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[Note.—When we commenced the Solutions, it was our intention

to obtain a specimen paper of questions on each of the subjects brought under the notice of the masters of these classes. That such papers would have been interesting and valuable, there can be little doubt, since the express object of the establishment of the classes is the preparation of masters for undergoing the necessary examination for certificates. The limits of our work, however, will not admit of the accomplishment of our wishes in in this respect; still, it is hoped the few specimen questions here appended will not be destitute of all the advantages such recognised papers would have afforded.]


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1. “ The weekly wages of 12 men and 3 boys amounts to £13 195. 6d., required the weekly wages of 15 men and 2 boys, if a man earns 10s. in the time that a boy earns 3s.”

As a man earns 10s. in the time that a boy earns 3s the value of a man's labour compared with that of a boy will be as 10 to 3.

Hence 1 boy does ®, and the 3 boys will do 2 the work of a man ; therefore the 2 boys will perform % x 2 of a man's work.

The fractions L and } may, therefore, be substituted for, as being equivalent to, the value of 3 boys' and 2


boys' labour, respectively, expressed in the terms of what a man can do. The question will resolve itself thus:

If the labour represented by 12%, cost £13 19s. 6d., what will be the cost of that represented by 15%-? To solve which we have,

1348 x 153 12% : 155 :: £13 19s. 6d. :

12 io
559 x 78 X 10 13 x 13

= £16 18s.
40 x
5 X 129

10 2. A boy in reducing 25 acres to poles, makes the mistake of multiplying by 4 and 28, instead of by 4 and 40; what divisor applied to his result will produce the right one ?

As the product 25 X 4 X 28 has the factor 28 instead of 40; it follows that it must now be divided by 28, and then multiplied by 40 ; i. e., (25 X 4 X 28 = 28) X 40. But since quantities may be multiplied in any order, we have, 25 X 4 X 28 X 40 = 28

= 25 X 4 X 28 X 8.

Here we find that the correct result will be produced by applying 1: = as a multiplier to the false result; · but the question requires the number applied to be a divisor. Therefore invert 10, and we have the divisor required ; thus,

25 X 4 X 28 = %= 25 X 4 X 28 x 1 l = 4000 poles.

3. A and B are travelling in the same direction, at a uniform speed, A being 11 miles in advance of B. If A should diminish his hourly rate by 120 yards, how should B at the same time alter his hourly rate to come up with A in 33 hours ?”

Had A continued at his original rate per hour, B would have had to increase his by 1760 x 1% = 3


880 yards per hour, in order to overtake A in 37 hours. But A decreases his speed by 120 yards per hour; therefore B will have to increase his 120 yards less than otherwise ; 880 120 = 760 yards per hour.

QUESTIONS IN MECHANICS. 1. “ An engine of 20 horse power has a duty of sixty millions ; required the number of bushels of coals it will consume in a day of 10 hours."

To comprehend this question, the import of “ the horse-power," and " the duty” of a steam-engine must be fully understood. The power indicates the relative work any particular engine will perform compared with what can be done in the same time by a horse; which, according to Watt, performs 33,000 units of work per minute. (See Tate's Exercises in Mechanics, Units of Work.)

The duty of an engine is the amount of work it performs by the agency of the steam evaporated by the consumption of one bushel of coals. The duty of engines is various, arising from many causes ; but experiments have shown that the same quantity of fuel evaporates a constant quantity of water. (See Tate's Mech., Steam Power.)

The horse-power of the engine in question is 20 ; therefore in a day of 10 hours it will perform

20 X 33,000 x 60 x 10 = 396,000,000 units of work. But during the consumption of one bushel of coals, the engine performs 60,000,000 units of work. 396,000,000

= 6*6 bushels of coals consumed 60,000,000 per day.

2. “ A train of 80 tons ascends a gradient having a rise of } in 100, with a uniform speed of 20 miles per hour. Required the horse-power, allowing friction at 7 lbs.


ton." Here there are two distinct forces acting in opposition to that of the source of motion, or the power of the engine-friction and gravitation. The former of these in effect is given 7 lbs. per ton ; i. e., one ton is moved over the space of one foot by the application of a force equal to that required to raise 7 lbs. 1 foot, in opposition to gravity; or, in other words, seven units of work are performed in moving a ton horizontally one foot. In the latter power-absorbing force, the whole load, 80 tons, is actually raised, in opposition to gravity, to the extent of one inch every 25 feet the train proceeds.

Now, to calculate the horse-power of the engine, we must find the space in feet passed over per minute by · the train; this multiplied by as many times 7 as there are tons in the train will give the units of work expended in opposition to the resistance of friction per minute ; and next find the perpendicular height in feet to which the train is raised per minute: this, multiplied by the weight of the train in pounds, will give the units of work expended in that space of time in opposition to the force of gravitation. The sum of these units, thus exhausted, must evidently be equal to the whole labouring power of the engine. To express that in horsepowers, we have only to divide it by the number of units a horse can perform in the same time, a minute. Thus,

20 X 5280 Space passed over per minute by train,

60 = 1760 feet.

.. 1760 X 7 X 80 = work due to friction per minute. Height the train is raised per minute, 1760 X }

5.86 feet. 100


.. 5.86 x 80 X 2240 = work due to gravity per minute.

Total work of the engine,

1760 X 7 X 20 + 5.86 X 80 X 2240 = 2,037,573:3 units.

2,037,573:3 .. Horse-power =


33,000 3. “In a condensing engine the length of a stroke is 6 feet; the steam is cut off at 2 feet of the stroke, the pressure of the steam 30 lbs. per square inch, and the elasticity of the vapour in the condenser is 3 lbs.

Required the work performed on one inch of the piston in each stroke.”

The work done by a steam-engine, working under high pressure, may be compared to the area of a rectangle; one side of which will represent the pressure of the steam, and the other the distance passed over by the piston or load. Thus, in the example before us, were the steam not cut off till after the completion of the stroke, the work done upon one inch of the piston during the stroke would be 30 x 6 180, which is the area of a rectangle whose sides are 30 and 6 respectively. But the rectangle will not apply when the steam is cut off at a certain point in the stroke, and the rest of it per ormed by the expansive power of the steam, except before the supply of steam is stopped ; because the pressure, 30 lbs. in this case, becomes successively less as the piston progresses. By Marriotte's law, the pressure at any point of the stroke may be ascertained :-" The decrease of the pressure of an elastic fluid is in the same ratio as the increase of volume."

Now, supposing that portion of the stroke after the steam is cut off to be divided into any number of equal parts, and by the above law, finding the pressure at each

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