176. If a weight of 1 lb. has to be raised through a height of 2 feet, or a weight of 2 lbs. through a height of 1 foot, it will be necessary to expend twice as much energy as would have raised a weight of 1 lb. through 1 foot, that is, 2 foot-pounds.
If a weight of 5 lbs. had to be raised from the floor up to a stool 3 feet high, how many units of energy would be required? To raise 5 lbs. through 1 foot requires 5 foot-pounds, and the process must be again repeated twice before the weight arrive at the top of the stool. For the whole operation 15 foot-pounds will therefore be necessary.
If 100 lbs. be raised through 20 feet, 100 foot-pounds of energy is required for the first foot, the same for the second, third, &c., up to the twentieth, making a total of 2,000 foot-pounds.
Here is a practical question for the sake of illustration. Which would it be preferable to hoist, by a rope passing over a single fixed pulley, a trunk weighing 40 lbs. to a height of 20 feet, or a trunk weighing 50 lbs. to a height of 15 feet? We shall find how much energy would be necessary in each case: 40 times 20 is 800; therefore in the first case the energy would be 800 foot-pounds. But 50 times 15 is 750; therefore the amount of work, in the second case, is only 750 lbs. Hence it is less exertion to carry 50 lbs. up 15 feet than 40 lbs. up 20 feet.
177. The rate of working of every source of energy, whether it lie in the muscles of men or other animals, in water-wheels, steam-engines, or other prime movers, is to be measured by the number of foot-pounds produced in the unit of time.
The power of a steam-engine is defined by its equivalent in horse-power. For example, it is meant that a steam-engine of 3 horse-power, could, when working for an hour, do as much work as 3 horses could do when working for the same time. The power of a horse is, however, an uncertain quantity, differing in different animals and not quite uniform in the same individual; accordingly the selection of this measure for the efficiency of the steam-engine is inconvenient. We replace it by a convenient standard horse-power, which is, however, a good deal larger than that continuously obtainable from any ordinary horse. A one horse-power steam-engine is capable of accomplishing 33,000 foot-pounds per minute.
178. We shall illustrate the numerical calculation of horse-power by an example: if a mine be 1,000 feet deep, how much water per minute would a 50 horse-power engine be capable of raising to the surface? The engine would yield 50 × 33,000 units of work per minute, but the weight has to be raised 1,000 feet, consequently the number of pounds of water raised per minute is
| 50 × 33,000 | = 1,650. |
| 1,000 |
179. We shall apply the principle of work to the consideration of the pulley already described (p. 90). In order to raise the weight of 14 lbs., it is necessary that the rope to which the power is applied should be pulled downwards by a force of 15 lbs., the extra pound being on account of the friction. To fix our ideas, we shall suppose the 14 lbs. to be raised 1 foot; to lift this load directly, without the intervention of the pulley, 14 foot-pounds would be necessary, but when it is raised by means of the pulley, 15, foot-pounds are necessary. Hence there is an absolute loss of ¹/₁₅th of the energy when the pulley is used. If a steam-engine of 1 horse-power were employed in raising weights by a rope passing over a pulley similar to that on which we have experimented, only ¹⁴/₁₅ths of the work would be usefully employed; but we find
| 33,000 × 14 | = 30,800. |
| 15 |