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95. Summary. Work is defined and measured by the product of force and the displacement in the direction of the force. If a force acts on a body, and the displacement has a direction different from that of the force, either (1) the force times a component of the distance or (2) a component of the force multiplied by the distance the body is moved is used for computing the work done. The component which is used is always parallel to the other factor.

The unit of work in the B. E. system is the foot-pound, and in the C. G. S. system it is the erg. Since the erg is a very small unit, the joule, which is equal to 10 ergs, is often used.

In order that a body may gain energy, work must be done on it. If the body does work, it must lose energy by an amount equal to the work which it does. The energy a body possesses is measured by the amount of work which could be done if that body were to lose all its energy. The energy which a body has may be potential or it may be kinetic. A body has kinetic energy when in motion, and the magnitude of the kinetic energy is always equal to mv2. The potential energy is measured either by the net work which is done in giving the body that energy or by the total work done when the body loses its potential energy. In the case of a weight lifted against gravitational forces the potential energy is equal to the product of the weight times the height it is lifted.

Other units used for work or energy are the calorie, the foot-ton, the kilowatt-hour, and the horse-power hour. A table giving the numerical relations of the different units of work is found in section 81.

Power is the rate at which work is done. A rate of work of 550 footpounds per second, or 33,000 foot-pounds per minute, is equal to 1 horse power. A rate of 1 joule per second is equal to 1 watt.

The relation between power, force, and velocity is given by equation (49),

Power = Fv.

Experimental evidence indicates that energy is never created nor destroyed, that the total amount of energy in the universe remains constant. In any system of isolated bodies (isolated in the sense that no energy can go into or out of it) the total energy is a constant. This principle is called the conservation of energy. The energy can change from one form to another, but the total amount in the isolated system will remain constant.

That the energy which we might use is becoming gradually less and less available is a fact of great economic importance. Energy "wasted" in friction is lost beyond recovery. Most of the energy obtained from coal and gasoline is lost forever.

It is important that everyone should understand the available sources of energy and have some idea of their practicability and duration. This question is one of continually increasing importance.

PROBLEMS

1. How much work is done in raising 40 cu. ft. of water to a tank from a well if the average height the water is raised is 30 ft.?

2. A mass of 1500 gm. is raised 200 cm. Compute the work in ergs.

3. If electrical energy costing 10 per kilowatt-hour is supplied to run a machine, how much would 10,000,000 ft.-lb. cost?

4. A force of 500 dynes acts for a distance of 2 m. on a mass of 1 kg. (a) How much acceleration will be given to the mass? (b) How much work will be done?

5. A laborer weighing 150 lb. carries a 10-pound hod containing 40 lb. of bricks up a vertical height of 20 ft. 5 times each hour. How much total work does he perform against gravity in an 8-hour day?

6. (a) How much work is required to roll a 150-pound barrel up an inclined plane 12 ft. long with an elevation of 4 ft.? (b) What is the direction of the least force required to roll the barrel up? (c) What is the magnitude of this force?

7. 100 lb. falls 30 ft. If all the energy could be converted into useful work, how much could be done?

8. A force of 30 lb. acting at an angle of 45° with the horizon drags a body a horizontal distance of 20 ft. Compute the work done.

9. (a) How much work is done in excess of that used in overcoming friction in driving a 3000-pound motor car 600 ft. up a 5 per cent grade (one that rises 5 ft. in each 100 ft. of road)? (b) What is the needed force?

10. A gun shoots a shell weighing a ton with a muzzle velocity of 2250 ft./sec. Compute the energy in foot-pounds.

11. A 3200-pound automobile, starting from rest, coasts down a hill 300 ft. long and 30 ft. high. If 50 per cent of the energy is lost in friction, how much speed should the car have at the bottom of the hill?

12. A constant force of 225 lb. acts on an automobile weighing 3200 lb. (a) Neglecting friction, what speed will the automobile attain on a level road in 10 sec. after the start? (b) How much work will be done in the 10 sec.? (c) How much kinetic energy will the car have at the end of the 10 sec.?

13. A mass of 400 gm. has a kinetic energy of 0.2 joules. (a) What is its velocity? (b) Through what distance must a force of 5000 dynes move in order to give it this energy?

