizontally, the only resistance to be overcome is the friction. When a team of horses pulls a loaded wagon, the only resistances which it must overcome are the friction between the wheels and the axles, and the resistance on the tires caused by the unevenness of the road. The work necessary to pump a certain amount of water is the weight of the water times the height through which it is lifted or pumped (plus, of course, the work lost in friction in the pipes). The work necessary to hoist a casting is the weight of the casting times the height to which it is lifted. The work done by a belt is the effective pull of the belt times the distance in feet which the belt travels. The work done in hoisting an elevator is the weight of the cage and of the load it carries times the height of the lift. Numerous other illustrations of work will suggest themselves to the student. 105. Power.-Power is the rate of doing work; that is, in calculating power the time required to do a certain number of footpounds of work is considered. If 10,000 lb. are lifted 7 ft. the work done is 70,000 foot-pounds, regardless of how long it takes. But, if one of two machines can do this in one-half the time that the other machine requires, then the first machine has twice the power of the second. = The engineer's standard of power is the Horse-power, which may be defined as the ability to do 33,000 foot-pounds of work per minute. The horse-power required to perform a certain amount of work is found by dividing the foot-pounds done per minute by 33,000. If an engine can do 1,980,000 foot-pounds in a minute, its horsepower would be 1,980,000÷33,000 60. An engine that can raise 66,000 lb. to a height of 10 ft. in 1 minute will do 66,000 lb. ×10 ft.=660,000 foot-pounds per minute, and this will equal 660000=20 horse-power. If another engine takes 4 minutes to do this same amount of work, it is only one-fourth as powerful; the work done per minute will be 660000=165,000 foot-pounds per minute; and its horse-power is 185000-5 horse-power. Example: 4 An electric crane lifts a casting weighing 3 tons to a height of 20 ft. from the floor in 30 seconds; what is the horse-power used? |