MACHINERY. 



The unit of work adopted in this country is the 

 amount of exertion necessary to overcome a pressure 

 of one pound through a space of a foot. Thus, a 

 hand does a unit of work when it lifts a pound- 

 weight a foot high ; or when it pulls in a foot of 

 the cord of a pulley, the tension of which is equal 

 to a pound-weight; or when it pushes with the 

 force of a pound against the lever of a capstan, 

 while the point against which it presses moves 

 through a foot. It follows that the amount of 

 units of work expended in any exertion of lifting, 

 pushing, or pulling, is found by multiplying the 

 number of pounds in the pressure by the number 

 of feet through which it is exerted. 10 Ibs. of coal 

 lifted 10 feet high is the same amount of work, 

 namely, 100 units, whether it is done in five 

 seconds or in fifty. But when we consider the 

 capacity of the agent for work, we require to take 

 the time into account. An agent or worker that 

 does 100 units of work in one second, has four 

 times the working-power of an agent that requires 

 four seconds to do 100 units. As a standard of 

 capacity for work, the power of a horse is taken. 

 From a number of experiments, Watt, the im- 

 prover of the steam-engine, estimated that a horse, 

 on an average, could do 33,000 units of work in 

 one minute ; that is, could lift 33,000 Ibs. i foot 

 high in I minute ; and this has been generally 

 assumed as the standard of working-power in all 

 mechanical agents. When a steam-engine, for 

 example, is said to be of 60 horse-power, the 

 meaning is, that it is capable of lifting 60 times 

 33,000 Ibs. or 1,980,000 Ibs. i foot high in i 

 minute ; or, which is the same thing, 19,800 Ibs. 

 joo feet high in i minute. 



Ex, How many horse-power does it require to 

 lift i ton of coals every five minutes from a pit 

 1000 feet deep? 2240 X 1000= 2,240,000, the 

 work to be done in 5 minutes. 2,240,000 -j- 5 = 

 448,000, the work to be done in i minute. .'. 448,000 

 -j- 33,000 (the work in one H. P.) = 13-6, the num- 

 ber of H. P. required. 



Work Accumulated in Moving Bodies. 



A heavy body in motion is capable of moving 

 other bodies. The force exerted to put it in 

 motion is all treasured up in it, and ready to be 

 exerted on any obstacle that may oppose it. In 

 other words, a moving body has accumulated in it 

 a power of doing the same amount of work as was 

 done upon it, abating what may have been lost by 

 friction or other obstacles ; and if we know the 

 velocity of a moving body and its weight, we are 

 able to calculate how many units of work have 

 been done upon it, and, therefore, how much work 

 it is capable of doing. 



The velocity being given in feet per second, and 

 the weight in pounds, the rule of calculation is as 

 follows : Square the -velocity, multiply by the weight, 

 and divide by 2 x 32^, or 64$. 



The reason of the rule may be thus shewn : 

 When a body begins to fall from a state of rest, it 

 descends i&rV feet in the first second, and it has 

 then acquired a velocity which, were gravity to 

 cease, would carry it on uniformly over 2 X i&rV, 

 or 32^ feet per second. The spaces descended 

 increase as the squares of the times that is, in 2 

 seconds the fall is 4 times i6Vi feet ; in 3 seconds, 

 9 times i6fV. The velocities, again, increase 

 simply as the times that is, at the end of 2 

 seconds the velocity is 2 times 32$ feet ; of 3 



seconds, 3 times 32^. Putting h for the height or 

 space fallen through in any number of seconds, v 

 for the velocity at the end of that time, / the 

 number of seconds, and g for 32!, the velocity 

 caused by gravity in one second, the relations of 

 these quantities are expressed in the following 

 formulas, which enable us to calculate any one of 

 them from the others. 

 h = i6& X t\ or (i.) h = \g X / 2 ; (2.) v=gy.t, 



If, in formula (i.), we substitute for i* its value in 



(3.) namely, -^ we get h = \g x ^, or (4.) h = . 

 o S^ l 



Knowing the velocity of a falling body, then, we 

 can find the space through which it must have 

 fallen in order to acquire it, by squaring the 

 velocity and dividing by 2 X 32^. The height thus 

 found is said to be the height due to that velocity. 



A falling body is urged by a pressure equal to 

 its own weight ; and, during a second, that pressure 

 acts upon it over 161*5 feet. Therefore, if the 

 body is a pound in weight, gravity has done upon 

 it in that time 16^ units of work ; and if it is 10 

 Ibs., the work done is 160}$ units. In other words, 

 the work done upon a falling body, and therefore 

 the work accumulated in it, is equal to the space 

 through which it has fallen multiplied by its weight 

 We may not know that space, but if we know the 



velocity, we can substitute for it its value, . 



Now, a cannon-ball, moving with a given 

 velocity, has the same force, whether it acquired 

 that velocity by falling from a certain height, or 

 by being shot from a cannon ; and the same rule 

 that enables us to calculate the units of work in it 

 in the former case, applies also in the latter. 

 Calling e the accumulated work, and w the weight 

 of the body, the rule given above may be expressed 



in a formula, thus : e = . 



V 



V* X IV IV IT . 



The expression , or X , is what was 



?-g 2 g 



formerly called the vis viva (living force) of a 

 moving body, but is now spoken of as its energy, 



Ex. i. A ram, weighing 5 cwt. or 560 Ibs. as 

 it strikes the top of a pile, has a velocity of 70 

 feet ; what is the work accumulated in it ? 



*__ = 42,653 units of work. 



2 X 32! 



Or we might proceed thus : The height from 

 which the ram must have fallen to acquire the 



70 2 



velocity of 70 feet, is = 76-166 feet. To raise 



2X 32^ 



the ram to this height required 76-166 X 560 = 

 42,653 units of work done upon it, and this work 

 it is capable of reproducing by its fall. 



Ex. 2 . A train weighing 100 tons has a velocity 

 of 50 feet a second, how far will it run before 

 stopping after the steam is shut off, supposing the 

 friction to be 8 Ibs. per ton, and that the train is 

 ascending an incline of i in 100? 



50 2 X 224,000 _ g 



2 X 32* 



in the train. Now, the resistance of friction is 

 8 X loo = 800 Ibs. and the resistance of the 

 incline is rta of the whole weight, or 2240 Ibs. 

 making together 3040 Ibs. To overcome this 

 resistance over I foot makes 3040 units of work, 



223 



> the work accu m u iated 