14. (a) What momentum will a force of 4000 dynes give in 5 sec. to a mass of 400 gm. which starts from rest? (b) How much kinetic energy will the mass now have? (c) How much work has been done by this force ?

15. A rifle weighing 7 lb. shoots a bullet, the weight of which is 0.0124 lb. (caliber 0.250), with a muzzle speed of 3000 ft./sec. (a) What is the momentum given to the bullet? (b) What is the kinetic energy in footpounds at the muzzle? (c) If no external force acted on the rifle, how much momentum would be given to it? (d) how much kinetic energy?

16. Find the horse power transmitted by a belt passing over a pulley, 4 ft. in diameter, which makes 300 revolutions per minute, the net force exerted by the belt on the pulley being 50 lb.

17. Prove that 1 horse-power hour is equal to 1,980,000 ft.-lb.

18. 30 tons of coal per hour are lifted 20 ft. by an elevator. What horse power is required?

19. Power is supplied at the rate of 25 watts. How many joules are supplied in 1 min. ?

20. It requires 10 H. P. to drive a certain car at 20 mi./hr. If the total retarding force is three times as great, how many horse power will be required to drive the car at 30 mi./hr.?

21. The retarding force on an automobile traveling at 30 mi./hr. is 200 lb. What horse power must be furnished to maintain that speed?

22. A man weighing 150 lb. runs upstairs, rising 20 ft. in 5 sec. Compute the horse power he develops.

23. An automobile requires 50 H. P. to drive it 60 mi./hr. What is the total retarding force?

24. A man skating at a rate of 100 yd. in 20 sec. pulls a sled with a force of 10 lb. Find the horse power exerted on the sled.

25. It requires 500 H. P. to draw a certain train at the speed of 30 mi./hr. What is the numerical value of the force?

26. How much more horse power must a 3000-pound motor car exert to travel at 30 mi./hr. up a grade that rises 5 ft. in each 100 ft. of road than to travel on a level road at the same rate?

27. A windmill raises half a cubic foot of water 20 ft. each minute. If half the energy is lost through friction, what horse power is developed by the windmill?

CHAPTER VIII

SIMPLE MACHINES

Simple machines, 96. The lever, 97. Mechanical advantage, 98. Types of levers, 99. The inclined plane, 100. Weighing-machines, 101. The hydraulic press, 102. The effect of friction, 103. The efficiency of machines, 104.

96. Simple machines. This chapter gives an account of the applications of the principles of work and energy to some extremely simple machines. While practical machines are usually complicated, they involve the same principles as the simple ones here described. If these principles are understood, it is not difficult to study out the details of the more practical types. It is more important to understand fully and clearly the general principles of machines than it is to understand all the intricate details of a few commercial machines. The student must not expect to learn machinery from this chapter, but he should learn some fundamental principles.

97. The lever. In a previous chapter (sect. 10) the simple lever was explained as an example of the law of moments of force. It was shown that when the force and the weight are in equilibrium,

or

Force X distance from axis = weight × distance from axis,

and, as in equation (1),

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This is called the law of the lever. A different method will now be used to deduce this law. The principle to be used is called the principle of work. It is based on the conservation of energy. The principle of work asserts that in the absence of friction the work supplied to any machine is equal to that done by the machine. Fig. 73 shows a lever which has been moved so that a weight w has

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The two triangles in Fig. 73 are similar; hence

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If this value of d/s is substituted in the preceding equation, equa

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This is the law of the lever, which we have thus derived by a different method, the principle of work.

The force F, as given by equation (1), is just sufficient to balance w. It is not a sufficient force to start w; that is, to accelerate it. If w is to be given an acceleration, more force is required and more work is done; for then the load will be given kinetic energy. The principle of work states that in frictionless machines

Energy supplied = work done,

or Applied force X distance moved=load × distance moved. In using this principle, distances must always be measured parallel to the forces (see section 83).

98. Mechanical advantage. The ratio of the weight to the applied force, w/F, is called the mechanical advantage. It is a measure of the gain in force produced by the machine. While no energy can be gained by the use of a machine, yet there can be a gain in

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