specific gravity of solids as well as liquids. B is a hollow ball, to which is attached a fine wire s supporting a dish c for receiving weights; proceeding from the under side of the ball, is the stirrup D carrying a heavy dish F for preserving the stability of the instrument when it floats, and for holding any solid body whose specific gravity is to be determined. The instrument is floated in pure water, and a weight of 1000 grains is put into the dish c; now the weight of the instrument is so adjusted that it sinks to about the middle of the fine stem; and a mark s is made at this point.

D

Fig. 9.

(1) To determine the specific gravity of a liquid. Place the instrument in the liquid and put weights into the dish c until the mark s on the stem sinks to the level of the surface of the liquid. These weights added to the weight of the instrument will be equal to the weight of the liquid displaced; but the weight of the instrument added to 1000 gr. is equal to the weight of an equal bulk of water; therefore the former sum divided by the latter will give the specific gravity of the liquid. Let the weight of the instrument w, grains, the weight put on the dish c=w grains, then we have

Weight of displaced water=w1+ 1000,

(2) To determine the specific gravity of a solid.

Place the instrument in water, and put the solid in the upper dish c; add weights to this dish until the mark s on the stem sinks to a level with the fluid; then these weights together with the weight of the body must be equal to 1000 grains. Let v weight of the body w=the weight added to the dish c; the

w+w2 = 1000,

.. w=1000-W2 ・ ・ ・

which gives the weight of the body.

(2),

Let the body be now placed in the lower dish F, and let weights be placed in the upper dish c until the mark s sinks to a level with the water; then these weights, together with the weight of the body in the water, are equal to 1000 grains. Let

w=the weight of the body in water; w3=the weight added to the upper dish c; then

Now having found the weight of body, eq. (2), and also its weight in water, eq. (3), we have, by taking the difference of these equations,

weight lost in water or w-w= W3 — W2•

Let s be put for the specific of the body; then

Example. Let w=300, w3=400; then by eq. (4) we have for the specific gravity of the body

21. The common Hydrometer has no dish at c, and the lower dish F is simply a heavy ball which serves to keep the instrument in a vertical position when floating in the liquid. There are no weights used with the instrument; but the upper stem s is graduated in such a manner as to enable the operator to ascertain the specific gravity of a liquid by the depth to which the instrument sinks in it.

Let v=the volume of the whole instrument; v=the volume included between any two consecutive graduations on the stem; n, n1=the number of divisions on the stem above the surface when the instrument is put into fluids whose specific gravities are s and s, respectively; then

vol. fluid displaced in the 1st case=v―nv,

.. wt. fluid displaced in the 1st case =s(v−nv),

but the weight of the fluid displaced in each case is the same, for the weight of the instrument remains the same,

..s(v-nv)=s, (v — n, v) ;

T

which gives the ratio of the specific gravities of the two liquids.

EXERCISES FOR THE STUDENT.

In the following exercises S. G. is put for "specific gravity," and the specific gravity of water is taken equal to unity.

1. A cubic foot of water weighs 1000 oz. Find the weight of a cubical block of stone whose side is 4 ft. and S. G. 1.25.

Ans. 80,000 oz.

2. Required the number of cubic feet contained in a body whose S. G. is s, and weight w oz.

Ans.

W 1000s

3. If s and s are the S. G. of two lumps of metal whose weights are w and w respectively; required the S. G. of the compound metal formed by fusing the lumps together, supposing no condensation of volume to take place.

Ans.

ss (w+w)

ws+ws

4. A body, weighing 20 gr., has a S. G. of 2·5; required its weight in water. Ans. 12 gr.

5. A body weighs w grs. in water and w, grs. in another liquid whose S. G. is s1; required the weight of the body.

6. A Nicholson's hydrometer weighs 250 grains, and requires 726 grains to sink it to the required depth in alcohol; required the S. G. of the alcohol.

Ans. 781.

7. If v and be put for the respective volumes of the lumps of metal in Example 3.; then it is required to find an expression for

the S. G. of the compound metal.

Ans.

vs+vs

V+v

8. If s, be put for the S. G. of the compound metal in Ex

22. To find the depth to which a rectangular piece of wood

will sink in a fluid..

Let ABCD represent a transverse section of the body, EF being the plane of floatation. Put b=AB, a=AD, l=the length of the body, w= the weight of cubic foot of the body, and x= ED the depth of immersion, then

Weight of the body=ablw,

weight of the displaced fluid=bxl×62·5;

.. bxl×62·5=ablw;

B

Fig. 10.

23. A barge (supposed for the sake of simplicity to be of a rectangular shape) is 7 ft. long, b ft. broad, and a ft. deep, outside measure. The thickness of the planking is e ft., and the weight of a cubic foot of the timber is w lbs. To what depth, d, will the barge sink when loaded with w lbs.?

Content of the timber, v=vol. exterior solid-vol. interior

=abl-(a-e)(b-2e)(1-2e)... (1).

For the sake of conciseness we shall put v for this expression, giving the volume of the timber.

Whole weight of the floating body, w1=wt. timber+wt. load

To determine the load necessary to sink the barge we have by making d=a in eq. (3)

Obs. The irregular form of boats as they are usually constructed, renders it difficult to calculate their volume; as a toler

7

ably near approximation to the truth, eminent surveyors take 12 of the rectangular displacement, as determined in the foregoing

investigation, for the true displacement in boats of an ordinary Hence we have, making v=0 in eq. (4),

curvature.

which is the weight in lbs. requisite to sink a vessel to a given depth a, the breadth of the vessel at that depth being b, and the length 1.

24. A globe of given diameter floats on water; it is required to find the weight of the displaced water, when the depth of immersion is given.

Let r=CK=CD the radius of the sphere; a eK the depth of immersion; w, the weight of displaced water in lbs. Here the displaced water has the form of a segment of a sphere, viz. ADK. Now we have by mensuration (see the Author's "Calculus," p. 200.)

If the we have

A

N

Fig. 11.

Content segment ADK=τa2(r—a);

.. w1 =πa2 (ra) × 62·5... (1).

B

ACK=0 be given; then for the value of a in eq. (1)

a=eK=CK+ce=r(1—cos 0) . . . (2).

To find the Buoyancy of Pontoons.

25. Pontoons are portable boats, with a covering of baulks and planks, &c., for forming floating bridges over rivers, &c. They are now usually made of tin in the shape of a cylinder with hemispherical ends.

B

N

1. Let NB represent a cylindrical pontoon with plane ends; ABD the plane of the water line parallel to the axis of the cylinder which passes through C; CN a vertical line cutting AD in e. To find the load w requisite to sink the pontoon to a given depth.

Put A the area ADK the surface of immersion; l=AB the length of the cylinder; then we have

vol. water displaced=AXI... (1);

Fig. 12.

K

D