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in the corresponding situations, the relation of their measures may be expressed by proportional equation, according to Def. 2.

3. If three variable physical quantities are fo connected, that one of them is increased or diminished, in proportion as both the others are increased or diminished; or, if the magnitudes of one of them in any two situations, have a ratio, which is compounded of the ratios of the magnitudes of the other two, in the corresponding fituations; the relation of the measures of these three may be expressed by a proportional equation, according to Def. 3.

4. In like manner may the relations of other combinations of physical quantities be expressed, according to Def. 4.

And when these proportional equations are obtained, by reasoning with regard to them, according to the preceding propositions, new relations of the physical quantities may be deduced.

2. Examples of Physical Problems. The use of algebra, in natural philosophy, may be properly illustrated by some X

examples

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examples of physical problemys. The solution of such problems inust be derived from known physical laws, which, though ultimately founded on experience, are here afsumed as principles, and reasoned upon mathematically. The experiments by which the principles are ascertained admit of various degrees of accuracy; and on the degree of physical accuracy in the principles will depend the physical accuracy of the conclufions mathematically deduced from them. If the principles are inaccurate, the conclusions must, in like manner, be innaccurate; and, if the limits of inaccuracy in the principles can be ascertained, the correfponding limits, in the conclusions derived from them, may likewise be calculated.

EXAMPLE I.

Let a glafs tube, 30 inches (a) long, be filled

with mercury, excepting 8 inches (b); and let it be inverted, as in the Torricellian experiment, so that the 8 inches of common air may rise to the top: It is required to find at what þeighi the mercury will remain fufpended, the mercury

in the barometer

barometer being at that time 28 inches (d) high.

The solution of this problem depends upon the following principles :

1. The pressure of the atmosphere is measured by the column of mercury in the barometer; and the elastic force of the air, in its natural state, which resists this prefsure, is therefore measured by the same column.

2. In different states, the elastic force of the air is reciprocally as the spaces which it occupies.

3. In this experiment, the mercury which remains suspended in the tube, together with the elastic force of the air in the top of it, being a counterbalance to the pressure of the atmosphere, may therefore be expressed by the column of mercury in the barometer.

Let the mercury in the tube be x inches, the air in the top of it occupies now the space ax; it occupied formerly b inches, and its elastic force was d inches of mercury: Now, therefore, the force must be

(a

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aX

bd (a-x:6::d :) inches (2.). Therefore

bd (3.)* 7 =d. This reduced, and putting a+d=2m, the equation is x*—2mx =bd-ad.

This resolved gives x=m+mo+bd-ad.
In numbers -

X=44 or 14. One of the roots 44 is plainly excluded in this case, and the other 14 is the true an[wer. If the column of mercury x, suspended in the tube, were a counterbalance to the pressure of the atmosphere, expressed by the height of the barometer d, together with the measure of the elastic force of b inches of common air in the space x-a, that is, bd

bd if x=dt

-=d, the equation will be the fame as before, and the root 44 would be the true answer. But the experiment in this question does not admit of such a supposition.

Ε Χ Α Μ Ρ L E The distance of the earth and moon (d,) and

their quantities of matter (t, 1,) being given, to find the point of equal attraction between them.

Let

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a

II.

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Let the distance of the point from the earth be x, its distance from the moon will be therefore dx. But gravitation is as the matter dire&tly, and as the square of the distance inversely; therefore the earth's attraction is as eas and the moon's attrac

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tion is as da12. But these are here equal ; therefore,

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dyt This equation reduced gives x

vt + VT Or mult. numerator and de-7

dt--dntl nominator by Vt - VS

t-i In round numbers, let d=60 semidiame ters of the earth, t=40,1=1, then x=52 femidiameters nearly.

There is another point beyond the moon at which the attractions are equal, and it would be found by putting the square root of 1 to be xd, which, in this case, would be a positive

dt+dytl quantity; and then x=.

72 nearly. If the quantities had been multiplied

before

tel.

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